where ε is called the absolute permittivity of force between charges by a factor of εr, its the medium. The force between the same two relative permittivity. charges placed in free space or vacuum at distance r is given by, While using Eq. (10.5) we assume that the medium is homogeneous, isotropic and Fvac = 1 § q1q 2 · --- (10.3) infinitely large. 4SH 0 ¨© r2 ¸¹ 10.4.3 Definition of Unit Charge from the Coulomb’s Law: Dividing Eq. (10.3) by (10.2) The force between two point charges q1 Fvac = 1 § q1q 2 · =H and q2, separated by a distance r in free space, is Fmed 4SH 0 ¨© r2 ¸¹ H0 written by using Eq. (10.2), 1 § q1q 2 · 1 q1q 2 q1q 4SH ©¨ r2 ¸¹ 4SH 0 r2 r2 F u 9 u 109 u 2 ε ε0 = 8.85 Í10-12 C2 N-1 m-2 diTehleectrraictiocoεn0stainst the relative permittivity If q1= q2= 1C and r = 1.0 m or of the medium and is denoted by εr or K. Then F = 9.0Í109 N K or Hr H Fvac --- (10.4) From this, we define, coulomb (C) the unit H0 Fmed of charge in SI units. Thus, One coulomb is the amount of charge which, when placed at a distance of one (i) εr is the ratio of absolute permittivity of a metre from another charge of the same medium to the permittivity of free space. magnitude in vacuum, experiences a force of 9.0 × 109 N. This force is a tremendously large (ii) εr is the ratio of the force between two point force realisable in practical situations. It is, charges placed a certain distance apart in therefore, necessary to express the charge in free space or vacuum to the force between smaller units for practical purpose. Subunits the same two point charges when placed at of coulomb are used in electrostatics. For the same distance in the given medium. example, micro-coulomb (10-6 C, µC), nano- coulomb (10-9 C, nC) or pico-coulomb. (10-12 C, εr is a dimensionless quantity. pC) are normally used units. (iii) εr is also called specific inductive capacity. The force between two point charges q1 Do you know ? and q2 placed at a distance r in a medium of relative permittivity εr, is given by Force between two charges of 1.0 C each, separated by a distance of 1.0 m is 9.0×109 N F 1 q1q 2 --- (10.5) or, about 10 million metric tonne. A normal For w4SaHt0eHrr, r2 (10.4) truck-load is about 10 metric tonne. So, this force is equivalent to about one million truck- εr = 80 then from Eq. loads. A tremendously large force indeed ! Fvac Hr 80 Fwater Fwater Fvac Example 10.2: Charge on an electron is 80 1.6×10-19 C. How many electrons are required This means that when two point charges to accumulate a charge of one coulomb? are placed some distance apart in water, the Solution: 1.6Í10-19 C = 1 electron § 1 · th 1 ©¨ 80 ¸¹ 1.6 u1019 force between them is reduced to of the ?1C electrons force between the same two charges placed at 0.625u1019 6.25u1018 electrons the same distance in vacuum. 6.25×1018 electrons are required to Thus, a material medium reduces the accumulate a charge of one coulomb. 192
It is now possible to measure a very Here, q1 = q2 = +1.6×10-19 C, r = 10-15m small amount of current in otto-amperes which measures flow of single electron. ? Fe 9 u109 (1.6 u1019 ) (1.6 u1019 ) (1015 )2 10.4.4 Coulomb’s Law in Vector Form: 9 u1.6 u1.6 u101 N As shown in Fig 10.4, q1 and q2 are two Fe 2.3u102 N force --- (10.8.a) similar point charges situated at points A and B. between the The gravitational r12 is the distance of separation between them. protons is given by F21 denotes the force exerted on q2 by q1 Fg = G m1m 2 r2 1 q1q 2 F21 4SH 0 u u r21 --- (10.6) 6.674 u1011 u1.67 u1027 u1.67 u1027 2 (1015 )2 r 21 Fg 1.86 u1034 N --- (10.8.b) Comparing 10.8. (a) and 10.8.(b) Fe 2.30 u102 N = 1.23u1036 Fg 1.86 u1034 N 1 Fig. 10.4: Coulomb’s law in vector form Thus, the electrostatic force is about 36 orders of magnitude stronger than the gravitational force. r21 is the unit vector along AB , away Comparison of gravitational and from B. Similarly, the force F12 exerted on q1 electrostatic forces: by q2 is given by Similarities F12 1 u q1q 2 u r12 --- (10.7) 1. Both 1forces obey inverse square law : 4SH 0 r12 2 F ∝ r2 2. Both are central forces : act along the r12 is the unit vector along BA , away from line joining the two objects. A. F12 acts on q1at A and is directed along BA, Differences away from A. The unit vectors r12 and r21 are 1. Gravitational force between two objects oppositely directed i.e., r12 r21 hence, is always attractive while electrostatic F 21 = - F12 force between two charges can be either attractive or repulsive depending Thus, the two charges experience force on the nature of charges. 2. Gravitational force is about 36 of equal magnitude and opposite in direction. orders of magnitude weaker than the electrostatic force. These two forces form an action- reaction pair. As F21 and F12 act along the line joining 10.5 Principle of Superposition: the two charges, the electrostatic force is a central force. The principle of superposition states that Example 10.3: Calculate and compare the when a number of charges are interacting, electrostatic and gravitational forces between two protons which are 10-15 m apart. Value of the resultant force on a particular charge is G = 6.674×10-11 m3 kg-1 s-2 and mass of the proton is 1.67×10-27 kg given by the vector sum of the forces exerted by individual charges. Consider a number of point charges q1, q2, q3 ------- kept at points A1, A2, A3--- as shown Solution: The electrostatic force between the in bFyigq. 21i0s.5F.12T.hTehfeorvcaelueexeorfteFd12onis the charge q1 calculated protons is given by Fe 1 q1q2 4SH0 r2 193
by ignoring the presence of other charges. F AC 2µC 4 cm 3µC B Similarly, we find F13 , F14 etc, one at a time, F AB A 3 cm using the coulomb’s law. θ F14 F15 F13 Fig. a: Position of charges. 4µC q2 q1 C A2 A1 Solution : Given, F12 AB = 4.0 cm, BC = 3.0 cm q3 q5 ? AC = 42 32 5.0 cm A3 q4 A5 Magnitude of force F AB on A due to B is, A4 § 1 · 2 u106 u 3u106 Fig. 10.5: Principle of superposition. F AB ¨ 4SH 0 ¸ (4 u102 )2 Total force F1 on charge q1 is the vector © ¹ sum of all such forces. 9 u109 u 6 u 1012 16 u104 F1 F12 F13 F14 ... = 3.37u10 1 ª q1q q1q 3 º = 33.7 N 4SH 0 « r13 2 ...»» , 2 r12 r13 This force acts at point A and is directed ¼» along BA (Fig. (b)). « r12 2 «¬ where r12 , r13 etc., are unit vectors directed to F q1 from q2, q3 etc., and r12, r13, r14,etc., are the F AC distances from q1 to q2, q3 etc respectively. F AB A Let there be N point charges q1, q2,q3 Fig. b: Forces acting at point A. etc., qN. The force F exerted by these charges Magnitude of force F AC on A due to C is, on a test charge q0 can be written using the summation notation Σ as follows, F AC § 1 · 2 u106 u 4 u106 Ftest F1 F2 F3 FN --- (10.9) ¦N = 1 ¦N q0qn r 0 n --- (10.10) ¨ 4SH 0 ¸ (5u102 )2 4SH 0 r02n © ¹ Fn n =1 9 u109 u 8.0 u1012 n =1 Where r0n is a unit vector directed from the 25 u 104 nth charge to the test charge qo and r0n is the 72 25 separation between them, r0n = r0n r0n u10 28.8 N Can you tell? This force acts at point A and is directed along CA . (Fig. 10.6.(b)) Three charges, q each, are placed at the vertices of an equilateral triangle. What will F = F AB + F AC be the resultant force on charge q placed at Magnitude of resultant force is, the centroid of the triangle? F = ¬ªFA2C FA2B 2FAC FAB cosT º¼ 1/2 Example 10.4: Three charges of 2µC, 3µC and 4µC are placed at points A, B and C 59.3 N respectively, as shown in Fig. a. Determine the force on A due to other charges. 194
Direction of the resultant force is 16.9° electric field of a charge is also a vector and is directed along the direction of the coulomb north of west. (Fig. c) force, experienced by a test charge. N The magnitude of electric field at a distance F r from a point charge Q is same at all points on F AC the surface of a sphere of radius r as shown in Fig. 10.6. Its direction is along the radius of W 16.9° A the sphere, pointing away from its centre if the F AB charge is positive. Fig. c: Direction of the resultant force. E 10.6 Electric Field: Space around a charge Q gets modified so E Q that when a test charge is brought in this region, 4SH0r 2 it experiences a coulomb force. This region + around a charged object in which coulomb force is experienced by another charge is called electric field. Fig. 10.6: Electric field due to a point charge (+Q). Mathematically, electric field is defined as the force experienced per unit charge. Let Q and SI unit of electric intensity is newton per q be two charges separated by a distance r. coulomb (NC-1). Practically, electric field is expressed in volt per metre(Vm-1). This is The coulomb force between them is given discussed in article 10.6.2. by F 1 Qq r , where, r is the unit vector 4SH 0 r2 Dimensional formula of E is, along the line joining Q to q. E = F [LMT 2 ] q0 [IT] Therefore, electric field due to charge Q is given by, E = [LMT 3 I1] E F Q r --- (10.11) 10.6.1 Electric Field Intensity due to a Point q 4SH0r 2 Charge in a Material Medium: The coulomb force acts across an empty space (vacuum) and does not need any Consider a point charge q placed at point O intervening medium for its transmission. in a medium of dielectric constant K as shown in Fig. 10. 7. The electric field exists around a charge irrespective of the presence of other charges. Since the coulomb force is a vector, the q0 A precise definition of electric field is: k Electric field is the force experienced by a q test charge in presence of the given charge at the given distance from it. Fig. 10.7: Field in a material medium. Consider the point P in the electric field of E lim F point charge q at distance r from it. A test charge qo0 q q0 placed at the point P will experience a force which is given by the Coulomb’s law, Test charge is a positive charge so small in magnitude that it dose not affect the surroundings of the given charge. 195
F 1 q q0 r 4SH 0 K r2 where r is the unit vector in the direction of Fig. 10.9 (a): force i.e., along OP. uniform electric field. By the definition of electric field intensity E F 1 q r q0 4SH 0 K r2 Fig. 10.9 (b): non uniform The direction of E will be along OP when electric field. q is positive and along PO when q is negative. 10.6.2 Practical Way of Calculating Electric The magnitude of electric field intensity in Field a medium is given by A pair of charged parallel plates is E 1q --- (10.12 ) arranged as shown in Fig. 10.10. The electric 4SH0K r2 field between them is uniform. A potential difference V is applied between two parallel For air or vacuum K = 1 then plates separated by a distance ‘d’. The electric field between them is directed from plate A to E = 1 q plate B as shown. 4SH 0 r2 The coulomb force between two charges and electric field E of a charge both follow the inverse square law, (F∝1/r2, E∝1/r2) Fig. 10.8. F∝1/r2, E∝1/r2 A B E Fig 10.10: Electric field or between two parallel plates. F r A charge +q placed between the plates experiences a force F due to the electric field. If Fig. 10.8: Variation of Coulomb force/ we have to move the charge against the direction Electric field due to a point charge. of field, i.e., towards the positive plate, we have to do some work on it. If we move the charge 1. Uniform electric field: A uniform electric +q from the negative plate B to the positive field is a field whose magnitude and plate A, the work done against the field is direction is same at all points. For example, W = Fd; where ‘d’ is the separation between the field between two parallel plates. Fig 10.9.a plates. The potential difference V between the two plates is given by 2. Non uniform electric field: A field whose magnitude and direction is not the same at W = Vq, but W = Fd all points. For example, field due to a point charge. In this case, the magnitude of field ∴Vq = Fd ∴ F/q = V/d = E is same at distance r from the point charge in any direction but the direction of the ∴ Electric field can be defined as field is not same. Fig 10.9.b E = V/d --- (10.13) This is the commonly used definition of electric field. 196
Example 10.5: Gap between two electrodes of magnitudes and are opposite in direction, the spark-plug used in an automobile engine is 1.25 mm. If the potential of 20 V is applied EA = - EC . EA + EC = 0 . Thus, the field at P is across the gap, what will be the magnitude of electric field between the electrodes? only to the charge at B and can be written as Ep EB 2 u106 4SH0 (BP)2 Solution: Ep 2 u106 u 9 u109 EV (5 / 2 )2 d 2 u 9 u103 u 2 20V V 25 1.25u103 m m E 16 u103 1.6 u104 36 u 103 25 This electric field is sufficient to ionize 1.44 u1015 NC-1 along BP the gaseous mixture of fuel compressed in the To calculate BP cylinder and ignite it. 5 BP (BA) cos (45)q 2 Can you tell? Example 10.7: A simplified model of hydrogen atom consists of an electron revolving about a Why a small voltage can produce a proton at a distance of 5.3×10-11m. The charge reasonably large electric field? on a proton is +1.6×10-19 C. Calculate the intensity of the electric field due to proton at Example 10.6: Three point charges are placed this distance. at the vertices of a right isosceles triangle as shown in the Fig. a. What is the magnitude and Solution: direction of the resultant electric field at point P E q 4SHor 2 which is the mid point of its hypotenuse? q 1.6 u1019 C C +10 µC EA EB r 5.3u1011 m 5cm P 1 9.0 u109 Nm2C 1 4SH o EC 1.6 u1019 +2 µC +10 µC E (5.3u1011 )2 u 9.0 u109 5.1u1011 NC 1 B 5cm A The force between electron and proton in Fig (a): Position of charges. hydrogen atom can be calculated by using the EB electric field. We have, E F F qE q EA ? F = -1.6×10-19 C× 5.1×1011 NC-1 = -8.16×108 N. P This force is attractive. EC Using the Coulomb's law, Fig (b): Electric field at point P. F 1 q1q2 Solution: Electric field is the force an a unit 4SH0 r 2 positive charge, the fields at P due to the charges at A, B and C are shown in the Fig. b. EA is the 9.0 u109 Nm2C 1 u (1.6 u 1019 C ) u (1.6 u 1019 C ) field at P due to charge at A and EC is the field (5.3u1011 m)2 at P due to charge at C. Since P is the midpoint of AC and the fields at A and C are equal in 8.6 u108 N Knowing electric field at a point is useful to estimate the force experienced by a charge at that point. 197
10.6.3 Electric Lines of Force: (6) Electric lines of force are crowded in a Michael Faraday (1791-1867) introduced region where electric intensity is large. the concept of lines of force for visualising (7) Electric lines of force are widely separated electric and magnetic fields. An electric line of from each other in a region where electric force is an imaginary curve drawn in such a intensity is small way that the tangent at any given point on this curve gives the direction of the electric field (8) The lines of force of an uniform electric at that point. See Fig.10.11. If a test charge is field are parallel to each other and are placed in an electric field it would be acted upon equally spaced. by a force at every point in the field and will The lines of force are purely a geometric move along a path. The path along which the unit positive charge moves is called a line of construction which help us visualise the nature force. of electric field in a region. The lines of force have no physical existence. Fig. 10.11: Electric line of force. A line of force is defined as a curve such Fig. 10.13 (a): Lines that the tangent at any point to this curve gives of force due to positive the direction of the electric field at that point. charge. The density of field lines indicates the strength of electric fields at the given point in Fig. 10.13 (b): Lines of space. Figure 10.12. force due to negative charge. (High field) Fig. 10.13 (c): Lines of force due to opposite charge. (Low field) Fig. 10.12: density Fig. 10.13; (d): of field lines and Lines of force strength of electric due to similar field. charge. Characteristics of electric lines of force Fig. 10.13 (e): Lines of force terminate (1) The lines of force originate from a on a conductor. positively charged object and terminate on a negatively charged object. Fig. 10.13 (f): Intensity of a electric field is more (2) The lines of force neither intersect nor meet at point A and less at B. each other, as it will mean that electric field More lines cross the area has two directions at a single point. at A and less at the same area at B. (3) The lines of force leave or terminate on a Fig. 10.13: The lines of force due to various conductor normally. geometrical arrangement of electrical charges. (4) The lines of force do not pass through Can you tell? conductor i.e. electric field inside a conductor is always zero, but they pass Lines of force are imaginary, can they through insulators. have any practical use? (5) Magnitude of the electric field intensity is proportional to the number of lines of force per unit area of the surface held perpendicular to the field. 198
