physics part 1

ALLENÒ CBSE 45 RACE SOLUTIONS CHAPTER-1 : ELECTRIC CHARGES & FIELDS 1. (ii) 2. (iii) 3. (iii) 4. (iv) 5. (i) 6. Dipole moment is defined as the product of magnitude of either charge & distance between them. pr = q .2 ar 7. N .m2 C 8. No, because Gauss law is valid only where the inverse square law is followed. 9. q 6e0 10. Zero 11. (i) On glass rod : positive charge and on silk : negative charge node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx (ii) On ebonite rod : negative charge and on wool : positive charge. www.notesdrive.com 12. When q1q2 < 0, if implies that if one charge is positive, the other must be negative, Hence, when q1q2 < 0, the force between the two charges will be attractive in nature. 13. The charged particle may or may not move along an electric line of force. If the charged particle was initially at rest, it will move along an electric line of force. In case the charged particle had some initial velocity making certain angle with a line of force, then its resultant path will not be along the line of force. 14. If a Gaussian surface does not include any net charge, there can be electric field on the surface of the Gaussian surface but parallel to the surface. However, it will require that there should be some source charge outside the Gaussian surface. 15. The magnitude of electric field due to an infinite plane sheet of charge is independent of the distance of the observation point from the sheet of charge. EXERCISE - V (RACE) CHAPTER-2 : ELECTROSTATIC POTENTIAL & CAPACITANCE For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false 1. Assertion : Electrons move away from a low potential to high potential region. Reason : Because electrons have negative charge. (i) a (ii) b (iii) c (iv) d E

46 Physics ALLENÒ 2. Assertion : The capacity of a given conductor remains same even if charge is varied on it. Reason : Capacitance depends upon nearly medium as well as size & shape of conductor. (i) a (ii) b (iii) c (iv) d 3. Assertion : Conductor having equal positive charge and volume, must also have same potential. Reason : Potential depends only charge and volume of conductor. (i) a (ii) b (iii) c (iv) d 4. Assertion : A metallic shield in form of a hollow shell may be built to block an electric field. Reason : In a hollow spherical shell, the electric field inside it is zero at every point. (i) a (ii) b (iii) c (iv) d www.notesdrive.com 5. Assertion : When charges are shared between any two bodies no charge is really lost, some loss node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx of energy does occur. Reason : Some energy disappears in the form of heat. (i) a (ii) b (iii) c (iv) d 6. What is the electrostatic potential due to an electric dipole at an equitorial point? 7. Define the term electostatic potential energy. 8. Why is the electrostatic potential constant through out the volume of the conductor? 9. What happens to the capacitance of a capacitor when a dielectric slab is placed between its plates? 10. Define the term dielectric strength of a dielectric. 11. Can electric potential at a point be zero while electric intensity at that point is not zero? 12. The potential energy of an electric dipole placed in a uniform electric field is zero. What is the orientation of vectors pr and uEr? 13. What is the amount of work done in moving a 100 C charge between two points 5 cm a part on an equi-potential surface? 14. Show that the electric field is always directed perpendicular to an equi-potential surface. 15. In parallel plate capacitor, the capacitance increases from 4mF to 80mF , on introducing a dielectric medium between the plates. What is the dielectric constant of the medium? E

ALLENÒ CBSE 47 RACE SOLUTIONS CHAPTER-2 : ELECTROSTATIC POTENTIAL & CAPACITANCE 1. (i) 2. (i) 3. (iv) 4. (i) 5. (i) 6. Zero 7. It is the amount of work done to bring a charge from infinity to its position in the system without node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx any acceleration. www.notesdrive.com 8. Electric field inside the conductor = - E= dV Þ dV = 0 dr dr \\ V = Constant 9. The introduction of dielectric increases the capacitance. 10. Dielectric strength is equal to the maximum value of electric field upto which a dielectric material can withstand without its breakdown. 11. Yes. The electric potential on the perpendicular bisector of an electric dipole is zero while electric intensity is not zero. 12. The potential energy of an electric dipole displaced through from the direction of uniform electric field r is given by U = - pE cos q. The potential energy of the dipole will be zero when q E = 90°, i.e. dipole is at right angles to the direction of field Er . 13. Zero 14. We know that W = E dl Cos q,but work done on equipotential surface is zero. E dl cos q = 0 but E ¹ 0 , dl ¹ 0 ,it means cos q = 0 then q = 90° Therefore E ^ dl . 15. K = Cm =20 Co E

48 Physics ALLENÒ EXERCISE - VI (MOCK TEST) CHAPTER-1 : ELECTRIC CHARGES & FIELDS 1. How does electric field at a point charge with distance r from an infinite thin sheet of charge? [1] 2. Define dielectric constant of a medium. What is it’s S.I. unit ? [1] 3. Draw equipotential surfaces for an electric dipole. [1] 4. What is electric line of force ? What is its importance ? [2] 5. Electric field intensity within a conductor is always zero. Why ? [2] 6. Fig. shows tracks of three charged particles crossing a uniform electrostatic field with same velocities along horizontal. Give the signs of the three charges. Which particle has the highest charge to mass ratio ? [2] www.notesdrive.com7. Can two balls having same kind of charge on them attract each other ? Explain.[2] node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx 8. Find the electric field intensity at any point on the axis of a uniformly charged ring or loop. What happens, if ring is far away from the point ? [3] 9. Explain the terms : (i) quantization of charge (ii) conservation of charge (iii) additive nature of charge. [3] 10. State Coulomb's law in electrostatics. Express it in vector form. What is importance of expressing it in vector form ? [3] 11. Explain the terms electric dipole and dipole moment. Derive a relation for the intensity of electric field at an equatorial point of an electric dipole. [5] 12. Concept of Electric field Electric field is an elegant way of characterizing the electrical environment of a system of charges. Electric field at a point in the space around a system of charges tells you the force a unit positive test charge would experience if placed at that point (without disturbing the system). Electric field is a characteristic of the system of charges and is independent of the test charge that you place at a point to determine the field (i) Which of the following statement is correct? The electric field at a point is (A) Always continuous. (B) Continuous if there is a charge at that point. (C) Discontinuous only if there is a negative charge at that point (D) Discontinuous if there is a charge at that point E

ALLENÒ CBSE 49 (ii) The force per unit charge is known as... (A) Electric Current (B) Electric Flux (C) Electric Field (D) Electric potential. (iii) The SI unit of electric field is... (B) V/m (A) N/C (D) both (A) and (B) (C) V m (iv) A proton of mass 'm' placed in electric field region remains stationary in air then magnitude of electric field is... (A) mge (B) mg/e (C) e/mg (D) e2g/m2 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx(v) Electric field is ...(B) Scalar (A) Vector (D) none of the above www.notesdrive.com(C) tensor (MOCK TEST) CHAPTER-2 : ELECTROSTATICS POTENTIAL & CAPACITANCE 1. Define the term electostatic potential energy. 2. A point charge +Q is placed at point O as shown in the figure. Is the potential difference VA – VB positive, negative or zero ? +Q O• A• B• 3. Write a relation for polarisation ( Pr) of a dielectric material in the presence of an external electric field Er . 4. Work done in moving a test charge between two points in an electric field is independent of the path followed comment ? 5. Explain the term electrostatic shielding. A 6. If a point charge +q is taken first from A to C and then from C to B of a circle C +q drawn with another point charge +q as centre [fig.], then along which path more work will be done? B 7. Derive an expression for the total work done in rotating an electric dipole through an angle in a uniform electric field? 8. The distance between the plates of a parallel plate capacitor is d. A metal plate of thickness (d/2) is placed between the plates. What will be the effect on the capacitance? E

50 Physics ALLENÒ 9. Two identical plane metallic surfaces A and B are kept parallel to each other in air separated by a distance of 1.0 cm as shown in the figure. Surface A is given a positive potential of 10V and the outer surface of B is earthed. (a) What is the magnitude and direction of uniform electric field between point Y and Z? What is the work done in moving a change of 20 C from point X to Y?www.notesdrive.com (b) Can we have non-zero electric potential in the space, where electric field strength is zero? node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 02_Unit-1.docx 10. (a) An air capacitor is given a charge of 2 C raising its potential to 200 V. If on inserting a dielectric medium, its potential falls to 50 V, what is the dielectric constant of the medium? (b) A conducting stab of thickness 't' is introduced without touching between the plates of a parallel plate capacitor separated by a distance d (t<d). Derive an expression for the capacitance of a capacitor? 11. Electrostatic potential of a charged body represents the degree of electrification of the body. It determines the flow of electric charge between two charged bodies placed in contact with each other. The charge always flows from a body of higher potential to another body at lower potential. The flow of charge stops as soon as the potential of the two bodies become equal. Now answer the followings. (i) A uniform electric field of 100 N/C exists vertically upward direction. The decrease in electric potential as one goes up through a height of 5cm is (A) 20V (B) 120 V (C) 5V (D) zero (ii) Work done to bring a unit positive charge unaccelerated from infinity to a point inside electric field is called (A) electric field (B) electric potential (C) capacitance (D) electric flux (iii) Electric potential at a distance of 27 cm from a point charge 9 µC, is given by (A) 200 V (B) 100 V (C) 300 V (D) 50 V (iv) Two charges –6µC and 10µC are placed at distant 18 cm apart. Electric potential at the midpoint joining these two will be (A) 4 × 105V (B) 4× 104V (C) 4× 103V (D) 4× 106V E

