JEE-Physics and the time period, T= 2 = 2 m i.e., time period (or frequency) is independent of speed of particle qB and radius of the orbit. Time period depends only on the field B and the nature of the particle, i.e., specific charge (q/m) of the particle. This principle has been used in a large number of devices such as cyclotron (a particle accelerator), bubble-chamber (a particle detector) or mass-spectrometer etc. • Case : The charged particle is moving at an angle to the field : ( 0°, 90° or 180°). Resolving the velocity of the particle along and perpendicular to the field.The particle moves with constant velocity v cos along the field ( no force acts on a charged particle when it moves parallel to the field). And at the same time it is also moving with velocity v sin perpendicular to the field due to which it will describe a circle (in a plane perpendicular to the field) Radius of the circular path r = m(v sin ) and Time period 2r 2m qB T= v sin = qB So the resultant path will be a helix with its axis parallel to the field B as shown in fig. v sin Helical Path r v cos B p The pitch p of the helix = linear distance travelled in one rotation p=T (vc os)= 2 m ( v cos ) qB Example An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV enters a region with uniform magnetic field of 0.15 T. Determine the radius of the trajectory of the electron if the field is – (a) Transverse to its initial velocity(b) Makes an angle of 30° with the initial velocity [Given : me = 9 × 10–31 kg] Solution 1 2eV 2 1.6 1019 2 103 = 8 × 107 m/s mv2 = eV v = m= 3 9 1031 2 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 b gmv 9 1031 8 3 107 (a) Radius r1 = qB = 1.6 1019 0.15 = 10–3 m = 1mm mv sin 1 (b) Radius r2 = qB = r1sin =1× sin30° =1 × 2 = 0.5 mm MOTION OF CHARGED PARTICLE IN COMBINED ELECTRIC AND MAGNETIC F I E L D S Let a moving charged particle is subjected simultaneously to both electric field E and magnetic field B . The moving charged particle will experience electric force Fe qE and magnetic force Fm =q ( v B) . Net force on the charge particle \"Lorentz-force\" F q(E v B) Depending on the direction of and various situation are possible and the motion in general v,E B is quite complex. Case : v , E and B all the three are collinear : E 29
JEE-Physics E v v' qq B As the particle is moving parallel or antiparallel to the field. The magnetic force on it will be zero particle a and only electric force will act So, acceleration of the F qE mm Hence, the particle will pass through the field following a straight line path (parallel to the field) with change in its speed. In this situation speed, velocity, momentum and kinetic energy all will change without change in direction of motion as shown in figure above. Case : , and are mutually perpendicular : v E B qv E v Fe q Fm B If in this situation direction and magnitude of E and B are such that F Resultant force F Fe Fm 0 a m 0 Then as shown in fig., the particle will pass through the field with same velocity F =F i.e. qE=qvB E em v B Example A beam of protons is deflected sideways. Could this deflection be caused by (i) a magnetic field (ii) an electric field? If either possible, what would be the difference? Solution Yes, the moving charged particle (e.g. proton, -particles etc.) may be deflected sideway either by an electric or by a magnetic field. (i) The force exerted by a magnetic field on the moving charged particle is always perpendicular NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 to direction of motion, so that no work is done on the particle by this magnetic force. That is the magnetic field simply deflects the particle and does not increase its kinetic energy. (ii) The force exerted by electric field on the charged particle at rest or in motion is always along the direction of field and the kinetic energy of the particle changes. Example A neutron, a proton, an electron an -particle enter a region of constant magnetic field with equal velocities. The magnetic field is along the inwards normal to the plane of the paper. The tracks of the particles are shown in fig. Relate the tracks to the particles. C ×××× B× × × × A× × × × × × × ×D 30 E
JEE-Physics Sol. Force on a charged particle in magnetic field F =q (v B) For neutron q=0, F=0 hence it will pass undeflected i.e., tracks C corresponds to neutron. If the particle is negatively charged, i.e. electron. F e(v B) It will experience a force to the right; so track D corresponds to electron. If the charge on particle is positive. It will experience a force to the left; so both tracks A and B corresponds to positively charged particles (i.e., protons and -particles). When motion of charged particle perpendicular to the magnetic field the path is a circle with radius mv rm m 4m m m m m r= qB q q 2e while q p e i.e. and as q q p So r > r p track B to -particle and A corresponds to proton. Example An electron does not suffer any deflection while passing through a region. Are you sure that there is no magnetic field? Is the reverse definite? Solution If electron passing through a certain region does not suffer any deflection, then we are not sure that there is no magnetic field in that region. This is due to that electron suffers no force when it moves parallel or antiparallel to magnetic field. Thus the magnetic field may exist parallel or antiparallel to the direction of motion of electron. The reverse is not true since an electron can also be deflected by the electric field. Example In a chamber, a uniform magnetic field of 8 × 10–4 T is maintained. An electron with a speed of 4.0 × 106m/s enters the chamber in a direction normal to the field. (a) Describe the path of the electron. (b) What is the frequency of revolution of the electron? (c) What happens to the path of the electron if it progressively loses its energy due to collisions with the atoms or molecules of the environment ? Solution mv 9.1 1031 4 106 (a) The path of the electron is a circle of radius r= Be = 1.6 1019 8 104 = 2.8 × 10–2m NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 The sense of rotation of the electron in its orbit can be determined from the direction of the centripetal force. F = – e ( v × B ) . So, if we look along the direction of B , the electron revolves clockwise. (b) the frequency of revolution of the electron in its circular orbit eb 1.6 1019 8.0 104 f= = Hz = 22.4 MHz 2 m 2 9.1 1031 (c) Due to collision with the atomic consistent of the environment, the electron progressively loses its speed. If the velocity vector of the electron remains in the same plane of the initial circular orbit after collisions, the radius of the circular orbit will decrease in proportion to the decreasing speed. However, in general, the velocity of the electron will not remain in the plane of the initial orbit after collision. In that case, the component of velocity normal to B will determine the radius of the orbit, while the component of velocity parallel to B remains constant. Thus, the path of the electron, between two collisions is, in general, helical. But an important fact must be noted: the frequency of orbital revolution remains the same, whatever be the speed of the electron. E 31
JEE-Physics Example A beam of protons with velocity 4 × 105 m/s enters a uniform magnetic field of 0.3 tesla at an angle of 60° to the magnetic field. Find the radius of the helical path taken by the proton beam. Also find the pitch of helix. Mass of proton=1.67× 10–27kg. Solution mv sin Radius of helix r= qB (component of velocity to field is vsin) (1.67 1027 )(4 105 ) 3 vsin =2 2 r (1.6 1019 )0.3 = 3 × 10–2m = 1.2cm vcos B p 2 r Again, pitch p=vcos × T (where T= v sin ) v cos 2r cos 60 2 (1.2 102 ) = 10–2m=4.35cm p= v sin = 4.35× sin 60 Example The region betwen x = 0 and x = L is filled with uniform, steady magnetic field B0kˆ . A particle of mass m, positive charge q and velocity v0ˆi travels along X-axis and enters the region of magnetic field. Neglect the gravity throughout the question. (a) Find the value of L if the particle emerges from the region of magnetic field with its final velocity at an angle 30° to its initial velocity. (b) Find the final velocity of the particle and the time spent by it in the magnetic field, if the magnetic field now extends upto 2.1 L. Solution (a) The particle is moving with velocity v0ˆi , perpendicular to magnetic field B0kˆ . Hence the particle will move along x=0 x=L mv0 D A NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 a circular arc OA of radius r = qB0 r 30° Let the particle leave the magnetic field at A. 30° mv0 C 2qB0 From CDA, sin60°= AD L L = r = CA r rsin30°= 2 L (b) As the magnetic field extends upto 2.1 L i.e., L > 2r, y v0i so the particle completes half cycle before leaving the magnetic field, as shown in figure. The magnetic field is always perpendicular to velocity vector, therefore the magnitude of velocity will remain the same. r m v0i Final velocity = v0 (ˆi ) =– v0ˆi Time spent in magnetic field= v0 = qB0 32 E
JEE-Physics Example A uniform magnetic field with a slit system as shown in fig. is to be used as a momentum filter for high energy charged particles. With a field of B tesla it is found that the filter transmits -particle each of energy 5.3 MeV. The magnetic field is increased to 2.3 B tesla and deuterons are passed into the filter. What is the energy of each deuteron transmitted by the filter ? Source Detector Solution In case of circular motion of a charged particle in a magnetic field E = r2q 2B2 k 2m b g b g b g b gSo according to the given problem (E ) = b g b gk r2 2e 2 B2 r 2 e 2 2.3B 2 (E k )D 2.3 2 4 (E k ) 22 2 2 4m and (E ) = 2 2m i.e. kD (5.3) 5.29 (E ) = × = 14.02 MeV kD 2 Example A charged sphere of mass m and charge q starts sliding from rest on a vertical fixed circular track of radius R from the position as shown in figure . There exists a uniform and constant horizontal magnetic field of induction B. Find the maximum force exerted by the track on the sphere. qm B NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 S o l . Magnetic force on sphere F = qvB (directed radially outward) RO m N – mg sin – qvB = mv2 N = mv2 + mg sin + qvB RR Fm N R mgsin mg Hence, at = /2 we get N = 2mgR + mg + qB 2gR = 3mg + qB 2gR max R Example An electron gun G emits electrons of energy 2keV travelling in the positive x-direction. The electrons are required to hit the spot S, where GS = 0.1m, and the line GS makes an angle of 60° with the x-axis, as shown in the figure. A uniform magnetic field B parallel to GS exists in the region outside the electron gun. Find the minimum value of B needed to make the electron hit S. BS 60° x G E 33
