Xam Idea Physics - Class 12 Term 1 and 2 Question Bank

Long Answer Questions [5 marks] Q. 1. State and explain Biot-Savart law. Use it to derive an expression for the magnetic field produced at a point near a long current carrying wire. [CBSE 2019 (55/3/1)] Ans. Biot-Savart law: Suppose the current I is flowing in a conductor and there is a small current element ‘ab’ of length ∆l. According to Biot-Savart the magnetic field (∆B) produced due to this current element at a point P distant r from the element is given by TB ? ITl sini or TB = n I Tlsini ...(i) r2 4r r2 n where 4r is a constant of proportionality. It depends on the medium between the current element and point of observation (P). µ is called the permeability of medium. Equation (i) is called Biot-Savart law. The product of current (I) and length element (∆l) (i.e., I ∆l) is called the current element. Current element is a vector quantity, its direction is along the direction of current. If the conductor be placed in vacuum (or air), then µ i×s r1e0p–l7acweedbebry/aµm0;pwehree-rme eµt0reis(coarllnedewthtoenp/aemrmpeeraeb2i)li.ty of free space (or air). In S.I. system µ0= 4π n0 Thus 4r = 10–7 weber/ampere × metre As in most cases the medium surrounding the conductor is air, therefore, in general, Biot-Savart law is written as n0 ITlsini TB = 4r r2 The direction of magnetic field is perpendicular to the plane containing current element and the line joining point of observation to current element. So in vector form the expression for magnetic field takes the form \"\" \" n0 IT l # r TB = 4r r3 Derivation of formula for magnetic field due to a current carrying wire using Biot-Savart law: Consider a wire EF carrying current I in upward direction. The point of observation is P at a finite distance R from the wire. If PM is perpendicular dropped from P on wire; then PM = R. The wire may be supposed to be formed of a large number of small current elements. Consider a small element I CD of length δl at a distance l from M. Let ∠CPM = φ and ∠CPD = δ φ, ∠ PDM = θ The length δl is very small, so that ∠PCM may also be taken equal to θ. The perpendicular dropped from C on PD is CN. The angle formed between element I \" and \" (= \" (π – θ). Therefore according to Biot-Savart law, the magnetic field due to dI r CP) is \" current element I dI at P is dB = n0 I dlsin(r – i) = n0 I dlsini ...(i) 4r r2 4r r2 198 Xam idea Physics–XII

But in ∆ CND, sin i = sin(+CDN) = CN = r dz CD dl or δl sin θ = r δφ ∴ From equation (i) dB = n0 Ird z = n0 I dz ...(ii) 4r r2 4r r Again from fig. cos z = R & r= R r cos z From equation (ii) d B = n0 I cos z dz ...(iii) 4r R If the wire is of finite length and its ends make angles α and β with line MP, then net magnetic field (B) at P is obtained by summing over magnetic fields due to all current elements, i.e., B = y–ab n0 I cosz dz = n0 I y–ab cos z dz 4r R 4rR n0 I n0 I 4r R 6sin z@a–b = 4r R 6sin a – sin(–b)@ i.e., B = n0 I (sina+ sinb) 4r R This is expression for magnetic field due to current carrying wire of finite length. If the wire is of infinite length (or very long), then α= β ⇒ π/2 ∴ B = n0 I asin r + sin r k = n0 I 61 + 1@ or B = n0 I 4r R 2 2 4r R 2rR Q. 2. (i) State Biot-Savart Law. Using this law, find an expression for the magnetic field at the centre of a circular coil of N-turns, radius R, carrying current I. [CBSE 2019 (55/3/1)] (ii) Sketch the magnetic field for a circular current loop, clearly indicating the direction of the field. [CBSE (F) 2010, Central 2016] Ans. (i) Biot-Savart Law: Refer to above question Magnetic field at the centre of circular loop: Consider a circular coil of radius R carrying current I in anticlockwise direction. Say, O is the centre of coil, at which magnetic field is to be computed. The coil may be supposed to be formed of a large number of current elements. Consider a small current element ‘ab’ of length ∆l. According to Biot Savart law the magnetic field due to current element ‘ab’ at centre O is TB = n0 I Tl sin i 4r R2 where θ is angle between current element ab and the line joining the element to the centre O. Here θ =90° because current element at each point of circular path is perpendicular to the radius. Therefore magnetic field produced at O, due to current element ab is n0 TB = 4r I Tl R2 According to Maxwell’s right hand rule, the direction of magnetic field at O is upward, perpendicular to the plane of coil. The direction of magnetic field due to all current elements is the same. Therefore the resultant magnetic field at the centre will be the sum of magnetic fields due to all current elements. Thus Moving Charges and Magnetism 199

B = /TB = / n0 I Tl = n0 I 2 / Tl 4r R2 4r R But ∑∆l = total length of circular coil =2πR (for one-turn) ∴ B = n0 I .2r R or B = n0 I 4r R2 2R If the coil contains N–turns, then ∑ ∆l =N. 2π R B = n0 I . N.2r R or B = n0 NI 4r R2 2R Here current in the coil is anticlockwise and the direction of magnetic field is perpendicular to the plane of coil upward; but if the current in the coil is clockwise, then the direction of magnetic field will be perpendicular to the plane of coil downward. (ii) Magnetic field lines due to a circular current loop: Q. 3. (i) Derive an expression for the magnetic field at a point on the axis of a current carrying circular loop. [CBSE (AI) 2013; (F) 2010; 2019 (55/3/1)] (ii) Two co-axial circular loops L1 and L2 of radii 3 cm and 4 cm are placed as shown. What should be the magnitude and direction of the current in the loop L2 so that the net magnetic field at the point O be zero? Ans. (i) Magnetic field at the axis of a circular loop: Consider a circular loop of radius R carrying current I, with its plane perpendicular to the plane of paper. Let P be a point of observation on the axis of this circular loop at a distance x from its centre O. Consider a small element of length dl of the coi\"l at point A. The magnitude of the magnetic induction dB at point P due to this element is given by dB = n0 I dlsin a ...(i) 4r r2 The direction of dB is perpendicular to the plane containing dl and r and is given by right hand screw rule. As the angle between I dl and r and is 90°, the magnitude of the magnetic induction dB is given by, dB = n0 I dlsin90o = n0 I dl . ...(ii) 4r r2 4r r2 If we consider the magnetic induction produced by the whole of the circular coil, then by symmetry the components of magnetic induction perpendicular to the axis will be cancelled out, while those parallel to the axis will be added up. Thus the resultant magnetic induction B at axial point P is along the axis and may be evaluated as follows: 200 Xam idea Physics–XII

The component of dB along the axis, dB x = n0 I dl sin a ...(iii) 4r r2 But sin a = R and r = (R 2 + x 2)1/2 r ` dB x = n0 I dl . R = n0 IR dl = n0 IR dl ..(iv) 4r r2 r 4rr 3 4r (R2 + x2)3/2 Therefore the magnitude of resultant magnetic induction at axial point P due to the whole circular coil is given by n0 IR n0 IR B = y 4r (R2 + x2)3/2 dl = 4r (R2 + x 2)3/2 y dl But y dl =length of the loop = 2πR ...(v) Therefore, B= 4r n0 IR 2) 3/2 (2rR) (R2 + x B = Bx it = n0 IR2 it. <At centre, x = \" = n0 I F 2 (R 2 + x 2)3/2 2R 0, B If the coil contains N turns, then n0 NIR2 B = 2 (R 2 + x 2)3/2 tesla. ...(vi) (ii) The magnetic field B= n0 NIa2 2 (a 2 + x 2)3/2 Here N= 1, a1 = 3 cm, x1 = 4 cm, I1 =1 A ∴ Magnetic field at O due to coil L1 is B1 = n0 # 1 # (3 # 10–2) 2 = n0 (9 # 10–4) 27(3 # 10–2)2 + (4 # 10–2)2A3/2 2 # 125 # 10–6 Magnetic field at O due to coil L2 is B2 = 27(4 n0 # I2 (4 # 10–2) 2 # 10–2)2 + (3 # 10–2)2A3/2 Here a2 = 4cm cm x2 =3 = n0 I2 # 16 # 10–4 2# 125 # 10–6 For zero magnetic field at O, the currents I1 and I2 should be in same direction, so current I2 should be in opposite directions and satisfy the condition B1 =B2 i.e., = n0 # 9 # 10–4 = n0 I2 # 16 # 10–4 & I2 = 9 A 2 # 125 # 10–4 2 # 125 # 10–4 16 Q. 4. (a) A straight thick long wire of uniform circular cross-section of radius ‘a’ is carrying a steady current I. The current is uniformly distributed across the cross-section. Use Ampere’s circuital law to obtain a relation showing the variation of the magnetic field (Br) inside and outside the wire with distance r, (r ≤ a) and (r > a) of the field point from the centre of its cross-section. What is the magnetic field at the surface of this wire? Plot a graph showing the nature of this variation. Moving Charges and Magnetism 201