ALLENÒ CBSE 51 UNIT-II CHAPTER-3 : CURRENT ELECTRICITY 1. ELECTRIC CURRENT (I) The rate of flow of electric charge across any cross-section is called electric current. (a) Instantaneous electric current I= dq dt (b) Average electric current Iav = Dq Dt 2. CURRENT DENSITY (J) Current flowing per unit area through any cross-section is called current density. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docxI AJ=IÞI= Jr.Ar=JA cos q A www.notesdrive.com 3. DRIFT VELOCITY (Vd) Average velocity with which electrons drift from low potential end to high potential end of the conductor (vd). Drift velocity is given by vr d = - et Er (in terms of applied electric field) m Vd = I (in terms of current through the conductor) neA I = neAvd where A is the area of cross-section and “Avd” represents the rate of flow. Mobility : The term vd is called mobility of charge carriers, represented by m = vd = et . E E m SI unit of mobility = m2/Vs 4. OHM’S LAW I = V where R = 1 l = rl where r (resistivity) = 1 R s A A s (conductivity ) Hence according to Ohm’s law when R is constant I µ V Þ I ~V curve is a straight line (at constant temperature). • Resistance of a conductor is given by R = rl = ml A ne2tA where r is resistivity. Its units is W-m. • Resistivity of a conductor, m (where m is mass of electron, n is number density of free r = ne2t electrons, t is average relaxation time). E

52 Physics ALLENÒ Variation of resistance with length: R = r l A A l (a) If a wire is cut to alter its length, then area remains same. \\Rµl (b) If a wire is stretched or drawn out or folded, area varies but volume remains constant. \\ R µ l2 Variation with area of cross-section or thickness (a) If area is increased / decreased but length is kept same.www.notesdrive.com \\ node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docxRµ1orRµ1(r = radius / thickness) A r2 (b) If area is increased/decreased but volume remains same. R µ 1 or R µ 1 A2 r4 5. TEMPERATURE DEPENDENCE OF RESISTANCE AND RESISTIVITY (1) For conductor For conductor, resistance and resistivity increases while increasing the temperature rt = r0(1+at), where ‘a’ is temperature coefficient of resistivity. As R µ r Þ R = R0(1 + at) r (R0 is the resistance at reference temperature) At temperature t1, R1 = R0 (1 + at1) At temperature t2, R2 = R0 (1 + at2) Þ a = R2 - R1 O For nichrome wire T R0 (t2 - t1 ) Resistivity rT of copper as a function of temperature T E

ALLENÒ CBSE 53 (2) For Semiconductor r T Temperature dependence of resistivity for a typical semiconductor. For Semiconductors resistance and resistivity decreases while increasing the temperature. (3) For Alloy Alloys (like nicrome and manganin etc) exibit a very weak depence of resistivity with temperature. 6. ELECTRIC ENERGY & ELECTRIC POWER Electric Energy : node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx For a charge q, moving with potential V, the electric energy is expressed as qV or Vit. www.notesdrive.com When a current is passed through a resistor energy is wasted in over coming the resistances of the wire. This energy is converted into heat (Heating effect of electric current). W= VIt = I2 Rt = V2 t R SI unit = joule (J) Commercial unit of electrical energy consumption : 1 unit of electrical energy = kilowatt hour = 1 kWh = 3.6 × 106 joules. Electric power : The energy liberated per second in a device is called its power. The electrical power P delivered by an electrical device is given by P = VI, where V = potential difference across device & I = current. If the current enters the higher potential point of the device then power is consumed by it (i.e. acts as load) . If the current enters the lower potential point then the device supplies power (i.e. acts as source). Power consumed by a resistor P = I2R = VI = V2 R SI unit = watt (W) 7. CELL • EMF (E) : The potential difference across the terminals of a practical cell when no current is being drawn from it. • Internal Resistance (r) : The opposition of flow of current inside the cell. It depends on (i) Distance between electrodes ­ Þ r­ (ii) Area of electrodes ­ Þ r¯ (iii) Concentration of electrolyte ­ Þ r­ (iv) Temperature ­ Þ r¯ E

54 Physics ALLENÒ Series grouping +E2 – r2 +E3– r3 n cells +E1 – r1 i R (a) Eequivalent = E1 + E2 + E3 + ....... En (b) requivalent = r1 + r2 + r3 + ...... rn (c) Current i = å Ei å ri + R (d) If all cells have equal emf E and equal internal resistance r then i = nE nr + R www.notesdrive.com Cases : node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx (i) If nr >> R Þ i = E (ii) If nr << R Þ i = nE r R E1 r1 Parallel Grouping E2 r2 E1 + E2 + E3 + ...... E3 r3 r1 r2 r3 i n cell (a) E equivalent = 1 1 1 r1 + r2 + r3 + ........ (b) requivalent = 1 1 1 r1 r2 1 + + r3 + ...... R (c) If all cells have equal emf. E and internal resistance r then Eequivalent = E requivalent = r Þ current i = r E n n +R Mixed combination Total number of identical cell in this circuit is nm. If n cells connected in a series and there are m such branches in the circuit than the internal resistance of the cells connected in a row = nr Er Er Er Er Er Er Er Er Er R E

ALLENÒ CBSE 55 Total internal resistance of the circuit 1 = 1 + 1 + ....upto m times rnet nr nr (Q There are such m rows) rnet = nr m Total e.m.f. of the circuit = total e.m.f. of the cells connected in a row ET = nE I = Enet = nE R + rnet R + nr m Condition for maximum power transfer : req = R e q 8. KIRCHHOFF’S LAWS (i) Junction Rule : It is based on conservation of charge. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx I I1 www.notesdrive.comI = I1 + I2 I2 (ii) Loop Rule : For any closed loop, total rise in potential + total fall in potential = 0. R +V – iR = 0 i V It is based on conservation of energy. 9. WHEAT STONE BRIDGE When current through the galvanometer is zero (null point or balance point) then P = R . Q S When PS > QR, VC < VD & PS < QR, VC > VD or PS = QR Þ products of opposite arms are equal. Potential difference between C & D at null point is zero . The null point is not affected by resistance of G & E. It is not affected even if the positions of G & E are interchanged. C PQ AG B RS D R E E

56 Physics ALLENÒ 10. IMPORTANT POINTS (i) A current flows through a conductor only when there is an electric field within the conductor because the drift velocity of electrons is directly proportional to the applied electric field. (ii) A metal has a resistance and gets often heated by flow of current because when free electrons drift through a metal, they make occasional collisions with the lattice. These collisions are inelastic and transfer energy to the lattice as internal energy. (iii) An ammeter is always connected in series whereas a voltmeter is connected in parallel because an ammeter is a low-resistance galvanometer while a voltmeter is a high-resistance galvanometer. NCERT IMPORTANT QUESTIONS CHAPTER - 3 : CURRENT ELECTRICITY 1. A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10–7 m2, and its resistance is measured to be 5.0W. What is the resistivity of the material at the temperature of the experiment? www.notesdrive.com Sol. l = 15m node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx A = 6.0 x 10–7 m2 R = 5.0 W r=? ρ = RA = 5.0 ´ 6.0 ´10-7 l 15 Þ ρ = 30 ´ 10 -7 = 2 × 10–7 W - m 15 2. A battery of emf 10 V and internal resistance 3W is connected to a resistor. If the current in the circuit is 0.5A,what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed? Sol. Electric Current I = E R+r 0.5 = 10 Þ 0.5 R + 1.5 = 10 R+3 0.5 R = 8.5 Þ R = 17W Terminal voltage V = I R = (0.5) (17) = 8.5 volt 3. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10–4 °C–1. Sol. R1 = 230 = 71.88 W and R2 = 230 = 82.14 W 3.2 2.8 Using, a = R2 - R1 , we get R1DT E