JEE-Physics S o l . The velocity of the electrons emitted by electron gun along x-axis, is v = 2EK E= 1 m mv2 2 The velocity of the electron can be resolved into two components v cos and v sin , parallel and perpendicular to the magnetic field respectively. Due to component v cos electron will move in the direction of magnetic field with constant speed v cos but due to component v sin , it will move on a circular path in the plane to magnetic field. Hence electron will move on a spiral path. As electrons are required to hit the spot S, hence distance travelled by electron in one time period along the direction of magnetic field must be just equal to GS. The electrons may also hit the spot S after two or more time periods but minimum value of B is required As per above discussion, G S = Distance travelled along the direction of magnetic field in one time period 2m 2E 1 qB B = m × cos × 2m × q GS 2m 2EK m b g= (v cos ) × T = v cos × qB = × cos × = 2 2 1.6 10 16 cos 60 2 3.14 9.1 10 31 = 4.68 × 10–3 T × 9.1 10 31 1.6 10 19 0.1 MOVING COIL GALVANOMETER A galvanometer is used to detect the current and has moderate resistance. N iS Principle. When a current carrying coil is placed in a magnetic field, it experiences a torque given NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 by = NiAB sin where is the angle between normal to plane of coil and direction of magnetic field. In actual arrangement the coil is suspended between the cylindrical pole pieces of a strong magnet. The cylindrical pole pieces give the field radial such that sin =1 (always). So torque =NiAB If C is torsional rigidity (i.e., restoring couple per unit twist of the suspension wire), then for deflection of coil =C. In equilibrium we have external couple = Restoring couple i.e.C=NiAB or NAB i i.e., i C In words the deflection produced is directly proportional to current in the coil. NAB The quantity = is called the current sensitivity of the galvanometer. Obviously for greater iC sensitivity of galvanometer the number of turns N, area of coil A and magnetic field B produced by pole pieces should be larger and torsional rigidity C should be smaller. That is why the suspension wire is used of phosphor bronze for which torsional rigidity C is smaller. 34 E
JEE-Physics CON VERSION OF GALVANOMETER INTO A MMETER An ammeter is a low resistance galvanometer; used to measure current directly in amperes and is always connected in series with the circuit. To convert a galvanometer into ammeter, a low resistance, called shunt, is connected in parallel to the galvanometer as shown in figure. S i a ig G b i Let i be the current in galvanometer for its full scale deflection and G the resistance of galvanometer. g Let i is the range of ammeter and i the current in shunt S. Then potential difference across a and s b is V = i G = i S. ...(i) ab g S At junction a, i=i + i i.e., i =i–i Sg Sg Therefore from (i) i G = (i–i )S or i (S+G) = iS i.e., i = S i ...(ii) g gg g SG This is the working equation for conversion of galvanometer into ammeter. Here i < i. g igG . If i << i, S= ig G From (ii) shunt required S= i ig i g 1 11 SG The resistance of ammeter R so formed is given by = + R = ....(iii) A RA G S A SG Note: Equation (ii) may also be used to increase the range of given ammeter. Here G will be resistance of given ammeter, S shunt applied, i its initial range and i the new range desired. g NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 CON VERSION OF GALVANOMETER INTO VOLTMETER A voltmeter is a high resistance galvanometer and is connected between two points across which potential difference, is to be measured i.e., voltmeter is connected in parallel with the circuit. To convert a galvanometer into voltmeter, a high resistance R in series is connected to the galvanometer. ig ig G R If V is range of voltmeter, then i = V or resistance in series V R = ig G ...(i) g RG This is working equation for conversion of galvanometer into voltmeter. The resistance of voltmeter so formed is R = R+G ...(ii) V Note: Equation (i) may also be used to increase the range of voltmeter. If V is initial range and V 0 is new range of voltmeter, then i = V0 V g G RG E 35
JEE-Physics SOME WORKED OUT EXAMPLES Example#1 Current i = 2.5 A flows along the circle x2 + y2 = 9 cm2 (here x & y in cm) as shown. Magnetic field at point (0, 0, 4 cm) is (A) 36 10 7 T kˆ (B) 36 107 T kˆ (C) 9 10 7 T kˆ (D) 9 10 7 T kˆ 5 5 Solution Ans. (A) Magnetic field on the axis of a circular loop B 0 2iR 2 10 7 2 2.5 32 104 kˆ 9 1 0 5 T kˆ 36 1 0 7 T kˆ 4 25 R 2 z2 3/2 125 106 Example#2 There are constant electric field E0ˆj & magnetic field Bkˆ present between plates P and P'. A particle of mass m is projected from plate P' along y axis with velocity v. After moving on the curved path, it passes through 1 point A just grazing the plate P with velocity v. The magnitude of impulse (i.e. F t p ) provided by magnetic 2 force during the motion of particle from origin to point A is :– y B0 E0 v2 Am P v1 x NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 m P' (A) m|v –v | (B) m v 2 v 2 (C) mv (D) mv 21 1 2 1 2 Solution Ans. (D) Electric force is only responsible for the change in momentum along y–axis. Therefore impulse provide by magnetic force is J = mv . B2 Example#3 Three identical charge particles A, B and C are projected perpendicular to the uniform magnetic field with velocities v , v and v (v < v < v ) respectively such that T , T and T are their respective time period 12 31 2 3 12 3 of revolution and r , r and r are respective radii of circular path described. Then :- 12 3 (A) r1 r2 r3 (B) T < T < T (C) r1 r2 r3 (D) r = r = r T1 T2 T3 123 T1 T2 T3 123 36 E
Solution JEE-Physics Ans. (C) T= 2m mv r v &r= qB qB T Example#4 An infinitely long straight wire is bent as shown in figure. The circular portion has a radius of 10 cm with its center O at a distance r from the straight part. The value of r such that the magnetic field at the center O of the circular portion is zero will be :- 10cm O r 10 20 1 5 (A) cm (B) cm (C) cm (D) cm 5 Solution Ans. (A) B circular loop B wire 0I 0 I r 10 cm 2 10 2 r Example#5 Two cylindrical straight and very long non magnetic conductors A and B, insulated from each other, carry a current I in the positive and the negative z–direction respectively. The direction of magnetic field at origin is y A B x (A) ˆi (B) ˆi (C) ˆj (D) – ˆj Solution x Ans. (C) y AB BA BB NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 Example#6 The magnetic force between wires as shown in figure is :- iL x I (A) 0 iI2 n xL (B) 0 iI2 n 2x L (C) 0iI n x L (D) None of these 2 2x 2 2x 2 x E 37
JEE-Physics Solution Ans. (C) Magnetic field at dr, B = 0 I iL 2 r dr0 I Force on small element at a distance r of wire of length L is dF = i(drr) 2xr F 0i I xL dr 0i I n x L I x r 2 x 2 Example#7 A wire carrying a current of 4A is bent in the form of a parabola x2 + y = 16 as shown in figure, where x and y are in meter. The wire is placed in a uniform magnetic field B 5 kˆ tesla. The force acting on the wire is y x (A) 80 ˆj N (B) – 80 ˆj N (C) 160 ˆj N (D) 160 ˆj N Ans. (C) Solution I 4 8ˆi 5 kˆ 1 6 0 ˆj N F B Example#8 A conducting coil is bent in the form of equilateral triangle of side 5 cm. Current flowing through it is 0.2 A. The magnetic moment of the triangle is :- (A) 3 × 10–2 A–m2 (B) 2.2 × 10–4 A–m2 (C) 2.2 × 10–2 A–m2 (D) 3 × 10–4 A–m2 Solution Ans. (B) Magnetic moment of current carrying triangular loop M = IA M 0.2 1 5 10 2 3 5 102 2 2 = 2.2 × 10–4 A-m2 Example#9 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 A disc of radius r and carrying positive charge q is rotating with angular speed in a uniform magnetic field B about a fixed axis as shown in figure, such that angle made by axis of disc with magnetic field is . Torque applied by axis on the disc is Disc B qr2B sin Fixed axis qr2B sin (A) , clockwise (B) , anticlockwise 38 2 4 qr2B sin qr2B sin (C) , anticlockwise (D) , clockwise 2 4 E
Solution JEE-Physics Ans. (D) M q M q mr2 qr2B sin clockwise M B L 2m 2m 2 4 Example#10 A particle of mass m and charge q is thrown from origin at t = 0 with velocity 2 i 3 j 4k units in a region with m uniform magnetic field 2 i units. After time t = qB , an electric field E is switched on, such that particle moves on a straight line with constant speed. E may be (A) –8ˆj 6kˆ units (B) 6ˆi 9kˆ units (C) 12ˆj 9kˆ units (D) 8ˆj 6kˆ units Solution Ans. (A) At t = n , 2ˆi 3ˆj 4kˆ ; For net force to be zero 8ˆj 6kˆ qB v qv B qE 0 E v B Example#11 A particle of specific charge q 1010 C kg–1 is projected from the origin along the positive m x–axis with a velocity of 105 ms–1 in a uniform magnetic field 2 103 kˆ tesla. Choose correct alternative(s) B (A) The centre of the circle lies on the y–axis (B) The time period of revolution is 10–7 s. 5 (C) The radius of the circular path is mm (D) The velocity of the particle at t = 1 107 s is 105 ˆj m / s 4 Solution Ans. (A,B,C,D) B F q , 105 ˆi , Force will be in y–direction y v v B 2 103 kˆ t= T 4 Motion of particle will be in xy plane Time period T 2m 2 107 s qB 1010 2 103 Radius of path mv 105 5 10 3 m 5 mm = qB 1010 2 10 3 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 T 10 7 x At t = s. Velocity of particle will be in +y direction. t=0 44 Example#12 y x A circular current carrying loop of radius R is bent about its diameter by 90° and I z I placed in a magnetic field B B0 (ˆi ˆj) as shown in figure. (A) The torque acting on the loop is zero (B) The magnetic moment of the loop is IR 2 ˆi ˆj 2 (C) The angular acceleration of the loop is non zero. (D) The magnetic moment of the loop is IR 2 ˆi ˆj 2 E 39