(b) Calculate the ratio of magnetic field at a point a above the surface of the wire to that at a a 2 value of the field of this wire? point 2 below its surface. What is the maximum  [CBSE Delhi 2010; Chennai 2015] Ans. (a) Magnetic field due to a straight thick wire of uniform cross-section: Consider an infinitely long cylindrical wire of radius a, carrying current I. Suppose that the current is uniformly distributed over whole cross-section of the wire. The cross-section of wire is circular. Current per unit cross-sectional area. fieild=art Iae2xtern al …(i) > a): We consider Magnetic points (r a circular path of radius r (> a) passing through external point P concentric with circular cross-section of wire. By symmetry the strength of magnetic field at every point of circular path is same and the direction of magnetic field is tangential to path at every point. So \" line integral of magnetic field B around the circular path y \" .d\"l= y B dl cos 0o = B 2rr B Current enclosed by path = Total current on circular cross-section of cylinder = I By Ampere’s circuital law \" .d\"l= y n ×current enclosed by path B & B 2rr = n0 ×I & B = n0 I 2rr This expression is same as the magnetic field due to a long current carrying straight wire. This shows that for external points the current flowing in wire may be supposed to be concerned at the axis of cylinder. Magnetic Field at Internal Points (r < a) : Consider a circular path of radius r (<a), passing through internal point Q concentric with circular cross-section of the wire. In this case the assumed circular path encloses only a path of current carrying circular cross-section of the wire. ∴ Current enclosed by path = current per unit cross-section × cross Q section of assumed circular path = i #rr2 = d Ir2 r R I 2 a2 Oa ra n ×r r = 2 ∴ By Ampere’s circuital law \" \" y = n0 # current closed by path B .dl &B = n0 Ir & B.2rr = n0 # Ir 2 2ra2 a2 Clearly, magnetic field strength inside the current carrying wire is directly proportional to distance of the point from the axis of wire. At surface of cylinder r = a, so magnetic field at surface of wire n0 I BS = 2ra (maximum value) The variation of magnetic field strength (B) with distance (r) from the axis of wire for internal and external points is shown in figure. 202 Xam idea Physics–XII

(b) BOutside = n0 I = n0 I a = n0 I 2rr 2raa + 2 3ra k B inside = n0 Ir = n0 I (a/2) = n0 I 2ra2 2ra2 4ra ∴ B outside = 4 B inside 3 n0 I Maximum value of magnetic field is at the surface given by BS = 2ra . Q. 5. Using Ampere’s circuital law find an expression for the magnetic field at a point on the axis of a long solenoid with closely wound turns. [CBSE (F) 2010, 2019(55/2/1)] Ans. Magnetic field due to a current carrying long solenoid: A solenoid is a long wire wound in the form of a close- packed helix, carrying current. To construct a solenoid a large number of closely packed turns of insulated copper wire are wound on a cylindrical tube of card-board or china clay. When an electric current is passed through the solenoid, a magnetic field is produced within the solenoid. If the solenoid is long and the successive insulated copper turns have no gaps, then the magnetic field within the solenoid is uniform; with practically no magnetic field outside it. The reason is that the solenoid may be supposed to be formed of a large number of circular current elements. The magnetic field due to a circular loop is along its axis and the current in upper and lower straight parts of solenoid is equal and opposite. Due to this the magnetic field in a direction perpendicular to the axis of solenoid is zero and so the resultant magnetic field is along the axis of the solenoid. If there are ‘n’ number of turns per metre length of solenoid and I amperes is the current flowing, then magnetic field at axis of long solenoid B = µ0 nI If there are N turns in length l of wire, then n = N or B = n0 NI l l Derivation: Consider a symmetrical long solenoid having number of turns per unit length equal to n. Let I be the current flowing in the solenoid, then by right hand rule, the magnetic field is parallel to the axis of the solenoid. Field outside the solenoid: Consider a closed path abcd. Applying Ampere’s law to this path y \" \" = n × 0 (since net current enclosed by path is zero) B. dl As dl ≠ 0 ∴ B = 0 This means that the magnetic field outside the solenoid is zero. \" Field inside the solenoid: Consider a closed path pqrs The line integral of magnetic field B along path pqrs is \"\" B\" .d\"l+ yqr \"\" \"\" \"\" ypqrs = ypq + yrs + ysp B. dl B.dl B.dl B.dl ...(i) \"\" For path pq, B and dl are along the same direction, ` ypq B\".d\"l= y B dl = Bl (pq = lsay) \"\" For paths qr and sp, B and d l are mutually perpendicular. Moving Charges and Magnetism 203

` yqr \"\" = ysp \" \" = y B dl cos 90o = 0 B.dl B. dl For path rs, B = 0 (since field is zero outside a solenoid) \"\" ` yrs =0 B.dl In view of these, equation (i) gives ` y& \"\" = ypq \"\" = Bl ...(ii) pqrs B.dl B.dl By Ampere’s law y \"\" = n0 # net current enclosed by path B.dl ∴ Bl = µ0 (nl I) ∴ B = µ0 nI This is the well known result. Q. 6. Using Ampere’s circuital law, derive an expression for the magnetic field along the axis of a toroidal solenoid. [CBSE (AI) 2013] OR (a) State Ampere’s circuital law. Use this law to obtain the expression for the magnetic field inside an air cored toroid of average radius ‘r’, having ‘n’ turns per unit length and carrying a steady current I. (b) An observer to the left of a solenoid of N turns each of cross section area ‘A’ observes that a steady current I in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic moment m = NIA. [CBSE Delhi 2015] Ans. Magnetic field due to a toroidal solenoid: A long solenoid shaped in the form of closed ring is called a toroidal solenoid (or endless solenoid). Let n be the number of turns per unit length of toroid and I the current flowing through it. The current causes the magnetic field inside the turns of the solenoid. The magnetic lines of force inside the toroid are in the form of concentric circles. By symmetry the magnetic field has the same magnitude at each point of circle and is along the tangent at every point on the circle. (i) For points inside the core of toroid Consider a circle of radius r in the region enclosed by turns of toroid. Now we apply Ampere’s circuital law to this circular path, i.e., \"\" y = n0 I ...(i) B.d l y \"\" = y Bdl cos 0 =B.2rr B.d l Length of toroid = 2πr N = Number of turns in toroid = n(2πr) and current in one-turn=I ∴ Current enclosed by circular path =(n 2πr). I ∴ Equation (i) gives B 2πr=µ0 (n 2πrI) ⇒ B= µ0 nI (ii) For points in the open space inside the toroid: No current flows through the Amperian loop, so I = 0 y \"B.d\"l= n0 I = 0 & B inside = 0 (iii) For points in the open space exterior to the toroid : The net current entering the plane of the toroid is exactly cancelled by the net current leaving the plane of the toroid. y \"B.d\"l= 0 & B exterior = 0 For observer, current is flowing in clockwise direction hence we will see magnetic field lines 204 Xam idea Physics–XII

going towards south pole. The solenoid can be regarded as a combination of circular loops placed side by side, each behaving like a magnet of magnetic moment IA, where I is the current and A area of the loop. These magnets neutralise each other except at the ends where south and north poles appear. Magnetic moment of bar magnet = NIA Q. 7. (a) Explain with the help of a labelled diagram construction, principle and working of a cyclotron stating clearly the functions of electric and magnetic fields on a charged particle. Derive an expression for time period of revolution and cyclotron frequency. Show that it is independent of the speed of the charged particles and radius of the circular path. [CBSE (AI) 2009, CBSE Delhi 2011, 2014, CBSE 2019 (55/2/1)] (b) What is resonance condition? How is it used to accelerated charged particles? (c) Also find the total KE attained by the charged particle. OR (a) Draw a schematic sketch of a cyclotron. Explain clearly the role of crossed electric and magnetic field in accelerating the charge. Hence derive the expression for the kinetic energy acquired by the particles. (b) An α–particle and a proton are released from the centre of the cyclotron and made to accelerate. (i) Can both be accelerated at the same cyclotron frequency? Give reason to justify your answer. (ii) W hen they are accelerated in turn, which of the two will have higher velocity at the exit slit of the dees? [CBSE (AI) 2013] Ans. (a) Cyclotron: The cyclotron, devised by Lawrence and Livingston, is a device for accelerating charged particles to high speed by the repeated application of accelerating potentials. Construction: The cyclotron consists of two flat semi - circular metal boxes called ‘dees’ and are arranged with a small gap between them. A source of ions is located near the mid-point of the gap between the dees (fig.). The dees are connected to the terminals of a radio frequency oscillator, so that a high frequency alternating potential of several million cycles per second exists between the dees. Thus dees act as electrodes. The dees are enclosed in an insulated metal box containing gas at low pressure. The whole apparatus is placed between the poles of a strong electromagnet which provides a magnetic field perpendicular to the plane of the dees. Working: The principle of action of the apparatus is shown in figure. The positive ions produced from a source S at the centre are accelerated by a dee which is at negative potential at that moment. Due to the presence of perpendicular magnetic field the ion will move in a circular path inside the dees. The magnetic field and the frequency of the applied voltages are so chosen that as the ion comes out of a dee, the dees change their polarity (positive becoming negative and vice-versa) and the ion is further accelerated and moves with higher velocity along a circular path of greater radius. The phenomenon is continued till the ion reaches at the periphery of the dees where an auxiliary negative electrode (deflecting plate) deflects the accelerated ion on the target to be bombarded Moving Charges and Magnetism 205

Role of electric field. Electric field accelerates the charge particle passing through the gap. Role of magnetic field As the accelerated charge particle enters normally to the uniform magnetic field, it exerts a magnetic force in the form of centripetal force and charge particle moves on a semicircular path of increasing radii in each dee (D1 or D2) alternatively. Expression for period of revolution and frequency: Suppose the positive ion with charge q moves in a dee with a velocity v then, mv2 qvB = r or r= mv ...(i) qB where m is the mass and r the radius of the path of ion in the dee and B is the strength of the magnetic field. The angular velocity ω of the ion is given by, ~= v = qB [from (i)] ...(ii) r m The time taken by the ion in describing a semi-circle, i.e., in turning through an angle π is, t = r = rm ...(iii) ~ Bq Thus the time is independent of the speed of the ion i.e., although the speed of the ion goes on increasing with increase in the radius (from eq. i) when it moves from one dee to the other, yet it takes the same time in each dee. From eq. (iii) it is clear that for a particular ion, m being known, B can be calculated for q producing resonance with the high frequency alternating potential. Significance: The applied voltage is adjusted so that the polarity of dees is reversed in the same time that it takes the ion to complete one half of the revolution. Now for the cyclotron to work, the applied alternating potential should also have the same semi-periodic time (T/2) as that taken by the ion to cross either dee, i.e., T rm 2 = t = qB ...(iv) or T = 2rm ...(v) qB This is the expression for period of revolution. Obviously, period of revolution is independent of speed of charged particle and radius of circular path. ∴ Frequency of revolution of particles o = 1 = qB T 2rm 206 Xam idea Physics–XII