ALLENÒ CBSE 57 DT = R2 - R1 i.e. T2 - T1 = R2 - R1 R1a R1a or T2 = R2 - R1 + T1 R1a T2 = 82.14 - 71.88 + 27 71.88 ´ 1.7 ´ 10-4 = 27 + 840 = 867 °C 4. A storage battery of emf 8.0 V and internal resistance 0.5W is being charged by a 120 V dc supply using a series resistor of 15.5W. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit? node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx Sol. During charging, www.notesdrive.com V= E + I ( r + R ) I = V-E = 120 -8 = 112 = 7A r+R 0.5 + 15.5 16 V +8V– 0.5W +– 15.5W 120V Terminal Voltage, V = E + Ir V = 8 + 7 × 0.5 = 11.5 volt 5. The number density of free electrons in a copper conductor estimated is 8.5 × 1028 m–3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10–6 m2 and it is carrying a current of 3.0 A. Sol. Using vd = I , We get neA 3 = 1.1 × 10–4 m s–1 1.6 ´10-19 ( )( )( )vd = 8.5´1028 2 ´10-6 Time taken, t= l 3 = 2.72 × 104 s vd = 1.1´10-4 E

58 Physics ALLENÒ 6. Explain the term 'drift velocity' of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of 'drift velocity'. Sol. Drift velocity is defined as the average velocity with which the electrons drift towards the positive terminal under the effect of applied electric field. Relation between electric current and drift velocity :- Let n be the number of free electron per unit volume of conducting wire. Number of free electrons related to small element = nAdx Number of free charge carrier related to small element (q) = (nAdx)e Current E Vd A I = dq dt I V I = neAwww.notesdrive.comædxö çè dt ÷ø node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx I = neAVd Vd Þ drift velocity 7. State Kirchhoff's rules. Explain briefly how these rules are justified. Sol. Kirchhoff's Laws: 1. Kirchhoff's Current Law (KCL) 2. Kirchhoff's Voltage Law (KVL) 1. Kirchhoff's Current Law (KCL) or Junction Rule It states that the sum of the currents entering any junction must equal to the sum of the electric currents leaving that junction. I1 i.e. SI = 0. O I3 (I1) + (I4) + (–I2) + (–I3) = 0 or I1 + I4 = I2 + I3 I2 This law is based upon conservation of charge. I4 SIGN CONVENTION Sign of currents flowing towards junction O are taken as positive and vice versa. 2. Kirchhoff's Voltage Law (KVL) Or Loop Rule: It states that the algebraic sum of all voltage drop in a closed electrical circuit must be zero i.e. SV = 0 A R1 R2 B In loop ABCD +I V1 V2 + V1 + V2 + E2 – E1 = 0 V1 + V2 = E1 – E2 – E1 – E2 This law is based upon conservation of energy. C DI E

ALLENÒ CBSE 59 8. Use Kirchhoff's rules to obtain conditions for the balance condition in a Wheatstone bridge. Sol. Let the current in galvanometer be Ig and resistance of galvanometer is Rg. Apply Kirchhoff's voltage law in loop ABDA- R1 B (I1–Ig) Ig R2 –I1R1 – IgRg + I2R3 = 0 ……(1) A I1 Apply KVL in loop BCDB - Rg G C –(I1 - Ig)R2 + (I2 + Ig)R4 + IgRg = 0 ……(2) I2 K2 R4 In balanced Wheat Stone Bridge (Ig = 0) R3 D (I2+Ig) From eq. (1) & (2) I I I1R1 = I2 R3 ……(3) I1R2 = I2R4 ……(4) eq. (3) (4) node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx +– K1 E www.notesdrive.com R1 = R3 This is the condition of balanced Wheat Stone Bridge. R2 R4 9. Using the concept of drift velocity of charge carriers in a conductor, deduce the relationship between current density and resistivity of the conductor. Sol. Relation between drift velocity and current is given by - I = neAVd I = neA æ et ö E ì vd = æ et ö E çè m ÷ø ï èç m ø÷ í îït Þ relaxation time I = æ ne2tE ö A çè m ø÷ From definition of current density J = I Þ J = æ ne2 t ö E Q ne2 t = s = 1 A èç m ÷ø m r J=1E r J = sE E

60 Physics ALLENÒ PREVIOUS YEARS QUESTION EXERCISE-I ONE MARK QUESTIONS : 1. Define the term conductivity of a conductor. On what factors does it depend ? Sol. Conductivity is the reciprocal of resistivity. It is defined as the conductance for a conductor of unit length and unit area of cross section. The conductance of a substance depends mainly on two factors, which are : Temperature and nature of material s = 1 = ne2t r m 2. Define the term ‘relaxation time’ in a conductor. Sol. Relaxation time is the time interval between two successive collisions of electrons in a conductor, when current flows. 3. Define mobility of a charge carrier. What is its relation with relaxation time ? Sol. Mobility of a charge carrier is defined as the drift velocity of the charge carrier per unit electric field i.e m = vd/E = et/m 4. Two wires one of copper and other of manganin have same resistance and equal length. Which wire is thicker and why ? Sol. Manganin is an alloy of Cu with manganese and nickel. Since the manganese and nickel have resistivity greater than copper, the pure copper has lower resistivity as compared to alloy manganin. For the same resistance and equal length manganin wire is thicker than copper. www.notesdrive.com [Q r µ A] R = r l node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docxA EXERCISE-II TWO MARK QUESTIONS : 1. The plot of the variation of potential difference across a combination of three identical cells in series, versus current is shown below. What is the emf and internal resistance of each cell ? 6V V 0 1A I Sol. V = Terminal voltage across cell combination Terminal voltage across a cell can be obtained by subtracting potential drop across internal resistance of the cell from the emf of the cell. E

ALLENÒ CBSE 61 V= E – Ir When I = 0, Þ V = E From graph it is found that when I = 0 , V = 6 V Þ E = 6 V As I = 1A, at V = 0 from graph As V = E – Ir Þ 0 = 6 – I × r Þ 6 = 1× r r=6W . As 3 identical cell are connected in series so emf and internal resistance of each cell is 2V and 2 W respectively . 2. A metal rod of square cross sectional area A having length l has current I flowing through it and a potential difference of V volt is applied across it’s ends (figure I). Now the rod is cut parallel to its length into two identical pieces and joined as shown in figure II, What potential difference must be maintained across the length 2l so that the current in the rod is still I ? node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx Sol. www.notesdrive.com V Al I ll I Case-I Case-II V = IR Vnew = I × Rnew V = I ´ rl1 Vnew = I ´ rL2 A1 A2 Now L2 = 2l1, A2 = A1/2 V 11 Vnew = 2 ´ 2 Vnew = 4 V Now 4V potential difference must be maintained across the length 2l to maintain same current in rod. 3. When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10–4 m/s. If the electron density in the wire is 8 × 1028 m–3, calculate the resistivity of the material of wire. Sol. Given : V = 5V Length of wire l = 0.1m vd = 2.5 × 10–4 m/s Electron density n = 8 × 1028 m–3 We know the drift velocity and current are related by the formula i = neAvd ........... (1) E

62 Physics ALLENÒ Where n is electron density, e is charge on electron, A is area of cross section and Vd is drift velocity. Also , i = V/R and R = rl/A where r is resistivity, l is length of conductor So i = VA / rl ........... (2) comparing equation (1) and (2) we get, VA / rl = neAvd or r = V / nelvd Put values of all, r = 5 / 0.1 × 8 × 1028 × 1.6 × 10–19 × 2.5 × 10–4 r = 1.56 × 10–5 ohm-meter 4. Two electric bulbs P and Q have their resistances in the ratio of 1 : 2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs. www.notesdrive.com Sol. Given RP 1 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docxRQ=2 but P = I2R and current remain same in series, so PP = RP 1 PQ RQ =2 5. A battery of emf 12 V and internal resistance 2 W is connected to a 4 W resistor as shown in the figure. (a) Show that a voltmeter when placed across the cell and across the resistor, in turn, gives the same reading. (b) To record the voltage and the current in the circuit, why is voltmeter placed in parallel and ammeter in series in the circuit ? V 2W 12V Sol. (a) 4W A When voltmeter across cell I = E/(R + r ) = 12 /(2 + 4) = 12/6 = 2 A V1 = E – Ir = 12 – (2 × 2 ) = 8 V When voltmeter across resistor V2 = ER /(r + R) = (12 × 4)/(4 + 2) = 12 × 4/6 = 8 V So V1 = V2 E

ALLENÒ CBSE 63 (b) A voltmeter is used to measure the potential difference across two points in a circuit .since in parallel connection the voltage in the branches remains same , and also the resistance of the voltmeter is very high due to which a very small current flow through the voltmeter so it is connected in parallel to measure the voltage . An ammeter is used to measure the current flowing through a component/circuit. Since, current remains same in series connections and also the resistance of an ammeter is very small due to which it doesn't affect the current to be measured . So an ammeter is connected in series to measure current. 6. Two cells of emfs 1.5 V and 2.0 V having internal resistance 0.2 W and 0.3 W respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell. Sol. Given E1 = 1.5 V & r1 = 0.2 W E2 = 2 V & r2 = 0.3 W node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx 1.5V 0.2W E1r2 + E2r1 1.5´.3 + 2 ´.2 0.85 www.notesdrive.comEquivalent emf =r1 + r2= 0.5 = 0.5 = 1.7V Equivalent resistance = r1r2 = .2 ´ .3 = .06 = 0.12 W 2V 0.3W r1 + r2 .2 + .3 .5 req = 0.12 W EXERCISE-III THREE MARK QUESTIONS 1. (i) Derive an expression for drift velocity of free electrons. (ii) How does drift velocity of electrons in a metallic conductor vary with increase in temperature ? Explain Sol. (i) Drift velocity : Drift velocity is defined as the average E velocity with which free electrons in a conductor get drifted Vd A in a direction opposite to the direction of the applied electric V field. I When conductor is subjected to an electric field E, each electron experience a force. Fr = -eEr and acquires an acceleration Fr -eEr a = m = m .....(i) Here m = mass of electron, e = charge, E = electric field. The average time difference between two consecutive collisions is known as relaxation time of electron. t = t1 + t2 + ...... + tn .....(ii) n As v = u + at (from equations of motion) E