JEE-Physics Solution Ans. (A,B) IR2 M B M 2 ˆi ˆj ; 0 Example#13 A current–carrying ring is placed in a magnetic field. The direction of the field is perpendicular to the plane of the ring– (A) There is no net force on the ring. (B) The ring will tend to expand. (C) The ring will tend to contract. (D) Either (B) or (C) depending on the directions of the current in the ring and the magnetic field. Solution Ans. (A,D) Net force = 0 and ring will tend to expand/contracts depending on I & B. Example#14 In the given figure, B is magnetic field at P due to shown segment AB of an infinite current carrying wire. A loop is taken as shown in figure. Which of the following statement(s) is/are correct. (B) B= 0i 2 r (A) B d 0i (C) Magnetic field at P will be tangential (D) None of these Solution Ans. (C) Magnetic field at P is tangential. Example#15 to 17 Two charge particles each of mass ‘m’, carrying charge +q and connected with each other by a massless inextensible string of length 2L are describing circular path in the plane of paper, each with speed v qB0L (where B is constant) about their m0 centre of mass in the region in which an uniform magnetic field B exists into the plane of paper as shown in figure. Neglect any effect of electrical & gravitational forces. 1 5 . The magnitude of the magnetic field such that no tension is developed in the string will be NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 (A) B 0 (B) B (C) 2B (D) 0 2 0 0 1 6 . If the actual magnitude of magnetic field is half to that of calculated in part (i) then tension in the string will be 3 q 2 B 2 L q 2 B 2 L 2 q 2 B 2 L 0 0 0 (A) (B) zero (C) (D) 4m 2m m 3 q 2 B 2 L 0 17. Given that the string breaks when the tension is T = . Now if the magnetic field is reduced to such a 4m value that the string just breaks then find the maximum separation between the two particles during their motion (A) 16 L (B) 4L (C) 14L (D) 2L 40 E
JEE-Physics Solution q+ T + 2q 15. Ans. (B) qvB mv2 T + qvB = L T = 0; q qB0 L B = mv2 B = B m L 0 16. Ans. (C) + T qvB 0 mv2 m(qB0L )2 q(qB0L ) B0 q 2 B 2 L m2R m2 0 T+ ; T 2R 2m qv(B0/2) 17. Ans. (C) Maximum reperation = 2R + (2R – 2L) 2L v V R mv2 = R =16L T + qvB = B B /4 = 4L maximum seperation – 2L = 14L R 0 Example#18 to 20 Curves in the graph shown give, as functions of radial distance r, the magnitude B of the magnetic field inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross sections of the wires. The wires are far from one another. B a bc 1 8 . Which wire has the greatest radius? d (D) d (A) a (B) b r (C) c 1 9 . Which wire has the greatest magnitude of the magnetic field on the surface? (D) d (A) a (B) b (C) c NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 2 0 . The current density in wire a is (B) less than in wire c (A) greater than in wire c (D) not comparable to that in wire c due to lack of information (C) equal to that in wire c Solution 18. Ans. (C) Inside the cylinder : B 0 I r ...(i) R 2R 2 I Outside the cylinder B 0 I ...(ii) r 2 r r 1 Inside cylinder B r and outside B r So from surface of cylinder nature of magnetic field changes. Hence it is clear from the graph that wire ‘c’ has greatest radius. E 41
JEE-Physics 19. Ans. (A) Magnitude of magnetic field is maximum at the surface of wire ‘a’. 20. Ans. (A) 0 I dB 0 I I Inside the wire B(r) 2 R2 r ; dr 2 R2 i.e. slope current density. R 2 It can be seen that slope of curve for wire a is greater than wire c. Example#21 Column-I gives some current distributions and a point P in the space around these current distributions. Column- II gives some expressions of magnetic field strength. Match column-I to corresponding field strength at point P given in column-II Column – I Column – II (A) A conducting loop shaped as regular hexagon of side x, (P) 30i carrying current i. P is the centroid of hexagon 32x (B) A cylinder of inner radius x and outer radius 3x, carrying (Q) 30i current i. Point P is at a distance 2x from the axis of x the cylinder (C) Two coaxial hollow cylinders of radii x and 2x, each carrying (R) 0i current i, but in opposite direction. P is a point at distance 2x 1.5x from the axis of the cylinders (D) Magnetic field at the centre of an n-sided regular (S) 0i Solution polygon, of circum circle of radius x, carrying current 3x i, n , P is centroid of the polygon. (T) Zero Ans. (A) (Q) ; (B) (P) ; (C) (S) ; (D) (R) For A : BP 6 4 0i 60 sin 30 sin 30 30i P x sin x x 60° x 0 3 i 30i NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 8 3 2 x For B: BP 2x P 2 2x x 3x For C : BP 0i 0i 3x 1.5x P 2 1.5 x x 3x For D : If n , n sided polygon circle so B = 0i 2x 42 E
JEE-Physics Example#22 A very small current carrying square loop (current I) of side 'L' is placed in y-z plane with centre at origin of the coordinate system (shown in figure). In column–I the coordinate of the points are given & in column–II magnitude of strength of magnetic field is given. Then .y P2(0, a, 0) .P3(a, a, 0) .I 0x P1(a, 0, 0) z Column I Column II (A) At point O (0, 0, 0) (P) 0 I L2 2 a 3 (B) At point P (a, 0, 0) (here a > > L) (Q) 2 20I 1 L (C) At point P (0, a, 0) (here (a > > L) (R) 0 5 I L2 2 16 a3 (D) At point P (a, a, 0) (here a > > L) (S) 0 I L2 3 4a3 (T) 0 5 IL2 4 a 3 2 . Ans. (A) (Q); (B) (P); (C) (S); (D) (R) For A : B P1 4 0 I sin sin 2 20 I 4 4 L 4L / 2 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 2kM 2 0 IL2 0 IL2 4 For B: B P2 r 3 a 3 2a 3 kM 0 IL2 0 IL2 r3 4 4 a3 For C : B P3 a3 kM 0 IL2 0 5IL2 For D : B P4 r 3 4 1 6 a 3 1 3 cos2 where cos 1 3 1 2 2 B P4 3 1 a2 E 43
JEE-Physics MAGNETISM GEO-MAGNETISM (BY DR. WILLIUM GILBERT) The branch of physics which deals with the study of earth's magnetic field is called geomagnetism. Important definitions : ( a ) Geographic axis : It is a straight line passing through the geographical poles of the earth. It is also called axis of rotation or polar axis of the earth. (b ) Geographic Meridian (GM) : It is a ver- Geographical tical plane at any place which passing through geographical axis of the earth. axis or polar axis Geographical North pole Magnetic south pole (c) Geographic equator : It is a great circle Geographical equator on the surface of the earth, in a plane perpendicular to the geographic axis. All the points on the geographic equator are at Magnetic north pole equal distance from the geographic poles. Magnetic equator Geographical South pole A great plane which passes through geographic equator and perpendicular to the geographic axis called geographic equatorial plane. This plane cuts the earth in two equal parts, a part has geographic north called northen hemisphere (NHS) and another part has geographic south called southern hemi sphere (SHS). ( d ) Magnetic axis : It is a straight line passing through magnetic poles of the earth. It is inclined to the geographic axis at nearly 17°. (e) Magnetic Meridian (MM) : (i) It is a vertical plane at North NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 any place which passing through magnetic axis of the earth. East (ii) It is a vertical plane at any place which passing through axis of free suspended bar magnet or magetic needle. West Magnetic meridian (iii) It is a vertical plane at any place which contains all South the magnetic field lines of earth of that place. ( f ) Magnetic equator : It is a great circle on the surface of the earth, in a plane perpendicular to the magnetic axis. All the points on the magnetic equator are at equal distance from the magnetic poles. MAIN ELEMENTS OF EARTH'S MAGNETIC FIELD Angle of declination () GM At a given place the acute angle between geographic meridian and the MM magnetic meridian is called angle of declination, i.e. at a given place it is the angle between the geographical north south direction and the direcation indicated by a agnetic compass needle in its equillibrium. 96 E
JEE-Physics (i) Angle of dip () GM (ii) It is an angle which the direction of resultant magnetic field of the BH horizontal earth substends with the horizontal line in magnetic meridian at the line E given place. B BV It is an angle which the axis of freely suspended magnetic needle MM for NHS (up or down) substends with the horizontal line in magnetic meridian at a given place. In northen hemi sphere, north pole of freely suspended magnetic needle will dip downwards i.e. towards the earth surface. In southern hemi sphere, south pole of freely suspended magnetic needle will dip downwards i.e. towards the earth surface. Dip circle : Angle of dip at a place is measured by the instrument called 'Dip-circle' in which a 90 0 Horizontal S 0° magnetic needle is free to rotate in vertical plane. 0 Dip About its horizontal axis. The ends of the needle EB 90º move over a vertical scale graduated in degree. Horizontal component of earth magnetic field (B ) Direction of earth's magnetic H field Horizontal component of earth magnetic field at a given place is the component of resultant mag- netic field of the earth along the horizontal line in magnetic meridian. B = Bcos and B = B sin .......(1) H v so that BV and B = B 2 B 2 .......(2) tan = BH H V At magnetic poles = 90° B = 0 and only B exist HV At magnetic equator = 0° B = 0 and only B exist VH decides the plane in which magnetic field lies at any place, () and () decides the direction of magnetic field and () and (B ) decides the magnitude of the field. H Apparent angle of dip (') : When the plane of vertical scale of dip circle is in the magnetic meridian, the BH' = B cos needle rest in the direction of earth's magnetic field. The angle made by the H needle with the horizontal is called true dip or actual dip. If the plane of NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 vertical scale of dip circle not kept in magnetic meridian, then the needle will GM not indicate the correct direction of earth magnetic field. BH In this situation the angle made by the needle with the horizontal is called the BV' =BV apparent angle of dip.Suppose the dip circle is set at an angle to t h e B MM magnetic meridian. Effective horizontal component in this plane will be B cos H and no effect on vertical component B V Apparent angle of dip B'V BV tan ' tan tan ' = B'H tan ' = B H cos cos For a vertical plane other than magnetic meridian > 0 cos < 1 tan' > tan ' > , so apparent angle of dip is always more than actual angle of dip at any place. tan For a vertical plane perpendicular to magnetic meridian = 90° tan' = cos 90 = ' = 90° , so in a plane perpendicular to magnetic meridian dip needle becomes just verticle. 97