This frequency is called the cyclotron frequency. Clearly the cyclotron frequency is independent of speed of particle. (b) Resonance condition: The condition of working of cyclotron is that the frequency of radio frequency alternating potential must be equal to the frequency of revolution of charged particles within the dees. This is called resonance condition. (c) Expression for KE attained If R be the radius of the path and vmax the velocity of the ion when it leaves the periphery, then in accordance with eq. (ii) vmax = qBR ...(vi) m The kinetic energy of the ion when it leaves the apparatus is, KE = 1 mvm2 ax = q2 B2 R2 ...(vii) 2 2m When charged particle crosses the gap between dees it gains KE = qV In one revolution, it crosses the gap twice, therefore if it completes n-revolutions before emerging the dees, the kinetic energy gained = 2nqV ...(viii) Thus KE = q2 B2 R2 = 2nqV 2m Q. 8. (a) Consider a beam of charged particles moving with varying speeds. Show how crossed electric and magnetic fields can be used to select charged particles of a particular velocity? (b) Name another device/machine which uses crossed electric and magnetic fields. What does this machine do and what are the functions of magnetic and electric fields in this machine? Where do these field exist in this machine? Write about their natures. [CBSE South 2016] \"\" Ans. (a) If we adjust the value of E and B such that magnitude of the two forces are equal, then total force on the charge is zero and the charge will move in the fields undeflected. This happen when E qE = Bqv or v= B (b) Name of the device: Cyclotron It accelerates charged particles or ions. l  Electric field accelerates the charged particles.   Magnetic field makes particles to move in circle. l  Electric field exists between the Dees.   Magnetic field exists both inside and outside the dees. l  Magnetic field is uniform.   Electric field is alternating in nature. Q. 9. Derive an expression for the force experienced by a current carrying straight conductor placed in a magnetic field. Under what condition is this force maximum? Ans. Force on a current carrying conductor on the basis of force on a moving charge: Consider a metallic conductor of length L, cross- sectional area A placed in a uniform magnetic field B and its length makes an angle θ with the direction of magnetic field B. The current in the conductor is I. According to free electron model of metals, the current in a metal is due to the motion of free electrons. When a conductor is placed in a magnetic field, the magnetic field exerts a force on every free-electron. The sum of forces acting on all electrons is the net force acting onneAtvhde conductor. If vd is the drift velocity of free electrons, then ...(i) current I = Moving Charges and Magnetism 207

where n is number of free electrons per unit volume. magnetic force on each electron =evd B sin θ ...(ii) \"\" Its direction is perpendicular to both vd and B Volume of conductor V = AL Therefore, the total number of free electrons in the conductor = nAL Net magnetic force on each conductor F = (force on one electron) × (number of electrons) = (evdB sin θ) . (nAL) = (neAvd). BL sin θ Using equation (i) F=IBL sin θ ...(iii) ∴ F=ILB sin θ This is the general formula for the force acting on a current carrying conductor. In vector form \" = I \" # \" ...(iv) F L B Force will be maximum when sin θ = 1 or θ = 90°. That is when length of conductor is perpendicular to magnetic field. Q. 10. Two long straight parallel conductors carry 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. [CBSE Delhi 2016] OR Derive an expression for the force per unit length between two long straight parallel current carrying conductors. Hence define SI unit of current (ampere). [CBSE (AI) 2009, 2010, 2012, Patna 2015] Ans. Suppose two long thin straight conductors (or wires) PQ and RS are placed parallel to each other in vacuum (or air) carrying currents I1 and I2 respectively. It has been observed experimentally that when the currents in the wire are in the same direction, they experience an attractive force (fig. a) and when they carry currents in opposite directions, they experience a repulsive force (fig. b). Let the conductors PQ and RS carry currents I1 and I2 in same direction and placed at separation r. Consider a current–element ‘ab’ of length ∆L of wire RS. The magnetic field produced by current-carrying conductor PQ at the location of other wire RS B1 = n0 I1 ...(i) 2r r According to Maxwell’s right hand rule or right hand palm rule number 1, the direction of B1 will be perpendicular to the plane of paper and directed downward. Due to this magnetic field, each element of other wire experiences a force. The direction of current element is perpendicular to the magnetic field; therefore the magnetic force on element ab of length ∆L TF = B1 I2 TL sin 90o = n0 I1 I2 TL 2r r ∴ The total force on conductor of length L will be n0 I1 I2 n0 I1 I2 F = 2r r / TL = 2r r L ∴ Force acting per unit length of conductor f = F = n0 I1 I2 N /m ...(ii) L 2r r According to Fleming’s left hand rule, the (a) (b) direction of magnetic force will be towards PQ i.e., the force will be attractive. 208 Xam idea Physics–XII

On the other hand if the currents I1 and I2 in wires are in opposite directions, the force will be repulsive. The magnitude of force in each case remains the same. Definition of SI unit of Current (ampere): In SI system of fundamental unit of current ‘ampere' is defined assuming the force between the two current carrying wires as standard. The force between two parallel current carrying conductors of separation r is n0 I1 I2 f = F = 2rr N/m L If I1 =I2 = 1 A, r = 1 m, then n0 f = 2r = 2×10–7 N/m Thus 1 ampere is the current which when flowing in each of parallel conductors placed at separation 1 m in vacuum exert a force of 2× 10–7 on 1 m length of either wire. Q . 11. Derive an expression for torque acting on a rectangular current carrying loop kept in a uniform magnetic field B. Indicate the direction of torque acting on the loop. [CBSE Delhi 2013; (F) 2009, 2019 (55/1/1)] OR \"\" Deduce the expression for the torque x acting on a planar loop of area A and carrying \" current I placed in a uniform magnetic field B. If the loop is free to rotate, what would be its orientation in stable equilibrium? [CBSE Ajmer 2015] Ans. Torque on a current carrying loop: Consider a rectangular loop PQRS of length l, breadth b suspended in a uniform magnetic field B . The length of loop = PQ = RS= l and breadth QR = SP = b. Let at any instant the normal to the plane of loop make an angle θ with the direction of magnetic field B and I be the current in the loop. We know that a force acts on a current carrying wire placed in a magnetic field. Therefore, each side of the loop will experience a force. The net force and torque acting on the loop will be determined by the forces acting on all sides of the loop. Suppose that the forces on sides PQ, QR, RS and SP are F1 ,F 2 , F 3 and F 4 respectively. The sides QR and SP make angle (90°– θ) with the direction of magnetic field. Therefore each of the forces F 2 and F 4 acting on these sides has same magnitude F′ = Blb sin (90°– θ) = Blb cos θ\". Acco\"rding to Fleming’s loepftpohsaitnedburtuthleeirthlienefoorfcaecstioFn2isansadmF4e. are equal and Therefore these forces cancel each other i.e., the resultant of F 2 and F 4 is zero. The sides PQ and RS of current loop are perpendicular to the magnetic field, therefore the magnitude of each of forces F1 and F 3 is F=IlB sin 90°=IlB According to Fleming’s left hand rule the forces F1 and F 3 acting on sides PQ and RS are equal and opposite, but their lines of action are different; therefore the resultant force of F1 and F 3 is zero, but they form a couple called the deflecting couple. When the normal to plane of loop makes an angle with the direction of magnetic field the perpendicular distance between F1 and F3 is b sin θ. ∴ Moment of couple or Torque, τ = (Magnitude of one force F) × perpendicular distance =(BIl). (b sin θ)=I (lb) B sin θ But lb = area of loop =A (say) Moving Charges and Magnetism 209

∴ Torque, τ = IAB sin θ If the loop contains N-turns, then τ = NI AB sin θ \" In vector form = \"\" x NIA # B The magnetic dipole moment of rectangular current loop = M = NIA \" \" \" ∴ = # x M B Direction of torque is perp\"end\"icular to direction of area of loop as well as the direction of magnetic field i.e., along IA # B. The current loop woul\"d be in stable equilibrium, if magnetic dipole moment is in the direction of the magnetic field (B) . Q. 12. (i) What is the relationship between the current and the magnetic moment of a current carrying circular loop? (ii) Deduce an expression for magnetic dipole moment of an electron revolving around a nucleus in a circular orbit. Indicate the direction of magnetic dipole moment. Use the expression to derive the relation between the magnetic moment of an electron moving in a circle and its related angular momentum. [CBSE (AI) 2010; (F) 2009] (iii) A muon is a particle that has the same charge as an electron but is 200 times heavier than it. If we had an atom in which the muon revolves around a proton instead of an electron, what would be the magnetic moment of the muon in the ground state of such an atom? Ans. (i) Relation between current and magnetic moment: Magnetic moment, for a current carrying coil is M = IA For circular coil of radius r, A=πr2 M= I. πr2 (ii) Magnetic moment of an electron moving in a circle: Consider an electron revolving around a nucleus (N) in circular path of radius r with speed v. The revolving electron is equIiv=aleTent to electric current where T is period of revolution = 2r r v I = e = ev 2rr/v 2rr ...(i) Area of current loop (electron orbit), A = πr2 Magnetic moment due to orbital motion, ev evr Ml = IA = 2rr (2rr2) = 2 ...(ii) This equation gives the magnetic dipole moment of a revolving electron. The direction of magnetic moment is along the axis. Relation between magnetic moment and angular momentum Orbital angular momentum of electron is mLa=ssmoefver lectron, ...(iii) Dwhiveidreinmge (ii) by (iii), we get Ml = evr/2 =2me 2e meLe Magnetic L me vr Ml = moment …(iv) This is expression of magnetic moment of revolving electron in terms of angular momentum of electron. In vector form \" = – e \" ...(v) 2me Ml L (iii) Magnetic moMmle=nt–o2fmemnu. oLn in the ground state: 210 Xam idea Physics–XII