64 Physics ALLENÒ The drift velvrodci=tyvrv1d+isvr2de+nf.i.n..e+d as – vr n vr d = ( ur1 + ur 2 + .... + ur n ) + a (t1 + t2 + .... + tn ) n vr d = 0 + a (t1 + t2 + .... + tn ) n (Q average thermal velocity = 0) \\ vrd = 0 + at eEr vr d = - æ m ö t Þ vrd = æ et ö E ç ÷ èç m ÷ø è ø www.notesdrive.com (ii) According to drift velocity expression, relaxation time is the time interval between node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx successive collisions of an electron. On increasing temperature, the electrons move faster and more collisions occur more quickly. Hence, relaxation time decreases with increase in temperature which implies that drift velocity also decreases with temperature. 2.(a) Define the term 'conductivity' of a metallic wire. Write it's SI unit. (b) Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field Er . Sol. (a) The conductivity of a material equals the reciprocal of the resistance of its wire of unit length and unit area of cross section. [Alternatively : The conductivity (s) of a material is the reciprocal of its resistivity (r)] (Also accept s = 1 ) r Its SI unit is æ1ö or ohm–1m–1 or (mho m–1) or siemen m–1 çè ohm - metre ÷ø (b) The acceleration, r = - e ur a m E r vr d eE The average drift velocity, vd, is given by = - m t (t = average time between collisions or average relaxation time) If n is the number of free electrons per unit volume, then current I is given by I = neA |vd | I = neA æ et ö E Þ I = æ ne2t ö EA èç m ÷ø ç m ÷ è ø But I = | j | A (j = current density) We, therefore, get |j|= ne2 t E , The term ne2 t is conductivity. \\ s = ne2t m m m Þ J = sE E

ALLENÒ CBSE 65 3. Two identical cells of emf 1.5 V each joined in parallel to supply energy to an external circuit consisting of two resistances of 7 W each joined in parallel. A very high resistance voltmeter reads the terminal voltage of cells to be 1.4 V. Calculate the internal resistance of each cell. Sol. A high resistance voltmeter means that no current flow through the voltmeter (practically very less current). When two batteries are connected in parallel, then E eq = E1r2 + E2r1 1.5V r1 + r2 Here r1 = r2 = r 1.5V E1 = E2 = 1.5V (given) V 7W Eeq = 1.5´ r +1.5´ r 7W 2r Eeq = 1.5 V node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx Now R1 = 7W ù given www.notesdrive.comR2 = 7W ûú So 1 = æ 1 + 1 ö W Req çè 7 7 ø÷ Req = 7 = 3.5W 2 Q I = terminal voltage equivalent resistance V = terminal voltage = 1.4 (given) = voltmeter reading So I = 1.4 = 0.4 A 3.5 Now V = Eeq – I × req 1.4 = 1.5 – 0.4 × req 0.4 × req = 0.1 req = 0.25 W As req = r/2 æ 1 1 1ö ççèQ req = r + r ÷ø÷ So r of each cell = 0.5W V 4. (a) The potential difference applied across a given 12 V 2 W R=4W resistor is altered so that the heat produced per second increases by a factor of 9. By what factor does the applied potential difference change? (b) In the figure shown, an ammeter A and a resistor of 4 A W are connected to the terminals of the source. The emf of the source is 12 V having an internal resistance of 2 W. Calculate the voltmeter and ammeter readings. E

66 Physics ALLENÒ Sol. (a) H= V2 t (initially) ...(1) R After altering potentialdifference,we can write, H' = V '2 t R Q H' = 9H we get, 9H = V'2 t ...(2) R Solving (1) & (2) we get, V' = 3V V 12v 2W (b) A www.notesdrive.com R=4W node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx Total current 'I' in the circuit, I = E r = 4 R 2 = 2A R+ + Now potential difference across battery is- V + Ir = E V + 2 × 2 = 12 V = 8volt 5. 30W E F I1 I3 C 40V 20W A 40W I2 BD 80V Use Kirchhoff’s rules, calculate the current in the arm AC of the given circuit. Sol. Applying Kirchhoff’s junction rule at node A I3 = I1 + I2 .....(1) Applying Kirchhoff’s KVL in loop EFCAE –30 I1 + 40 – 40 I3 = 0 .....(2) 3I1 + 4I3 = 4 Applying KVL in loop EFDBE –30 I1 + 20 I2 – 80 = 0 .....(3) –3I1 + 2I2 = 8 E

ALLENÒ CBSE 67 from eq (1) we put the value of I3 in eq (2) 3I1 + 4(I1 + I2) = 4 7I1 + 4I2 = 4 .....(4) from eq (3) and (4) we get I1 = –12/13 A Putting I1 in eq (4) we get I2 = 34 A 13 Then from eq(1) we get I3 = 22/3 A 6. Find the magnitude and direction of current in 1 W resistor in the given circuit 6V P 2W node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx 9V www.notesdrive.com 1W 3W 3W Q Sol. Applying KVL in loop PRSQP R I1 6V P I2 2W M I2 – I1 9V 1W Q N S 3W 3W 6 – 3I1 + (I2 – I1) = 0 ....(1) 6 – 4I1 + I2 = 0 4I1 – I2 = 6 Applying KVL in loop MPQNM 9 – 2I2 – (I2 – I1) – 3I2 = 0 9 – 6I2 + I1 = 0 – I1 + 6I2 = 9 ....(2) On solving eq (1) and (2) we get I2 = 42 A 23 I1 = 45 A 23 Net current through 1W resistor = I1 – I2 = 3 A from Q to P 23 E

68 Physics ALLENÒ EXERCISE-IV FIVE MARK QUESTIONS : 1. (i) Plot a graph showing variation of voltage v/s the current drawn from the cell. How can one get information from this plot about the emf of the cell and it’s internal resistance ? (ii) Two cell of emf’s E1 and E2 and internal resistance r1 and r2 are connected in parallel. Obtain the expression for the emf and internal resistance of a single equivalent cell that can replace this combination ? Sol. (i) The terminal potential difference across the cell, V V = E – Ir or V = –rI + E E Comapring the above relation with the equation of a straight line i.e. y = mx + c, it follows that graph between I (along x axis) and V (along y axis) will be a straight line having slope equal to –rwww.notesdrive.com and making intercept equal to E on y axis. Thus we get node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docxO Isc I information about EMF of cell and its internal resistance from this plot. Isc ® Short ckt. Current (ii) + E1, r1 I1 V I º I I A I B1 B2 C A C I2 + – E2, r2 I = I1 + I2 ....(1) V = V(B1) – V(B2) V = Potential difference across terminal B1 and B2 V = E1 – I1r1 – for first cell V = E2 – I2r2 – for second cell and I = I1 + I2 as per above eq(1) So I= E1 - V + E2 - V Þ æ E1 + E2 ö - V æ 1 + 1 ö r1 r2 ç r1 r2 ÷ ç r1 r2 ÷ è ø è ø ( )So V= E1r2 + E2r1 - I(r1r2 ) (r1 + r2) r1 + r2 So V = Eeq – Ireq Eeq = E1r2 + E2r1 Þ req = r1r2 r1 + r2 r1 + r2 1 11 Þ Eeq = E1 + E2 Now req = r1 + r2 req r1 r2 E