JEE-Physics Example At a certain place, the horizontal component of earth's magnetic field is 3 times of the vertical component. What the angle of dip at that place. Solution BV BV 1 B= 3 B , tan = B H = 3B v = 3 = tan30° = 30° H V Example A compass needle of magnetic moment is 60 A-m2 pointing towards geographical north at a certain place where the horizontal component of ear th's magnetic field is 40T, experiences a torque 1.2 × 10–3 N-m. What is the declination of that place. Solution Sm N N 1.2 10 3 1 WE MM = MBsin sin = MB = 24 10 4 = = 30° 2 GM Nm S Example If the dip circle is set at 45° to the magnetic meridian, then the apparent dip is 30°. Calculate the true dip. Solution F Itan HG KJtan' = cos ; tan = tan'cos = tan30°cos45° = 3 × 2 ; tan = 6 = tan–11 1 11 6 Example A magnetic needle is free to rotate in a vertical plane and that plane makes an angle of 600 with magnetic 2 meridian. If the needle stays in a direction making an angle of tan–1 with the horizontal direction, what 3 would be the actual dip at that place ? Solution ( 2 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 tan–1 , tan = tan'cos ' = 3 = 60°) 2 21 1 tan = tan(tan–1 3 )cos60° tan = 3 × 2 = 3 = 30° Example A 1-meter long narrow solenoid having 1000 turns is placed in magnetic meridian. Find the current in the solenoid which neutralises the earth’s horizontal field of 0.36 oersted at the centre of the solenoid. Solution The magnetic field intensity at the centre of solenoid is H = ni = 1000 i A/m = 4i ovested ( 1 amp/meter = 4 × 10–3 oersted) Since it neutralises the earth’s field of 0.36 oersted, it is equal and opposite to the earth’s field. 4i = 0.36 0.36 i = 4 3.14 = 0.0286 ampere = 28.6 milli-ampere or 28.6 mA 98 E
JEE-Physics Example If 1 and 2 are angles of dip in two vertical planes at right angle to each other and is true dip then prove cot2 = cot21 + cot22. Solution If the vertical plane in which dip is 1 subtends an angle with meridian than other vertical plane in which dip is 2 and is perpendicular to first will make an angle of 90 – with magnetic meridian. If 1 and 2 are apparent dips than tan 1 BV tan 2 BH BV ) BV BH cos cos(90 BH sin cot2 1 cot2 2 1 1 B 2 cos2 B 2 sin2 B 2 B cos 2 cot2 (tan 1 )2 H H H B sin (tan 2 )2 B 2 B 2 V V So cot21 + cot22 = cot2 Example Considering earth as a short bar magnet show that the angle of dip is related to magnetic latitude as tan = 2tan Solution For a magnetic dipole the field components at point P (r, ) are given as B Br Br µ0 2M cos B µ0 M sin P 4 r3 4 r3 Sr tan B V Br µ0 2M cos 4 r3 B H B 4 r3 µ0 M sin N or tan = – 2 cot From figure So tan 2 cot or tan = 2tan 2 2 Example At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of the magnetic meridian points 60° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 direction and magnitude of the earth’s field at the location. Solution From formula, BH = B cos B= BH = BH sec = 0.16 × 2 = 0.32 G. cos The earth’s field is 0.32 G, in direction 60° upwards from horizontal, in a plane (magnetic meridian) 12° West of geographical meridian. Example A dip circle shows an apparent dip of 60° at a place where the true dip is 45°. If the dip circle is rotated through 90° what apparent dip will it show? Solution Let 1 and 2 be apparent dip shown by dip circle in two perpendicular positions then true dip is given by cot2 = cot21 + cot22 or cot2 45° = cot2 60° + cot22 or cot2 2 2 or cot2 = 0.816 giving 2 = 51° 3 E 99
JEE-Physics Applications of Geo-magnetism :- (Based on B ) C H PA Tangent galvanometer :- S It is an instrument which can detect/measures electric currents. It is also called moving magnet galvanometer. T1 B Principle :- It is based on 'tangent law' T2 Construction :- (i) It consists of a circular coil of a large number of turns of insulated copper wire wound over a vertical circular frame. (ii) A small magnetic compass needle is pivoted at the centre of vertical circular coil. This needle can rotate freely in a horizontal plane. Tangent law :- If a current is passed through the vertical coil, then magnetic field produced at its centre is perpendicular to the horizontal component of earth's magnetic field since coil is in magnetic meridian. So in the effect of two crossed fields (B B ) compass needle comes in equillabrium according to Coil 2(ACW) H 0 tangent law. BF N M Torque on needle due to (B ) – 1 = MB sin( 90–) 1(CW) 0 0 90- Torque on needle due to (B ) – 2 = MBHsin W E H B0 At equilibrium condition of needle net torque on it is zero S 1 2 0 ; | 1| | 2 | x Coil MB0sin(90–) = MBHsin B0cos = BHsin B =B sin ; B =B tan where B = Bcos , angle of dip. 0H 0H H cos F I2B HR 0NI = B tan HG JKso I = 0N tan 2R H I = K tan , so for this galvanometer I tan The electric current is proportional to the tangent of the angle of deflection Reduction factor : is a constant for the given galvanometer at given place. The reduction factor of a tangent galvanometer is numerically equal to the current required to produce a NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 deflection of 45° in it. = 45° I = K tan(45°) ; I = K SI unit of 'K' ampere Sensitivity : A tangent galvanometer is both sensitive and accurate if the change in its deflection is large for a given fractional change in current. = K tan or d = K sec2 d d d 2d or d sin 2 d sin cos sin 2 2 d = (d)max if sin 2 1 sin so 2 4 The tangent galvanometer has maximum sensitivity when = 45°. 100 E
JEE-Physics Example Why is a short magnetic needle used in tangent galvanometer ? Solution It is kept short so that the field arround it is uniform. Example Why does a tangent galvanometer fail to work at magnetic poles of the earth? Solution Because horizontal component of earth's field (B ) is zero at the magnetic poles. H Example A cell of an emf of 2V and internal resistance of 0.5 is sending current through a tangent galvanometer of resistance 4.5. If another external resistance of 95 is introduced, the deflection of galvanometer is 45°. Calculate the raduction factor of galvanometer. Solution E 22 I = r R R ' = 0.5 4.5 95 = 100 ampere From I = K tan reduction factor K= I 2 tan = 100 tan 45 = 0.02 ampere Example Two tangent galvanometers A and B have their number of turns in the ratio 1 : 3 and diameters in the ratio 1:2 (a) Which galvanometer has greater reduction factor (b) Which galvanometer shown greater deflection, when both are connected in series to a d.c. source. Solution NA 1 RA 1 ; Reduction factor K= KA RA NB 1 3 3 K (a) NB = 3 , RB = KB = RB × = × = A > K NA 2 1 2 B 2 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 (b) From I = Ktan , (I = same in series combination) K > K tanA < tanB A < B A B VIBR ATION M AGNETOMETER : SN It is an instrument used to compare the horizontal components of mirror magnetic field of earth of two different places, to compare magnetic fields and magetic moments of two bar magnets. It i s also cal led oscillation magnetometer. Principle : This device works on the principle, that whenever a freely suspended bar magent horizontal component in earth magnetic field (B ) is slightly disturbed from its equilibrium position H then, it will experience a torque and executes angular S.H.M. *Rotation is possible only in horizontal plane. E 101
JEE-Physics Angular S.H.M of magnetic dipole :- When a dipole is suspended in a uniform magnetic field it will align itself parallel to field. Now if it is given a small angular displacement about its equilibrium position. The restoring torque acts on it : = – M B sin I = – MB sin = – MB , ( sin ~ ) H H H = MB H (– ) = 2 (–) 2 MB H BH N II I S The time period of angular S.H.M. T 2 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 MBH M = magnetic moment of bar magnet I = moment of inertia of bar magnet about its geometric axis Comparision of magnetic moments of magnets of the same size Let the two magnets of same size have moment of inertia I and magnetic moments M and M . Suspend the 12 two given magnets turn by turn in the metal stirrup of the vibration magnetometer and note the time period in each case. II Then T = 2 M1B and T = 2 M2B 1 2 T1 M2 or M1 T22 Dividing, T2 = M1 M 2 T12 Since T and T are known therefore the ratio M2 can be determined. 12 M1 Comparision of magnetic moments of magnets of different sizes Let the two magnets have moments of inertia I and I and magnetic moments S N 12 S N M and M respectively. Place the two given magnets one upon the other as (a) sum combination 12 shown in Fig. (a). This combination is called sum combination'. It has moment of inertia (I + I ) and magnetic moment (M + M ). Put this combination in the SN 12 12 NS (b) difference combination magnetometer and set it into oscillations. The time period T is determined. 1 I1 I2 .........(1) T1 = 2 (M1 M 2 )B Now, the two magnets are placed as shown in Fig. (b). This combination is called 'difference combination'. It has moment of inertia (I + I ) and magnetic moment (M – M ). This combination is put in the magneto meter 12 12 and its time period T is determined. 2 T = 2 I1 I2 2 (M1 M 2 )B ......(2) Dividing, T1 M1 M 2 [from equation (1) and (2)] T2 = M1 M 2 knowing T and T , we can determine M1 12 M2 . 102 E
JEE-Physics Comparision of earth's magnetic field at two different places Let the vibrating magnet have moment of inertia I and magnetic moment M. Let it be vibrated in places where earth's magnetic field is B H1 and B H2 II Then, T = 2 M B H1 and T = 2 MBH2 1 2 T and T are determined by placing magnetometer at two different places, turn by turn. 12 T1 B H2 T12 B H2 B 2 cos 2 B1 T22 cos 2 Dividing, T2 = B H 1 or T22 = B H1 = B1 cos 1 B 2 = T12 cos 1 Knowing T, T and 1 2 the ratio B1 can be determined. 1 2 B2 Example Magnetic moments of two identical magnets are M and 2M respectively. Both are combined in such a way that their similar poles are same side. The time period in this is case 'T '. If polrity of one of the magnets 1 is reversed its period becomes 'T ' then find out ratio of their time periods respectively. 2 Solution T1 T2 IM IM SN SN SN NS I 2M I 2M M = 2M + M = 3M M = 2M–M = M system system I = 2I I = 2I system system I (I same, BH same) T 1 ; T = 2 MB H system M NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 T1 M 2 M1 T2 M1 3M 3 Example A magnet is suspended in such a way when it oscillates in the horizontal plane. It makes 20 oscillations per minute at a place where dip angle is 30° and 15 oscillations per min at a place where dip angle is 60°. Find the ratio of the total earth's magnetic field at the two places. Solution 1 MBH f2 = 1 . MB cos I and M are same in given cases 42 I f = 2 I B1 f12 cos 2 20 20 cos 60 16 B 2 = f22 × cos 1 = 15 15 × cos 30 = 9 3 E 103