10.7 Electric Flux: imagine a small charge +q present at a point O inside closed surface. Imagine an infinitesimal As discussed previously, the number of area dA of the given irregular closed surface. lines of force per unit area is the intensity of the electric field E . =AreNa uenmcbloesrinogf lines of force ... E the lines of force - (10.14) Fig. 10.15: Gauss' law. The magnitude of electric field intensity at point P on dS due to charge +q at point O is, Fig. 10.14: Flux through area S. E = 1 § q · Number of lines of force = (E).(Area) 4SH 0 ©¨ r2 ¸¹ When the area is inclined at an angle θ with the direction of electric field, Fig. 10.14, the electric The direction of E is away from point O. flux can be calculated as follows. Let θ be the angle subtended by normal drawn and Let tvheectoarngdleS between electric field E to area dS and the direction of E. Electric flux, area be θ, then the electric flux dφ, passing through area dS, = Ecosθ dS passing through area dS is given by q cosT dS dφ = (component of dS along E ).(area S 4SH0r 2 of ) d § q ·§ dS cosT · dφ = E (dS cos θ) ¨ 4SH 0 ¸ ¨© r2 ¹¸ © ¹ dφ = EdS cos θ dφ = E .d S --- (10.15) dI § q · dZ --- (10.17) ¨ 4SH 0 ¸ Total flux through the entire surface © ¹ ) ³ dI ³ E.dS E.S --- (10.16) where, dZ = dS cosT is the solid angle s r2 The SI unit of electric flux can be calculated subtended by area dS at point O. using, Total electric flux, φ , crossing the given closed surface can be obtEained by integrating ) E S =(V/m) m2 =Vm Eq. (10.17) over its area. Thus, 10.8 Gauss' Law: Karl Friedrich Gauss (1777-1855) one )E ³ dI ³ E.ds ³ q dZ q ³ dZ of the greatest mathematician of all times, s 4SH 0 4SH 0 formulated a law expressing the relationship between the electric charge and its electric field But ³ dZ 4S solid angle subtended by which is called the Gauss’ law. Gauss' law is analongous to Coulomb’s law in the sense that it entire closed surface at point O too expresses the relationship between electric field and electric charge. Gauss' law provides Total flux q ( 4S ) equivalent method for finding electric intensity. 4SH 0 It relates values of field at a closed surface and the total charges enclosed by that surface. ³)E E ds q / H0 --- (10.18) s Consider a closed surface of any shape which encloses number of positive electric This is true for every electric charge charges (Fig. 10.15). To prove Gauss’ theorem, enclosed by a given closed surface. Total flux due to charge q1, over the given closed surface = + q1/ε0 199
Total flux due to charge q2, over the given Example 10.8: A charge of 5.0 C is kept at the closed surface = + q2/ε0 centre of a sphere of radius 1 m. What is the flux passing through the sphere? How will this value Total flux due to charge qn, over the given change if the radius of the sphere is doubled? closed surface = + qn/ε0 Solution: Flux per unit area is given by Eq. Positive sign in Eq. (10.18) indicates that 10.16. the flux is directed outwards, away form the charge. If the charge is negative, the flux will According to Gauss law, the total flux be is directed inwards as shown in Fig 10.16 (b). If a charge is outside the closed surface the through the sphere ) ³ E.ds , where the net flux through it will be zero Fig 10.16 (c). integration is over the surface of the sphere. As the electric field is same all over the sphere i.e. | E | = constant and the direction of E as well as that of ds is along the radius, we get flux = ) = | E | 4S R2 E q 9 u109 u 5.0 C 4SH 0 r 2 (1.0 m)2 Fig. 10.16 (a): Flux due to positive charge. E 9u109 u 5 4.5u1010 NC-1 I ES ?flux = 4.5u1010 u 4S (1)2 5.65u1011Vm Thus the total flux is independent of radius. Fig. 10.16 (b): Flux due to negative charge. E ∝ 1/r2, and area ∝ r2. This can also be seen from Gauss' law, where the net flux crossing a closed surface is equal to q/ε0 where q is the net charge inside the closed surface. As the charge Fig. 10.16 (c): Flux due to charge outside a inside the sphere is unchanged, the flux passing closed surface is zero. through a sphere of any radius is the same. Thus, if the radius of the sphere is increased According to the superposition principle, by a factor of 2, the net flux passing through the total flux φ due to all charges enclosed its surface remains unchanged. As shown in within the given closed surface is Fig. 10.17, same number of lines of force cross both the surfaces. The total flux is independent of shape of the closed surface because Eq. 10.18 does not involve any radius. = q1 q2 q3 qn i=n qi Q H0 H0 H0 H0 H0 = ¦) E + + + ---- + Hi=1 0 Statement of Gauss' law The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ε0 where Q is the total charge within the surface. Fig. 10.17: Flux is independent of the shape and size of closed surface. Gauss' law is applicable to any closed surface of regular or irregular shape. 200
Do you know ? dipole moment p . Gaussian surface Dipole mpo=mqe(n2tlis) de fined as All the lines of force originating from a point charge penetrate an imaginary --- (10.19) three dimensional surface. The total flux ΦE = q/ε0. The same number of lines of force A dipole moment is a vector whose will cross the surface of any shape. The total flux through both the surfaces is the magnitude is q (2l) and the direction is from same. Calculating flux involves calculating the negative to the positive charge. The unit of ³ E ds , hence it is convenient to consider a dipole moment is Columb-meter (Cm) or Debye regular surface surrounding the given charge distribution. A surface enclosing the given (D). 1D = 3.33×10-30 Cm. If two charges +e and charge distribution and symmetric about it is a Gaussian surface. -e are separated by 1.0A0, the dipole moment is For example. if we have a point charge 1.6×10-29 Cm or 4.8 D. For example, a water the Gaussian surface will be a sphere. If the charge distribution is linear, the Gaussian molecule has a permanent dipole moment of surface would be a cylinder with the charges distributed along its axis. Gaussian surface Natural dipole: offers convenience of calculating the The water molecule is non-linear, i.e., integral ³ E ds . the two hydrogen atoms and one oxygen atom are not in a straight line. The two hydrogen- Remember that a Gaussian surface oxygen bonds in water molecule are at an is purely imaginary and does not exist angle of 105°. The positive charge of a water physically. molecule is effectively concentrated on the hydrogen side and the negative charge on 10.9 Electric Dipole: the oxygen side of the molecule. Thus, the positive and negative charges of the water A pair of equal and opposite charges molecule are inherently separated by a small separated by a finite distance is called an electric distance. This separation of positive and dipole. It is shown in Fig. (10.18). negative charges gives rise to the permanent dipole moment of a water molecule. Fig. 10.18: Electric dipole. x-y axial line, Molecules of water, P-Q equatorial line. ammonia, sulphur di- oxide, sodium chloride Line joining the two charges is called the etc. have an inherent dipole axis. A line passing through the dipole separation of centers of positive and negative charges. Such molecules are called axis is called axial line. A line passing through polar molecules. the centre of the dipole and perpendicular to the Polar molecules are the molecules in which the center of positive charge and axial line is called the equatorial line as shown the negative charge is naturally separated. in Fig. 10.18. Molecules such as H2, Cl2, CO2 CH4 and many others have their positive and negative Strength of a dipole is measured in terms charges effectively centered at the same point and are called non-polar molecules. of a quantity called the dipole moment. Let q be the magnitude of each charge and 2l be Non-polar molecules are the the distance from negative c(h2al r)geis to positive molecules in which the center of positive charge. Then the product q called the charge and the negative charge is one and the same. They do not have a permanent electric dipole. When an external electric field is applied to such molecules the centers of positive and negative charge are displaced and a dipole is induced. 201
6.172×10-30 Cm or 1.85 D. Its direction is from Example 10.9: An electric dipole of length 2.0 oxygen to hydrogen. See box on Natural dipole. cm is placed with its axis making an angle of 30° with a uniform electric field of 105 N/C. as 10.9.1 Couple Acting on an Electric Dipole in shown in figure. If it experiences a torque of a Uniform Electric Field: Consider an electric dipole placed in a 10 3 Nm, calculate the magnitude of charge on the dipole. uniform electric field E. The axis of electric dipole makes an angle θ with the direction of electric field as shown in Fig. 10.19 a. P Solution: Given W 10 3, Nm, E 105 N / C, Fig. 10.19 (a): Dipole in uniform electric field. 2l 2.0 u102 m, T 300 W qE2l sinT 10 3Nm q105 N / C 2.0 u102 m § 1 · ¨© 2 ¹¸ ? q = 10 3 3 u102 C 103 = 1.73u102 C Fig. 10.19 (b): Couple acting on a dipole. 10.9.2 Electric Intensity at a Point due to an Figure 10.19. b shows the couple acting on Electric Dipole: an electric dipole in uniform electric field. Case 1 : At a point on the axis of a dipole. The force acting on charge - q at A is Consider an electric dipole consisting of F A = - qE in the direction opposite to that two charges -q and +q separated by a distance 2l as shown in Fig. 10.20. Let P be a point at a of E and the force acting on charge +q at B distance r from the centre C of the dipole. The is F B = + qE in the direction of E . Since F A electric intensity Ea at P due to the dipole is the vector sum of the field due to the charge - q at = - F B , the two equal and opposite forces A and + q at B. separated by a distance form a couple. Moment of the couple is called torque and is defined by W d u F where, d is the perpendicular distance b..e.MtwaegennittuhdeetwofoTeoqruqaulea=nd opposite forces. Magnitude of force × Perpendicular distance ... Torque on the dipole = W BP u qE Fig. 10.20: Electric field of a dipole along ... τ = qE2lsinθ its axis. --- (10.20) Electric field intensity at P due to the b..u. tτ p q u 2l --- (10.21) charge -q at A = pEsinθ 1 ( +ql))2PPQPDD In vector form W p u E --- (10.22) = EA 4SH 0 ( r If θ =90° sin θ =1, then τ = pE When the axis of electric dipole is where PD is unit vector directed along PD perpendicular to uniform electric field, torque Electric intensity at P due to charge +q at B of the couple acting on the electric dipole is 1 q 4SH 0 -l maximum, i.e., τ = pE. It θ = 0 then τ = 0, this is = EB ( r ) 2PPPQQQ the minimum torque on the dipole. Torque tends to align the dipole axis along the direction of where PQ is a unit vector directed along PQ . electric field. 202
The magnitude of EB is greater than that of where PQ is the unit vector directed along PQ E A . (Because BP < AB) or BP Electric field at PE Aisatnhde,EsuB m of E A and E B Resultant field Ea at P on the axis, due to the dipole is ∴ Eeq = E A + E B Ea = EB + EA Consider ∆ ACP The magnitude of Ea is given by (AP)2 = (PC)2 + (AC)2 = r2 + l2 = (BP)2 | Ea | 1 ªq q º ? EA 1q --- (10.25) 4SH 0 «¬ (r l)2 (r l)2 »¼ 4SH0 (r 2 l 2 ) --- (10.26) | Ea |= q ª r2 + l2 2lr - r2 2lr l2 º EB 1 q 4SH 0 « (r2 l 2 )2 » 4SH0 (r2 l 2 ) ¬ ¼ | Ea |= 2(2lq)r EA = EB 4SH0 (r 2 l 2 )2 The resultant of fields E A and EB acting But 2lq = p, the dipole moment at point P can be calculated by resolving these | Ea |= 1 2 pr vectors E A and EB along the equatorial line 4SH 0 (r2 l 2 )2 and along a direction perpendicular to it. --- (10.23) Ea , is directed along PQ, which is the direction of the dipole moment p i.e. from the Fig. 10.21 (b): Components of negative to the positive charge, parallel to the the field at point P. axis. If r>> l, l2 can be neglected compared to r2, Ea = 1 2p --- (10.24) 4SH 0 r3 The field will be along the direction of the Consider Fig. 10.21 (b). Let the y-axis coincide with the equator of the dipole x-axis dipole moment p . will be parallel to dipole axis, as shown. The Case 2: At a point on the equatorial line. As origin is at point P. shown in Fig. 10.21 (a) The y-components of EA and EB are EAsinθ Fig. 10.21 (a): and EBsinθ respectively. They are equal in magnitude but opposite in direction and cancel r Electric field of a each other. There is no contribution from them dipole at a point on towards the resultant. the equatorial line. The x-components of EA and EB are EAcosθ and EBcosθ respectively. They are of equal Electric field at point P due to charge -q at magnitude and are in the same direction A is: E A 1 (q) PA ∴| Eeq | EA cosT EB cosT --- (10.27) 4SH 0 (A P)2 By using Eq. 10.25 and 10.26 where PA is the unit vector direction along PA . Eeq = 2EAcosT Similarly, Electric field at P due to charge + q at 2 § 4SH q l 2 ) · l ¨ (r 2 ¸ r2 l2 1 (q) © 0 ¹ 4SH 0 (BP)2 B is: E B PQ 2ql 4SH0 (r2 l 2 )3/2 203
If r>>l then 2 is very small compared to r2 (b) Surface charge density (σ) l Eeq 1p 1p --- (10.28) Suppose a charge q is uniformly distributed 4SH0 (r2 )3/2 4SH0 r3 over a surface of area A . As shown in Fig. 10.23, The direction of this field is along - p (anti- then the surface charge density σ is defined as parallel to p ) as shown in Fig. 10.21 (c). V q --- (10.30) A Fig. 10.21 (c): SI unit of σ is (C/m2) Electric field at point P is anti- For example, charge distributed uniformly on a thin disc or a synthetic cloth. If the charge parallel to p . is not distributed uniformally over the surface of a conductor, then charge dq on small area Comparing Eq. 10.28 and 10.24 we find element dA can be written as dq = σ dA. that the electric intensity at an axial point is twice that at a point on the equatorial position, lying at the same distance from the centre of the dipole. 10.10 Continuous Charge Distribution: A system of charges can be considered as Fig. 10.23: Surface charge. a continuous charge distribution, if the charges (c) Volume charge density (ρ) are located very close together, compared to their distances from the point where the Suppose a charge q is uniformly distributed intensity of electric field is to be found out. throughout a volume V, then the volume charge The charge distribution is continuous density ρ is defined as the charge per unit in the sense that, a system of closely spaced charges is equivalent to a total charge which volume. is continuously distributed along a line or a surface or a volume. To find the electric field q --- (10.31) due to continuous charge distribution, we define U following terms for different types of charge distribution. V S.I. unit of ρ is (C/m3) For example, charge on a solid plastic sphere or a solid plastic cube. (a) Linear charge density (λ). If the charge is not distributed uniformaly over the volume of a material, then charge dq As shown in Fig. 10.22 charge q is over small volume element dV can be written uniformly distributed along a liner conductor of as dq = ρ dV. length l. The linear charge density λ is defined as, O q --- (10.29) Fig. 10.24:Volume charge. l Electric field due to a continuous charge SI unit of λ is (C/m). distribution can be calculated by adding electric fields due to all these small charges. For example, charge distributed uniformly Can you tell? on a straight thin rod or a thin nylon thread. If The surface charge density of Earth is the charge is not distributed uniformly over the σ = -1.33 nC/m2. That is about 8.3×109 electrons per square meter. If that is the length of thin conductor then charge dq on small case why don't we feel it? element of length dl can be written as dq = λdl Fig. 10.22: Linear charge. 204