In Bohr’s theory, value of angular momentum L in ground state is L= h 2r e h eh ∴ Ml = 2mn # 2r = 4rmn = eh = 1 eh = 1 # 1.6 # 10–19 # 6.63 # 10–34 4r (200 me) 200 4rme 200 4 # 3.14 # 9.1 # 10–31 = 4.64 # 10–26 Am2 Q. 13. Draw the labelled diagram of a moving coil galvanometer. Prove that in a radial magnetic field, the deflection of the coil is directly proportional to the current flowing in the coil. [CBSE (F) 2012] OR (a) Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and working. (b) Answer the following: (i) Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer? (ii) Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Explain, giving reason. [CBSE (AI) 2014] OR Explain, using a labelled diagram, the principle and working of a moving coil galvanometer. What is the function of (i) uniform radial magnetic field, (ii) soft iron core? Define the terms (i) current sensitivity and (ii) voltage sensitivity of a galvanometer. Why does increasing the current sensitivity not necessarily increase voltage sensitivity?  [CBSE Allahabad 2015] Ans. • Moving coil galvanometer: A galvanometer is used to detect current in a circuit. Construction: It consists of a rectangular coil wound on a non-conducting metallic frame and is suspended by phosphor bronze strip between the pole-pieces (N and S) of a strong permanent magnet. A soft iron core in cylindrical form is placed between the coil. One end of coil is attached to suspension wire which also serves as one terminal (T1) of galvanometer. The other end of coil is connected to a loosely coiled strip, which serves as the other terminal (T2). The other end of the suspension is attached to a torsion head which can be rotated to set the coil in zero position. A mirror (M) is fixed on the phosphor bronze strip by means of which the deflection of the coil is measured by the lamp and scale arrangement. The levelling screws are also provided at the base of the instrument. The pole pieces of the permanent magnet are cylindrical so that the magnetic field is radial at any position of the coil. Moving Charges and Magnetism 211

Principle and working: When current (I) is passed in the coil, torque τ acts on the coil, given by τ =NIAB sin θ where θ is the angle between the normal to plane of coil and the magnetic field of strength B, N is the number of turns in a coil. A current carrying coil, in the presence of a magnetic field, experiences a torque, which produces proportionate deflection. i.e., Deflection, θ ∝ τ (Torque) When the magnetic field is radial, as in the case of cylindrical pole pieces and soft iron core, then in every position of coil the plane of the coil, is parallel to the magnetic field lines, so that θ =90­° and sin 90°=1. The coil experiences a uniform coupler. Deflecting torque, τ = NIAB If C is the torsional rigidity of the wire and is the twist of suspension strip, then restoring torque = C θ For equilibrium, deflecting torque = restoring torque i.e. NIAB = C θ ∴ i= NAB I ...(i) i.e. C θ∝I Deflection of coil is directly proportional to current flowing in the coil and hence we can construct a linear scale. Importance (or function) of uniform radial magnetic field: Torque for current carrying coil in a magnetic field is τ = NIAB sin θ In radial magnetic field sinθ = 1, so torque is τ = NIAB This makes the deflection (θ) proportional to current. In other words, the radial magnetic field makes the scale linear. • The cylindrical, soft iron core makes the field radial and increases the strength of the magnetic field, i.e., the magnitude of the torque. Sensitivity of galvanometer : Current sensitivity: It is defined as the deflection of coil per unit current flowing in it. Sensitivity, SI = c i m = NAB ...(ii) I C Voltage sensitivity: It is defined as the deflection of coil per unit potential difference across its ends i.e., SV = i = NAB , ...(iii) V Rg .C where Rg is resistance of galvanometer. Clearly for greater sensitivity number of turns N, area A and magnetic field strength B should be large and torsional rigidity C of suspension should be small. Dividing (iii) by (ii) SV = 1 & SV = 1 SI SI G G Clearly the voltage sensitivity depends on current sensitivity and the resistance of galvanometer. If we increase current sensitivity then it is not certain that voltage sensitivity will be increased. Thus, the increase of current sensitivity does not imply the increase of voltage sensitivity. Q. 14. With the help of a circuit, show how a moving coil galvanometer can be converted into an ammeter of a given range. Write the necessary mathematical formula. Ans. Conversion of galvanometer into ammeter An ammeter is a low resistance galvanometer and is connected in series in a circuit to read current directly. 212 Xam idea Physics–XII

The resistance of an ammeter is to be made as low as possible so that it may read current without any appreciable error. Therefore to convert a galvanometer into ammeter a shunt resistance. (i.e., small resistance in parallel) is connected across the coil of galvanometer. Let G be the resistance of galvanometer and Ig the current required for full scale deflection. Suppose this galvanometer is to converted into ammeter of range I ampere and the value of shunt required is S. If Is is current in shunt, then from fig. I = Ig +IS ⇒ IS =(I – Ig) ...(i) Also potential difference across A and B (VAB)=IS. S = Ig . G Substituting value of IS from (i), we get or (I – Ig) S = Ig G or IS – Ig S = Ig G or IS = Ig (S+G) S or Ig = S + G I ...(ii) ...(iii) i.e. required shunt, S = GIg I–Ig This is the working equation of conversion of galvanometer into ammeter. The resistance (RA) of ammeter so formed is given by 1 = 1 + 1 or 1 = S+G & RA = SG RA S G RA SG S+G If k is figure of merit of the galvanometer and n is the number of scale divisions, then Ig= nk. Out of the total main current I amperes, only a small permissible value Ig flows through the galvanometer and the rest IS = (I – Ig) passes through the shunt. Remark: An ideal ammeter has zero resistance. Q. 15. A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a resistance R1 in series with the coil. If a resistance R2 is connected in series with it, then it can measure upto V/2 volts. Find the resistance, in terms of R1 and R2, required to be connected to convert it into a voltmeter that can read upto 2 V. Also find the resistance G of the galvanometer in terms of R1 and R2. [CBSE Delhi 2015] Ans. Let Ig be the current through galvanometer at full deflection To measure V volts, V = Ig (G + R1) ...(i) V volts, V = Ig (G + R2) ...(ii) 2 2 2 V volts, 2 V = Ig (G + R3) ...(iii) To measure for conversion of range dividing (i) by (ii), G + R1 2 = G + R2 & G = R1 –2R2 Putting the value of G in (i), we have Ig = V & Ig = V R1 – 2R2 + R1 2R1 –2R2 Substituting the value of G and Ig in equation (iii), we have 2V = V (R1 –2R2 + R3) 2R1 –2R2 4R1 – 4R2 = R1 – 2R2 + R3 R3 = 3R1 – 2R2 Moving Charges and Magnetism 213

Self-Assessment Test Time allowed: 1 hour Max. marks: 30 1. Choose and write the correct option in the following questions. (3 × 1 = 3) (i) A current loop in a magnetic field (a) can be in equilibrium in two orientations, both the equilibrium states are unstable. (b) can be in equilibrium in two orientations, one stable while the other is unstable. (c) experiences a torque whether the field is uniform or non uniform in all orientations. (d) can be in equilibrium in one orientation. (ii) Two circular coils 1 and 2 are made from the same wire but the radius of the first coil is twice that of the second coil. What ratio of the potential difference (in volt) should be applied across them, so that the magnetic field at their centres is the same? (a) 2 (b) 3 (c) 4 (d) 6 (iii) Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is (a) 40 W (b) 25 W (c) 250 W (d) 500 W 2. Fill in the blanks. (2 × 1 = 2) (i) Ampere's law is to Biot-Savart law, what Gauss's law is to ______________. (ii) In reality the turns of the toroidal coil form a ______________ and there is always a small magnetic field external to the toroid. 3. Using the concept of force between two infinitely long parallel current carrying conductors, define one ampere of current 1 4. Define one tesla using the expression for the magnetic force acting on a particle of charge 'q' moving with velocity v in a magnetic field B . 1 5. A beam of electrons projected along +x-axis, experiences a force due to a magnetic field along the +y-axis. What is the direction of the magnetic field? 1 x e z y 6. A point charge is moving with a constant velocity perpendicular to a uniform magnetic field as shown in the figure. What should be the magnitude and direction of the electric field so that the particle moves undeviated along the same path? 2 Y B v +q X 214 Xam idea Physics–XII