JEE-Physics Example Time period of thin rectangular bar magnet of vibration magnetometer is 'T'. If it is broken into two equal parts than find out time period of each part at the same place. (a) Along its length (b) Perpendicular to its length Solution I ml2 Time period of thin rectangular bar magnet T = 2 MB H (here I = ) 12 2 A 2A Case I perpendicular to its axis : N S N SN S for each part mass and length both become half moment of inertia I' I magnetic moment M' M 8 8 Time period of each part T' I/8 T [B same] M/2 = H 2 If bar magnet broken into 'n' equal parts perpendicular to its axis then time period of each part becomes T T´ = NS n Case - II parallel to its axis :- for each part: mass become half and length remain same moment of inertia I' I magnetic moment M' M 2 2 (A' = A/2) Time period of each part T' I/2 (A' = A/2) M/2 If bar magnet cuts into 'n' equal parts parallel to its axis then time period of each part remain equal to time NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 period original magnet Example A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. T he time period of osci llation in a horizontal magnetic field is 25/4 sec. One of the magnets is removed and if the other magnet oscillates in the same field, calculate the time period. Solution Magnetic moment is a vector quantity. If the magnetic moments of the two magnets are M each then, the net magnetic moment when the magnets are placed perpendicular to each other, is 2I Meff. = M 2 M 2 M 2 and the moment of inertia is 2I. so T = 2 M 2H When one of the magnets is withdrawn, the time period is T' = 2 I MH T' 1 or T' = T 51 = 2 sec T 21/2 21/ 4 24 4 104 E
JEE-Physics NEUTRAL POINT It is a point where net magnetic field is zero. At this point magnetic field of bar magnet or current carrying coil or current carrying wire is just neutralised by magnetic field of earth. (B ) H A compass needle placed at this neutral point can set itself in any direction. Location of Neutral Points : N (a) When N–pole of magnet directed towards North :- Two neutral points N symmetrically located on equatorial line of magnet. Let distance of each BH BH BH neutral point from centre of magnet is 'y' then W y B =B E eq H B= 0 . M nutral y nutral H 4 (y2 2 )3 / 2 point point Bequatorial Bequ 0 M 4 y3 S S . =B (If y >>> ) nutral point N H BH (b) When S–pole of magnet directed towards North :- S Baxis Two neutral points symmetrically located on the axial line of magnet. Let BH distance of each neutral points from centre of the magnet is x, then x 0 . 2Mx W E 4 (x2 2 )2 x B = B BH axis H 0 2M =B (If x >>> ) N BH 4 H nutral point S Baxis x3 (c) If magnet is held vertically on the board, then only one neutral point is obtained on the horizontal board. N N nutral point BH N S BH S nutral point S When N-pole on the board When S-pole on the board NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 Example The magnetic field at a point x on the axis of a small bar magnet is equal to the field at a point y on the equator of the same magnet. Find the ratio of the distances of x and y from the centre of the magnet. Solution B =B 0 2M = 0 M 2 1 x3 2 x axis equatorial 4 x3 4 y3 x 3 = y 3 y 3 = 1 y = 21/3 Example A coil of 0.1 m radius and 100 turns placed perpendicular magnetic meridian. When current of 2 ampere is flow through the coil then the neutral point is obtained at the centre. Find out magnetising field of earth. Solution Magnetic field at centre of coil B = 0NI = µ H ( H = H) C e 2R 0 C Magnetising field of earth NI 100 2 He= 2R = 2 0.1 = 1000 A/m E 105
JEE-Physics Example A short magnet of moment 6.75 A–m2 produces a neutral point on its axis. If the horizontal component of earth's magnetic field 5 x 10–5 Wb/m2, Calculate the distance of the neutral point from the centre. Solution F I F I2KM 1/3 2 10 7 6.75 1/3 2KM GH KJ HG JKBH = d3 d = B H = 5 10 5 = 0.3 m = 30 cm MISCELLENEOUS Magnetic field of long Bar magnet r+ S 2 N B2 B1 r (i) At Axial position :– Magnetic field at point 'P' due to north pole B1 = 0 m (away from north pole) 4 (r – )2 Magnetic field at point 'P' due to south pole B2 = 0 m (towards north pole) 4 (r )2 Net magnetic field at point 'P' LM OP LM OP= 0m Baxis = B1 – B2 , ( B1 > B2) MN PQ NM QP4 11 = 0m 4r – 4 (r2 – 2 )2 (r – )2 (r )2 B axis 0 2Mr , where M = m (2) 4 (r 2 – 2 )2 B1 If magnet is short r >> , then B axis ~ 0 2M 4 r3 P B equatorial (ii) At equatorial position :– Magnetic field at point 'P' due to north pole :– 0 m r2+ 2 B2 r2+ 2 4 2 ..... (1) (along NP line) r r2 2 HF IKB1= S ON Magnetic field at point 'P' due to south pole :– 2 0 m NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 4 2 ..... (2) r2 2 FH KIB2= (along PS line) From equation (1) & (2) B1 = B2 = 0 · r2 m 2 = B (Let) 4 Net magnetic field at point 'P' Beq = 2 B cos = 2· 0 m [where cos = 4 (r2 2 ) cos, (r2 2 ) ] = 2· 0 m 4 (r2 2 ) (r2 2 ) B eq. 0 (r 2 M 2 , where M = m(2) 4 2 )3 If magnet is short r >> , then B eq. ~ 0 M 4 r3 106 E
JEE-Physics Magnetic shielding If a soft iron ring is placed in magnetic field, most of the lines are found to pass through the ring and no lines pass through the space inside the ring. The inside of the ring is thus protected against any external magnetic effect. This phenomenon is called magnetic screening or shielding and is used to protect costly wrist–watches and other instruments from external magnetic fields by enclosing them in a soft–iron case or box. B=0 B=0 Iron ring in a field Super conductor in a field ( i ) Super conductors also provides perfect magnetic screening due to exclusion of lines of force. This effect is called 'Meissner effect' (ii) Relative magnetic premeability of super conductor is zero. So we can say that super conductors behaves like perfect dimagnetic. Magnetic map This map represents basic three elements of earth magnetic field at all position of earth, and following conclu- sions are normally used for magnetic survey of earth's magnetic field. Lines joining Position of same Isogonic lines, Agonic lines Positions of zero Isoclinic lines , Aclinic lines, mag. equator Positions of same Isodynamic lines Adynamic line or magnetic axis of earth Positions of zero Positions of same B H Positions of zero B H Dipole - Dipole Interactions : S.No. Relative position of dipoles Magnetic force (Fm) (a) NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 (b) r M2 0 . 6M1M 2 (along r) (c) M1 Fm 4 r4 Fm M1 r M2 0 . 3M1M 2 (along r) 4 r4 Fm Fm 0 . 3 M1M 2 (perpendicular to r) M1 Fm 4 r4 Fm M2 r E 107
JEE-Physics MAGNETIC MATERIALS IMPORTANT DEFINATIONS AND RELATIONS Magnetising field or Magnetic intensity (H) Field in which amaterial is placed for magnetisation, called as magnetising field. magnetic field permeability of free space Magnetising field = B0 = 0 H SI Unit H : ampere/meter Intensity of magnetisation ( I ) When a magnetic material is placed in magnetising field then induced dipole moment per unit volume of M that material is known as intensity of magnetisation = V I SI Unit : ampere/meter = = ampere meter2 ] [ M IA meter3 V V Magnetic susceptibility ( m) m = I [It is a scalar with no units and dimensions] H Physically it represent the ease with which a magnetic material can be magnetised A material with more m , can be change into magnet easily. Magnetic permeability µ µ = B m = total magnetic field inside the material H magnetising field It measures the degree to which a magnetic material can be penetrated (or permeated) by the magnetic field lines SI Unit of µ : µ = B m Wb / m2 Wb H A = H H A/m A m A m m [ = L I weber henry – ampere] NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 Relative permeability µr = 0 It has no units and dimensions. Relation between permeability and susceptibility When a magnetic material is placed in magnetic field for magnetisation then total magnetic field in material B0 , where = induced field. = µ0 ; = µ0 Bm B0 Bi Bi B0 H Bi I F II HG JK = µ0 + µ0 = µ0( + ) = µ0 1 H Bm H I Bm H I H GFH KJIB I H 1 µ = µ0(1 + m) = µ0 H µr = 1+ m for vacuum = 0, ( µr = 1) m at STP for air = 0.04 ( at S.T.P. for air µr = 1.04) m 108 E
JEE-Physics CL ASSIFICATION OF MAGNETIC MATERIALS On the basis of magnetic properties of the materials [as magnetisation intensitily (I), Susceptibility ( ) and relative permeability (r)] Faraday devide these materials in three classes – PROPERTIES DIAMAGNETIC PARAMAGNETIC FERROMAGNETIC Cause of magnetism Orbital motion of electrons Spin motion of electrons Formation of domains Substance placed in uni- Poor magnetisation in op- Poor magnetisation in Strong magnetisation in form magnetic field. posite direction. same direction. same direction. HereBm < B0 Here Bm > B0 Here Bm >>> B0 M MM I – H curve I Small, negative, var- I Small, positive, varies I very large, positive & m – T curve ies linearly with field linearly with field varies non–linearly with field I I I H H H m small, negative & small, positive & m very large, positive temperature independent m mT° varies inversely with temp. & temp. dependent 1 (Curie law) m 1 (Curie Weiss m T T TC law) (for T > TC) (TC = Curie temperature) m m m NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 TT T TC TC(Iiron)=770°C or 1043K µr (µ < µ0) 1 > µr > 0 2 > µr > 1 (µ > µ0) µr >>> 1 (µ >>> µ0) Magnetic moment of single Atoms donot have any Atoms have permanent Atoms have permanent atom permanent magnetic megnetic moment which megnetic moment which moment are randomly oriented. are organised in domains. (i.e. in absence of external magnetic field the magnetic moment of whole material is zero) E 109