Do you know ? iii. One can get static shock if charge transferred is large. Static charge can be useful Static charges can be created whenever iv. Dust or dirt particles gathered on computer or TV screens can catch static there is a friction between an insulator and charges and can be troublesome. other object. For example, when an insulator like rubber or ebonite is rubbed against Precautions against static charge a cloth, the friction between them causes i. Home appliances should be grounded. electrons to be transferred from one to the ii. Avoid using rubber soled footwear. other. This property of insulators is used iii. Keep your surroundings humid. (dry air in many applications such as Photocopier, Inkjet printer, Panting metal panels, can retain static charges). Electrostatic precipitation/separators etc. Static charge can be harmful Internet my friend i. When charge transferred from one body 1. h t t p s : / / w w w. p h y s i c s c l a s s r o o m . to other is very large sparking can take com>class place. For example lightning in sky. ii. Sparking can be dangerous while 2. https://courses.lumenlearing.com>elect refuelling your vehicle. 3. https://www.khanacademy.org>science. 4. https://www.topper.com>guides>physics ExercisesExercises 1. Choose the correct option. (A) 1:4 (B) 1:2 i. A positively charged glass rod is brought (C) 1:1 (D) 1:16 close to a metallic rod isolated from iv. Two charges of 1.0 C each are placed ground. The charge on the side of the one meter apart in free space. The force metallic rod away from the glass rod will between them will be be (A) 1.0 N (B) 9×109 N (A) same as that on the glass rod and equal (C) 9×10-9 N (D) 10 N in quantity v. Two point charges of +5 µC are so (B) opposite to that on the glass of and placed that they experience a force of equal in quantity 80×10-3 N. They are then moved apart, (C) same as that on the glass rod but lesser so that the force is now 2.0×10-3 N. The in quantity distance between them is now (D) same as that on the glass rod but more (A) 1/4 the previous distance in quantity (B) double the previous distance ii. An electron is placed between two (C) four times the previous distance parallel plates connected to a battery. If (D) half the previous distance the battery is switched on, the electron vi. A metallic sphere A isolated from ground will is charged to +50 µC. This sphere is (A) be attracted to the +ve plate brought in contact with other isolated (B) be attracted to the -ve plate metallic sphere B of half the radius of (C) remain stationary sphere A. The charge on the two sphere (D) will move parallel to the plates will be now in the ratio iii. A charge of + 7 µC is placed at the centre (A) 1:2 of two concentric spheres with radius 2.0 (C) 4:1 (B) 2:1 (D) 1:1 cm and 4.0 cm respectively. The ratio of vii. Which of the following produces uniform the flux through them will be electric field? 205
(A) point charge (B) linear charge (C) two parallel plates iii. Four charges of + 6×10-8 C each are (D) charge distributed an circular any placed at the corners of a square whose viii. Two point charges of A = +5.0 µC and B = -5.0 µC are separated by 5.0 cm. sides a are 3 cm each. Calculate the A point charge C = 1.0 µC is placed at 3.0 cm away from the centre on the resultant force on each charge and show perpendicular bisector of the line joining the two point charges. The charge at C its direction on a digram drawn to scale. will experience a force directed towards [Ans: 6.89×10-2 N] (A) point A (B) point B iv. The electric field in a region is given by (C) a direction parallel to line AB (D) a direction along the perpendicular E = 5.0 kN/C. Calculate the electric flux bisector Through a square of side 10.0 cm in the 2. Answer the following questions. following cases i. What is the magnitude of charge on an (a) the square is along the XY plane electron? ii. State the law of conservation of charge. [Ans: = 5.0×10-2 Vm] iii. Define a unit charge. iv. Two parallel plates have a potential (b) The square is along XZ plane difference of 10V between them. If the [Ans: Zero] plates are 0.5 mm apart, what will be the strength of electric charge. (c) The normal to the square makes an v. What is uniform electric field? vi. If two lines of force intersect of one angle of 450 with the Z axis. point. What does it mean? vii. State the units of linear charge density. [Ans: 3.5×10-2 Vm] viii. What is the unit of dipole moment? ix. What is relative permittivity? v. Three equal charges of 10×10-8 C respectively, each located at the corners of a right triangle whose sides are 15 cm, 20 cm and 25 cm respectively. Find the force exerted on the charge located at the 90° angle. [Ans: 4.59.×10-3 N] vi. A potential difference of 5000 volt is applied between two parallel plates 5 cm a part a small oil drop having a charge of 9.6 ×10-19 C falls between the plates. Find (a) electric field intensity between the plates and (b) the force on the oil drop. [Ans: (a) 1.0.×105 N/C 3. Solve numerical examples. (b) 9.6.×10-14 N] i. Two small spheres 18 cm apart have vii. Calculate the electric field due to a charge equal negative charges and repel each of -8.0×10-8 C at a distance of 5.0 cm other with the force of 6×10-3 N. Find the from it. total charge on both spheres. [Ans: -2.88×10-2 N/C] [Ans: q = 2.938×10-7 C] *** ii. A charge +q exerts a force of magnitude -0.2 N on another charge -2q. If they are separated by 25.0 cm, determine the value of q. [Ans: q = 0.8333 µC] 206
11. Electric Current Through Conductors Can you recall? 1. Do you recall that the flow of charged particles in a conductor constitutes a current? 2. An electric current in a metallic conductor such as a wire is due to flow of electrons, the negatively charged particles in the wire. 3. What is the role of the valence electrons which are the outermost electrons of an atom? 11.1 Introduction: the plane P from time t to t + ∆t, i.e. during the time interval ∆t. Then the current is given by The valence electrons become de-localized when a large number of atoms come together I (l) = lim 'q --- (11.2) in a metal. These are the conduction electrons Here, the 'ctouror'ent t is expressed as the limit or free electrons constituting an electric current when a potential difference is applied across the of the ratio ∆q/ ∆t as ∆t tends to zero. conductor. The current during lightening could be 11.2 Electric current : as high as 10,000 A, while the current in the house hold circuit could be of the order of a few Consider an imaginary gas of both amperes. Currents of the order of a milliampere negatively and positively charged particles. (mA), a microampere (µA) or a nanoampere Fig. 11.1 shows the negatively and positively (nA) are common in semiconductor devices. charged particles flowing randomly in various directions across a plane P. In a time interval 11.3 Flow of current through a conductor : t, let the amount of positive charge flowing in the forward direction be q+ and the amount of A current can be generated by positively negative charge flowing in the forward direction or negatively charged particles. In an be q-. electrolyte, both positively and negatively charged particles take part in the conduction. In a metal, the free electrons are responsible for conduction. These electrons flow and generate a net current under the action of an applied electric field. As long as a steady field exists, the electrons continue to flow in the form of a steady current. Such steady electric fields are generated by cells and batteries. Fig. 11.1: Flow of charged particles. Do you know ? Thus the net charge flowing in the forward Sign convention : The direction of the direction is q = q+- q-. For a steady flow, this current in a circuit is drawn in the direction quantity is proportional to the time t. The ratio in which positively charged particles would S qtI uisnidteo fifntehdeacsIurt=hreeqntct uirsreanmt pIe. re (A), t-h-a- t(1o1f.t1h)e move, even if the current is constituted by charge and time is coulomb (C) and second (s) the negatively charged particles, electrons, respectivly. which move in the direction opposite to that Let I be the current varying with time. Let the electric field. We use this as a convention. ∆q be the amount of net charge flowing across 11.4 Drift speed : Imagine a copper rod with no current flowing through it. Fig 11.2 shows the schematic of a conductor with the free electrons 207
in random motion. There is no net motion of t = VLd --- (11.4) these electrons in any direction. If electric field is applied along the length of the copper rod, From the Eq. (11.1), and Eq. (11.3), the current and a current is set up in the rod, these electrons still move randomly, but tend to 'drift' in a I= q = n A Le = n AVd e --- (11.5) particular direction. Their direction is opposite Hence t L / Vd to that of the applied electric field. I J Direction of electric field : Direction of an Vd = nAe = ne --- (11.6) electric field at a point is the direction of the where J = I/A is current density. J is uniform force on the test charge placed at that point. over the cross sectional area A of the wire. Its The electrons under the action of the unit is A/m2 applied electric field drift with a drift speed tVhde. Here, J = I --- (11.7) The drift speed in a copper conductor is of A order of 10-4 m/s-10-5 m/s, whereas the electron From Eq. (11.6), random speed is of the order of 106 m/s. J =(ne)V d --- (11.8) Fig. 11.2: Free electrons in random motion For electrons, ne is negative and J and V d inside the conductor. have opposite directions, V d is the drift velocity. How is the current through a conductor Example 11.1: A metallic wire of diameter related to the drift speed of electrons? Figure 0.02m contains 1028 free electrons per cubic 11.3 shows a part of conducting wire with its meter. Find the drift velocity for free electrons, free electrons having the drift speed Vd in the having an electric current of 100 amperes direction opposite to the electric field E . flowing through the wire. (Given : charge on electron = 1.6 × 10-19C) Solution: Given e = 1.6 × 10-19 C n = 1028 electrons/m3 D = 0.02m r = D/2 = 0.01m I = 100 A Vd = J = n I ne Ae where A is the cross sectional area of the wire. Fig. 11.3: Conducting wire with the applied A = πr2 = 3.142 × (0.01)2 electric field. = 3.142 × 10-4m2 It is assumed that all the electron move with 100 the same drift speed Vthdeacnrdosthsaste,ctthioencu(Arr)eonft I is Vd = 3.142 u10-4 u10281.6 u 10-19 the same throughout the 10249 wire. Consider the length L of the wire. Let n be 5.027 the number of free electrons per unit volume of the wire. Then the total number of electrons in Vd = 10-3 × 0.1989 = 1.9 × 10-4 m/s the length L of the conducting wire is nAL. The Example 11.2: A copper wire of radius 0.6 mm carries a current of 1A. Assuming the current to total charge in the length L is, be uniformly distributed over a cross sectional area, find the magnitude of current density. q = n A L e --- (11.3) where e is the electron charge. This is total charge that moves through any cross section of the wire in a certain time interval t, 208
Solution: Given Reciprocal of resistance is called conductance. r = 0.6 mm = 0.6 ×10-3 m I = 1A C = 1 --- (11.10) J=? R Area of copper wire = πr2 The unit of conductance is siemens or (Ω)-1 = 3.142 × (0.6)2 × 10-6 Example 11.3: A Flashlight uses two 1.5V batteries to provide a steady current of 0.5A in = 3.142 × 0.36 × 10-6 the filament. Determine the resistance of the glowing filament. = 1.1311 × 10-6 m2 J = I 1 Solution: A 1.1311u106 V 3 J = 0.884 × 106 A/m2 R = I 0.5 6.0: 11.5 Ohm’s law : The relationship between the current ∴ Resistance of the glowing filament is 6.0 Ω. through a conductor and applied potential difference was first discovered by German Physical origin of Ohm’s law : scientist George Simon Ohm in 1828 AD. This relationship is known as Ohm’s law. We know that electrical conduction in a conductor is due to mobile charge carriers, the It states that “The current I through electrons. It is assumed that these conduction a conductor is directly proportional to the electrons are free to move inside the volume potential difference V applied across its two of the conductor. During their random motion, ends provided the physical state of the conductor electrons collide with the ion cores within the is unchanged”. conductor. It is assumed that electrons do not collide with each other. These random motions The graph of current versus potential average to zero. On the application of an difference across the conductor is a straight line electric field E , the motion of the electrons is a as shown in Fig. 11.4 combination of the random motion of electrons due to collisions and that due to the electric field Ideal Ohm's law E . The electrons drift under the action of the Fig. 11.4: I-V curve for a conductor. field E and move in a direction opposite to the direction of the field E . Consider an electron of mass m subjected to an electric field E . The force experienced by the electron will be F = e E . The acceleration experienced by the electron will then be In general, I ∝ V a = eE --- (11.11) m or V= I R or R =V , --- (11.9) I where R is a proportionality constant and is The type of collision the conduction electrons undergo is such that the drift velocity called the resistance of the conductor. The unit attained before the collision has nothing to do with the drift velocity after the collision. After of resistance is ohm (Ω), the collision, the electron will move in random direction, but will still drift in the direction 1Ω = 1volt 1ampere If potential difference of 1volt across opposite to E . a conductor produces a current of 1ampere Let τ be the average time between two successive collisions. Thus on an average, the through it, then the resistance of the conductor is 1Ω. 209
electrons will acquire a drift speed Vd = aτ , vacuum tubes, junction diodes, thermistors etc. where a is the acceleration given by Eq (11.11). Resistance R for such non-linear devices at a particular value of the potential difference V is Also, at any given instant of time, the average given by, drift speed of the electron will also be Vd = aτ . R = lim 'V = dV From Eq. (11.11), 'Io0 'I dI --- (11.16) eEW Vd =aW m --- (11.12) where ∆V is the potential difference between the two values of potential From the Eq. (11.6) and Eq. (11.12), Vd =J = eEτ --- (11.13) V 'V to V+ 'V , ne m 2 2 which gives and ∆I is the corresponding change in the § m · current. ¨© e 2 nW ¸¹ E = J --- (11.14) 11.7 Electrical Energy and Power: or, E = ρJ, where ρ is the resistivity of the Consider a resistor AB connected to a material and cell in a circuit shown in Fig. 11.6 with current m flowing from A to B. The cell maintains a ne2W U --- (11.15) potential difference V between the two terminals of the resistor, higher potential at A and lower For a given material, m, n, e2 and τ will at B. Let Q be the charge flowing in time ∆t be constant and ρ will also be constant, ρ through the resistor from A to B. The potential is independent of E , the externally applied difference V between the two points A and B, is electric field. equal to the amount of work W, done to carry a 11.6 Limitations of the Ohm’s law: unit positive charge from A to B. It is given by Ohm’s law is obeyed by various materials V = W , W = VQ --- (11.17) and devices. The devices for which potential Q difference (V) versus current (I) curve is a straight line passing through origin, inclined to V-axis, are called linear devices or ohmic devices (Fig. 11.4). Resistance of these devices is constant. Several conductors obey the Ohms law. They follow the linear I-V characteristic. Fig. 11.6: A simple circuit with a cell and a resistor. The cell provides this energy through the charge Q, to the resistor AB where the work is Fig. 11.5: I-V curve for non-Ohmic performed. When the charge Q flows from the devices. higher potential point A to the lower potential point B, i.e. through a decrease in potential of The devices for which the I-V curve is not value V, its potential energy decreases by an a straight line as shown in Fig. 11.5 are called amount non-ohmic devices. They do not obey the Ohm’s law and the resistance of these devices ∆U = QV = I ∆tV --- (11.18) is a function of V or I; e.g. liquid electrolytes, where I is current due to the charge Q flowing in time ∆t. Where will this energy go? By the principle of conservation of energy, it is 210