ALLENÒ CBSE 69 2. (i) Define the term drift velocity. (ii) On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend? (iii) Why alloys like constantan and manganin are used for making standard resistors ? Sol. (i) Drift velocity is defined as the average velocity with which the electrons drift towards the positive terminal under the effect of applied electric field. (ii) We know that the current flowing through the conductor is : I = nAevd E A \\ I = neA æ eEt ö e– e– èç m ÷ø I node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx l Using E=V www.notesdrive.com l +– æ eV ö æ ne2 At ö 1 V èç ml ÷ø ç ml ÷ R I = neA t = è V = V ø I µ V ® Which is ohm’s law Where R = ml is constant for particular conductor at a particular temperature and is called nAe2t the resistance of the conductor. R = æ m t ö l = rl Þ r = æ m t ö èç ne2 ÷ø A A èç ne2 ø÷ Where r is the specific resistance or resistivity of the material of the wire. It depends on number of free electron per unit volume and temperature. (iii) Alloys like constantan and manganin are used for making standard resistors, because : (a) They have high value of resistivity and (b) Temperature coefficient of resistance is negligible. 3. Wheatstone bridge is an arrangement of four resistances P, Q, R and S connected as shown in the figure. Their values are so adjusted that the galvanometer G shows no deflection. The bridge is then said to be balanced when this condition is achieved. In the setup shown here, the points B and D are at the same potential and it can be shown that P /Q = R/S This is called balancing condition. If any three resistances arc known, the fourth can be found. The practical form of Wheatstone bridge is slide wire bridge or Meter Bridge. Using this the unknown resistance can be determined as S = æ100 - l ö ´ R , where l is the balancing length of the Meter Bridge. çè l ø÷ E

70 Physics ALLENÒ B P Q C A I1 Rg Ig I2 G R K2 S ID I +– K1 E (i) In a Wheatstone bridge circuit, P = 5 Ohm, Q = 6 Ohm, R = 10 Ohm and S = 5 Ohm. What is the value of additional resistance to be used in series with S, so that the bridge is balanced? (A) 9 Ohm (B) 7 ohm (C) 10 Ohm (D) 5 Ohm Ans. (B) 7 ohmwww.notesdrive.com (ii) A Wheatstone bridge consisting of four arms of resistance P, Q, R, S is most sensitive when node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx (A) all the resistance are equal (B) all the resistance are unequal (C) the resistance P and Q are equal but R >> P and S >> Q (D) the resistance P and Q are equal but R << P and S << Q Ans. (A) all the resistance are equal (iii) The percentage error in measuring resistance with a meter bridge can be minimized by adjusting the balancing point close to (A) 0 (B) 20 cm (C) 50 cm (D) 80 cm Ans. (C) 50 cm (iv) In meter bridge experiment, the ratio of left gap resistance is 2 : 3. The balance point from left is (A) 20 cm (B) 50 cm (C) 40 cm (D) 60 cm Ans. (C) 40 cm EXERCISE - V (RACE) For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false 1. Assertion : Total current entering a circuit is equal to that leaving it by Kirchhoff's law. Reason : It is based on conservation of charge. (i) a (ii) b (iii) c (iv) d 2. Assertion : Terminal potential of a cell is always less than its emf. Reason : Potential drop due to internal resistance of cell increases the terminal potential difference. (i) a (ii) b (iii) c (iv) d E

ALLENÒ CBSE 71 3. Assertion : Potentiometer measures the correct value of emf of a cell. Reason : No current flow through the cell being balnanced at null point of potentiometer. (i) a (ii) b (iii) c (iv) d 4. Assertion : The connecting wires are made of copper. Reason : The electrical conductivity of copper is high. (i) a (ii) b (iii) c (iv) d 5. Assertion : There is no current in the metal in the absence of electric field. Reason : Motion of free electron are randomly. (i) a (ii) b (iii) c (iv) d 6. Define the term relaxation time in a conductor. 7. Write the SI unit of conductivity. 8. Sate the principle of working of a potentiometer? 9. If n cells each of emf e and internal resistance r are connected in series, then what will be the total emf and total internal resistance of the combination? 10. A wire is streched so as to change its length by 0.1%, then what will be the percentage increase node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx in its resistance? www.notesdrive.com 11. Write the condition under which the p.d. between the terminals of a battery and its emf is equal. 12. A cell of emf 2 V and internal résistance 0.1 is connected to a 3.9 external resistance. What will be the p.d. across the terminals of the cell? 13. What is meant by the sensitivity of a potentiometer? 14. A current of 4A is flowing in a copper wire. If the same current is allowed to flow through another copper wire of double the radius, what is the effect on the drift velocity of free electrons? 15. When electrons drift in a metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction? RACE CHAPTER-3 (SOLUTIONS) 1. (i) 2. (iv) 3. (i) 4. (i) 5. (i) 6. Relaxation time is the time internal between two successive collisions of electrons in a conductor, when current flows. 7. 1 or ohm–1 meter–1 ohm - meter 8. The potential difference across any two points of current carrying wire, having uniform cross section area & material of the potentiometer is directly proportional to the length between the two points i.e. V µ l. 9. eeq = ne req = nr 10. 0.2% R µ l2 Þ DR = 2Dl Rl 11. When the battery is in an open circuit or when no current is drawn from battery. 12. I = r E = 0.5 A V = E – IR = 1.95 V +R 13. A potentiometer said to be sensitivity if it can measure very small p.d. or it shows large change in balancing length for small change in p.d. 14. I = n e A vd Since e is a constant and n and I are the same, vd µ 1/A. As the area of cross-section is increased four times, therefore, the drift velocity will be one-fourth that in the first wire. 15. No. As the collision is a random process, the electrons cannot get deflected in the same direction. E

72 Physics ALLENÒ EXERCISE - VI (MOCK TEST) 1. Why are alloys used for making standard resistance coils ? [1] 2. The variation of potential difference V with length l in case of two potentiometer wires P and Q is as shown. Which one of these will you prefer for comparing emfs of two primary cells and why ? [1] P VQ l 3. What is advantage of using thick metallic strips to join wires in a potentiometer ? www.notesdrive.com [1] 4. Two heated wires of same dimensions are first connected in series and then in parallel with a node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx source of supply. What will be the ratio of heat produced in the two cases? [2] 5. Two electric bulbs A and B are marked 220V, 40 W and 220V, 60W respectively. Which one has a higher resistance? [2] 6. Why do we prefer potentiometer to measure the emf of cell than a voltmeter? [2] 7. Show the resistance of a conductor is given by R= ml . [3] ne2tA 8. Potential difference V is applied across the ends of copper wire of length (l) and diameter D. What is the effect on drift velocity of electrons if (1) V is doubled (2) l is doubled (3) D is doubled [3] 9. What is drift velocity? Derive expression for drift velocity of electrons in a conductor in terms of relaxation time of electrons? [3] 10.(a) Using the principle of wheat stone bridge describe the method to determine the specific resistance of a wire in the laboratory. Draw the circuit diagram and write the formula used ? (b) In a whetstone bridge experiment, a student by mistake, connects key (K) in place of galvanometer and 11. Relation between V, H and r of a cell Emf of a cell is the potential difference between two electrodes of the cell when no current is drawn from the cell. Internal resistance is the resistance offered by the electrolyte of a cell when the electric current flows through it. The internal resistance of a cell depends upon the following factors: (i) Distance between the electrodes (ii) Nature and temperature of the electrolyte (iii) Nature of electrodes (iv) Area of electrodes. E

ALLENÒ CBSE 73 For a freshly prepared cell, the value of internal resistance is generally low and goes on increasing as the cell is put to more use. The potential difference between the two electrodes of a cell in a closed circuit is called terminal potential difference and its value is less than the emf of the cell during discharging and more then emf of the cell during charging of the cell in a closed circuit. It can be written as V = E - Ir or V = E + Ir r IE I R (i) The terminal potential difference of two electrodes of a cell is equal to emf of the cell when node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx(A) I ¹ 0(B) I = 0(C) both (A) and (B) (D) neither (A) and (B) www.notesdrive.com(ii) A cell of emf E and internal resistance r gives a current of 0.5 A with an external resistance of 12 Ohm and a current of 0.25 A with an external resistance of 25 Ohm. What is the value of internal resistance of the cell ? (A) 5 Ohm (B) 1 Ohm (C) 7 Ohm (D) 3 Ohm (iii) If external resistance connected to a cell has beer, increased to 5 times, the potential difference across the terminals of the cell increases from 10 V to 30 V. Then the emf of the cell is (A) 30 V (B) 60 V (C) 50 V (D) 40 V (iv) During the charging of cell the correct relation is (A) E = V + Ir (B) E = V – Ir (C) V = E + Ir (D) V = E – Ir E

74 Physics ALLENÒ IMPORTANT NOTES ______________________________________________________________________www.notesdrive.com ______________________________________________________________________ ______________________________________________________________________ node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 03_Unit-2.docx ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ E

ALLENÒ CBSE 75 UNIT-III : MAGNETIC EFFECT OF CURRENT & MAGNETISM CHAPTER-4 : MOVING CHARGES & MAGNETISM 1. BIOT-SAVART LAW (BSL) According to the Biot Savart Law, the magnitude of magnetic field dBr is proportional to the current I, the element length drl and inversely proportional to the square of the distance r. Its direction is perpendicular to the plane containing drl and rr . Field at point P due to current element dB = µ0 Idl sin q b cw ×d®B 4p r2 Id®l P m0 = 10-7 Tm / Anode06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docxq ®r 4p www.notesdrive.com O m0 = permeability of free space (air/vacuum) ®r = OOP®P Vector form :- a dBr = µ0 Iduurl ´ ˆr = µ0 Iduurl ´ rr 4p r2 4p r3 2. APPLICATION OF BIOT SAVART LAW (a) Finite length wire : BP = µ0I (sin q1 + sin q2 ) 4pd I M q2 P q1 d (b) Infinite length wire: BP = µ0I 2pd I q2=90° MP q1=90° d E