JEE-Physics PROPERTIES DIAMAGNETIC PARAMAGNETIC FERROMAGNETIC Behaviour of substance in It moves from stronger to It moves with week force Nonuniform magnetic field weaker magnetic field. from weaker magnetic field to stronger magnetic Strongly attract from field. weaker magnetic field to stronger magnetic field. , When rod of material is It becomes perpendicular If there is strong magnetic Weak magnetic field be- suspended between poles to the direction of external field in between the poles tween magnetic poles can of magnet. magnetic field. then rod becomes parallel made rod parallel to field to the magnetic field. direction. Magnetic moment of NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 substance in presence of Value M is ver y le ss and Value M is low but i n M is very high and in opposite to H . direction of H . direction of H . external magnetic field Examples Bi, Cu, Ag, Pb, H2O, Hg, Na, K, Mg, Mn, Sn, Fe,Co, Ni all their alloys, H2, He, Ne, Au, Zn, Sb, NaCl, Diamond.( May be Pt, Al, O2 Fe3O4 Gd, Alnico, etc. found in solid, liquid or (May be found in solid, liq- (Normally found only in sol- gas). uid or gas.) ids) (crystalline solids) 110 E
JEE-Physics Magnetic susceptibility of some Elements at 300 K Diamagnetic Paramagnetic Substance Substance Bismuth – 1.66 × 10–5 Aluminium 2.3 × 10–5 Copper – 9.8 × 10–6 Calcium 1.9 × 10–5 Diamond – 2.2 × 10–5 Chromium 2.7 × 10–4 Gold – 3.6 × 10–5 Lithium 2.1 × 10–5 Lead – 1.7 × 10–5 Magnesium 1.2 × 10–5 Mercury – 2.9 × 10–5 Niobium 2.6 × 10–5 Nitrogen (STP) – 5.0 × 10–9 Oxygen (STP) 2.1 × 10–6 Silver – 2.6 × 10–5 Platinum 2.9 × 10–4 Silicon – 4.2 × 10–6 Tungsten 6.8 × 10–5 MAGNETIC HYSTERESIS B (+)ve saturation a Only Ferromagnetic materials show magnetic hysteresis, when Ferromagnetic material is placed in external magnetic field for b magnetisation then B increases with H non-linearly along Oa. c BR f H Hc If H is again bring to zero then it decreases along path ab. Due to lagging behind of B with H this curve is known as hysteresis e curve.[Lagging of B behind H is called hysteresis] d (-)ve saturation Cause of hysteresis : By removing external magnetising field (H = 0), the magnetic moment of some domains remains aligned in the applied direction of previous magnetising field which results into a residual magnetism. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 Residual magnetism (ob) = Br retentivity remanence Retentivity of a specimen is a measure of the magnetic field remaining in the ferromagnetic specimen when the magnetising field is removed. Coercivity (oc) : Coercivity is an measure of magnetising field required to destroy the residual magnetism of the ferromagnetic specimen. Ferromagnetic materials Soft magnetic materials Hard magnetic materials Low retentivity, low coercivity and High retentivity, high coercivity small hysteresis loss. and large hysteresis loss suitable for making electromagnets, suitable for permanent magnet cores of transformers etc. Ex. Soft iron, Ex. Steel, Alnico (used in magnetic shielding) E 111
JEE-Physics HYSTERESIS LOSS B (i) The area of hystersis loop for a ferromagnetic material is equal Soft to the energy loss per cycle of magnetisation and demagnetisation magnetic material z zper unit volume. H H WH = B.dH 0 I.dH Hard (ii) Its value is different for different materials. magnetic (iii) The work done per cycle per unit volume of material is equal to material the area of hysteresis loop. B VAnt Total energy loss in material WH = V A n t joule = J calorie i.e WH = volume of material × area of hysteresis curve × frequency × time. M MM B0 B0 B0 fig.(a) fig.(b) fig.(c) The materials of both (a) and (b) remain strongly magnetized when B0 is reduced to zero. The material of (a) is also hard to demagnetize, it would be good for permanent magnets. The material of (b) magnetizes and demagnetizes more easily, it could be used as a computer memory material. The material of (c) would be useful for transformers and other alternating-current devices where zero hysteresis would be optimal. Example Obtain the earth’s magnetisation. Assume that the earth’s field can be approximated by a giant bar magnet of magnetic moment 8.0 × 1022 Am2. The earth’s radius is 6400 km. Solution The earth’s radius R = 6400 km = 6.4 × 106 m Magnetisation is the magetic moment per unit volume. Hence, M 8.0 1022 3 24.0 104 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 I = 4 R3 = 4 (6.4 106 )3 = 4 262.1 = 72.9 Am–1 3 Example A solenoid of 500 turns/m is carrying a current of 3A. Its core is made of iron which has a relative permeability of 5000. Determine the magnitudes of the magnetic intensity, magnetization and the magnetic field inside the core. Solution The magnetic intensity H = ni = 500 m–1 × 3A = 1500 Am–1 and r = 5000 so = r – 1 = 5000 – 1 = 4999 ~_ 5000 and = 5000 0 The magnetisation I = H = 7.5 × 106 Am–1 The magnetic field B = 5000 0 H = 5000 × 4 × 10–7 × 1500 = 9.4 T Example A domain in ferromagnetic iron is in the form of a cube of side length l µm. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The molecular mass of iron is 55 g/mole and its density is 7.9 g/cm3. Assume that each iron atom has a dipole moment of 9.27 × 10–24 Am2. 112 E
JEE-Physics Solution The volume of the cubic domain V = (10–6)3 = 10–18 m3 = 10–12 cm3 mass = volume × density = 7.9 × 10–12 g An Avogadro number (6.023 × 1023) of iron atoms has a mass of 55g. The number of atoms in the domain N = 7.9 10 12 6.023 1023 = 8.65 × 1010 atoms 55 The maximum possible dipole moment Mmax is achieved for the (unrealistic) case when all the atomic moments are perfectly aligned. Mmax = (8.65 × 1010) × (9.27 × 10–24) = 8.0 × 10–13 Am2 The consequent magnetisation is Imax M max 8.0 10 13 = 8.0 × 105 Am–1 = domain volume = 10 18 Example Relation between permeability and magnetising field H for a sample of iron is 0.4 12 10 4 ) henery/ =( H meter. where unit of H is A/m. Find value of H for which magnetic induction of 1.0 Wb/m2 can be produce. Solution Magnetic induction for medium B = H 1 = (0.4 12 10 4 ) H H 1 0.4 1 = 0.4 + 12 × 10–4 H H = 12 10 4 = 500 A/m Example When a rod of magnetic material of size 10 cm × 0.5 cm × 0.2 cm is located in magnetising field of 0.5 × 104 A/m then a magnetic moment of 5 A-m2 is induced in it. Find out magnetic induction in rod. Solution F IM GHF IKJ GH JKTotal magnetic induction B = 0 (I + H) = 0 M H I V V GFH IKJ= 4 × 10–75 10 6 0.5 10 4 = 6.28 Wb/m2 Example A rod of magnetic material of cross section 0.25 cm2 is located in 4000 A/m magnetising field. Magnetic flux passes through the rod is 25 × 10–6 Wb. Find out for rod (i) permeability (ii) magnetic susceptibility (iii) magnetisation NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\03.Magnetism theory.p65 Solution (i) Magnetic flux 25 10 6 = BA B = /A = 0.25 10 4 = 1 wb/m2 B1 B = H = H = 4000 = 2.25 × 10–4 wb/A-m (ii) 2.5 10 4 1 = 199 – 1 = 198 ( r 4 10 7 ) m = r – 1 = 0 – 1 = 0 (iii) Magnetization I = m H = 198 × 4000 = 7.92 × 105 A/m E 113
MATHEMATICAL TOOLS Mathematics is the language of physics. It becomes easier to describe, understand and apply the physical principles, if one has a good knowledge of mathematics. MATHEMATICAL TOOLS Differentiation Integration Vectors To solve the problems of physics Newton made significant contributions to Mathematics by inventing differentiation and integration. Cutting a tree with a blade Cutting a string with an axe APPROPRIATE CHOICE OF TOOL IS VERY IMPORTANT RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 1
1. FUNCTION Function is a rule of relationship between two variables in which one is assumed to be dependent and the other independent variable, for example : e.g. The temperatures at which water boils depends on the elevation above sea level (the boiling point drops as you ascend). Here elevation above sea level is the independent & temperature is the dependent variable e.g. The interest paid on a cash investment depends on the length of time the investment is held. Here time is the independent and interest is the dependent variable. In each of the above example, value of one variable quantity (dependent variable) , which we might call y, depends on the value of another variable quantity (independent variable), which we might call x. Since the value of y is completely determined by the value of x, we say that y is a function of x and repre- sent it mathematically as y = f(x). Here f represents the function, x the independent variable & y is the dependent variable. x f f(x) Input Ouput (Domain) (Range) All possible values of independent variables (x) are called domain of function. All possible values of dependent variable (y) are called range of function. Think of a function f as a kind of machine that produces an output value f(x) in its range whenever we feed it an input value x from its domain (figure). When we study circles, we usually call the area A and the radius r. Since area depends on radius, we say that A is a function of r, A = f(r) . The equation A = r2 is a rule that tells how to calculate a unique (single) output value of A for each possible input value of the radius r. A = f(r) = r2 . (Here the rule of relationship which describes the function may be described as square & multiply by ). If r = 1 A = if r = 2 A = 4 if r = 3 A = 9 The set of all possible input values for the radius is called the domain of the function. The set of all output values of the area is the range of the function. We usually denote functions in one of the two ways : 1. By giving a formula such as y = x2 that uses a dependent variable y to denote the value of the function. 2. By giving a formula such as f(x) = x2 that defines a function symbol f to name the function. Strictly speaking, we should call the function f and not f(x), y = sin x. Here the function is sine, x is the independent variable. Example 1. The volume V of a ball (solid sphere) of radius r is given by the function V(r) = (4 / 3) (r)3 Solution : The volume of a ball of radius 3m is ? V(3) = 4 / 3(3)3 = 36 m3 . Example 2. Suppose that the function F is defined for all real numbers r by the formula F(r) = 2(r – 1) + 3. Solution : Evaluate F at the input values 0, 2, x + 2, and F(2). In each case we substitute the given input value for r into the formula for F : F(0) = 2(0 – 1) + 3 = – 2 + 3 = 1 ; F(2) = 2(2 – 1) + 3 = 2 + 3 = 5 F(x + 2) = 2(x +2 – 1) + 3 = 2x + 5 ; F(F(2)) = F(5) = 2(5 – 1) + 3 = 11. RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 2