converted into some other form of energy. 11.8 Resistors: In the limit as ∆t 0, Resistors are used to limit the current following through a particular path of a circuit. dU --- (11.19) Commercially available resistors are mainly of dt = I.V two types : Carbon resistors and Wire wound Here, dU the time rate of transfer of resistors. High value resistors are mostly carbon energy datndisispgoiwveenr, by, resistors. They are small and inexpensive. The values of these resistors are colour coded to dU mark their values in ohms. The colour coding P = dt = I.V --- (11.20) is standardized by Electronic Industries Association (EIA). One such resistor is shown We can also say that this power is in Fig. 11.7. transferred by the cell to the resistor or any other device in place of the resistor, such as a motor, a rechargeable battery etc. Because of the presence of an electric Fig. 11.7: Carbon composition resistor. field, the free electrons move across a resistor Colour code: and there would be an increase in their kinetic energy as they move. When the electrons collide Colours 1st 2nd Multiplier Tolerance with the ion cores the energy gained by them is shared among the ion cores. Consequently, digit digit vibrations of the ions increase, resulting in heating up of the resistor. Thus, some amount Black 00 ×100 of energy is dissipated in the form of heat in a resistor. The energy dissipated in time interval Brown 1 1 ×101 ±1% ∆t is given by Eq. (11.18). The energy dissipated per unit time is actually the power dissipated Red 22 ×102 ±2% and is given by Eq. (11.20). Using Eq. (11.20), and using Ohm’s law, V=IR, Orange 3 3 ×103 ∴P =V 2 =I2R --- (11.21) Yellow 4 4 ×104 R Green 5 5 ×105 It is the power dissipation across a resistor Blue 66 ×106 which is responsible for heating it up. For Violet 7 7 ×107 example, the filament of an electric bulb heats Gray 88 ×108 up to incandescence, radiating out heat and White 9 9 ×109 light. For Gold ×10-1 ±5% Example 11.4 : An electric heater takes 6A For Silver × 10-2 ±10% current from a 230V supply line, calculate the power of the heater and electric energy No colour - ±20% consumed by it in 5 hours. Easy Bytes: Solution : Given I = 6A, V = 230V Finding it difficult to memorize the colour We know that, code sequence? No need to worry, we have a P = I × V = (6A) (230V) = 1380 W one liner which will help you out “B. B. Roy in P = 1.38 kW Great Britain has Very Good Wife” Energy consumed = Power × time = (1.38 kW) × (5 h) B B R O Y G B V GW = 6.90kWh (1.0 Kwh = 1 unit of power) = 6.9 units of electrical energy. This funny one liner makes it easy to recall the sequence of digits and multipliers. 211
In the four band resistor colour code Because of series combination, the supply illustrated in the above table, the first three voltage between two resistors R1 and R2 is V1 bands (closest together) indicate the value in adfRlni2ov.dwiid.seVe.d2t,hinarronseudesgrptihheecestthiccvueoermlrryeebnsiiantsnarteodtimrotnaRh,ie1nssausnptahdpmelyetshavecmourlerterasiegninsettaoilsrIl ohms. The first two bands indicate two numbers the resistors. and third band often called decimal multiplier. The fourth band separated by a space from the three value bands, (so that you know which end to start reading from), indicates tolerance of the resistor. Example i. Colour code of resistor is Yellow Violet Orange Gold Value : 4 7 103 ±5% i.e. 47×103 = 47000Ω = 47kΩ ±5% The value of the resistor is 47kΩ ±5% Fig. 11.9: Series combination of two ii. From given values of resistor; find the colour resistors R1 and R2. bands of this resistor According to Ohm’s law, 330Ω = 33×10 3 3 101 R 1 = V1 , R2 = V2 --- (11.22) I I Orange Orange Brown tolerance band 11.8.1 Rheostat: Total voltage V=V1+V2 --- (11.23) A rheostat shown in Fig. 11.8 is an From equation .... (11.22) and (11.23) adjustable resistor used in applications that require adjustment of current or resistance in we write an electric circuit. The rheostat can be used to adjust potential difference between two points V = II(RR1 +R2 ) --- (11.24) in a circuit, change the intensity of lights and ∴V = --- (11.25) control the speed of motors, etc. Its resistive s element can be a metal wire or a ribbon, carbon films or a conducting liquid, depending upon Thus the equivalent resistance of the series the application. In hi-fi equipment, rheostats are used for volume control. circuit Rs = R1+R2 When a number of resistors are connected in series, the equivalent resistance is equal to the sum of individual resistances. For n number of resistors, i =n ∑Rs = R1+ R2 + R2+...........+Rn= Ri -- (11.26) i =1 II. Parallel Combination of Resistors: In the parallel combination, the resistors are connected in such a way that the same Fig . 11.8: Rheostat. voltage is applied across each resistor. 11.8.2 Combination of Resistors: A number of resistors are said to be connected in parallel if all of them are connected I. Series combination of Resistors: between the same two electrical points each having individual path as shown in Fig. 11.10. In series combination of resistors, these are connected in single electrical path as shown in In parallel combination the total current I Fig 11.9. Hence the same electric current flows is divided into I1 and I2 as shown in the circuit through each resistor in a series combination. 212
diagram Fig.11.10, whereas voltage V across Example 11.5: Calculate i) total resistance and them remains the same, ii) total current in the following circuit. R1 = 3Ω, R2 = 6Ω, R3 = 5Ω, V = 14V Fig. 11.10 : Two resistors in parallel Circuit diagram combination. Solution: I = I1+ I2 -- (11.27) i) Total resistance = R = R +R T P3 wcuhrererentIf1lioswciunrgretnhtrofulogwhiRng2. through R1 and I2 is R1 R2 3u6 RP = R1 + R2 9 2: When Ohm’s law is applied to R RToT=ta2l + 5 = 7Ω = 7Ω 1 Resistance V= I1R1 i.e. I1 = V --- (11.28a) ii) Total current : law applied to R2 R1 I = V 14V Ohm’s RT 7: V =I2R2 i.e. I2 =V --- (11.28b) R2 I = 2A 11.9 Specific Resistance (Resistivity): From Eq. (11.27) and Eq. (11.28), ∴ I =V +V , At a particular temperature, the resistance R1 R2 of a given conductor is observed to depend on the nature of material of conductor, the area of If, I = V , its cross-section, and its length. Rp It is found that resistance R of a conductor V =V +V , of uniform cross section is Rp R1 R2 i. directly proportional to its length l, i.e. R∝ l ∴ 1 = 1 + 1 , --- (11.29) ii. inversely proportional to its area of Rp R1 R2 cross section A, wcohmerbeinRaptiiosnth. e equivalent resistance in parallel i.e. R∝ l A If n irnespisatroarlslelR, t1h, eRe2q,uiRv3a..l.e.n..t...r,esRisntanacree connected From i and ii of the combination is given by R =ρ l --- (11.31) A 1 +1 +............ 1 n 1 R1 R2 Rn i =1 R ∑1= +1 = --(11.30) where ρ is a constant of proportionality and R3 Rp it is called specific resistance or resistivity Thus when a number of resistors are of the material of the conductor at a given connected in parallel, the reciprocal of the temperature. equivalent resistance is equal to the sum of the From Eq. (11.31), we write reciprocals of individual resistances. U RA --- (11.32) l 213
SI unit of resistivity is ohm-meter. Conductivity : Reciprocal of resistivity is Resistivity of a conductor is numerically the called conductivity of a material, σ = (ρ)-1. resistance per unit length, and per unit area of cross-section of material of the conductor. SI unit of σ is: (Ωm)-1 i.e. siemens/meter (Sm-1) i.e. when, R = 1Ω, A =1m2 and l = 1m, then, ρ =1Ωm Table 11.1 : Resistivity of various materials Material Resistivity ρ Material Resistivity ρ (Ω.m) (Ω.m) Conductors Semiconductors Silver 1.59 × 10-8 Carbon 3.5 × 10-5 Copper 1.72 × 10-8 Germanium 0.5 Gold 2.44 × 10-8 Silicon 3 × 104 Aluminium 2.82 × 10-8 Tungsten 5.6 × 10-8 Insulators 1011-1013 Iron 9.7 × 10-8 Glass 1011-1015 Mercury 95.8 × 10-8 Mica 1013-1016 Nichrome (alloy) 100 × 10-8 Rubber (hard) 1016 Teflon 3 × 108 Wood (maple) Example 11.6: Calculate the resistance per Again, the SI unit of ρ is metre, at room temperature, of a constantan unit(E ) = V/m = V m=Ω.m (alloy) wire of diameter 1.25mm. The resistivity unit(J ) A/m 2 A of constantan at room temperature is 5.0 × 10-7 Ωm. In terms of conductivity σ of a material, from Solution: ρ = 5.0 × 10-7 Ωm d = 1.25 × 10-3 m (11.33), r = .625 × 10-3 m J 1 E VE --- (11.34) U Cross-sReRceetsisoiissnttaiavlniActyerepUaer=mπRerltA2er = Rl tFhoer a particular resistor, we had (Eq. 11.9) given by resistance R R = V I i.e. RU 5 u 107 Compare this with the above Eq (11.33). l A (0.625u103 )2 u 3.142 11.10 Va riation of Resistance with R l 0.41 :m1 Temperature: Resistivity of a material varies with temperature. It is a property of material. Fig. 11.11 shows the variation of resistivity of ∴ Resistance per metre= 0.41 Ωm-1 copper as a function of temperature (K). It can be seen that the variation is linear over a certain Resistivity ρ is a property of a material, range of temperatures. Such a linear relation can be expressed as, while the resistance R refers to a particular object. Similarly, the electric field E at a point is specified in a material with the potential difference across the resistance, and the current ρ = ρ0[1+ α(T - T0)], --- (11.35) density J in a material instead of the current I where T0 is the chosen reference temperature and in the resistor. Then for an isotropic material, ρfo0rienxtahme prlees,isTtoivciatyn at the chosen temperature, be 0 oC. U E or E U J ---- (11.33) J 214
Fig. 11.11: Resistivity as a function of R = 2.5 × 1.32 temperature (K). T In the above Eq. (11.35), RT = 3.3Ω D U U0 R - R0 --- (11.36) Superconductivity : U0 (T -T0 ) R0 (T -T0 ) We know that the resistivity of a metal decreases as the temperature decreases. Here, α is called the temperature coefficient In case of some metals and metal alloys, the resistivity suddenly drops to zero at a of resistivity. Table (11.1) shows the resistivity of particular temperature (Tc). This temperature is called critical temperature, for example, some of the metals. The temperature coefficient mercury loses its resistance completely to zero at 4.2K. of resistance is defined as the increase in Superconductivity can be harnessed so as to be useful for mankind. It is already in use in resistance per unit original resistance at the obtaining very high magnetic field (a few Tesla) in superconducting magnet. These chosen reference temperature, per degree rise in magnets are used in research quality NMR spectrometers. For its operation, the current temperature. The unit of α is oC-1 or oK-1 (per carrying coils are required to be kept at a temperature less than the critical temperature degree celcius or per degree kelvin). of the coil material. From Eq. (11.36) 11.11 Electromotive Force (emf): When charges flow through a conductor, R = R [1+ α (T - T )] --- (11.37) 00 a potential difference has to be established between the two ends of the conductor. For a For small difference in temperatures, steady flow of charges, this potential difference is required to be maintained across the two ends 1 dR of the conductor, the terminals. There is a device D R0 dT --- (11.38) that does so by doing work on the charges, thereby maintaining the potential difference. Do you know ? Such a device is called an emf device and it provides the emf ε. The charges move in the Here, the temperature difference is more conductor owing to the energy provided by the important than the temperature alone. emf device. The device supplies this energy Therefore, as the sizes of degrees on the through the work it does. Celsius scale and the Absolute scale are identical, any scale can be used. You must have used some of these emf Example 11.7: A piece of platinum wire has devices. Power cells, batteries,Solar cells, fuel resistance of 2.5 Ω at 0o C. If its temperature cells, and even generators, are some examples coefficient of resistance is 4 ×10-3/oC. Find the of emf devices familiar to you. resistance of the wire at 80o C. Solution: Fig. 11.12: Circuit with emf device. R0= 2.5 Ω α = 0.004/oC T - 0 = T = 80o C RT = R0(1+ αT ) RT = 2.5 (1+ 0.004 × 80) = 2.5 (1 + 0.32) 215
Fig. 11.12 shows a circuit with an emf from the potential difference across its two device and a resistor R. Here, the emf device terminals (V). keeps the positive terminal (+) at a higher electric potential than the negative terminal (-). V= ε - (I) (r) --- (11.40) The negative sign is due to the fact that the The emf is represented by an arrow from current I flows through the emf device from the the negative terminal to the positive terminal of a device such as a Voltaic cell. When the circuit negative terminal to the positive terminal. is open, there is no net flow of charge carriers within the device. When connected in a circuit, By the application of Ohm’s law Eq. (11.9), there is a flow of carriers from one terminal to the other terminal inside the emf device. V = IR The positive charge carriers move towards the positive terminal which acts as cathode inside Hence IR = ε - Ir --- (11.41) the emf device. Thus the positive charge carriers move from the region of lower potential energy, Or to the region of higher potential energy which is cathode inside the emf device. Here, the energy I = ε --- (11.42) source is chemical in nature. In a Solar cell, it is R +r the photon energy in the Solar radiation. Thus, the maximum current that can be drawn from the emf device is when R = 0, i.e. Imax = ε --- (11.43) r This is the maximum allowed current from an emf device (or a cell). This decides the Now suppose that a charge dq flows maximum current rating of a cell or a battery. through the cross section of the circuit (Fig. 11.12), in time dt. 11.12 Cells in Series: It is clear that the same amount of charge In a series combination, cells are connected in single electrical path, such that the positive dq flows throughout the circuit, including the terminal of one cell is connected to the negative terminal of the next cell, and so on. The terminal emf device. It enters the negative terminal (low voltage of battery/cell is equal to the sum of voltages of individual cells in series, as shown potential terminal) and leaves the positive in Fig 11.13 a. terminal (higher potential terminal). Hence, Figure shows two 1.5V cells in series. This combination provides total voltage of 3.0V the device must do work dw on the charge dq, (1.5×2). so that it moves in the above manner. Thus we define the emf of the emf device. ε = dw --- (11.39) dq The SI unit of emf is joule/coulomb (J/C). In an ideal device, there is no internal Fig. 11.13 (a): Cells in parallel. resistance to the motion of charge carriers. The emf of the device is then equal to the potential difference across the two terminals of the device. In a real emf device, there is an internal resistance to the motion of charge carriers. Fig. 11.13 (b): Cells in parallel. If such a device is not connected in a circuit, The equivalent emf of n number of cells there is no current through it. In that case the in series combination is the algebraic sum of emf is equal to the potential difference across their individual emf. The equivalent internal the two terminals of the emf device connected resistance of n cells in a series combination is in a circuit, there is no current through it. If a the sum of their individual internal resistance. current (I) flows through an emf device, there is V = I . ri --- (11.44) an internal resistance (r) and the emf (ε) differs ¦ H i ¦ i i 216
• Advantages of cells in series. Considering Eq. (11.49) and Eq. (11.50) we can (i) The cells connected in series produce a write, larger resultant voltage. H = H1r2 H 2 r1 r1 +r2 (ii) Cells which are damaged can be easily eq identified, hence can be easily replaced. r1r2 r1 +r2 11.13 Cells in parallel: req i.e. 1 11 Consider two cells which are connected req in parallel. Here, positive terminals of all the r1 r2 cells are connected together and the negative terminals of all the cells are connected together. Heq H1 H 2 In parallel connection, the current is divided --- (11.51) aFmigo. n1g1.1th3eb.bCraonncshiedseri.peo. iIn1tsanBd1 Ia2nadsBs2hohwavninign req r1 r2 potenFtoiarlsthVeB1fairnsdt cVeBl2l, respectively. For n number of cells connected in parallel with the potential difference emf ε1, ε2, ε3, ......, εn and internal resistance r1, r2, r3, ......, rn across its terminals is, 1 1 1 1 1 V = VB1 - VB2 = ε1 - I1r1 --- (11.45) req r1 r2 r3 ........... rn --- (11.52) ? I1 = H1 -V --- (11.46) and H eq H1 H2 ............. Hn Point B1 and B2r1are connected exactly similarly req r1 r2 rn to the second cell. Hence, considering the second cell we write, --- (11.53) VWCo=emkVbnBio1nw-inVgtBh2tah=te Iεl2a=-stII1t2+hrr2eI ;2eIe2quaHt2iro2nVs, --- (11.47) Substitution of emfs should be done algebraically by considering proper ± signs ? I = H1 V H2 V --- (11.48) according to polarity. r1 r1 r2 r2 • Advantages of cells in parallel : For cells § H1 H2 · V § 1 1 · connected in parallel in a circuit, the circuit ¨ r1 r2 ¸ ¨ r1 r2 ¸ will not break open even if a cell gets © ¹ © ¹ damaged or open. Thus, V § 1 1 · § H1 H2 · I • Disadvantages of cells in parallel : The ¨ r1 r2 ¸ ¨ r1 r2 ¸ voltage developed by the cells in parallel © ¹ © ¹ connection cannot be increased by increasing number of cells present in circuit. § r1 +r2 · H1r2 H 2r1 11.14 Types of Cells: ¨ r1r2 ¸ r1r2 ? V © ¹ I Electrical cells can be divided into several categories like primary cell, secondary cell, fuel cell, etc. V= H1r2 H 2r1 I r1r2 ---(11.49) A primary cell cannot be charged again. It r1 + r2 r1 + r2 can be used only once. Dry cells, alkaline cells are different examples of primary cells. Primary cells are low cost and can be used easily. If we replace the cells by a single cell But these are not suitable for heavy loads. Secondary cells are used for such applications. connected between points B1 and B2 with the The secondary cell are rechargeable and can be emf εeq and the internal resistance req as in Fig. reused. The chemical reaction in a secondary (11.13b), cells is reversible. Lead acid cell, and fuel cell then, V = εeq - Ireq --- (11.50) 217