7. (a) Obtain the conditions under which an electron does not suffer any deflection while passing through a magnetic field. (b) Two protons P and Q moving with the same speed pass through the magnetic fields B1 and B 2 respectively, at right angles to the field directions. If B 2 > B1 , which of the two protons will describe the circular path of smaller radius? Explain. 2 8. 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 when they carry currents equal to I and 3 I respectively. 2 √3 9. (a) Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field. (b) A proton and a deuteron having equal momenta enter in a region of uniform magnetic field at right angle to the direction of the field. Depict their trajectories in the field. 2 10. (a) Depict the magnetic field lines due to a circular current carrying loop showing the direction of field lines. (b) A current I is flowing in a conductor placed along the x-axis as shown in the figure. Find the magnitude and direction of the magnetic field due to a small current element dl lying at the origin at points (i) (0, d, 0) and (ii) (0, 0, d). 3 y I x O dl z-axis 11. A proton, a deuteron and an alpha particle, are accelerated through the same potential difference and then subjected to a uniform magnetic field B , perpendicular to the direction of their motions. Compare (i) their kinetic energies, and (ii) if the radius of the circular path described by proton is 5 cm, determine the radii of the paths described by deuteron and alpha particle. 3 12. State the principle of a moving coil galvanometer. Explain its working and obtain the expression for the deflection produced due to the current passed through the coil. Define current sensitivity. 3 13. (a) State and explain the law used to determine magnetic field at a point due to a current element. Derive the expression for the magnetic field due to a circular current carrying loop of radius r at its centre. Moving Charges and Magnetism 215

(b) A long wire with a small current element of length 1 cm is placed at the origin and carries a current of 10 A along the X-axis. Find out the magnitude and direction of the magnetic field due to the element on the Y-axis at a distance 0.5 m from it. OR (a) Derive the expression for the magnetic field due to a current carrying coil of radius r at a distance x from the centre along the X-axis. (b) A straight wire carrying a current of 5 A is bent into a semicircular arc of radius 2 cm as shown in the figure. Find the magnitude and direction of the magnetic field at the centre of the arc. 5 Answers 1. (i) (b) (ii) (c) (iii) (c) 2. (i) Coulomb's law (ii) helix 11. (i) 1 : 1 : 2 (ii) 5 2 cm, 5 2 cm 13. (b) 4 × 108 T OR (b) 7.85 × 10–5 T zzz 216 Xam idea Physics–XII

Magnetism and Chapter –5 Matter 1. Magnetic Dipole Moment of a Current Loop and Revolving Electron Magnetic dipole moment of a magnet is given as , M = m 2l, where m is pole strength, 2l is separation between poles. Its SI unit is ampere (metre)2 abbreviated as Am2. Magnetic dipole moment of a current loop is M = NIA The direction of M is perpendicular of the plane of loop and given by right hand thumb rule. Magnetic dipole moment of a revolving electron ev evr = IA = 2rr ×rr2 = 2 where v is velocity, r is radius of orbit e M= 2me L amp m2 where L = mevr is angular momentum of revolving electron. 2. Magnetic Field Intensity due to a Magnetic Dipole Magnetic field intensity at a general point having polar coordinates (r, θ) due to a short magnet is given by B = n0 M 1 + 3cos2i 4r r3 where M is the magnetic moment of the magnet. Special Cases (i) At axial point θ = 0, n0 Baxis = 4r 2M r3 (ii) At equatorial point θ = 90° n0 Beqt. = 4r M r3 3. Gauss’s law in magnetism The net magnetic flux through any closed surface is zero. y B.ds = 0 4. Earth’s Magnetism The earth’s magnetic field may be approximated by a magnetic dipole lying at the centre of earth such that the magnetic north pole Nm is near geographical north pole Ng and its magnetic south pole Sm is near geographical south pole Sg. In reality, the north magnetic pole behaves like the south pole of a bar magnet inside the earth and vice versa. The magnitude of earth's magnetic field at earth's surface is about 4×10–5 T. Magnetism and Matter 217

5. Elements of Earths' Magnetic Field Earth’s magnetic field may be specified completely by three quantities called the elements of earth's magnetic field. These quantities are (i) Angle of declination (a): It is the angle between geographical meridian and the magnetic meridian planes. (ii) Angle of dip (θ) : It is the angle made by resultant magnetic field Be with the horizontal. The angle of dip is 0° at magnetic equator and 90° at magnetic poles. Angle of dip is measured by dip circle. It is also called as magnetic inclination (iii) Horizontal component (H) of earth's magnetic field (Be) H = Be cos q …(i) Vertical component of Be is V = Be sin q …(ii) ∴ Be = H2 + V2 …(iii) V and tan i = H …(iv) 6. Important Terms in Magnetism (i) Magnetic permeability (µ): It is the ability of a material to allow magnetic lines of force to pass B through it and is equal to n= H , where B is the magnetic field strength and H is the magnetic field intensity. B n B0 n0 The relative magnetic permeability nr = = where µ0 is the permeability of free space and B0 is the magnetic field strength in vacuum. (ii) Intensity of magnetisation (M ): It is defined as the magnetic moment per unit volume of a magnetised material. Its unit is Am–1. i.e., fieMld=inmVtensity (H): It is the magnetic field used for magnetisation of a material. (iii) Magnetising If I is the current in the solenoid, then magnetising field intensity H=nI, where n = number of turns per metre. Its unit is Am–1. (iv) Magnetic susceptibility: It is defined as the intensity of magnetisation per unit magnetising field, i.e., |m = M H It has no unit. It measures the ability of a substance to take up magnetisation when placed in a magnetic field. 218 Xam idea Physics–XII

7. Classification of Magnetic Materials Magnetic materials may be classified into three categories : (i) Diamagnetic substances: These are the substances in which feeble magnetism is produced in a direction opposite to the applied magnetic field. These substances are repelled by a strong magnet. These substances have small negative values of susceptibility χ and positive low value of relative permeability µr, i.e., –1 # |m 1 0 and 0 # nr1 1 The examples of diamagnetic substances are bismuth, antimony, copper, lead, water, nitrogen (at STP) and sodium chloride. (ii) Paramagnetic substances: These are the substances in which feeble magnetism is induced in the same direction as the applied magnetic field. These are feebly attracted by a strong magnet. These substances have small positive values of M and and relative permeability greater than 1, i.e., χ µr 0 1 |m 1 f, 1 1 nr 1 1 + f where ε is a small positive number. The examples of paramagnetic substances are platinum, aluminium, calcium, manganese, oxygen (at STP) and copper chloride. (iii) Ferromagnetic substances: These are the substances in which a strong magnetism is produced in the same direction as the applied magnetic field. These are strongly attracted by a magnet. These substances are characterised by large positive values of M and and values of much greater than 1, eg. Iron, cobalt, nickel and alloy like alnico. χ µr i.e., |m 22 1, nr 22 1 Distinction between Dia–, Para– and Ferromagnetics Property Diamagnetic Paramagnetic Ferromagnetic Remark (i) Magnetic B < B0 B > B0 B >> B0 iBn0diuscmtioangninetifcree induction B space (ii) Intensity of small and small and very high and m is magnetic magnetisation negative positive positive moment m small and small and M = V negative positive (iii) Magnetic very high and susceptibility positive |= M H (iv) Relative nr < 1 nr > 1 tnhre>or>de1r (of permeability the nr = n thousands) n0 8. Curie Law It states that the magnetic susceptibility of paramagnetic substances is inversely proportional to absolute temperature, i.e., |m ? 1 & | = C where C is called Curie constant T T 9. Curie Temperature When temperature is increased continuously, the magnetic susceptibility of ferromagnetic substances decrease and at a stage the substance changes to paramagnetic. The temperature of transition at which a ferromagnetic substance changes to paramagnetic is called Curie temperature. It is denoted by TC . It is different for different materials. In paramagnetic phase the susceptibility is given by C |m = T – TC Magnetism and Matter 219

10. Diamagnetism is universal properties of all substances but it is weak in para and ferromagnetic substances and hence difficult to detect. 11. Electromagnets and Permanent Magnets Electromagnets are made of soft iron which is characterised by low retentivity, low coercivity and high permeability. The hysteresis curve must be narrow. The energy dissipated in magnetisation and demagnetisation is consequently small. Permanent magnets are made of steel which is characterised by high retentivity, high permeability and high coercivity. They can retain their attractive property for a long period of time at room temperatures. Selected NCERT Textbook Questions Magnetism Q. 1. A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10–2 N–m. What is the magnitude of magnetic moment of the magnet? Ans. Given, B = 0.25 T, τ = 4.5 × 10–2 N-m, θ = 30° We have τ = mB sin θ & Magnetic moment m= x = 4.5×10–2 = 4.5×10–2 = 0.36 A–m2 B sin i 0.25× sin 30° 0.25×0.5 Q. 2. A short bar magnet of magnetic moment m = 0.32 JT –1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (i) stable and (ii) unstable equilibrium? What is the potential energy of the magnet in each case? Ans. Given m = 0.32 JT –1, B = 0.15 T Potential energy of magnet in magnetic field U = – mB cos θ (i) In stable equilibrium the potential energy of magnet is the minimum; so cos θ =1 or θ = 0° Thus in stable equilibrium position, the bar magnet is so aligned that its magnetic moment is along the direction of magnetic field (θ = 0°). Um = – mB = – 0.32 × 0.15 = – 4.8 × 10–2 J (ii) In unstable equilibrium, the potential energy of magnet is the maximum. Thus in unstable equilibrium position, the bar magnetic is so aligned that its magnetic moment is opposite to the direction of the magnetic field, i.e., cos θ = – 1 or q =180°. In this orientation potential energy, Umax=+mB = + 4.8 × 10–2 J. Q. 3. (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 solenoid acts like a bar magnet. What is the associated magnetic moment? (b) If the solenoid is free to turn about the vertical direction in an external uniform horizontal magnetic field at 0.25 T, what is the magnitude of the torque on the solenoid when its axis makes an angle of 30° with the direction of the external field. Ans. (a) If solenoid is suspended freely, it stays in N-S direction. The polarity of solenoid depends on the sense of flow of current. If to an observer looking towards an end of a solenoid, the current appears anticlockwise, the end of solenoid will be N-pole and other end will be S-pole. Magnetic moment, m = NIA = 800 × 3.0 × 2.5 × 10–4 = 0.60 A-m2 (b) Torque on solenoid τ = mB sin θ      = 0.60 × 0.25 sin 30°
 = 0.60 × 0.25 × 0.5 = 7.5 ×10–2 N–m 220 Xam idea Physics–XII