76 Physics ALLENÒ (c) Magnetic field at centre of current carrying circular loop : I(ACW) O B0 = µ0I R 2R Magnetic field at centre of current carrying circular coil (N>1) B0 = µ0NI , where N ® number of turns 2R (d) Magnetic field at an axial point of current carrying circular coil : N Bx = µ0NIR2 I 2(x2 + R2 )3/2 O P x axis [where sinq = R ] Rq Bx x2 + R2 x Bx – x curve for circular coil www.notesdrive.com B0(max) Point of inflextion node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docxPoint of curvature change P Q P&Q Point of zero curvature R –R/2 O +R/2 –x (x = 0) +x (centre) 3. AMPERE'S CIRCUITAL LAW (ACL) It states that line integral of magnetic field is equal to m0 times the total current passing through the surface Ñò Br · r = m0 I dl 4. APPLICATION OF ACL (A) Field at a axial point of solenoid : For finite length :- BP = m0nI (cos q1 - cos q2 ) 2 Angle q1 and q2 both measured in same sense from the axis of the solenoid to end vectors. For Infinite length :- Bin = m0 ni n = N = Total turn L Total length Bout = 0 q1 q2 P E

ALLENÒ CBSE 77 (B) Field inside toroid : B = m0 nI , where n = N/2pRm, turn density mean radius Rm = R1 + R2 2 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx(C) Solid cylindrical wire : www.notesdrive.comI B O 32 1 R Cross section view R r Point Result (1) r > R Bout = µ0I (2) r = R BS = µ0I 2pr 2pR (3) r < R Bin = µ0Ir (4) r = 0 Baxis = 0 2pR2 5. MAGNETIC FORCE ON MOVING CHARGE IN MAGNETIC FIELD Vector from Frm = q (vr ´ Br ext ) Always éêëêFFrrmm ^ vr ù Br ext ú ^ ûú ®v ®Fm CW (+q, m) B® ext Magnitude form : Fm = qvBsinq q=90° (v^ B) Þ Fm = qvB (max) q=0° or 180° Þ Fm = 0 (min) E

78 Physics ALLENÒ 6. MOTION OF CHARGE IN UNIFORM MAGNETIC FIELD (CIRCULAR PHATH ) (vr ^ Br, q=90°) qvB = mv2 r ×B v O r Fm (a) Radius of circular path : r= mv , where P = mv = 2mEK = 2mqVacc qB (b) Time period : T= 2pm qB (c) Kinetic energy of charge : EK ( qBr )2 2m www.notesdrive.com = Motion of charge in uniform magnetic field at any angle except 0° or 180° or 90° node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx (Helical path) (a) Radius of helical path : r= mv sin q qB (b) Time period : T= 2pm qB (c) Pitch of helix :P = (vcosq) T, where T = 2pm qB r r Combined effect of E & B on moving charge Electromegnetic or Lorentz force r rr FL = Fe + Fm Þ FrL = qEr + q (vr ´ Br ) 7. MAGNETIC FORCE ON CURRENT CARRYING WIRE (OR CONDUCTOR) (a) Straight wire :- Frm = I(Lr ´ Br ext/ uniform ) L I Bext N (b) L Frm = I(Lr ´ Br ext/ uniform ) Arbitrary wire :- ®L N I Bext L 8. Magnetic force between two long parallel wires 12 I1 I2 F = µ0I1I2 N/m l 2pd éparallel currents Þ Attraction ù d êëantiparallel currents Þ Repulsionúû E

ALLENÒ CBSE 79 9. MAGNETIC TORQUE ON A CLOSED CURRENT CIRCUIT When a plane closed current circuit is placed in uniform magnetic field , it experience a zero net force, but experience a torque given by tr = NI Ar ´ Br = Mr ´ Br = BINA sin q where Ar = area vector outward from the face of the circuit where the current is anticlockwise, = magnetic induction of the uniform magnetic field. Mr = magnetic moment = NI Ar Unit of magnetic moment = Am2 Note : This expression can be used only if Br is uniform. 10. MOVING COIL GALVANOMETER (MCG) Principle : When current carrying coil is placed in a magnetic field it experiences a torque. t = NI AB ...(1) Scale node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx Pointer Permanent Magnet www.notesdrive.com Coil NS Pivot Soft iron Core Uniform radial magnetic field It consists of a plane coil of many turns suspended in a radial magnetic field. When a current is passed in the coil it experiences a torque which produces a twist in the suspension. This deflection is directly proportional to the torque t = Kf ...(2) By equation (1) & (2) \\ NIAB = Kf; I = æ K øö÷ f ; èç NAB K = elastic torsional constant of the suspension I = Cf ; C = K NAB C Þ Galvanometer constant So I µ f tµI E

80 Physics ALLENÒ Galvanometer : An instrument used to measure strength of current by measuring the deflection of coil due to torque produced by a magnetic field. tµiµq A galvanometer can be converted into ammeter & voltmeter with desired range as below. Current Sensitivity (CS) It is defined as the deflection per unit current. CS = f / I = NAB K Voltage Sensitivity (VS) It is defined as deflection per unit voltage. VSwww.notesdrive.com=f/V= f / IR = æ NAB ö èç KR ÷ø node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx 11. AMMETER : It is a modified form of suspended coil galvanometer, it is used to measure current. A shunt (small resistance) is connected in parallel with galvanometer to convert into ammeter . Uses of shunt : Ig Rg (i) To prevent G coil due to over loading. (ii) To increase the range of Ammmeter. (iii) To convert G in to Ammeter. S = IgRg IS S I - Ig where Rg = galvanometer resistance Ig = Maximum current that can flow through the galvanometer . I = Maximum current that can be measured using the given ammeter . An ideal ammeter has zero resistance. Ammeter resistance = RA = SR g S+ Rg 12. VOLTMETER : A high resistance is put in series with galvanometer. It is used to measure potential difference. V = Ig (Rg + R) R = V - Rg Ig Ig Rg R Voltmeter Resistance : RV = Rg + R V0 R ® ¥ , Ideal voltmeter E

ALLENÒ CBSE 81 CHAPTER-5 : MAGNETISM AND MATTER 1. MAGNETIC FIELD LINES The magnetic field lines are imaginary lines to present mg. field. The magnetic field line of a magnet form continuous close loops and tangent to the field line at a given point represents the direction of the net magnetic field at the point. 2. MAGNETIC MOMENT OF REVOLVING ELECTRON If an electron is revolving with T as a time period then equivalent current is given I = e magnetic T moment is given as ml = IA = Ipr2 = eVr 2 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx mr l = -eLr L = angular momentum www.notesdrive.com2me NOTE: The ratio of magnetic moment to angular momentum called gyromagnetic ratio. Bohar magnetron = 9.27 × 10–24 Am2 3. GAUSS'S LAW FOR MAGNETISM It states that the net magnetic flux through any closed surface is zero fB = S B·DS = 0 4. MAGNETIC PROPERTIES OF MATERIALS w Intensity of magnetisation I = M/V w Magnetic induction B = µH = µ0(H + I) w Magnetic permeability µ= B H w Magnetic susceptibility cm = I = µr – 1 H 5. CLASSIFICATION OF MAGNETIC MATERIALS Diamagnetic Paramagnetic Ferromagnetic –1£ c < 0 0£c<e c >> 0 £ µr < 1 mr >>1 m < m0 1 < µr < 1 + e µ > µ0 m >> m0 E

82 Physics ALLENÒ NCERT IMPORTANT QUESTIONS CHAPTER-4 : MOVING CHARGES & MAGNETISM 1. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A. Sol. Given i1 = 8A, i2 = 5A, a = 4 × 10–2 m, l = 0.10m, F = ? F= m0i1i2 l= 4p ´10-7 ´ 8´ 5´ .10 = 2´10-5 N 2pa 2p ´ 4 ´10-2 Nature of force Þ Attractive 2. A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre. www.notesdrive.com Sol. Using , B= m0N I ,we get node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docxl B= 4p´10-7 ´ 400 ´ 5´8 = 8p × 10–3 T 80 ´10-2 3. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil? Sol. Using, t = NIAB sin q t = 20 × 12 × (10 × 10 × 10–4) × 0.8 × sin 30° =2.4 × 0.8 × 1/2 = 0.96 N-m 4. A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid? Sol. The number of turns per unit length is, n= 500 = 1000 turns/ m , I = 5A 0.5 B = µ0n I = 4p × 10–7 × 103 × 5 = 6.28 × 10–3 T 5. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T? Sol. i = 8A, q = 30º, B = 0.15T, F/l = ? F = i l B sin q F/l = i B sin q = 8 × 0.15 sin 30º = 1.20 × 1/2 = 0.6 N/m E