are some examples of secondary cells. Lead ii. Current in each resistor : acid battery is used widely in vehicles and other applications which require high load currents. Total current I in the circuit is, Solar cells are secondary cells that convert Solar energy into electrical energy. I = ε = 15 = 2.1A RT + r 6 +1 Fuel cells vehicles (FCVs) are electric vehicles that use fuel cells instead of lead acid Consider resistors between A and B. batteries to power the vehicles. Hydrogen is used as a fuel in fuel cells. The by- product 4reΩsisrtLeoseritstIo1rsbeantdheI2cbuerrethnet through one of the after its burning is water. This is important current in the other in terms of reducing emission of greenhouse gases produced by traditional gasoline fueled that iIs1,× I 4==II2f×ro4m symmetry of the two arms. vehicles. The hydrogen fuel cell vehicles are B1ut 2 thus more environment friendly. + I2 = I = 2.1A Example 11.8: A network of resistors is I1 connected to a 15 V battery with internal ∴ I1 = I2 =1.05A resistance 1 Ω as shown in the circuit diagram. that is, the current in each 4Ω resistor is 1.05A, Calculate the current in 1Ω resistor between B and C (i) The equivalent resistance, (ii) Current in each resistor, would be 2.1A. (iii) Voltage drops VAB, VBC and VDC. Now, consider the resistances between C 1Ω D and D rΩesirsetLsoiers.ttoIr3sbaentdheI4cubreretnhte through one of the 6 current in the other I3 × 6 = I4 × 6 ∴ curIr3e=ntI4in= 1.05A That is, each 6 Ω resistor is 1.05A r =1 Ω iii. Voltage drop across BC is VBC 15 V V = I × 1 = 2.1 × 2.1 = 2 V Solution : BC i) Equivalent Resistance (ReqR) C=DR=AB66++×R66BC+3R:DC Voltage drop across CD is VCD RAB = VCD = I × RCD = 2.1 × 3 =6.3V 4×4 [Note : Total voltage drop across AD is 4+4 (4.2 V+2.1V+6.3 V) =12.6 V, while its emf is 15 V. The loss of the voltage is 2.4 V]. 2:, Internet my friend RBC = 1Ω https://www.britannica.com/science/ RT= Req = 2 + 1 + 3 = 6 Ω superconductivityphysics ∴ Equivalent Resistance is 6 Ω 218
ExercisesExercises 1. Choose correct alternative number of cells. The number of rows should be i) You are given four bulbs of 25 W, 40 W, 60 W and 100 W of power, all operating (A) 2 (B) 4 at 230 V. Which of them has the lowest resistance? (C) 5 (D) 100 (A) 25 W (C) 40 W viii) Five dry cells each of voltage 1.5 V are connected as shown in diagram (C) 60 W (D) 100 W ii) Which of the following is an ohmic conductor? (A) transistor (B)vacuum tube What is the overall voltage with this arrangement? (C) electrolyte (D) nichrome wire iii) A rheostat is used (A) 0V (B) 4.5V (A) to bring on a known change of (C) 6.0V (D) 7.5V resistance in the circuit to alter the current 2. Give reasons / short answers (B) to continuously change the resistance in any arbitrary manner and there by alter i) In given circuit diagram two resistors are the current connected to a 5V supply. (C) to make and break the circuit at any instant (D) neither to alter the resistance nor the current iv) The wire of length L and resistance R is stretched so that its radius of cross-section a] Calculate potential difference across is halved. What is its new resistance? the 8Ω resistor. (A) 5R (B) 8R b] A third resistor is now connected (C)4R (D) 16R in parallel with 6Ω resistor. Will the potential difference across the 8Ω resistor v) Masses of three pieces of wires made of the larger, smaller or the same as before? the same metal are in the ratio 1:3:5 and Explain the reason for your answer. their lengths are in the ratio 5:3:1. The ratios of their resistances are ii) Prove that the current density of a metallic conductor is directly proportional to the (A) 1:3:5 (B) 5:3:1 drift speed of electrons. (C) 1:15:125 (D) 125:15:1 3. Answer the following questions. vi) The internal resistance of a cell of emf i) Distinguish between Ohmic and non- 2V is 0.1Ω it is connected to a resistance ohmic substances; explain with the help of 0.9Ω. The voltage across the cell will of example. be ii) DC current flows in a metal piece of non- (A) 0.5 V (B) 1.8 V uniform cross-section. Which of these quantities remains constant along the (C) 1.95 V (D) 3V conductor: current, current density or drift speed? vii) 100 cells each of emf 5V and internal resistance 1Ω are to be arranged so as to produce maximum current in a 25Ω resistance. Each row contains equal 219
4. Solve the following problems. vii) A silver wire has a resistance of 4.2 Ω i) What is the resistance of one of the rails of at 27° C and resistance 5.4 Ω at 100° C. a railway track 20 km long at 20° C? The Determine the temperature coefficient of cross section area of rail is 25 cm2 and the resistance. rail is made of steel having resistivity at [Ans: 3.91×10-3/°C] 20° C as 6×10-8 Ω m. viii) A 6m long wire has diameter 0.5 mm. Its [Ans: 0.48 Ω] resistance is 50 Ω. Find the resistivity and ii) A battery after a long use has an emf conductivity. 24 V and an internal resistance 380 Ω. [Ans: 1.636×10-6Ω/m, 6.112×105m/Ω] Calculate the maximum current drawn from the battery? Can this battery drive ix) Find the value of resistances for the starting motor of car? following colour code. [Ans: 0.063 A] 1. Blue Green Red Gold iii) A battery of emf 12 V and internal [Ans: 6.5 kΩ ± 5%] resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, 2. Brown Black Red Silver [Ans: 1.0 kΩ ± 10%] a] Calculate resistance of resistor. 3. Red Red Orange Gold b] Calculate terminal voltage of the [Ans: 2.2 kΩ ± 5%] battery when the circuit is closed. 4. Orange White Red Gold [Ans: a) 21 Ω, b) 10.5 V] [Ans: 3.9 kΩ ± 5%] iv) The magnitude of current density in a 5. Yellow Violet Brown Silver copper wire is 500 A/cm2. If the number of free electrons per cm3 of copper is [Ans: 4.70 kΩ ± 10%] 8.47×1022 calculate the drift velocity of the electrons through the copper wire x) Find the colour code for the following (charge on an e = 1.6×10-19 C) value of resistor having tolerance ± 10% a) 330Ω b) 100Ω c) 47kΩ [Ans: 3.69×10-4 m/s] d) 160Ω e) 1kΩ v) Three resistors 10 Ω, 20 Ω and 30 Ω are xi) A current 4A flows through an automobile connected in series combination. headlight. How many electrons flow through the headlight in a time 2hrs. i] Find equivalent resistance of series combination. [Ans : 1.8 ×1023] ii] When this series combination is xii) The heating element connected to 230V connected to 12V supply, by neglecting the value of internal resistance, obtain draws a current of 5A. Determine the potential difference across each resistor. amount of heat dissipated in 1 hour (J = 4.2 J/cal.). [Ans: i) 60 Ω, ii) 2 V, 4 V, 6 V] [Ans : 985.7 kcal] vi) Two resistors 1k Ω and 2k Ω are connected in parallel combination. *** i] Find equivalent resistance of parallel combination ii] When this parallel combination is connected to 9 V supply, by neglecting internal resistance calculate current through each resistor. [Ans: i) 0.66 kΩ, ii) 9 mA, 4.5 mA] 220
12. Magnetism Can you recall? force? 4. If you freely hang a bar magnetic horizontally, 1. What is a bar magnet? 2. What are the magnetic lines of force? in which direction will it become stable? 3. What are the rules concerning the lines of quadrupole. In this Chapter the main focus 12.1 Introduction: The history of magnetism dates back will be on elementary aspects of magnetism and to earlier than 600 B.C., but it is only in the terrestrial magnetism. twentieth century that scientists began to understand it and developed technologies based 12.2 Magnetic Lines of Force and Magnetic on this understanding. William Gilbert (1544- 1603) was the first to systematically investigate Field: the phenomenon of magnetism using scientific method. He also discovered that Earth is a weak You have studied properties of electric lines magnet. Danish physicist Hans Oersted (1777- 1851) suggested a link between electricity and of force earlier in the Chapter on electrostatics. magnetism. James Clerk Maxwell (1831-1879) proved that electricity and magnetism represent In a similar manner, magnetic lines of force different aspects of the same fundamental force field. originate from the north pole and end at the In electrostatics you have learnt about the south pole of a bar magnet. The magnetic relationship between the electric field and force due to electric charges and electric dipoles. lines of force of a magnet have the following Analogous concepts exist in magnetism except that magnetic poles do not exist in isolation, properties: and we always have a magnetic dipole or a i) The magnetic lines of force of a magnet Do you know ? or a solenoid form closed loops. This is in Some commonly known facts about magnetism. contrast to the case of an electric dipole, (i) Every magnet regardless of its size and where the electric lines of force originate shape has two poles called north pole and south pole. from the positive charge and end on (ii) If a magnet is broken into two or more pieces then each piece behaves like an the negative charge, without forming a independent magnet with somewhat weaker magnetic field. complete loop (see Fig. 12.4). Thus isolated magnetic monopoles do not exist. The search for magnetic ii) The direction of the net magnetic field B monopoles is still going on. (iii) Like magnetic poles repel each other, at a point is given by the tangent to the whereas unlike poles attract each other. (iv) When a bar magnet/ magnetic needle is magnetic line of force at that point in the suspended freely or is pivoted, it aligns itself in geographically North-South direction of line of force. direction. iii) The number of lines of force crossing per unit area decides the magnitude of the magnetic field B . iv) The magnetic lines of force do not intersect. This is because had they intersected, the direction of magnetic field would not be unique at that point. Try this You can take a bar magnet and a small compass needle. Place the bar magnet at a fixed position on a paper and place the needle at various positions. Noting the orientation of the needle, the magnetic field direction at various locations can be traced. Density of lines of force i.e., the number of lines of force per unit area normal to the surface 221
around a particular point determines the strength Equator:- A line passing through the centre of the magnetic field at that point. The number of a magnet and perpendicular to its axis is of lines of force is called magnetic flux (φ ). SI called magnetic equator. The plane containing unit of magnetic flux (φ ) is weber (Wb). For a all equators is called the equatorial plane. The specific case of uniform magnetic field which locus of points, on the equatorial plane, which is normal to the finite area A, the magnitude of are equidistant from the centre of the magnet magnetic field strength B at a point in the area is called the equatorial circle. The popularly A is given by known ‘equator’ in Geography is actually Magnetic Field = magnetic flux an ‘equatorial circle’. Such a circle with any diameter is an equator. area i. e. B = φA --- (12.1) Magnetic length (2l):- It is the distance SI unit of magnetic field (B) is expressed between the two poles of a magnet. as weber/m2 or Tesla. 1 Tesla = 104 Gauss. 5 However, magnetic lines are only a crude Magnetic length (2l) = 6 × Geometric length way of representing magnetic field. It is a --- (12.2) pictorial representation of the strength of the magnetic field (B). It is better defined in terms 12.3.1 Magnetic field due to a bar magnet of Lorentz force law which you will learn in std at a point along its axis and at a point XII. along its equator: 12.3 The Bar magnet: A bar magnet is said to have magnetic pole Fig. 12.2 (a): Magnetic field at a point strength +qm and - qm at the north and south along the axis of the magnet. poles, respectively. The separation of magnetic poles inside the magnet is 2l. As the bar magnet Beq Beq P0 m has two poles, with equal and opposite pole 4S r3 strength, it is called a magnetic dipole. This is analogous to an electric dipole. The magnetic Fig. 12.2 (b): Magnetic dipole moment, therefore, becomes m = qm.2l field along the equatorial ( 2l is a vector from south pole to north pole) point. in analogy with the electric dipole moment. SI unit of pole strength (q ) is A m. Consider a bar magnet of dipole length 2l m and magnetic dipole moment m as shown in Fig. 12.2 (a). We will now find magnetic field at SI unit of magnetic dipole moment m is A m2. a point P along the axis of the bar magnet. Axis:- It is the line passing through both the poles of a bar magnet. Obviously, there is only Let r be the distance of point P from the one axis for a given bar magnet. centre O of the magnetic dipole. Fig. 12.1: Bar magnet OS = ON = l ... NP SP r2 l2 We now use the electrostatic analogy to obtain the magnetic field due to a bar magnet at a large distance r >> l. Consider the electric field due to an electric dipole with a dipole moment p. 222
The Electrostatic Analogue: compared with charge q in electrostatics. As suggested by Maxwell, electricity Accordingly, we can write the equivalent and magnetism could be studied analogously. physical quantities in electrostatics and The pole strength (qm) in magnetism can be magnetism as shown in table 12.1. Table 12.1: The Electrostatic Analogue Quantity Electrostatics Magnetism Basic physical quantity Electrostatic charge Magnetic pole Field Electric Field E Magnetic Field B Constant 1 P0 Dipole moment 4SH 0 4S Force p = q ( ) amlo=ngqSm ( 2l ) (bar magnet) Energy (In external field) 2l N pole of a dipole Coulomb’s law along (-ve) (+ve) charge Axial field for a short F =qE F = qm B dipole U = - p.E U = -m.B F § 1 · q1q 2 No analogous law as ¨ 4SH 0 ¸ r2 magnetic monopoles © ¹ do not exist 2p along p P0 2m 4SH 0 r 3 4S r3 Equatorial field for a short p opposite to p P0 m dipole 4SH 0 r 3 4S r3 You have studied the electric field due to an Similarly, the equatorial magnetic field electric dipole of length 2l (p = 2ql) at a distance Beq P0 m 4S r3 r along the dipolar axis (Eq. 10.24) which is --- (12.4) given by, Negative sign shows that the direction of Ea 1 2 p , r !! l Beq is opposite to m . 4SH 0 r For the same distance from centre O of a 3 The electric field on the equator (Eq. 10.28) bar magnet, is antiparallel to p and is given by Baxis = 2Beq --- (12.5) E eq 1 p , r !! l 12.3.2 Magnetic field due to a bar magnet at 4SH 0 r3 an arbitrary point: Using the analogy given in Table 12.1, minoimtsFeimngt.a1gm2n.e3wtiScithhfoiewcledsn.atrMbeaaargtmneOati.gcnPemtiosomfamennyatgpnmoeitniicst we can thus write the axial magnetic field of a bar magnet at a distance r, r >> l, 2l being the creosmolpvoenden(atsboaulot ntgherceanntrde paoelforpnthegenrdmi,catuhglenaerpto)toiinntrtPo. length of bar magnet, For the component mcosθ Ba P0 2m --- (12.3) is an axial point. 4S r3 223