Q. 4. A bar magnet of magnetic moment 1.5 JT –1 lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment (i) normal to the field direction? and (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)? Ans. (a) Work done in aligning a magnet from orientation θ1 to θ2 is given by W = U2 – U1 =– mB cos θ2 – (–mB cos θ1) =– mB (cos θ2 – cos θ1) …(i) (i) Here θ1 = 0°, θ2 =90° ∴ W = mB (cos 0° – cos 90°) = mB (1– 0) = mB = 1.5 × 0.22 = 0.33 J (ii) Here θ1 = 0°, θ2 =180° ∴ W = mB (cos 0° – cos 180°) =2mB = 2 × 1.5 × 0.22 = 0.66 J (b) Torque τ = mB sin θ In (i) θ = 90°, τ = mB sin 90° = mB = 1.5 × 0.22 = 0.33 N-m This torque tends to align the magnet along the direction of field direction. In (ii) θ = 180°, τ = mB sin 180° = 0 Q. 5. A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10–4 m2, carrying a current of 4.0 A is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid? (b) What are the force and torque on the solenoid if a uniform magnetic field of 7.5 × 10–2 T is set up at an angle of 30° with the axis of the solenoid? Ans. Given N =2000, A = 1.6 × 10–4 m2, I = 4.0 A (a) Magnetic moment of solenoid, m = NIA = 2000 × 4.0 × 1.6 × 10– 4 = 1.28 A-m2 (b) Net force on current carrying solenoid (or magnetic dipole) in uniform magnetic field is always zero. Given, B = 7.5 × 10– 2 T, θ = 30° Torque τ = mB sin θ   τ = 1.28 × 7.5 × 10–2 × sin 30°   = 1.28 × 7.5 × 10–2 × 0.5    = 4.8 × 10–2 N-m Q. 6. A short bar magnet has a magnetic moment of 0.48 JT–1. Give the magnitude and direction of the magnetic field produced by the magnet at a distance of 10 cm from the centre of magnet on (a) the axis, (b) equatorial lines (normal bisector) of the magnet. Ans. Given m = 0.48 JT–1, r = 10 cm = 0.10 m n0 2m (a) Magnetic field at axis, B1 = 4r r3        = (10–7) # 2 # 0.48 = 0.96 # 10–4 T (0.10)3      = 0.96 G along S-N direction (b) Magnetic field at equatorial line B2 = n0 m = 0.48 # 10–4 T 4r r3 = 0.48 G along N-S direction Magnetism and Matter 221

Q. 7. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60° and one of the fields has a magnitude of 1.2 × 10– 2 T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of other field? Ans. For equilibrium, the net torque on magnetic field must be zero. Therefore, the torques exerted by fields B1 and B2 on the dipole must be equal and opposite.      τ1= τ2 mB1 sin θ1 = mB2 sin θ2 ⇒ B2 = B1 sin i1 sin i2 Given B1 = 1.2 × 10– 2 T θ1 = 15°, θ2 = (60° – 15°) = 45° ∴ B2 = 1.2×10 –2 × sin 15° = 1.2×10 –2 × 0.2588 = 4.4×10–3 T sin 45° 0.7071 Earth’s Magnetism Q. 8. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at a place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place. (Given cos 22° = 0.927, sin 22° = 0.375). Ans. By definition, angle of dip θ = 22° Given H = 0.35 G or Be = H = 0.35 G We have H = Be cos θ cos i cos 22° or Be = 0.35 = 0.38 G 0.927 Q. 9. At a certain location in Africa, compass points 12° west of geographical north. The north top of magnetic needle of a dip circle placed in the plane of the magnetic meridian points 60° above the horizontal. The horizontal component of earth’s magnetic field is measured to be 0.16 gauss. Specify the direction and magnitude of earth’s magnetic field at the location. Ans. This problem illustrates how the three elements of earth’s field : angle of declination (α) angle of dip (θ) and horizontal component H; determine the earth’s magnetic field completely. Here angle of declination (α) = 12° Angle of dip θ = 60° and horizontal component, H = 0.16 gauss        = 0.16 × 10– 4 T If Be is the total earth’s magnetic field, then the relation between Be and H is H = Be cos θ gauss & Be = H = 0.16×10–4 = 0.16×10– 4 = 0.32 × 10 –4 T cos i cos 60° 0.5 Thus, the magnitude of earth’s field is 0.32 × 10– 4 T = 0.32 G and it lies in a vertical plane 12° west of geographical meridian making an angle of 60° (upwards) with the horizontal direction. Q. 10. A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographical meridian. The earth magnetic field at the location is 0.33 G and the angle of dip is zero. Locate the line of neutral points (Ignore the thickness of the cable). 222 Xam idea Physics–XII

Ans. Given Be = 0.33 G, I = 2.5 A Angle of dip, θ = 0 ∴ H = Be cos 0° = 0.33 G = 0.33 × 10–4 T, V = Be sin 0° = 0 Magnetic field due to current carrying cable Bc = n0 I 2rr The cable is perpendicular to magnetic meridian. For neutral point, the magnetic produced by cable must be equal and opposite to earth’s magnetic field, i.e., Bc = H & n0 I = H 2rr n0 I r = 2rH = 4r ×10–7 ×2.5 = 1.5×10–2 m =1.5 cm 2r×0.33×10– 4 That is the line of neutral points is parallel to cable at a distance 1.5 cm above the plane of paper. Q . 11. A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm. below and above the cable? Ans. Given Be = 0.39 G, θ = 35° (i) Below the Cable The magnetic field due to horizontal wires. B1 = 4× n0 I = 4× 4r×10– 7 ×1.0 = 2.0×10– 5 T 2rR 2r ×4×10– 2 = 0.2 × 10 – 4 T = 0.2 G This is directed along NS direction. The earth’s horizontal magnetic field is directed from south to north H = Be cos θ = 0.39 cos 35° = 0.39 × 0.82 = 0.32 G ∴  Net horizontal magnetic field BH = H – B1 = 0.32 – 0.2 = 0.12 G. Vertical component of earth’s magnetic field BV = Be sin θ = 0.39 sin 35° = 0.39 ×0.57 = 0.22 G Resultant magnetic field BR = B 2 + BV2 H       = (0.12)2 + (0.22)2 = 0.25 G Angle made by resultant field with horizontal and z = tan–1 BV = tan–1 d 0.22 n= tan–1 (1.8333) = 61.4° BH 0.12 (ii) Above the Cable : If point is above the cable the direction of magnetic field B1 will be along SN direction. So H and B1 will be added. ∴ BH = 0.32 + 0.2 =0.524        BV = 0.22 G ∴ Resultant magnetic field BR = B 2 + BV2 H          = (0.52)2 + (0.22)2 = 0.57 G and z = tan–1 BV = tan– 1 0.22 = tan–1 (0.4230) = 22.9° BH 0.524 Magnetism and Matter 223

Multiple Choice Questions [1 mark] Choose and write the correct option(s) in the following questions. 1. Magnetism in substances is caused by (a) orbital motion of electrons only (b) spin motion of electrons only (c) due to spin and orbital motions of electrons both (d) hidden magnets 2. A magnetic needle is kept in a uniform magnetic field. It experiences (a) a force and a torque (b) a force but not a torque (c) a torque but not a force (d) neither a torque nor a force 3. A magnetic needle is kept in a non-uniform magnetic field. It experiences (a) a force and a torque (b) a force but not a torque (c) a torque but not a force (d) neither a force nor a torque 4. A bar magnet of magnetic moment m is placed in a uniform magnetic field of induction B . The torque exerted on it is (a) m . B (b) –m . B (c) m × B (d) –m × B 5. A uniform magnetic field exists in space in the plane of paper and is initially directed from left to right. When a bar of soft iron is placed in the field parallel to it, the lines of force passing through it will be represented by (a) (b) (c) (d) 6. Points A and B are situated perpendicular to the axis of a 2 cm long bar magnet at large distances x and 3x from its centre on opposite sides. The ratio of the magnetic fields at A and B will be approximately equal to (a) 1: 9 (b) 2: 9 (c) 27: 1 (d) 9: 1 7. A paramagnetic sample shows a net magnetisation of 8 Am–1 when placed in an external magnetic field of 0.6 T at a temperature of 4 K. When the same sample is placed in an external magnetic field of 0.2 T at a temperature of 16 K, the magnetisation will be [ NCERT Exemplar] (a) 32 Am-1 (b) 2 Am-1 (c) 6 Am–1 (d) 2.4 Am–1 3 3 8. A toroid of n turns, mean radius R and cross-sectional radius a carries current I. It is placed on a horizontal table taken as X-Y plane. Its magnetic moment m [NCERT Exemplar] (a) is non-zero and points in the Z-direction by symmetry. (b) points along the axis of the toroid ( m = mφ). 1 at large distances outside the toroid. (c) is zero, otherwise there would be a field falling as r3 (d) is pointing radially outwards. 9. A long solenoid has 1000 turns per metre and carries a current of 1 A. It has a soft iron core of μr =1000. The core is heated beyond the Curie temperature, Tc , then [NCERT Exemplar] (a) the H field in the solenoid is (nearly) unchanged but the B field decreases drastically. (b) the H and B fields in the solenoid are nearly unchanged. (c) the magnetisation in the core reverses direction. (d) the magnetisation in the core diminishes by a factor of about 108. 224 Xam idea Physics–XII