ALLENÒ CBSE 83 6. The horizontal component of the earth's magnetic field at a certain place is 3.0 × 10–5 T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west; (b) south to north? Sol. F = I (lr ´ Br ) F = I l B sin q The force per unit length is f = F/ l = I B sin q (a) When the current is flowing from east to west, q = 90° Hence, f = I B Þ 1 × 3 × 10–5 = 3 × 10–5 N m–1 (b) When the current is flowing from south to north, node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx q = 0° Hence, f = 0 www.notesdrive.com Hence there is no force on the conductor. 7. Obtain an expression for magnetic field due to infinitely long straight current carrying conductor by using Ampere's law. Sol. Magnetic field due to infinitely long straight current carrying conductor : Apply ampere's law to find out magnetic field at point 'p' ®B d®l Ñò Br .drl = m0 å I P Ñò Bdl cos q = m0I ì q = 0º I r ïícos q = 1 0 îï å I = I ÑòB dl = m 0 I Þ B × 2pr = m0I amperian loop B = m0I . 2pr 8. Two long straight parallel conductors are carrying steady current I1 and I2 separated by a distance d. if the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere. Sol. We consider two long conductor X1Y1 and X2Y2 placed Y2 Y1 parallel to each other at a distance d apart. The current I1 I1 B2 P and I2 are flowing as shown in diagram . I2 X1 Magnetic field at point P (on the conductor X1Y1 due to Q FF d current I2 flowing through the long conductor X2Y2) is given by B1 B2 = µ0 2I2 / 4pd = µ0I2 X2 2pd E

84 Physics ALLENÒ According to right hand rule, the direction of the magnetic field B2 at a point P is perpendicular to the plane of paper and in inward direction. Now, the conductor X1Y1 carrying current I1 lies in the magnetic field B2 produced by the conductor X2Y2. Since F=BIl, the force experienced by the unit length of the conductor X1Y1 due to magnetic field B2 is given by F= B2× (I1×l) = m0I1I2 2pd or F = µ0 × 2 I1I2 4pd .......... (1) Applying Fleming's left hand rule,it follows that the force F on the conductor X1Y1 act in the plane of the paper and towards left. If we proceed in a similar manner then it can be proved that www.notesdrive.com the conductor X2Y2 experience an equal force in the plane of the paper but towards right . node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docxTherefore, the two parallel conductor carrying current in the same direction attracts each other. Definition of Ampere : Let I1 = I2= 1 A and d = 1 m Then from eq (1) we have that F = 2 × 10–7 N/m This means that one ampere of current is that much current which when flow through two infinitely long parallel conductors ,separated by one meter in free space, causes a force of 2×10-7 N per meter on each conductor. 9. How is a galvanometer converted into a voltmeter and an ammeter? Draw the relevant diagrams and find resistance of the arrangement in each case. Take resistance of galvanometer as G. Sol. Conversion of Galvanometer into Voltmeter : Voltmeter is used to measure p.d. so it is connected in parallel, in a circuit and it's resistance should be infinite in ideal case, thus in order to maximize resistance of galvanometer according to required range we connect a suitable resistance in series as shown : G RX Ig V If potential difference between the points to be measured = V and if galvanometer gives full- scale deflection, when current \"ig\" passes through it. Then, V = Ig(Rg + RX) Þ V = IgRg + IgRX Þ V – IgRg = IgRX RX = (V – IgRg) /Ig RX = V - Rg Ig Also, equivalent resistance of voltmeter : Rv = Rx + Rg E

ALLENÒ CBSE 85 Conversion of Galvanometer into Ammeter : Ammeter is a current measuring device so it's resistance should be zero in ideal case, thus in order to minimize it's resistance according to the required range, we connect a suitable resistance in parallel with it, which is called shunt 'S' as shown. Ig G Then current through shunt : IS = (I – Ig) Potential difference across the shunt : Vg = IgRg IS RS But VS = (I – Ig)RS I VS = Vg A RS(I – Ig) = IgRg RS = Ig Rg I - Ig node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docxThus equivalent resistanceG'=Rg .RS Rg + RS www.notesdrive.com 10. Write the formula of Biot-Savart's law in vector form. Obtain an expression of magnetic field on the axis of a current carrying circular loop. Draw necessary diagram. Sol. Vector form of Biot-Savart's law dlr ´ rˆ rr ( )dBr=m0I íìˆr = r 4pr 2 î Magnetic field at an axial point of a current carrying circular loop : dBr m0I dlr ´ rr ( )Þ = 4pr3 Magnetic field due to small element at point 'P' dB = m0Idl sin q ………..(1) 4pr2 r Angle between dl and rˆ is always 90° (q = 90°). Direction of magnetic field is perpendicular r to the plane of dl and rˆ . dB =m40pIrd2l {sin 90° =1 Total magnetic field at point P ò ò {2pR Baxis = 2pR µ0Idl ´ R sin f = R 0 4pr 2 r r Baxis = dBsin f Þ 0 E

86 Physics ALLENÒ m0IR 2 pR µ0 IR m 0 IR 2 4p r3 0 4p r3 2r3 òBaxis = dl Þ Baxis = ´ 2pR Þ Baxis = Baxis = m 0 IR 2 {r = (R2 + x2 )1/ 2 2 (R2 + x2 )3/2 Baxis = µ0NIR 2 2 (R2 + x2 )3/2 11. A rectangular coil of sides 'l' and 'b' carrying a current I is subjected to a uniform magnetic field B acting perpendicular to its plane. Obtain the expression for the torque acting on it. Sol. Consider a rectangular conducting loop (PQRS) of length l and breadth b placed in a uniform magnetic field. Let I be the current flowing in the loop in clockwise direction . Let at any instant the angle between the magnetic field and normal to the rectangular coil is q. www.notesdrive.com Force Far1ct=inIg(lro´n Brth)eÞarFm1 PQ of the loop, node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx= IBl(Directed inside the sheet of paper) aFSnoimdrciaelacFrrtlF2rya3=l,oafInoc(gtrlricn´teghBreaor)csntÞaintmhgFee2oalnr=irnmtIehB,eQhlaRernmacneRd(tSDhfoeoirryfeccetchtaeeFrnd4lcooeaolucptteis,anidcgheotonhtethheserh.eaTermthoeSfrePpfaooprfeetrho)enllyootwp oarfeoercqeusal,Fr1opapnodsiFrte2 act on the loop. F1 and F2 form a couple and try to rotate the loop clockwise. The magnitude of the torque (t) due to forces Fr1 and Fr2 is given by t = Magnitude of the either force × Perpendicular distance between forces t = IlB × b sin q {l × b = A, area of the loop In t= IAB sin qI(Ar ´ Br ) vector form tr = If the loop has N turns, then net torque acting on the loop is t = N I A B sin q t= MMr ´BBrsin q tr = E

ALLENÒ CBSE 87 CHAPTER-5 : MAGNETISM AND MATTER 1. A closely wound solenoid of 800 turns and area of cross section 2.5 × 10–4 m2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment? Sol. Using , M = NIA, we get M = 800 × 3 × 2.5 × 10–4 M= 0.60 A.m2 2. A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) Intensity of magnetisation (I), (c) B and (d) the magnetising current Im. Sol. (a) The field H is dependent of the material of the core, and is node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx H = ni = 1000 × 2.0 = 2 ×103 A/m. www.notesdrive.com (b) The magnetic field B is given by B = µr µ0 H = 400 × 4W ×10–7 æ Tm ö × 2 × 103 (A/m) çè A ø÷ = 1.0 T (c) Magnetisation is given by B = µoI + µoH I = (B – µ0H)/ µ0 I= B -H m0 I= 1.0 - 2 ´103 4p ´10-7 = 7.93 × 105 A/m (d) The magnetising current Im is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus B = µ0 n(I' + Im) Using I' = 2A, B = 1 T, we get 1.0 = 4p × 10–7 × 1000 (2 + Im) Im = 1.0 - 2= 796 - 2 4p ´10-4 = 794 A. E

88 Physics ALLENÒ 3. Prove that the magnetic moment of the electron revolving around a nucleus in an orbit of radius r with orbital speed v is equal to evr/2. Hence using Bohr's postulate of quantization of angular momentum, deduce the expression for the magnetic moment of hydrogen atom in the ground state. Sol. Here we consider an e- of mass me which is revolving with velocity v on a circular path of radius r. I = e ïìT Þ Time period T íîïT =2vp I = ev 2pr Magnetic moment of electron is given by µl = IA Þ µl = ev ´ (p r2 ) 2pr µl = evr ……(1) 2 www.notesdrive.com µl = e(m evr) {mevr = angular momentum = L} node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx2m e µl = eL 2me In vector form eLr uur = - 2me µl r r For (–ve) charges µl and L are is opposite direction for (+ve) charges µl and L are in same direction. µl = e Þ gyromagnetic ratio L 2m for electron, µl = 8.8 × 1010 C/kg L According to Bohr's theory, orbital angular momentum of electron is given by, L = nh/2p Magnetic moment of electron, µl = e ´ nh 2m e 2p µl = neh 4p me For n = 1 : µl = eh Þ 1.6 ´10-19 ´ 6.62 ´10-34 4p m 4 ´ 3.14 ´ 9.1´10-31 e µl = 9.27 × 10–24 A × m2 9.27´10-24 A´ m2 = 1 Bohr magneton E