Beq = P0 m 4S r3 107 u 0.5 5 u 108 0.625u105 Wb / m2 8 u 103 P (0.2)3 Fig. 12.3: Magnetic field at an arbitrary point. 12.4 Gauss' Law of Magnetism: ppeoripnetAnaldtsiotch,uelasrafomtorerd,itsththeaenpcoecionrmt.PpUoissninaegnntethqeumaretsosiurnilaθtsl The Gauss' law for electric field is known to you. It states that the net electric flux through a closed Gaussian surface is proportional to the net electric charge enclosed by the surface (Eq. (10.18)). The Gauss' law for magnetic fields states that the net magnetic flux Φ through a closed Gaussian surface is zero, i.e.,B of axial and equatorial fields, we get (Gauss' law for magnetic Ba Po 2m cosT --- (12.6) fields) 4S r3 The magnetic force lines of (a) bar magnet, (b) current carrying finite solenoid, directed along m cosθ and and (c) electric dipole are shown in Fig.12.4(a), 12.4(b) and 12.4(c), respectively. The curves Beq Po msinT --- (12.7) labelled (i) and (ii) are cross sections of three 4S r3 dimensional closed Gaussian surfaces. directed opposite to msinθ Thus, the magnitude of the resultant magnetic field B, at point P is given by B Ba 2 Beq2 ∴ B 4PSo rm3 >2 cosT @2 >sinT @2 ?B Po m 3cos2 T 1 --- (12.8) 4S r3 Lre.t α be the angle made by the direction of B with by using eq (12.6) and eq (12.7), Then, --- (12.9) of and m Fig. 12.4 (a): Bar magnet. The angle between directions B Fig. 12.4 (b): Current (I) carrying solenoid. is then T D . Example 12.1: A short magnetic dipole has magnetic moment 0.5 A m2. Calculate its magnetic field at a distance of 20 cm from the centre of magnetic dipole on (i) the axis (ii) the equatorial line (Given µ0 = 4 π Í10-7 SI units) Solution : m = 0.5 Am2 , r = 20 cm = 0.2 m Ba = P0 2m 107 u 2 u 0.5 1u107 4S r3 (0.2)3 8 u 103 1 u 104 1.25u105 Wb / m2 8 224
This fact clearly indicates that there is some magnetic field present everywhere on the Earth . This is called Terrestrial Magnetism. It is extremely useful during navigation. Magnetic parameters of the Earth are described below. The magnetic lines of force enter the Earth's surface at the north pole and emerge from the south pole. Unless and otherwise stated, the directions mentioned (South, North, etc.) are always, Geographic. Fig. 12.4 (c): Electric dipole. Fig. 12.5: Earth's magnetism. Magnetic Axis :- The Earth is considered to If we compare the number of lines of force be a huge magnet. Magnetic north pole (N) entering in and leaving out of the surface (i), it of the Earth is located below Antarctica while is clearly seen that they are equal. The Gaussian the south pole (S) is below north Canada. The surface does not include poles. It means that the straight line NS joining these two poles is called flux associated with any closed surface is equal the magnetic axis, MM'. to zero. When we consider surface (ii), in Fig. Magnetic equator :- A great circle in the plane 12.4 (b), we are enclosing the North pole. As perpendicular to magnetic axis is magnetic even a thin slice of a bar magnet will have North equatorial circle,AA'. It happens to pass through and South poles associated with it, the closed India near Thiruvananthapuram. Gaussian surface will also include a South pole. Geographic Meridian:- A plane perpendicular However in Fig. 12.4(c), for an electric dipole, to the surface of the Earth (vertical plane) the field lines begin from positive charge and perpendicular to geographic axis is geographic end on negative charge. For a closed surface meridian. (Fig.12.6) (ii), there is a net outward flux since it does Magnetic Meridian:- A plane perpendicular to include a net (positive) charge. According to the surface of the Earth (Vertical plane) and passing Gauss' law of electrostatics as studied earlier, through the magnetic axis is magnetic meridian. Direction of resultant magnetic field of the , where q is the positive charge Earth is always along or parallel to magnetic meridian. (Fig.12.6) enclosed. Thus, situation is entirely different Magnetic declination:- Angle between the geographic and the magnetic meridian at a from magnetic lines of force, which are shown place is called ‘magnetic declination’ (α). The declination is small in India. It is 0° 58′ west at in Fig. 12.4(a) and Fig. 12.4(b). Thus, Gauss' law Mumbai and 0041′ east at Delhi. Thus, at both these places, magnetic needle shows true North of magnetism can be written as . From the above we conclude that for electrostatics, an isolated electric charge exists but an isolated magnetic pole does not exist. In short, only dipoles exist in case of magnetism. 12.5 Earth’s Magnetism: It is common experience that a bar magnet or a magnetic needle suspended freely in air always aligns itself along geographic N-S direction. If it has a freedom to rotate about horizontal axis, it inclines with some angle with the horizontal in the vertical N-S plane. 225
accurately (Fig.12.6). (magnetic equator) B= BH along South to North, BV= 0 and φ =0 φ Magnetic maps of the Earth:- Fig. 12.6: Magnetic declination. Magnetic inclination or angle of dip (φ):- φ) Magnetic elements of the Earth (wBiHt,hαtiamned. Angle made by the direction of resultant vary from place to place and also magnetic field with the horizontal at a place is inclination or angle of dip at the place (Fig. The maps providing these values at different 12.7). locations are called magnetic maps. These are extremely useful for navigation. Magnetic maps drawn by joining places with the same value of a particular element are called Iso- magnetic charts. Lines joining the places of equal horizontal components (BH) are known as ‘Isodynamic lines’ Lines joining the places of equal declination (α) are called Isogonic lines. Lines joining the places of equal inclination or dip (φ) are called Aclinic lines. Example 12.2: Earth's magnetic field at the equator is approximately 4×10-5 T. Calculate Earth's dipole moment. (Radius of Earth = 6.4×106 m, µ0 = 4π×10-7 SI units) Solution: Given Fig. 12.7: Magnetic inclination. Beq = 4 ×10-5 T Earth’s magnetic field:- Magnetic force r = 6.4 ×106 m experienced per unit pole strength is magnetic field B at that place. It can be resolved in Assume that Earth is a bar magnet with N and components along the horizontal, BH and along S poles being the geographical South and North poles, respectively. The equatorial magnetic vertical, BV . The vertical component can be field due to Earth's dipole can be written as conveniently determined. The two components can be related with the angle of dip (φ) as, Beq P0m 4S r3 BH = B cosφ, BV = B sinφ m 4S Beq u r3 / P0 BV 4 u105 u (6.4 u106 )3 u107 BH tan I --- (12.10) 1.05u1020 A m2 Example 12.3: At a given place on tmheisEakretpht, a bar magnet of magnetic moment B2 BV 2 BH 2 ? B BV 2 BH 2 --- (12.11) horizontal in the East-West direction. P and Special cases fQieladreofthtehitswmo angenuettraalnpdoiBnHts due to magnetic is the horizontal 1) At the magnetic North pole, B = BV , component of the Earth's magnetic field. directed upward, BH = 0 and φ =900. (A) Calculate the angles dibreetcwtieoennofpmos.ition 2) At the magnetic south pole, B = BV ,- vectors of P and Q with the directed downward, BH = 0 and φ = 2700. (B) Points P and Q are 1 m from the centre of 3) Anywhere on the magnetic great circle the bar magnet and BH 3.5u105 T . Calculate 226
magnetic dipole moment of the bar magnet. (B) tan2T 2 Neutral point is that point where the resultant magnetic field is zero. ?sec2T 1 tan2T 1 2 3 Solution: oo(fpApm)oasAgitnseettsoieceBnfHieflardtotmhBe the figure, the ?cos2T 1 direction due to the bar 3 magnet is points P and Q. r 1m and B BH 3.5u105 T (Given) Also, T D = 900 at P and it is2700 at Q. we have, B P0 m 3cos2T 1 4S r3 ? m BH u r3 § P0 · ©¨ 4S ¸¹ 3cos2T 1 3.5u105 u13 107 u 3 1 1 3 ?m 350 247.5 A m2 2 Always remember: In this Chapter we have used B as a symbol for magnetic field. Calling it magnetic induction is unreasonable. We have used the words magnetic field which are used in spoken language. ExercisesExercises 1. Choose the correct option. iii) The horizontal and vertical component of magnetic field of Earth are same at i) Let r be the distance of a point on the some place on the surface of Earth. The axis of a bar magnet from its center. The magnetic dip angle at this place will be magnetic field at r is always proportional to (A) 30o (B) 45o (A) 1/r2 (B) 1/r3 (C) 1/r (C) 0o (D) 90o (D) not necessarily 1/r3 at all points iv) Inside a bar magnet, the magnetic field ii) Magnetic meridian is the plane lines (A) perpendicular to the magnetic axis of (A) are not present Earth (B) are parallel to the cross sectional area (B) perpendicular to geographic axis of of the magnet Earth (C) are in the direction from N pole to S (C) passing through the magnetic axis of pole Earth (D) are in the direction from S pole to N (D) passing through the geographic axis pole Earth 227
v) A place where the vertical components of ii) A magnet makes an angle of 45o with the Earth's magnetic field is zero has the angle horizontal in a plane making an angle of of dip equal to 30o with the magnetic meridian. Find the true value of the dip angle at the place. (A) 0o (B) 45o [Ans: tan-1 (0.866)] (C) 60o (D) 90o iii) Two small and similar bar magnets have vi) A place where the horizontal component magnetic dipole moment of 1.0 Am2 of Earth's magnetic field is zero lies at each. They are kept in a plane in such a way that their axes are perpendicular to (A) geographic equator each other. A line drawn through the axis of one magnet passes through the center (B) geomagnetic equator of other magnet. If the distance between their centers is 2 m, find the magnitude of (C) one of the geographic poles magnetic field at the mid point of the line joining their centers. (D) one of the geomagnetic poles vii) A magnetic needle kept nonparallel to the magnetic field in a nonuniform magnetic field experiences (A) a force but not a torque [Ans: 5 u107 T ] (B) a torque but not a force iv) A circular magnet is made with its north pole at the centre, separated from the (C) both a force and a torque surrounding circular south pole by an air a gap. Draw the magnetic field lines in the (D) neither force nor a torque gap. [The magnet is hypothetical magnet]. 2. Answer the following questions in brief. i) What happens if a bar magnet is cut into Draw a diagram to illustrate the magnetic two pieces transverse to its length/ along lines of force between the south poles of its length? two such magnets. ii) What could be the equation for Gauss' v) Two bar magnets are placed on a straight law of magnetism, if a monopole of pole line with their north poles facing each other strength p is enclosed by a surface? on a horizontal surface. Draw magnetic lines around them. Mark the position of 3. Answer the following questions in detail. any neutral points (points where there is no resultant magnetic field) on your diagram. i) Explain the Gauss' law for magnetic fields. ii) What is a geographic meridian. How does *** the declination vary with latitude? Where is it minimum? iii) Define the Angle of Dip. What happens to Internet my friend angle of dip as we move towards magnetic https://www.ngdc.noaa.gov pole from magnetic equator? 4. Solve the following Problems. i) A magnetic pole of bar magnet with pole strength of 100 A m is 20 cm away from the centre of a bar magnet. Bar magnet has pole strength of 200 A m and has a length 5 cm. If the magnetic pole is on the axis of the bar magnet, find the force on the magnetic pole. [Ans: 2.5×10-2N] 228
13. Electromagnetic Waves and Communication System Can you recall? 1. What is a wave? 4. What are Lenz's law, Ampere's law and 2. What is the difference between longitudinal Faraday's law? and transverse waves? 5. By which mechanism heat is lost by hot 3. What are electric and magnetic fields and bodies ? what are their sources? with the circuit changes. (4) Ampere’s law gives the relation between 13.1 Introduction : The information age in which we live the induced magnetic field associated with a loop and the current flowing through is based almost entirely on the physics of the loop. Maxwell (1831-1879) noticed electromagnetic (EM) waves. We are now a major flaw in the Ampere’s law for globally connected by TV, cellphone and time dependant fields. He noticed that the internet. All these gadgets use EM waves as magnetic field can be generated not only carriers for transmission of signals. Energy by electric current but also by changing from the Sun, an essential requirement for life electric field. Therefore in the year 1861, on Earth, reaches us by means of EM waves he added one more term to the equation that travel through nearly 150 million km of describing this law. This term is called empty space. There are EM waves from light the displacement current. This term is bulbs, heated engine blocks of automobiles, extremely important and the EM waves x-ray machines, lightning flashes, and some which are an outcome of these equations radioactive materials. Stars, other objects in our would not have been possible in absence milky way galaxy and other galaxies are known of this term. to emit EM waves. Hence, it is important for us As a result, the set of four equations to make a careful study of the properties of EM describing the above four laws is called waves. Maxwell’s equations. In 1888, H. Hertz (1857-1894) succeeded 13.2 EM wave: in producing and detecting the existence of EM There are four basic laws which describe waves. He also demonstrated their properties namely reflection, refraction and interference. the behaviour of electric and magnetic fields, In 1895, an Indian physicist Sir Jagdish the relation between them and their generation Chandra Bose (1858-1937) produced EM by charges and currents. These laws are as waves ranging in wavelengths from 5 mm to 25 follows. nm. His work, however, remained confined to (1) Gauss' law for electrostatics, which is laboratory only. In 1896, an Italian physicist G. Marconi essentially the Coulomb’s law, describes (1874-1937) became pioneer in establishing the relationship between static electric wireless communication. He was awarded the charges and the electric field produced by Nobel prize in physics in 1909 for his work in them. developing wireless telegraphy, telephony and (2) Gauss' law for magnetism, which is broadcasting. similar to the Gauss' law for electrostatics mentioned above, states that \"magnetic 13.2.1 Sources of EM waves: monopoles which are thought to be According to Maxwell’s theory, \"accelerated magnetic charges equivalent to the electric charges, do not exist\". Magnetic poles charges radiate EM waves\". Consider a charge always occur in pairs. oscillating with some frequency. This produces (3) Faraday’s law which gives the relation between electromotive force (emf) induced in a circuit when the magnetic flux linked 229
an oscillating electric field in space, which 3) produces an oscillating magnetic field which in turn is a source of oscillating electric field. Thus (Faraday’s law with Lenz’s law) varying electric and magnetic fields regenerate Here φm is the magnetic flux and the each other. integral is over a closed loop. Time varying Waves that are caused by the acceleration magnetic field induces an electromotive of charged particles and consist of electric force (emf) and hence, an electric field. The and magnetic fields vibrating sinusoidally at direction of the induced emf is such that the right angles to each other and to the direction change is opposed. of propagation are called EM waves or EM radiation. Figure 13.1 shows an EM wave 4) propagating along z-axis. The time varying electric field is along the x-axis and time varying (Ampere-Maxwell law) magnetic field is along the y-axis. and tHheereinµte0girsatlhies permeability of vacuum over a closed loop, I is the current flowing through the loop. φE is the electric flux linked with the circuit. Magnetic field is generated by moving Fig. 13.1: EM wave propagating along z-axis. charges and also by varying electric fields. 13.2.2 Characteristics of EM waves: Do you know ? 1) The electric and magnetic fields, E and B are always perpendicular to each other In 1865, Maxwell proposed that an and also to the direction of propagation oscillating electric charge radiates energy in the form of EM wave. EM waves are of the EM wave. Thus the EM waves are periodic changes in electric and magnetic fields, which propagate through space. transverse waves. Thus, energy can be transported in the form of EM waves. 2) The cross product E ×B gives the direction Maxwell’s Equations for Charges and in which the EM wave travels. E ×B also Currents in Vacuum gives the energy carried by EM wave. 3) The E and B fields vary sinusoidally and are in phase. 1) (Gauss’ law) 4) EM waves are produced by accelerated electric charges. Here E is the electric Tfiheledinatnedgrεa0l is 5) EM waves can travel through free space permittivity of vacuum. is the as well as through solids, liquids and over a closed surface S. The law states that gases. electric flux through any closed surface S is 6) In free space, EM waves travel with bedqyeusctahrlietboestsuhretfhateocteareldleailvteiicodtneridcbcebhtywareεge0e.nQGaianneunsescle’loclstarewidc velocity c, equal to that of light in free space. charge and electric field it produces. c 1 3u108 m / s , P0H 0 2) (Gauss’ law for magnetism). where µ0 (4π×10-7 Tm/A) is permeability and ε0 (8.85×10-12 C2/Nm2) is permittivity Here B is the magnetic field. The of free space. integral is over a closed surface S. The law states that magnetic flux through a closed 7) In a given material medium, the velocity surface is always zero, i.e., the magnetic field lines are continuous closed curves, (vm) of EM waves is given by vm 1 having neither beginning nor end. PH where µ is the permeability and ε is the 230