Example 3. A function (x) is defined as (x) = x2 + 3, Find 0) , (1), x2), (x+1) and 1)). Solution : (0) = 02 + 3 = 3 ; Example 4. Solution : (1) = 12 + 3 = 4 ; (x2) = (x2)2+3 = x4+3 (x+1) = (x + 1)2 + 3 = x2 + 2x + 4 ; ((1)) = 4) = 42+3 = 19 If function F is defined for all real numbers x by the formula F(x) = x2 . Evaluate F at the input values 0,2, x + 2 and F(2) F(0) = 0 ; F(2) = 22 = 4 ; F(x+2) = (x+2)2 ; F(F(2)) = F(4) = 42 = 16 2. TRIGONOMETRY 2.1 MEASUREMENT OF ANGLE AND RELATIONSHIP BETWEEN DEGREES AND RADIAN In navigation and astronomy, angles are measured in degrees, but in calculus it is best to use units called radians because of the way they simplify later calculations. B Let ACB be a central angle in a circle of radius r , as in figure. Then the angle ACB or is defined in radius as - Cr A Arc length AB = Radius = r If r = 1 then = AB The radian measure for a circle of unit radius of angle ACB is defined to be the length of the circular arc AB. Since the circumference of the circle is 2 and one complete revolution of a circle is 360º, the relation between radians and degrees is given by : radians = 180º Angle Conversion formulas 1 degree = ( 0.02) radian 180 1 radian 57 degrees 180 Radians to degrees : multiply by Example 5. (i) Convert 45º to radians. (ii) Convert rad to degrees. 6 Solution : (i) 45 • = rad 180 4 Example 6. Solution : (ii) 180 = 30º 6• Convert 30º to radians. 30º× = rad 180 6 RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 3
Example 7. Convert rad to degrees. Solution : 3 180 = 60º 3× Standard values (1) 300 = rad (2) 45o = rad (3) 60o = rad 6 4 3 (4) 90o = rad (5) 120o = 2 rad (6) 135o = 3 rad 2 3 4 (7) 150o = 5 rad (8) 180o = rad (9) 360o = 2 rad 6 (Check these values yourself to see that the satisfy the conversion formulae) 2.2. MEASUREMENT OF POSITIVE AND NEGATIVE ANGLES y x Negative Measure An angle in the xy-plane is said to be in standard position if its vertex lies at the origin and its initial ray lies along the positive x-axis (Fig.). Angles measured counterclockwise from the positive x-axis are assigned positive measures ; angles measured clockwise are assigned negative measures. y y x x 9 3 4 –4 2.3 SIX BASIC TRIGONOMETRIC FUNCTIONS The trigonometric function of a general angle are defined in terms of x, y, and r. Sine : sin = opp = y hyp r hyp r Cosecant : cosec = opp = y adj x hyp r Cosine : cos = hyp = r Secant : sec = adj = x opp y Cotangent : cot = adj x Tangent : tan = adj = x = opp y RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 4
VALUES OF TRIGONOMETRIC FUNCTIONS If the circle in (Fig. above) has radius r = 1, the equations defining sin and cos become cos = x, sin = y We can then calculate the values of the cosine and sine directly from the coordinates of P. Example 8. Find the six trigonometric ratios from given figure 5 4 3 Solution : opp 4 adj 3 sin = hyp = 5 ; cos = hyp = 5 ; Example 9 Solution : opp 4 hyp 5 tan = adj = 3 ; cosec = opp = 4 ; hyp 5 adj 3 sec = adj = 3 ; cot= opp = 4 Find the sine and cosine of angle shown in the unit circle if coordinate of point p are as shown. y 1 , 3 2 2 P 31 2 x 1 2 1 cos = x-coordinate of P = – 2 3 sin = y-coordinate of P = . 2 RULES FOR FINDING TRIGONOMETRIC RATIO OF ANGLES GREATER THAN 90° 2.4 Step 1 Identify the quadrant in which angle lies. Step 2 (a) If angle = (n ± ) where n is an integer. Then trigonometric function of (n ± ) = same trigonometric function of and sign will be decided by CAST Rule. RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 5
(b) If angle = (2n 1) where n is an integer. Then 2 trigonometric function of (2n 1) = complimentary trignometric function of 2 and sign will be decided by CAST Rule. Values of sin , cos and tan for some standard angles. Degree 0 30 37 45 53 60 90 120 135 180 Radians 0 / 6 37 /180 / 4 53 /180 / 3 / 2 2 / 3 3 / 4 sin 0 1/ 2 3 / 5 1/ 2 4 / 5 3 / 2 1 3 / 2 1/ 2 0 cos 1 3 / 2 4 / 5 1/ 2 3 / 5 tan 0 1/ 3 3 / 4 1 4 / 3 1/ 2 0 1/ 2 1/ 2 1 3 3 1 0 Example 10 Evaluate sin 120° Solution : sin 120° = sin (90° + 30°) = cos 30° = 3 Example 11 2 Solution 3 Aliter sin 120° = sin (180° – 60°) = sin 60° = 2 Evaluate cos 135° cos 135° = cos (90° + 45°) = – sin 45° = – 1 2 Example 12 Evaluate cos 210° Solution : cos 210° = cos (180° + 30°) = – cos30° = – 3 2 Example 13 Evaluate tan 210° Solution : tan 210° = tan (180° + 30°) = tan 30° = 1 3 RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 6
2.5 GENERAL TRIGONOMETRIC FORMULAS : 1. cos2 + sin2 = 1 2. cos(A + B) = cos A cos B – sin A sin B 1 + tan2 = sec2 . 1+ cot2 = cosec2 . sin( A + B) = sin A cos B + cos A sin B tan (A+B) = tan A tanB 3. sin 2 = 2 sin cos ; 1 tan A tanB cos2 = 1 cos 2 ; 2 cos 2 = cos2 – sin2 = 2cos2 – 1 = 1 – 2sin2 sin2 = 1– cos2 2 3. DIFFERENTIATION 3.1 FINITE DIFFERENCE The finite difference between two values of a physical quantity is represented by notation. For example : Difference in two values of y is written as y as given in the table below. y2 100 100 100 y1 50 99 99.5 y = y2 – y1 50 1 0.5 INFINITELY SMALL DIFFERENCE : The infinitely small difference means very-very small difference. And this difference is represented by ‘d’ notation instead of ‘’. For example infinitely small difference in the values of y is written as ‘dy’ if y2 = 100 and y1 = 99.99999999........ then dy = 0.000000...................00001 3.2 DEFINITION OF DIFFERENTIATION Another name for differentiation is derivative. Suppose y is a function of x or y = f(x) Differentiation of y with respect to x is denoted by symbol f ’(x) dy where f ’ (x) = dx dx is very small change in x and dy is corresponding very small change in y. NOTATION : There are many ways to denote the derivative of a function y = f(x). Besides f ’(x), the most common notations are these : y´ “y prime” or “y dash” Nice and brief but does not name the independent variable. dy “dy by dx” Names the variables and uses d for dx derivative. df “df by dx” Emphasizes the function’s name. dx “d by dx of f” d f(x) Emphasizes the idea that differentiation is dx an operation performed on f. Dxf “dx of f” A common operator notation. y “y dot” One of Newton’s notations, now common for time derivatives i.e. dy . dt f´(x) f dash x Most common notation, it names the independent variable and Emphasize the function’s name. RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 7
3.3 SLOPE OF A LINE It is the tan of angle made by a line with the positive direction of x-axis, measured in anticlockwise direction. Slope = tan ( In 1st quadrant tan is +ve & 2nd quadrant tan is –ve ) In Figure - 1 slope is positive In Figure - 2 slope is negative < 90° (1st quadrant) > 90° (2nd quadrant) 3.4 AVERAGE RATES OF CHANGE : Given an arbitrary function y = f(x) we calculate the average rate of change of y with respect to x over the interval (x , x + x) by dividing the change in value of y, i.e.y = f(x + x) – f(x), by length of interval x over which the change occurred. The average rate of change of y with respect to x over the interval [x, x + x] = y f(x x) f(x) x x Geometrically, y = QR = tan = Slope of the line PQ y+ y Q x PR therefore we can say that average rate of change of y with y respect to x is equal to slope of the line joining P & Q. y P R x 3.5 THE DERIVATIVE OF A FUNCTION x x + x y In triangle QPR tan = x We know that, average rate of change of y w.r.t. x is y = f(x x) – f(x) . x x If the limit of this ratio exists as x 0, then it is called the derivative of given function f(x) and is denoted as dy = lim f(x x) – f(x) f ’(x) = dx x x 0 3.6 GEOMETRICAL MEANING OF DIFFERENTIATION The geometrical meaning of differentiation is very much useful in the analysis of graphs in physics. To under- stand the geometrical meaning of derivatives we should have knowledge of secant and tangent to a curve Secant and tangent to a curve Secant : A secant to a curve is a straight line, which intersects the curve at any two points. y q Secant P x RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 8
Tangent:- A tangent is a straight line, which touches the curve at a particular point. Tangent is a limiting case of secant which intersects the curve at two overlapping points. In the figure-1 shown, if value of x is gradually reduced then the point Q will move nearer to the point P. If the process is continuously repeated (Figure - 2) value of x will be infinitely small and secant PQ to the given curve will become a tangent at point P . Therefore y = dy dx = tan x 0 x y+ y Q y y P R x x x + x Figure - 1 we can say that differentiation of y with respect to x,i.e. dy is dx Q dy y+y Q y equal to slope of the tangent at point P (x, y) or tan = dx y Q R (From fig. 1, the average rate of change of y from x to x + x is identical with the slope of secant PQ.) PQ x x x+x Figure - 2 3.7 RULES FOR DIFFERENTIATION RULE NO. 1 : DERIVATIVE OF A CONSTANT The first rule of differentiation is that the derivative of every constant function is zero. If c is constant, then dc = 0. dx Example 14 d (8) 0 , d 1 0 , d3 0 dx dx 2 dx RULE NO. 2 : POWER RULE If n is a real number, then d xn nxn1 . dx To apply the power Rule, we subtract 1 from the original exponent (n) and multiply the result by n. RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 9
Example 15 f x x2 x3 x4 .... f' 1 2x 3x2 4x3 .... d 1 d 1 Example 16 (i) = (x–1) = (–1)x–2 = x2 Example 17 dx x dx (ii) d4 = 4d (x–3) = 4(–3)x–4 = 12 . dx x4 dx x3 (a) d (x1/ 2 ) = 1 x 1/ 2 1 dx 2 = 2x | | Function defined for x 0 derivative defined only for x > 0 (b) d (x1/ 5 ) = 1 x–4/5 dx 5 | | Function defined for x 0 derivative not defined at x = 0 RULE NO. 3 : THE CONSTANT MULTIPLE RULE If u is a differentiable function of x, and c is a constant, then d (cu) = c du dx dx In particular, if n is a positive integer, then d (cxn ) = cn xn–1 dx Example 18 The derivative formula Example 19 d (3x2 ) = 3(2x) = 6x dx says that if we rescale the graph of y = x2 by multiplying each y–coordinate by 3, then we multiply the slope at each point by 3. A useful special case The derivative of the negative of a differentiable function is the negative of the function’s derivative. Rule 3 with c = –1 gives. d (u) = d (1 u) = –1 d (u) = d (u) dx dx dx dx RULE NO. 4 : THE SUM RULE The derivative of the sum of two differentiable functions is the sum of their derivatives. If u and v are differentiable functions of x, then their sum u + v is differentiable at every point where u and v are both differentiable functions is their derivatives. d (u v) = d [u (1)v] = du (1) dv du dv dx dx dx dx dx dx The Sum Rule also extends to sums of more than two functions, as long as there are only finitely many functions in the sum. If u , u ,.........u are differentiable at x, then so is u + u + ........+ u , and 12 n 12 n d (u1 u2 ..... un ) = du1 du2 ....... dun . dx dx dx dx RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 10