10. The magnetic field of Earth can be modelled by that of a point dipole placed at the centre of the Earth. The dipole axis makes an angle of 11.3° with the axis of Earth. At Mumbai, declination is nearly zero. Then, [NCERT Exemplar] (a) the declination varies between 11.3° W to 11.3° E. (b) the least declination is 0°. (c) the plane defined by dipole axis and Earth axis passes through Greenwich. (d) declination averaged over Earth must be always negative. 11. In a plane perpendicular to the magnetic meridian, the dip needle will be (a) vertical (b) horizontal (c) inclined equal to the angle of dip at that place (d) pointing in any direction 12. The meniscus of a liquid contained in one of the limbs of a narrow U-tube is placed between the pole-pieces of an electromagnet with the meniscus in a line with the field. When the electromagnet is switched on, the liquid is seen to rise in the limb. This indicates that the liquid is (a) ferromagnetic (b) paramagnetic (c) diamagnetic (d) non-magnetic 13. Electro-magnets are made of soft iron because soft iron has (a) small susceptibility and small retentivity (b) large susceptibility and small retentivity (c) large permeability and large retentivity (d) small permeability and large retentivity. 14. In a permanent magnet at room temperature [NCERT Exemplar] (a) magnetic moment of each molecule is zero. (b) the individual molecules have non-zero magnetic moment which are all perfectly aligned. (c) domains are partially aligned. (d) domains are all perfectly aligned. 15. If a magnetic substance is kept in a magnetic field, then which of the following substances is thrown out? (a) Paramagnetic (b) Ferromagnetic (c) Diamegnetic (d) Antiferromagnetic 16. Above Curie’s temperature ferromagnetic substances becomes (a) paramagnetic (b) diamagnetic (c) superconductor (d) no change 17. In the hysteresis cycle, the value of H needed to make the intensity of magnetisation zero is called (a) retentivity (b) coercive force (c) Lorentz force (d) none of the above 18. A permanent magnet attracts (b) only ferromagnetic substances (a) all substances (c) some substances and repels others (d) ferromagnetic substances and repels all others 19. Susceptibility is positive for (b) ferromagnetic substances (a) paramagnetic substances (c) non-magnetic substances (d) diamagnetic substances 20. If the horizontal and vertical components of earth’s magnetic field are equal at a certain place, the angle of dip is (a) 90° (b) 60° (c) 45° (d) 0° Magnetism and Matter 225

Answers 2. (c) 3. (a) 4. (c) 5. (b) 6. (c) 8. (c) 9. (a), (d) 10. (a) 11. (a) 12. (b) 1. (c) 14. (c) 15. (c) 16. (a) 17. (b) 18. (b) 7. (b) 20. (c) 13. (b) [1 mark] 19. (a), (b) Fill in the Blanks 1. The unit of magnetic dipole moment is _______________. 2. Diamagnetic substances when placed in a magnetic field, are magnetised in the direction _______________ to the magnetic field. 3. Paramagnetic materials when placed in a magnetic field are magnetised in the direction _______________ to the magnetic field. 4. The angle between the magnetic moment of a bar magnet and its magnetic field at an equatorial point is _______________. 5. The ability of a material to retain magnetism after removal of magnetizing field is called as _______________. 6. SI unit of magnetic pole strength is _______________. 7. Inside the body of a magnet the direction of magnetic field lines is from _______________. 8. For paramagnetic materials magnetic susceptibility is related with temperature as inversely proportional to _______________. 9. There is no effect of temperature on _______________ type of materials. 10. Ferromagnetism can be explained on the basis of formation of _______________ within the materials. Answers 1. Am2 2. opposite 3. parallel 4. 180° 5. retentivity 6. ampere-meter 8. T 9. diamagnetic 7. South pole to North pole 10. domain Very Short Answer Questions [1 mark] Q. 1. Where on the earth’s surface is the value of angle of dip maximum? OR [CBSE (AI) 2011] Where on the surface of earth is the angle of dip 90°? Ans. Angle of dip (90°) is maximum at magnetic poles. Q. 2. A magnetic needle, free to rotate in a vertical plane, orients itself vertically at a certain place on the Earth. What are the values of (i) horizontal component of Earth’s magnetic field and (ii) angle of dip at this place? [CBSE (F) 2012] Ans. (i) 0 (ii) 90° Q. 3. Where on the earth’s surface is the value of vertical component of earth’s magnetic field zero?  [CBSE (F) 2011] Ans. Vertical component of earth’s magnetic field is zero at magnetic equator. Q. 4. The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°. What is the value of vertical component of earth’s magnetic field at equator? [CBSE Delhi 2012] Ans. Zero 226 Xam idea Physics–XII

Q. 5. A small magnet is pivoted to move freely in the magnetic meridian. At what place on earth’s surface will the magnet be vertical? [CBSE (F) 2012] Ans. Magnet will be vertical at the either magnetic pole of earth. Q. 6. Which of the following substances are diamagnetic? Bi, Al, Na, Cu, Ca and Ni [CBSE Delhi 2013] Ans. Diamagnetic substances are (i) Bi (ii) Cu. Q. 7. What are permanent magnets? Give one example. [CBSE Delhi 2013] Ans. Substances that retain their attractive property for a long period of time at room temperature are called permanent magnets. Examples: Those pieces which are made up of steel, alnico, cobalt and ticonal. Q. 8. Mention two characteristics of a material that can be used for making permanent magnets. [CBSE Delhi 2010] Ans. For making permanent magnet, the material must have high retentivity and high coercivity (e.g., steel). Q. 9. Why is the core of an electromagnet made of ferromagnetic materials? [CBSE Delhi 2010] Ans. Ferromagnetic material has a high permeability. So on passing current through windings it gains sufficient magnetism immediately. Q. 10. The permeability of a magnetic material is 0.9983. Name the type of magnetic materials it represents. [CBSE Delhi 2011] Ans. µ is <1 and > 0, so magnetic material is diamagnetic. Q . 11. The susceptibility of a magnetic materials is – 4.2×10–6. Name the type of magnetic materials it represents. [CBSE Delhi 2011] Ans. Susceptibility of material is negative, so given material is diamagnetic. Q . 12. In what way is the behaviour of a diamagnetic material different from that of a paramagnetic, when kept in an external magnetic field? [CBSE Central 2016] Ans. A diamagnetic specimen would move towards the weaker region of the field while a paramagnetic specimen would move towards the stronger region. Q. 13. At a place, the horizontal component of earth’s magnetic field is B and angle of dip is 60°. What is the value of horizontal component of the earth’s magnetic field at equator? [CBSE Delhi 2017] Ans. Here, BH = B and δ = 60° We know that BH = BE cos δ B = BE cos 60° ⇒ BE = 2B At equator δ = 0° ∴ BH = 2B cos 0° = 2B Q. 14. What is the angle of dip at a place where the horizontal and vertical components of the Earth’s magnetic field are equal? [CBSE (F) 2012] Ans. We know BV = tand BH Given BV =BH then tan δ = 1 Angle of dip, δ = 45° Q. 15. The magnetic susceptibility of magnesium at 300 K is 1.2 × 105. At what temperature will its magnetic susceptibility become 1.44 × 105? [CBSE 2019 (55/2/1)] Ans. The susceptibility of a paramagnetic substance is inversely proportional to the absolute temperature. Magnetism and Matter 227

| ? 1 T |= C (where C is curie constant) T Here |1 = 1.2 × 105, T1 = 300 K |2 = 1.44 × 105, T2 = ? |1 = C & C = |1 T1 ...(i) T1 |2 = C ...(ii) T2 T2 = C = |1 T1 = 1.2 ×105 × 300 = 250 K |2 |2 1.44 ×105 Q. 16. The magnetic susceptibility χ of a given material is – 0.5. Identify the magnetic material.  [CBSE 2019 (55/2/1)] Ans. The susceptibility of material is – 0.5, which is negative. Hence, material is diamagnetic substance. Q. 17. Write one important property of a paramagnetic material. [CBSE 2019 (55/5/1)] Ans. It moves from weaker magnetic field towards stronger magnetic field. Q. 18. Do the diamagnetic substances have resultant magnetic moment in an atom in the absence of external magnetic field? [CBSE 2019 (55/5/1)] Ans. No, diamagnetic substances have no resultant magnetic moment in the absence of external magnetic field. Q. 19. How does the (i) pole strength and (ii) magnetic moment of each part of a bar magnet change if it is cut into two equal pieces transverse to length? Ans. When a bar magnet of magnetic moment (M = m2l ) is cut into two equal pieces transverse to its length, (i) the pole strength remains unchanged (since pole strength depends on number of atoms in cross-sectional area). (ii) the magnetic moment is reduced to half (since M ∝ length and here length is halved). Q. 20. A hypothetical bar magnet (AB) is cut into two equal parts. One part is now kept over the other, so that the pole C2 is above C1. If M is the magnetic moment of the original magnet, what would be the magnetic moment of the combination, so formed? Ans. The magnetic moment of each half bar magnet is M but oppositely 2 directed, so net magnetic moment of combination = M – M =0 (zero). 2 2 Short Answer Questions–I [2 marks] Q. 1. The susceptibility of a magnetic material is 2.6 × 10–5. Identify the type of magnetic material and state its two properties. [CBSE Delhi 2012] Ans. The material having positive and small susceptibility is paramagnetic material. 228 Xam idea Physics–XII