permittivity of the given medium. Do you know ? 8) The EM waves obey the principle of According to quantum theory, an electron, while orbiting around the nucleus superposition. in a stable orbit does not emit EM radiation 9) The ratio of the amplitudes of electric and even though it undergoes acceleration. It will emit an EM radiation only when it falls magnetic fields is constant at any point and from an orbit of higher energy to one of lower energy. is equal to the velocity of the EM wave. EM waves (such as X-rays) are | E0 |= c | B0 | or | E0 | = 1 --- (13.1) produced when fast moving electrons hit | B0 | P0 H 0 a target of high atomic number (such as molybdenum, copper, etc.). E0 and B0 are the amplitudes of E and An electric charge at rest has an B respectively. electric field in the region around it but has no magnetic field. When the charge moves, 10) As the electric field vector ( E0 ) is more it produces both electric and magnetic prominent than the magnetic field vector fields. If the charge moves with a constant velocity, the magnetic field will not change ( B0 ), it is responsible for optical effects with time and as such it cannot produce an EM wave. But if the charge is accelerated, due to EM waves. For this reason, electric both the magnetic and electric fields change with space and time and an EM wave is vector is called light vector. produced. Thus an oscillating charge emits an EM wave which has the same frequency 11) The intensity of a wave is proportional to as that of the oscillation of the charge. the square of its amplitude and is given by the equations IE 1 H E0 2 , IB 1 B02 --- (13.2) 2 2 P0 0 12) The energy of EM waves is distributed equally between the electric and magnetic fields. IE = IB both x- and y-axes. As per property (2) of EM Example 13.1: Calculate the velocity of EM waves, E ×B should be along the direction of waves in vacuum. propagation which is along the x- axis Solution: The velocity of EM wave in free Since (+ j ) ×(+ k ) = i , B is along the k , space is given by i.e., along the z-direction. 1 1 Thus, the amplitude of B = 3.2×10-8 T and c its direction is along the z-axis. (8.85 u 1012 C2 )(4S u107 T.m ) P0H 0 Nm 2 A c 3.00 u108 m / s Example 13.3: A beam of red light has an amplitude 2.5 times the amplitude of second Example 13.2: In free space, an EM wave of beam of the same colour. Calculate the ratio of frequency 28 MHz travels along the x-direction. the intensities of the two waves. The amplitude of the electric field is E = 9.6 V/m and its direction is along the y-axis. What Solution: Intensity ∝ (Amplitude)2 is amplitude and direction of magnetic field B? I2 ∝ (a)2 and I1 ∝ (2.5a)2 I1 (2.5a )2 Solution : We have, ? I2 a2 (2.5) 2 6.25 . |B| |E| 9.6 V / m c 3u108 m / s In an EM wave, the magnetic field and B 3.2 u108 T electric field both vary sinusoidally with x. For It is given that E is along y-direction and ay-waxaivseatnrdavBellailnognagltohnegzxa-axxisi,swhiatvhirnegfeErenacloentog Chapter 8, we can write Ey and Bz as the wave propagates along x-axis. The magnetic field B should be in a direction perpendicular to 231
Ey= E0 sin (kx-ωt) --- (13.2) Ey E0 sin 2S ªx t »¼º ¬« O and Bz= B0 sin (kx-ωt), --- (13.3) Q where E0 is the amplitude of the electric field Ey E0 sin 2S ªx 3 u 1010 t º «¬102 »¼ Ey and B0 is the amplitude of the magnetic field 2S is the propagation constant and λ B. k O Ey E0 sin 2S ¬ª100x 3u1010 tº¼ V / m z is the wavelength of the wave. ω = 2πυ is the Example 13.6: The magnetic field of angular frequency of oscillations, υ being the frequency of the wave. an = EM wave travelling along x-axis t is B k 4×10-4 sin (ωt - kx). Here B is in tesla, is Both the electric and magnetic fields in second and x is in m. Calculate the peak value attain their maximum (and minimum) values at the same time and at the same point in space, of electric force acting on a particle of charge 5 i.e., E and B oscillate in phase with the same µC travelling with a velocity of 5×105 m/s along frequency. the y-axis. Solution : Example 13.4: An EM wave of frequency B0= 4×10-4 T, q = 5 µC = 5×10-6 C 50 MHz travels in vacuum along the positive v = 5×105 m/s x-axis and E at a particular point, x and at a E0= cB0=(3×108) × (4×10-4) particular instant of time t is 9.6 j V/m. Find =12×104 N/C the magnitude and direction of B at this point Maximum electric force = qE0 = (5×10-6) (12×104) x and at time t. = 60×10-2 = 0.6 N Solution : B=E 9.6 3.2 u108 T 13.3 Electromagnetic Spectrum: c 3u108 The orderly distribution (sequential arrangement) of EM waves according to their As the wave propagates along +x axis and wavelengths (or frequencies) in the form of distinct groups having different properties E is along +y axis, direction of B will be along is called the EM spectrum (Fig. 13.2). The +z-axis i.e. B = 3.2×10-8 k T. properties of different types of EM waves are given in Table 13.1. Example 13.5: For an EM wave propagating along x direction, the magnetic field oscillates along the z-direction at a frequency of 3×1010 Hz and has amplitude of 10-9 T. a) What is the wavelength of the wave? b) Write the expression representing the corresponding electric field. Solution : a) O = c 3u108 m / s 102 m X 3u1010 / s b) E0= cB0= (3×108 m/s) × (10-9 T) = 0.3 V/m. Fig. 13.2: Electromagnetic spectrum. Since B acts along z-axis, E acts along y-axis. We briefly describe different types of EM Expression representing the oscillating electric waves in the order of decreasing wavelength (or field is increasing frequency). 13.3.1 Radio waves : Ey E0 sin (kx -Zt) Radio waves are produced by accelerated motion of charges in a conducting wire. The Ey E0 sin ª§ 2S · x (2SQ )t º «¬¨© O ¸¹ ¼» 232
frequency of waves produced by the circuit Notation used for high frequencies depends upon the magnitudes of the inductance 1 kHz = one kilo Hertz =1000 Hz = 103 Hz and the capacitance (This will be discussed 1 MHz = one mega Hertz = 106 Hz in XIIth standard). Thus, by choosing suitable 1 GHz = one giga Hertz =109 Hz values of the inductance and the capacitance, Notation used for small wavelengths radio waves of desired frequency can be 1 µm = one micrometer = 10-6 m produced. 1 Å= one angstrom = 10-10 m= 10-8 cm Properties : 1nm = one nanometer = 10-9 m 1) They have very long wavelengths ranging Uses : from a few centimetres to a few hundreds 1) Radio waves are used for wireless of kilometres. 2) The frequency range of AM band is 530 communication purpose. kHz to 1710 kHz. Frequency of the waves 2) They are used for radio broadcasting and used for TV-transmission range from 54 MHz to 890 MHz, while those for FM radio transmission of TV signals. band range from 88 MHz to 108MHz. 3) Cellular phones use radio waves to transmit voice communication in the ultra high frequency (UHF) band. Table 13.1: Properties of different types of EM waves Name Wavelength Frequency Generated By range in m range in Hz Gamma 6×10-13 to 1×10-10 5×1020 to 3×1018 a) Transitions of nuclear energy levels rays 1×10-11 to 3×10-8 3×1019 to 1×1016 b) Radioactive substances X-rays 3×10-8 to 4×10-7 1×1016 to 8×1014 a) Bombardment of high energy Ultraviolet 8×1014 to 4×1014 electrons (keV) on a high atomic (UV waves) 4×10-7 to 8×10-7 4×1014 to 1×1012 number target (Cu, Mg, Co) 8×10-7 to 3×10-4 Visible light 3×10-4 to 6×10-2 b) Energy level transitions of 6×10-4 to 1×105 innermost orbital electrons Infrared (IR) radiations Rearrangement of orbital electrons of Microwaves atom between energy levels. As in high Radio waves voltage gas discharge tube, the Sun and mercury vapour lamp, etc. Rearrangement of outer orbital electrons in atoms and molecules e.g., gas discharge tube Hot objects 1×1012 to 5×109 Special electronic devices such as 5×1011 to 8×1010 klystron tube Acceleration of electrons in circuits 13.3.2 Microwaves : they are incident. These waves were discovered of by H. 2) They can be detected by crystal detectors. Uses Hertz in 1888. Microwaves are produced by 1) Used for the transmission of TV signals. oscillator electric circuits containing a capacitor 2) Used for long distance telephone and an inductor. They can be produced by special vacuum tubes. communication. Properties 3) Microwave ovens are used for cooking. 1) They heat certain substances on which 4) Used in radar systems for the location of 233
distant objects like ships, aeroplanes etc, Do you know ? 5) They are used in the study of atomic and Stars and galaxies emit different types molecular structure. of waves. Radio waves and visible light can 13.3.3 Infrared waves pass through the Earth’s atmosphere and reach the ground without getting absorbed These waves were discovered by William significantly. Thus the radio telescopes Herschel (1737-1822) in 1800. All hot bodies and optical telescopes can be placed on the are sources of infrared rays. About 60% of ground. All other type of waves get absorbed the solar radiations are infrared in nature. by the atmospheric gases and dust particles. Thermocouples, thermopile and bolometers are Hence, the γ-ray, X-ray, ultraviolet, infrared, used to detect infrared rays. and microwave telescopes are kept aboard Properties artificial satellites and are operated remotely 1) When infrared rays are incident on any from the Earth. Even though the visible radiation reaches the surface of the Earth, object, the object gets heated. its intensity decreases to some extent due 2) These rays are strongly absorbed by glass. to absorption and scattering by atmospheric 3) They can penetrate through thick columns gases and dust particles. Optical telescopes are therefore located at higher altitudes. of fog, mist and cloud cover. Uses The Indian Giant Metrewave Radio 1) Used in remote sensing. Telescope (GMRT) near Pune is an important 2) Used in diagnosis of superficial tumours milestone in the field of Radio-astronomy. Also, Indian Astronomical Observatory and varicose veins. houses the Himalayan Chandra Telescope 3) Used to cure infantile paralysis and to (HCT), the 2 m optical-IR Telescope, which is situated at Hanle, Ladakh, at an altitude of treat sprains, dislocations and fractures. 4500 m. 4) They are used in Solar water heaters and Table 13.2: Wavelengths of colours in cookers. visible light 5) Special infrared photographs of the body Colour Wavelength called thermograms, can reveal diseased organs because these parts radiate less violet 380-450 nm heat than the healthy organs. 6) Infrared binoculars and thermal imaging blue 450-495 nm cameras are used in military applications for night vision. green 495-570 nm 7) Used to keep green house warm. 8) Used in remote controls of TV, VCR, etc yellow 570-590 nm 13.3.4 Visible light : It is the most familiar form of EM waves. orange 590-620 nm These waves are detected by human eye. Therefore this wavelength range is called the red 620-750 nm visible light. The visible light is emitted due to atomic excitations. 13.3.5 Ultraviolet rays : Properties : Ultraviolet rays were discovered by J. Ritter (1776-1810) in 1801. They can be 1) Different wavelengths give rise to different produced by the mercury vapour lamp, electric colours. These are given in Table 13.2. spark and carbon arc lamp. They can also be obtained by striking electrical discharge in 2) Visible light emitted or reflected from hydrogen and xenon gas tubes. The Sun is the objects around us provides us information most important natural source of ultraviolet about those objects and hence about the rays, most of which are absorbed by the ozone surroundings. layer in the Earth’s atmosphere. 234
Properties : Properties 1) They are high energy EM waves. 1) They produce fluorescence in certain 2) They are not deflected by electric and materials, such as 'phosphors'. magnetic fields. 2) They cause photoelectric effect. 3) X-rays ionize the gases through which 3) They cannot pass through glass but pass they pass. through quartz, fluorite, rock salt etc. 4) They have high penetrating power. 4) They possess the property of synthesizing 5) Their over dose can kill living plant and vitamin D, when skin is exposed to them. animal overdose tissues and hence are Uses : harmful. Uses 1) Ultraviolet rays destroy germs and bacteria 1) Useful in the study of the structure of and hence they are used for sterilizing crystals. surgical instruments and for purification 2) X-ray photographs are useful to detect of water. bone fracture. X-rays have many other medical uses such as CT scan. 2) Used in burglar alarms and security 3) X-rays are used to detect flaws or cracks in systems. metals. 4) These are used for detection of explosives, 3) Used to distinguish real and fake gems. opium etc. 4) Used in analysis of chemical compounds. 13.3.7 Gamma Rays (γ-rays) 5) Used to detect forgery. Discovered by P. Villard (1860-1934) in 1900. Gamma rays are emitted from the nuclei Do you know ? of some radioactive elements such as uranium, radium etc. 1. A fluorescent light bulb is coated from Properties with a powder inside and contains a gas; 1) They are highest energy EM waves. electricity causes the gas to emit ultraviolet (energy range keV - GeV) radiation, which then stimulates the tube 2) They are highly penetrating. coating to emit light. 3) They have a small ionising power. 2. The pixels of a television or computer 4) They kill living cells. screen fluoresce when electrons from an Uses electron gun strike them. 1) Used as insecticide disinfection for wheat 3. What we call 'visible light' is just the part and flour. of the EM spectrum that human eyes see. 2) Used for food preservation. Many other animals would define 'visible' 3) Used in radiotherapy for the treatment of somewhat differently. For instance, many cancer and tumour. animals including insects and birds, see 4) They are used to produce nuclear in the UV region. Natural world is full of reactions. signals that animals see and humans cannot. Many birds including bluebirds, budgies, 13.4 Propagation of EM Waves: parrots and even peacocks have ultraviolet You must have seen a TV antenna used to patterns that make them even more vivid to each other than they are to us. receive the TV signals from the transmitting tower or from a satellite. In communication 13.3.6 X-rays: using radio waves, an antenna in the transmitter German physicist W. C. Rontgen (1845- radiates the EM waves, which travel through space and reach the receiving antenna at the 1923) discovered X-rays in 1895 while studying other end. As the EM wave travels away from cathode rays (which is a stream of electrons the transmitter; the strength of the wave keeps emitted by the cathode in a vacuum tube). X-rays are also called Rontgen rays. X-rays are produced when cathode rays are suddenly stopped by an obstacle. 235
on decreasing. Several factors influence the propagation of EM waves and the path they follow. It is also important to understand the composition of the Earth’s atmosphere as it plays a vital role in the propagation of EM waves. Different layers of Earth’s atmosphere are shown in Fig. 13.3. Do you know ? Ionizing radiations : Fig. 13.4: Propagation of EM waves. Ultraviolet, X-ray and gamma rays 13.4.1 Ground (surface) wave: have sufficient energy to cause ionization When a radio wave from a transmitting i.e. they strip electrons from atoms and antenna propagates near surface of the Earth molecules lying along their path. The atoms so as to reach the receiving antenna, the wave lose their electrons and are then known as propagation is called ground wave or surface ions. Ionization is harmful to human beings wave propagation. because it can kill or damage living cells, or make them grow abnormally as cancers. In this mode, radio waves travel close to Fluorescent lamps are based on ionization of the surface of the Earth and move along its gas. Ionizing radiation is also used in various curved surface from transmitter to receiver. equipments in laboratory and industry. The radio waves induce currents in the Fig 13.3: Earth and atmospheric layers. ground and lose their energy by absorption. Different modes of propagation of EM Therefore, the signal cannot be transmitted over large distances. Radio waves having frequency waves are described below and are shown in less than 2 MHz (in the medium frequency band) Fig. 13.4. are transmitted by ground wave propagation. This is suitable for local broadcasting only. Do you know ? For TV or FM signals (very high frequency), X-rays have many practical applications ground wave propagation cannot be used. in medicine and industry. Because X-ray 13.4.2 Space wave: photons are of such high energy, they can penetrate several centimetres of solid matter When the radio waves from the transmitting and can be used to visualize the interiors of antenna reach the receiving antenna either materials that are opaque to ordinary light. directly along a straight line (line of sight) or after reflection from the ground or satellite or after reflection from troposphere, the wave propagation is called space wave propagation. The radio waves reflected from troposphere are called tropospheric waves. Radio waves with frequency greater than 30 MHz can pass through the ionosphere (60 km - 1000 km) after suffering a small deviation. Hence, these waves cannot be transmitted by space wave propagation except by using a satellite. Also, for TV signals which have high frequency, transmission over long distance is not possible by means of space wave propagation. The maximum distance over which a signal can reach is called its range. For larger TV 236