ALLENÒ CBSE 89 4. Establish different relations between magnetisation(I), magnetic intensity(H), magnetic susceptibility (c) and relative magnetic permeability(µr). Define the magnetic susceptibility (c). Sol. When current is passed through the solenoid a magnetic field is developed inside the solenoid. If a magnetic substance is kept in a magnetising field of intensity (H), then it gets magnetised. The total magnetic induction (B) is the sum of the magnetic induction (B0) in vacuum and magnetic induction (B1) due to magnetisation of material. B = B0 + BM B = µ0H + µ0I node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docxB = µ0 (H + I){I = cH } www.notesdrive.comB = µ0[H + cH] µH = µ0H [1 + c] µ = µ0 [1 + c] {µr = µ/µ0} µr =[1 + c] Magnetic susceptibility : It is a property which determine how easily a specimen can be magnetised. It is expressed as I cm = H Since I is magnetic moment per unit volume, units of I and H are same (Am–1). cm has no units and no. dimension. E

90 Physics ALLENÒ PREVIOUS YEARS QUESTION CHAPTER - 4 : MOVING CHARGES & MAGNETISM CHAPTER - 5 : MAGNETISM AND MATTER EXERCISE-I ONE MARK QUESTIONS : 1. Write the relation for the force acting on a charged particle q moving with velocity vr in the presence of a magnetic field Br . Sol. ur r ur F = q(v´ B) 2. A proton is accelerated through a potential difference V, subjected to a uniform magnetic field acting normal to the velocity of the proton. If the potential difference is doubled, how will the www.notesdrive.com radius of the circular path described by the proton in the magnetic field change? node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx Sol. r= mv = 2m (KE) qB qB = 2m(qV) qB rµ V Thus if V2 = 2V1 Þ r2 = 2r1 3. A proton and an electron travelling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency? Sol. We know that, frequency n= qB and me < < mp 2pm So electron will move with higher frequency 4. What can be the cause of helical motion of a charged particle ? Sol. When a charged particle moves in uniform external magnetic field, with velocity not perpendicular or parallel to the magnetic field, then charge particle experiences a force also along with torque and perform helical motion. 5. Write the underlying principle of a moving coil galvanometer. Sol. Moving coil galvanometer works on the principle that when a current carrying coil placed in a magnetic field, it experiences a torque so, it get deflected and by measuring deflection we can measure current. E

ALLENÒ CBSE 91 EXERCISE-II TWO MARK QUESTIONS : 1. Draw the magnetic field lines due to a current passing through a long solenoid. Sol. Q Q dl c a b P B P ××××××××××××××××××××××× 2. A rectangular coil of sides ‘l’ and ‘b’ carrying a current I is subjected to a uniform magnetic field Br acting perpendicular to its plane. Obtain the expression for the torque acting on it. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx (Refer to NCERT Q. No. 11) www.not Besdrive.com 3. Find the condition under which the charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select charged particles of a particular speed. (Er ) Sol. Consider a charge 'q' moving with velocity vr in the presence of both electricy field & magnetic field (Br ) experiences a force given as- E v ( )Fr = q Er + vr ´ Br = FrE + FrB FE Assume, Er & Br are ^ to each other & also ^ to the velocity of the particle. x Directions of electric force ( FrE ) & magnetic force ( FrB ) are just opposite. \\ Fr = q (E - vB) ˆj and magnetic force are equal then, net force on tzhe FB is zero & it particle if magnitudes of electric will move undeflected in the fields. qE =qvB or v = E / B The above condition is used to select charged particles of a particular velocity. EXERCISE-III THREE MARK QUESTIONS : 1. Draw the magnetic field lines due to a circular wire carrying current I. (Refer to NCERT Q. No. 11) Sol. Magnetic field due to circular wire carrying current I. i i E

92 Physics ALLENÒ 2. Two identical coils P and Q each of radius R are lying in perpendicular planes such that they have a common centre. Find the magnitude and direction of the magnetic field at the common centre of the two coils, if they carry currents equal to I and 3 I respectively Q PO Sol. y BP y BR Q I2 BP P x I1 BQ z www.notesdrive.com q x BQ node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx Given I1 = I (current in coil P) ; I2 = 3 I (current in coil Q) Magnetic field at the centre of circular current carrying coil, is given by B = m0I/2R So BP = m0I 2R BQ = 3m 0 I 2R net field at centre (BR) BR = B2P + B2Q = æ m0I ö2 + æ 3m0 I ö2 çè 2R ø÷ çèç 2R ÷÷ø So, BR = m0I , Direction tanq = BP = 1 R BQ 3 q = 30o 3. (a) Write the expression for the magnitude of the magnetic force acting on a charged particle moving with velocity v in the presence of magnetic field B. (b) A neutron, an electron and an alpha particle moving with equal velocities, enter a uniform magnetic field going in to the plane of the paper as shown. Trace their paths in the field and justify your answer. ×××××× a• × × × × × × n•× × × × × × e•× × × × × × E

ALLENÒ CBSE 93 Sol. (a) Force acting on a charged particle q which is moving with velocity v in magnetic field B is given by F = q(vr ´Br) The right hand rule gives the direction of the force .The direction of the force is perpendicular to the plane containing velocity vr and magnetic field Br . (b) A charge particle experience a force when it enters the magnetic field .Due to the presence of magnetic field, charged particle will move in a circular path because the force is perpendicular to the velocity of the charged particle , required centripetal force will be provided by magnetic force. Radius of the circular path in which the charged particle is moving is given by r = mv/qB since v and B are constant so radius of path of particle is proportional to their mass to charge ratio. Alpha-particle does trace circular path in anti-clockwise sense and its deviation will be in the direction of (vr X Br) . node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx Neutron does pass without any deviation as magnetic field does not exert force on neutral www.notesdrive.comparticle. Electron does, trace circular path in clockwise sense as its deviation is in the direction opposite to (vr X Br) with a smaller radius due to mass to charge. ×××× a n× × × × e× × × × 4. (a) State Biot – Savart law and express this law in the vector form.(Refer to Theory topic-1) (b) Two identical circular coils, P and Q each of radius R, carrying currents 1A and 3 A 5. (a) respectively, are placed concentric and perpendicular to each other lying in the XY and YZ (b) planes. Find the magnitude and direction of the net magnetic field at the centre of the coils. (c) (Refer to 3 marks PYQ.-2) Sol. (a) State Ampere’s circuital law. Use this law to find magnetic field due to straight infinite current carrying wire. (c) (Refer to NCERT Q. no. 7) How are the magnetic field lines different from the electrostatic field lines ? Ampere’s circuital law : The line integral of magnetic field over a closed loop is m0 times the total current threading through that loop Ñò Br × dlr = m0 (SI) The magnetic field lines form continuous closed loops, whereas electrostatic field lines never forms closed loop. E

94 Physics ALLENÒ 6. Two long straight parallel conductors are carrying steady current I1 and I2 separated by a distance d. if the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere. (Refer to CH.-4, NCERT Q. no. 8) 7. How is a galvanometer converted into a voltmeter and an ammeter ? Draw the relevant diagrams and find the resistance of the arrangement in each case. Take resistance of galvanometer as G. (Refer to CH.-4, NCERT Q. no. 9) 8. (a) A point charge q moving with speed v enters a uniform magnetic field B that is acting into the plane of the paper as shown. What is the path followed by the charge q and in which plane does it move? (b) How does the path followed by the charge get affected if its velocity has a component parallel to Br ? (c) If an electric field Er is also applied such that the particle continues moving along the original straight line path, what should be the magnitude and direction of the electric field Er ? Y ×B www.notesdrive.com vq node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx Sol. (a) X (b) Charge q moves in circular path . It does move in X-Y plane. (c) If velocity has component parallel to B then charge q moves in Helical path. vr = -vˆi [Q The particle is moving along negative x-direction] Br = -Bkˆ Q The magnetic field is perpendicular to the plane of the paper directed inwards i.e. negative z-direction. vr ´ Br = q ëé-vˆi ´ -Bkˆ ( ) ( )\\ Force acting due to magnetic field Fm = q ù û r = -qvBˆj ëéQˆi ´kˆ = -ˆjûù Fm Þ magnitude of Fm = qvB, Direction Þ For the undeflected motion of particle, ( )Fe = qE should be applied in y +ˆj direction E