Example 20 (a) y = x4 + 12x 4 (b) y = x3 + 3 x2 – 5x + 1 dy d (x4) d (12x) dy d x3 d 4 x2 d (5x) d (1) dx dx dx dx dx 3 dx dx dx = 4x3 + 12 48 = 3x2 + 3 .2x – 5 + 0 = 3x2 + 3 x – 5. Notice that we can differentiate any polynomial term by term, the way we differentiated the polynomials in above example. RULE NO. 5 : THE PRODUCT RULE If u and v are differentiable at x, then so is their product uv, and d (uv) = u dv v du . dx dx dx The derivative of the product uv is u times the derivative of v plus v times the derivative of u. In prime notation (uv)’ = uv’ + vu’. While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of two functions is not the product of their derivatives. For instance, dd dd dx (x.x) = dx (x2) = 2x, while dx (x) . dx (x) = 1.1 = 1. Example 21 Find the derivatives of y = (x2 + 1) (x3 + 3). Solution : From the product Rule with u = x2 + 1 and v = x3 + 3, we find d [( x 2 1)(x3 3)] = (x2 + 1) (3x2) + (x3 + 3) (2x) dx = 3x4 + 3x2 + 2x4 + 6x = 5x4 + 3x2 + 6x. Example can be done as well (perhaps better) by multiplying out the original expression for y and differenti- ating the resulting polynomial. We now check : y = (x2 + 1) (x3 + 3) = x5 + x3 + 3x2 + 3 dy dx = 5x4 + 3x2 + 6x. This is in agreement with our first calculation. There are times, however, when the product Rule must be used. In the following examples. We have only numerical values to work with. Example 22 Let y = uv be the product of the functions u and v. Find y’(2) if u’(2) = 3, u’(2) = –4, v(2) = 1, Solution : and v’(2) = 2. From the Product Rule, in the form y’ = (uv)’ = uv’ + vu’ , we have y’(2) = u(2) v’(2) + v(2) u’ (2) = (3) (2) + (1) (–4) = 6 – 4 = 2. RULE NO. 6 : THE QUOTIENT RULE If u and v are differentiable at x, and v(x) 0, then the quotient u/v is differentiable at x, d u v du u dv dx dx and = dx v v2 RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 11
Just as the derivative of the product of two differentiable functions is not the product of their derivatives, the derivative of the quotient of two functions is not the quotient of their derivatives. Example 23 t2 1 Solution : Find the derivative of y = t2 1 We apply the Quotient Rule with u = t2 – 1 and v = t2 + 1 : dy (t2 1).2t (t2 1).2t d u v(du / dt) u(dv / dt) dt = (t2 1)2 dt v v2 = 2t3 2t 2t3 2t 4t (t2 1)2 = (t2 1)2 . RULE NO. 7 : DERIVATIVE OF SINE FUNCTION d (sin x) cos x dx Example 24 (a) y = x2 – sin x : dy = 2x d (sin x) Difference Rule (b) dx dx Product Rule (c) Quotient Rule = 2x – cos x y = x2 sin x : dy d dx = x2 dx (sin x) + 2x sin x = x2 cos x + 2x sin x sin x dy x. d (sin x) sin x.1 y= : dx dx = x x2 = x cos x sin x . x2 RULE NO. 8 : DERIVATIVE OF COSINE FUNCTION d (cos x) sin x dx Example 25 (a) y = 5x + cos x dy d d dx = dx (5x) + dx (cosx) Sum Rule = 5 – sin x (b) y = sinx cosx dy d d dx = sinx dx (cosx) + cosx dx (sinx) Product Rule = sinx (– sinx) + cosx (cosx) = cos2 x – sin2 x RULE NO. 9 : DERIVATIVES OF OTHER TRIGONOMETRIC FUNCTIONS Because sin x and cos x are differentiable functions of x , the related functions sin x 1 tan x = cos x ; sec x = cos x RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 12
cos x 1 cot x = sin x ; cosec x = sin x are differentiable at every value of x at which they are defined. There derivatives. Calculated from the Quotient Rule, are given by the following formulas. d d dx (tan x) = sec2 x ; dx (sec x) = sec x tan x d d dx (cot x) = – cosec2 x ; dx (cosec x) = – cosec x cot x Example 26 Find dy / dx if y = tan x . Solution : d (tan x) = d sin x cos x d (sin x ) – sin x d (cos x) dx dx dx dx = cos x cos2 x cos x cos x – sin x (–sin x) = cos2 x = cos2 x sin2 x = 1 = sec2 x cos2 x cos2 x Example 27 (a) dd (b) dx (3x + cot x) = 3 + dx (cot x) = 3 – cosec2 x d 2 d d = (2cosec x) = 2 (cosec x) dx sin x dx dx = 2 (–cosec x cot x) = – 2 cosec x cot x RULE NO. 10 : DERIVATIVE OF LOGARITHM AND EXPONENTIAL FUNCTIONS d loge x 1 d ex ex dx x dx Example 28. y = ex . log (x) e dy d ex d dy ex dx . log (x) +. [log (x)] ex = ex . log (x) + dx dx e dx e x Example 29. d (sin t) Answer : dt cos t Example 30. d (cos t) Answer : dt sin t RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 13
Example 31 (a) d (b) dx cos 3x (c) d = – sin 3x dx 3x = –3 sin 3x d dx sin 2x d = cos 2x dx (2x) = cos 2x . 2 = 2 cos 2x d dt (A sin (t + ) d = A cos (t + ) dt (t + ) = A cos (t + ). . = A cos (t + ) Example 32 d 1 d d Example 33 = dx (3x – 2)–1 = – 1(3x – 2)–2 dx (3x – 2) dx 3 x – 2 3 = –1 (3x – 2)–2 (3) = – (3x – 2)2 d A cos(t ) dt = –A sin (t + ) RULE NO. 11 : RADIAN VS. DEGREES d sin(x°) = d sin x = cos x = cos(x°) . dx dx 180 180 180 180 3.8 DOUBLE DIFFERENTIATION If f is differentiable function, then its derivative f ’ is also a function, so f ’ may have a derivative of its own, denoted by (f ’ )’ = f ’’ . This new function f ’’ is called the second derivative of f because it is the derivative of the derivative of f . Using Leibniz notation, we write the second derivative of y = f (x) as d dy d2y dx dx dx2 Another notation is f ’’ (x) = D2 f (x) = D2f(x) INTERPRETATION OF DOUBLE DERIVATIVE We can interpret f ’’ (x) as the slope of the curve y = f ’(x) at the point (x, f ’(x)). In other words, it is the rate of change of the slope of the original curve y = f (x) . In general, we can interpret a second derivative as a rate of change of a rate of change. The most familiar example of this is acceleration, which we define as follows. RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 14
If s = s(t) is the position function of an object that moves in a straight line, we known that its first derivative represents the velocity v(t) of the object as a function of time : ds v (t) = s’ (t) = dt The instantaneous rate of change of velocity with respect to time is called the acceleration a(t) of the object. Thus, the acceleration function is the derivative of the velocity function and is therefore the second derivative of the position function : a (t) = v ’ (t) = s’’ (t) dv d2s or in Leibniz notation, a = dt = dt2 Example 34 : If f (x) = x cos x , find f ’’ (x). Solution : Using the Product Rule, we have dd f ’ (x) = x dx (cos x) + cos x dx (x) = – x sin x + cos x To find f ’’ (x) we differentiate f ’ (x) : d f ’’ (x) = dx (–x sin x + cos x) d dd = – x dx (sin x) + sin x dx (– x) + dx (cos x) = – x cos x – sin x – sin x = – x cos x – 2 sin x Example 35 : The position of a particle is given by the equation s = f (t) = t3 – 6t2 + 9t where t is measured in seconds and s in meters. Find the acceleration at time t. What is the acceleration after 4 s ? Solution : The velocity function is the derivative of the position function : s = f (t) = t3 – 6t2 + 9t ds v(t) = dt = 3 t2 – 12 t + 9 The acceleration is the derivative of the velocity function : d2s dv a (t) = dt2 = dt = 6t – 12 a(4) = 6(4) – 12 = 12 m/s2 APPLICATION OF DERIVATIVES 3.9 3.9.1 DIFFERENTIATION AS A RATE OF CHANGE dy dx is rate of change of ‘y’ with respect to ‘x’ : For examples : dx (i) v = dt this means velocity ‘v’ is rate of change of displacement ‘x’ with respect to time ‘t’ dv (ii) a = dt this means acceleration ‘a’ is rate of change of velocity ‘v’ with respect to time ‘t’ . RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 15
dp (iii) F = dt this means force ‘F’ is rate of change of momentum ‘p’ with respect to time ‘t’ . dL (iv) = dt this means torque ‘ ’ is rate of change of angular momentum ‘L’ with respect to time ‘t’ dW (v) Power = dt this means power ‘P’ is rate of change of work ‘W’ with respect to time ‘t’ dq (vi) = dt this means current ‘’ is rate of flow of charge ‘q’ with respect to time ‘t’ Example 36. The area A of a circle is related to its diameter by the equation A = D2. 4 How fast is the area changing with respect to the diameter when the diameter is 10 m? Solution : The (instantaneous) rate of change of the area with respect to the diameter is dA = 2D = D dD 4 2 When D = 10 m, the area is changing at rate (/2) 10 = 5 m2/m. This means that a small change D m in the diameter would result in a change of about 5 D m2 in the area of the circle. Example 37. Experimental and theoretical investigations revealed that the t (seconds) s (meters) Solution : distance a body released from rest falls in time t is proportional to t=0 the square of the amount of time it has fallen. We express this by 0 saying that t=2 5 10 1 t=3 15 s = gt2 , 20 25 2 30 where s is distance and g is the acceleration due to Earth’s 35 gravity. This equation holds in a vacuum, where there is no air 40 resistance, but it closely models the fall of dense, heavy objects 45 in air. Figure shows the free fall of a heavy ball bearing released from rest at time t = 0 sec. A ball bearing falling from rest (a) How many meters does the ball fall in the first 2 sec? (b) What is its velocity, speed, and acceleration then? (a) The free–fall equation is s = 4.9 t2. During the first 2 sec. the ball falls s(2) = 4.9(2)2 = 19.6 m, (b) At any time t, velocity is derivative of displacement : d v(t) = s’(t) = dt (4.9t2) = 9.8 t. At t = 2, the velocity is v(2) = 19.6 m/sec in the downward (increasing s) direction. The speed at t = 2 is speed = |v(2)| = 19.6 m/sec. d2s a = dt2 = 9.8 m/s2 3.9.2 MAXIMA AND MINIMA y Suppose a quantity y depends on another quantity x in a manner shown in the figure. It becomes maximum at x1 and minimum at x. At these points the tangent to the curve is parallel to the xaxis 2 and hence its slope is tan = 0. Thus, at a maximum or a minimum, dy x1 x2 x slope = dx = 0. RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 16
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