coverage, the height of the transmitting antenna Example 13.8: If the height of a TV transmitting should be as large as possible. This is the reason antenna is 128 m, how much square area can be why the transmitting and receiving antennas are covered by the transmitted signal if the receiving mounted on top of high rise buildings. antenna is at the ground level? (Radius of the Earth = 6400 km) Range is the straight line distance from the Solution: point of transmission (the top of the antenna) to the point on Earth where the wave will hit Range = d 2Rh while travelling along a straight line. Range is shown by d in Fig. 13.5. Let the height of the = 2(6400u103 )(128) transmitting antenna (AA') situated at A be h. B represents the point on the surface of the Earth = 16.384 u105 u103 at which the space wave hits the Earth. The triangle OA'B is a right angled triangle. From = 16.384 u108 ∆ OA' B we can write = 4.047u104 OA'2 = A'B2 + OB2 (R+h)2 = d2 + R2 40.470 km or R2 + h2 + 2Rh = d2 + R2 As h << R, we can ignore h2 and write Area covered = S d 2 3.14 u (40)2 d ≅ 2Rh The range can be increased by mounting = 5144 .58 k4m2 the receiver at a height h' say at a point C on the surface of the Earth. The range increases to d + Example 13.9: The height of a transmitting d' where d' is 2Rh ' Thus antenna is 68 m and the receiving antenna is Total range = d d ' 2Rh 2Rh ' at the top of a tower of height 34 m. Calculate the maximum distance between them for Fig. 13.5: Range of the signal (not to scale). satisfactory transmission in line of sight mode. Example 13.7: A radar has a power of 10 kW (radius of Earth = 6400 km) and is operating at a frequency of 20 GHz. It is located on the top of a hill of height 500 m. ht = 68 m, hr = 34m, R = 6400 km = 6.4×106m Calculate the maximum distance upto which it Solution: can detect object located on the surface of the Earth . (Radius of Earth = 6.4×106 m) dmax 2Rht 2Rhr Solution: = 2u 6.4 u106 u 68 2u 6.4 u106 u 34 Maximum distance (range) = d = 2Rh = 870 u103 435 u103 = 2u (6.4 u106 ) u 500 m = 29.5u103 20.9u103 = 8u104 80 km, 50.4 u103 m where R is radius of the Earth and h is the height of the radar above Earth’s surface. = 50.4 km 13.4.3 Sky wave propagation: When radio waves from a transmitting antenna reach the receiving antenna after reflection in the ionosphere, the wave propagation is called sky wave propagation. The sky waves include waves of frequency between 3 MHz and 30 MHz. These waves can suffer multiple reflections between the ionosphere and the Earth. Therefore, they can be transmitted over large distances. Critical frequency : It is the maximum value of the frequency of radio wave which can be reflected back to the Earth from the ionosphere when the waves are directed normally to ionosphere. 237
Skip distance (zone) : It is the shortest distance place and the receiver at another place. The from a transmitter measured along the surface communication channel is a passage through of the Earth at which a sky wave of fixed which signals transfer in between a transmitter frequency (if grater than critical frequency) will and a receiver. This channel may be in the form be returned to the Earth so that no sky waves of wires or cables, or may also be wireless, can be received within the skip distance. depending on the types of communication system. 13.5 Introduction to Communication System: Communicationisexchangeofinformation. There are two basic modes of communication: (i) point to point communication Since ancient times it is practiced in various and (ii) broadcast. ways e.g., through speaking, writing, singing, using body language etc. After the discovery In point to point communication mode, of electricity in the late 19th century, human communication takes place over a link between a communication systems changed dramatically. single transmitter and a receiver e.g. Telephony. Modern communication is based upon the In the broadcast mode there are large number of discoveries and inventions by a number of receivers corresponding to the single transmitter scientists like J. C. Bose (1858-1937), S. F. B. e.g., Radio and Television transmission. Morse (1791-1872), G. Marconi (1874-1937) 13.5.2 C ommonly used terms in electronic and Alexander Graham Bell (1847-1922) in the 19th and 20th centuries. communication system: Following terms are useful to understand In the 20th century we could send messages any communication system: over large distances using analogue signals, 1) Signal :- The information converted into cables and radio waves. With the advancements electrical form that is suitable for transmission of digitization technologies, we can now is called a signal. In a radio station, music and communicate with the entire world almost in speech are converted into electrical form by a real time. microphone for transmission into space. This electrical form of sound is the signal. A signal The ability to communicate is an important can be analog or digital as shown in Fig. 13.7. feature of modern life. We can speak directly to others all around the world and generate vast (a) (b) amount of information every day. Fig 13.7: (a) Analog signal. (b) Digital signal. Here we will briefly discuss how (i) Analog signal: A continuously varying communication systems work.Acommunication system is a device or set up used in transmission signal (voltage or current) is called and reception of information from one place to an analog signal. Since a wave is a another. fundamental analog signal, sound and 13.5.1 Elements of a communication system: picture signals in TV are analog in nature (Fig 13.7 a) There are three basic (essential) elements (ii) Digital signal :Asignal (voltage or current) of every communication system: a) Transmitter, that can have only two discrete values b) Communication channel and c) Receiver. is called a digital signal. For example, a square wave is a digital signal. It has two Fig. 13.6: Block diagram of the basic elements values viz, +5 V and 0 V. (Fig- 13.7 b) of a communication system. 2) Transmitter :- A transmitter converts the signal produced by a source of information In a communication system, as shown into a form suitable for transmission through a in Fig. 13.6, the transmitter is located at one channel and subsequent reception. 238
3) Transducer :- A device that converts one used to increase the range of a communication form of energy into another form of energy is system. These are shown in Fig. 13.8. called a transducer. For example, a microphone converts sound energy into electrical energy. Fig.13.8: Use of repeater station to increase Therefore, a microphone is a transducer. the range of communication. Similarly, a loudspeaker is a transducer which converts electrical energy into sound energy. Do you know ? 4) Receiver :- The receiver receives the message signal at the channel output, reconstructs it in To transmit a signal we need an antenna recognizable form of the original message for or an aerial. For efficient transmission and delivering it to the user of information. reception, the transmitting and receiving 5) Noise :- A random unwanted signal is called antennas must have a length at least λ/4 noise. The source generating the noise may be where λ is the wavelength of the signal. located inside or outside the system. Efforts should be made to minimise the noise level in a For an audio signal of 15kHz, the communication system. required length of the antenna is λ/4 which 6) Attenuation :- The loss of strength of the can be seen to be equal to 5 km. signal while propagating through the channel is known as attenuation. It occurs because the The highest TV tower in Rameshwaram, channel distorts, reflects and refracts the signals Tamilnadu, is the tallest tower in India and as it passes through it. is ranked 32nd in the world with pinnacle 7) Amplification :- Amplification is the height of 323 metre. It is used for television process of raising the strength of a signal, using broadcast by the Doordarshan. an electronic circuit called amplifier. 8) Range :- The maximum (largest) distance 13.6 Modulation: between a source and a destination up to which As mentioned earlier, an audio signal has the signal can be received with sufficient strength is termed as range. low frequency (< 20 KHz). Low frequency 9) Bandwidth :- The bandwidth of an electronic signals can not be transmitted over large circuit is the range of frequencies over which it distances. Because of this, a high frequency operates efficiently. wave, called a carrier wave, is used. Some 10) Modulation :- The signals in communication characteristic (e.g. amplitude, frequency or system (e.g. music, speech etc.) are low phase) of this wave is changed in accordance frequency signals and cannot be transmitted with the amplitude of the signal. This process over large distances. In order to transmit the is known as modulation. Modulation also signal to large distances, it is superimposed on helps avoid mixing up of signals from different a high frequency wave (called carrier wave). transmitters as different carrier wave frequencies This process is called modulation. Modulation can be allotted to different transmitters. Without is done at the transmitter and is an important the use of these waves, the audio signals, if part of a communication system. transmitted directly by different transmitters, 11) Demodulation :- The process of regaining would have got mixed up. signal from a modulated wave is called demodulation. This is the reverse process of Modulation can be done by modifying modulation. the (i) amplitude (amplitude modulation) 12) Repeater :- It is a combination of a (ii) frequency (frequency modulation), and (iii) transmitter and a receiver. The receiver receives phase (phase modulation) of the carrier wave in the signal from the transmitter, amplifies it and proportion to the amplitude or intensity of the transmits it to the next repeater. Repeaters are 239
signal wave keeping the other two properties susceptible to noise. This modulation is used forDisplacement same. Figure 13.9 (a) shows a carrier wave high quality broadcast transmission. and (b) shows the signal. The carrier wave is a high frequency wave while the signal is a Phase modulation (PM) is easier than low frequency wave. Amplitude modulation, frequency modulation. It is used in determining frequency modulation and phase modulation of the velocity of a moving target which cannot be carrier waves are shown in Fig. 13.9 (c), (d) and done using frequency modulation. (e) respectively. a Amplitude modulation (AM) is simple to implement and has large range. It is also b cheaper. Its disadvantages are that (i) it is not very efficient as far as power usage is concerned c (ii) it is prone to noise and (iii) the reproduced signal may not exactly match the original signal. d In spite of this, these are used for commercial broadcasting in the long, medium and short e wave bands. Time Frequency modulation (FM) is more Fig. 13.9: (a) Carrier wave, (b) signal (c) complex as compared to amplitude modulation AM (d) FM and (e) PM. and, therefore is more difficult to implement. However, its main advantage is that it Internet my friend reproduces the original signal closely and is less https://www.iiap.res.in/centers/iao ExercisesExercises 1. Choose the correct option. vi) The waves used by artificial satellites for i) The EM wave emitted by the Sun and communication purposes are responsible for heating the Earth’s (A) Microwave atmosphere due to green house effect is (B) AM radio waves (A) Infra-red radiation (B) X ray (C) FM radio waves (C) Microwave (D) Visible light (D) X-rays ii) Earth ’s atmosphere is richest in vii) If a TV telecast is to cover a radius of (A) UV (B) IR 640 km, what should be the height of (C) X-ray (D) Microwaves transmitting antenna? iii) How does the frequency of a beam of (A) 32000 m (B) 53000 m ultraviolet light change when it travels (C) 42000 m (D) 55000 m from air into glass? 2. Answer briefly. (A) No change (B) increases i) State two characteristics of an EM wave. (C) decreases (D) remains same ii) Why are microwaves used in radar? iv) The direction of EM wave is given by iii) What are EM waves? (B) E . B iv) How are EM waves produced? (A) E ×B v) Can we produce a pure electric or (C) along E (D) along B v) The maximum distance upto which TV magnetic wave in space? Why? transmission from a TV tower of height h vi) Does an ordinary electric lamp emit EM can be received is proportional to waves? (A) h1/2 (B) h vii) Why do light waves travel in vacuum (C) h3/2 (D) h2 whereas sound wave cannot? 240
viii) What are ultraviolet rays? Give two uses. iii) The speed of light is 3×108 m/s. Calculate the frequency of red light of wavelength of ix) What are radio waves? Give its two uses. 6.5×10-7 m. x) Name the most harmful radiation [Ans: υ = 4.6×1014 Hz] entering the Earth's atmosphere from the iv) Calculate the wavelength of a microwave outer space. of frequency 8.0 GHz. xi) Give reasons for the following: [Ans: 3.75 cm] (i) Long distance radio broadcast uses v) In a EM wave the electric field oscillates short wave bands. sinusoidally at a frequency of 2×1010 Hz. (ii) Satellites are used for long distance What is the wavelength of the wave? TV transmission. [Ans: 1.5×10-2 m] xii) Name the three basic units of any vi) The amplitude of the magnetic field part communication system. of a harmonic EM wave in vacuum is xiii) What is a carrier wave? B0= 5×10-7 T. What is the amplitude of the electric field part of the wave? xiv) Why high frequency carrier waves are used for transmission of audio signals? [Ans: 150V/m] xv) What is modulation? vii) A TV tower has a height of 200 m. xvi) What is meant by amplitude modulation? How much population is covered by TV xvii) What is meant by noise? transmission if the average population xviii) What is meant by bandwidth? density around the tower is 1000/km2? xix) What is demodulation? (Radius of the Earth = 6.4×106 m) xx) What type of modulation is required for [Ans: 8×106] television broadcast? viii) Height of a TV tower is 600 m at a given xxi) How does the effective power radiated place. Calculate its coverage range if the by an antenna vary with wavelength? radius of the Earth is 6400 km. What xxii) Why should broadcasting programs use should be the height to get the double different frequencies? coverage area? xxiii) Explain the necessity of a carrier wave [Ans: 87.6 km, 1200 m] in communication. ix) Atransmitting antenna at the top of a tower xxiv) Why does amplitude modulation give has a height 32 m and that of the receiving noisy reception? antenna is 50 m. What is the maximum xxv) Explain why is modulation needed. distance between them for satisfactory 2. Solve the numerical problem. communication in line of sight mode ? i) Calculate the frequency in MHz of a radio Given radius of Earth is 6.4×106 m. wave of wavelength 250 m. Remember [Ans: 45.537 km] that the speed of all EM waves in vacuum *** is 3.0×108 m/s. [Ans: 1.2 MHz] ii) Calculate the wavelength in nm of an X-ray wave of frequency 2.0×1018 Hz. [Ans: 0.15 nm] 241
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