Properties (i) They have tendency to move from a region of weak magnetic field to strong magnetic field, i.e., they get weakly attracted to a magnet. (ii) When a paramagnetic material is placed in an external field the field lines get concentrated inside the material, and the field inside is enhanced. Q. 2. The susceptibility of a magnetic material is –2.6 × 10–5. Identify the type of magnetic material and state its two properties. [CBSE Delhi 2012] Ans. The magnetic material having negative susceptibility is diamagnetic in nature. Properties: (i) This material has + ve but low relative permeability. (ii) They have the tendency to move from stronger to weaker part of the external magnetic field. Q. 3. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known at to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place. [CBSE Delhi 2011] Ans. Angle of dip, θ = 60° H = 0.4 G = 0.4 × 10–4 T If Be is earth’s magnetic field, then H = Be cos θ. & Be = H = 0.4 # 10–4 T = 0.4 # 10–4 T = 0.8 # 10–4 T = 0.8 G cos i cos 60° 0.5 Q. 4. A compass needle, free to turn in a vertical plane orients itself with its axis vertical at a certain place on the earth. Find out the values of (i) horizontal component of earth’s magnetic field and (ii) angle of dip at the place. [CBSE Delhi 2013] Ans. If compass needle orients itself with its axis vertical at a place, then (i) BH = 0 because BV = | B | BV (ii) tand = BH = 3 ⇒ Angle of dip d = 90°, Concept: It is possible only on magnetic north or south poles. Q. 5. Write two properties of a material suitable for making (a) a permanent magnet, and (b) an electromagnet. [CBSE (AI) 2017] Ans. (a) Two properties of material used for making permanent magnets are (i) High coercivity (ii) High retentivity (iii) High permeability (b) Two properties of material used for making electromagnets are (i) High permeability (ii) Low coercivity (iii) Low retentivity Q. 6. From molecular view point, discuss the temperature dependence of susceptibility for diamagnetism, paramagnetism and ferromagnetism. Ans. Diamagnetism is due to orbital motion of electrons developing magnetic moments opposite to applied field and hence is not much affected by temperature. Paramagentism and ferromagnestism is due to alignments of atomic magnetic moments in the direction of the applied field. As temperature increases, this alignment is disturbed and hence susceptibilities of both decrease as temperature increases. Magnetism and Matter 229

Q. 7. Consider the plane S formed by the dipole axis and the axis of earth. Let P be point on the magnetic equator and in S. Let Q be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angles at P and Q. Ans. In following figure: (i) P is in S (needle will point both north) Declination = 0 P is also on magnetic equator. ∴ Dip = 0° (ii) Q is on magnetic equator. ∴ Dip = 0° but declination = 11.3 Q. 8. What is the basic difference between the atom and molecule of a diamagnetic and a paramagnetic material? Why are elements with even atomic number more likely to be diamagnetic? Ans. Atoms/molecules of a diamagnetic substance contain even number of electrons and these electrons form the pairs of opposite spin; while the atoms/molecules of a paramagnetic substance have excess of electrons spinning in the same direction. The elements with even atomic number Z has even number of electrons in its atoms/molecules, so they are more likely to form electrons pairs of opposite spin and hence more likely to be diamagnetic. Short Answer Questions–II [3 marks] Q. 1. Depict the field-line pattern due to a current carrying solenoid of finite length. (i) In what way do these lines differ from those due to an electric dipole? (ii) Why can’t two magnetic field lines intersect each other? [CBSE (F) 2009] Field Field Ans. (i) Difference: Field lines of a solenoid form continuous current loops, while in the case of an electric dipole the field lines begin from a positive charge and end on a negative charge or escape to infinity. (ii) Two magnetic field lines cannot intersect because at the point of intersection, there will be two directions of magnetic field which is impossible. Q. 2. Explain the following: (i) Why do magnetic field lines form continuous closed loops? (ii) Why are the field lines repelled (expelled) when a diamagnetic material is placed in an external uniform magnetic field? [CBSE (F) 2011] Ans. (i) Magnetic lines of force form continuous closed loops because a magnet is always a dipole and as a result, the net magnetic flux of a magnet is always zero. (ii) When a diamagnetic substance is placed in an external magnetic field, a feeble magnetism is induced in opposite direction. So, magnetic lines of force are repelled. 230 Xam idea Physics–XII

Q. 3. (i) Mention two properties of soft iron due to which it is preferred for making an electromagnet. (ii) State Gauss’s law in magnetism. How is it different from Gauss’s law in electrostatics and why? [CBSE South 2016] Ans. (i) Low coercivity and high permeability (ii) Gauss’s Law in magnetism: The net magnetic flux through any closed surface is zero. y B.ds = 0 Gauss’s Law in electrostatics: The net electric flux through any closed surface is 1 times the net charge enclosed by the surface. f0 y E.ds = q f0 The difference between the Gauss’s law of magnetism and that for electrostatic is a reflection of the fact that magnetic monopole do not exist i.e., magnetic poles always exist in pairs. Q. 4. Show diagrammatically the behaviour of magnetic field lines in the presence of (i) paramagnetic and (ii) diamagnetic substances. How does one explain this distinguishing feature?          OR [CBSE (AI) 2014] Draw the magnetic field lines distinguishing between diamagnetic and paramagnetic materials. Give a simple explanation to account for the difference in the magnetic behaviour of these materials. [CBSE Bhubaneshwar 2015, Central 2016] Ans. • A paramagnetic material tends to move from weaker field to stronger field regions of the magnetic field. So, the number of lines of magnetic field increases when passing through it. Magnetic dipole moments are induced in the direction of magnetic field. Paramagnetic materials has a small positive susceptibility. • A diamagnetic material tends to move from stronger field to weaker field region of the magnetic field. So, the number of lines of magnetic field passing through it decreases. Magnetic dipole moments are induced in the opposite direction of the applied magnetic field. Diamagnetic materials has a negative susceptibility in the range (–1 ≤ χ < 0). Q. 5. Draw the magnetic field lines for a current carrying solenoid when a rod made of (a) copper, (b) aluminium and (c) iron are inserted within the solenoid as shown. [CBSE Sample Paper 2018] Ans. (a) When a bar of diamagnetic material (copper) is placed in an external magnetic field, the field lines are repelled or expelled and the field inside the material is reduced. NnsS Magnetism and Matter 231

(b) When a bar of paramagnetic material (Aluminium) is placed in an external field, the field lines gets concentrated inside the material and the field inside is enhanced. Ns nS (c) When a ferromagnetic material (Iron) is placed in an internal magnetic field, the field lines are highly concentrated inside the material. Ns nS Q. 6. In what way is Gauss’s law in magnetism different from that used in electrostatics? Explain briefly. The Earth’s magnetic field at the equator is approximately 0.4 G. Estimate the Earth’s magnetic dipole moment. Given: Radius of the Earth = 6400 km. [CBSE Patna 2015] Ans. As we know that y E . dS = 1 [q] Isolated positive or negative charge exists freely. So, Gauss’s law states that f0 Isolated magnetic poles do not exist. So, Gauss’s law states that y B . dS = 0 Magnetic field intensity at the equator is n0 B= 4r . m = 10–7 m R3 R3 ∴ m = 107 . BR3 = 107 × 0.4 × 10–4 × (6400 × 103)3 = 1.05 × 1023 Am2 Q. 7. A bar magnet of magnetic moment 6 J/T is aligned at 60° with a uniform external magnetic field of 0·44 T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii). [CBSE Examination Paper 2018] Ans. (a) Work done = mB(cos q1− cos q2) (i) i1 = 60°, i2 = 90° ∴ Work done = mB(cos 60°− cos 90°) = mB c 1 – 0 m = 1 mB 2 2 = 1 × 6 × 0.44 J =1.32 J 2 (ii) i1 = 60°, i2 =180° ∴ Work done = mB(cos 60°− cos 180°) = mB c 1 – (–1) m = 3 mB 2 2 = 3 × 6 × 0.44 J = 3.96 J 2 232 Xam idea Physics–XII

(b) Torque = | m × B |= mB sin i For i =180° and B = 0.44 T we have Torque = 6 × 0.44 sin 180°=0 Q. 8. (a) An iron ring of relative permeability mr has windings of insulated copper wire of n turns per metre. When the current in the windings is I, find the expression for the magnetic field in the ring. (b) The susceptibility of a magnetic material is 0.9853. Identify the type of magnetic material. Draw the modification of the field pattern on keeping a piece of this material in a uniform magnetic field. [CBSE Examination Paper 2019] Ans. (a) From Ampere’s circuital law, we have, &y B.dl = n0 nr Ienclosed ...(i) For the field inside the ring, we can write &y B.dl = &y Bdl = B.2rr (r = radius of the ring) Also, Ienclosed = (2rrn) I    ` B. 2rr = n0 nr . (n.2rr) I     [Using equation (i)] ` B= n0 nr .nI (b) The material is paramagnetic. The field pattern gets modified as shown in the figure below. Q. 9. (a) Show that the time period (T) of oscillations of a freely suspended magnetic dipole of I magnetic moment (m) in a uniform magnetic field (B) is given by T = 2r mB , where I is a moment of inertia of the magnetic dipole. (b) Identify the following magnetic materials: [CBSE 2019 (55/3/1)] (i) A material having susceptibility _|mi = – 0.00015 (ii)  A material having susceptibility _|mi = 10–5  Ans. (a) Let us consider a uniform magnetic field B exists in the region, in which a magnet of dipole moment m is placed. The dipole is making small angle i with the magnetic field. The torque acts on the magnet is given by x = –mB sin i (Restoring torque) = –mB i (  i in small) ...(i) Also the torque on dipole try to restore its initial position i.e., along the direction of magnetic field. (I = moment of inertia) In equilibrium d2 i dt2 I = –mB sin i ...(ii) Negative sign implies that restoring torque is in opposition to deflecting torque. d2 i = –mB i ...(iii) dt2 I ...(iv) Comparing with equation of angular SHM d2 i = – ~2 i dt2 We have ~2 = mB & ~ = mB I I Magnetism and Matter 233

⇒ 2r = mB & T = I T I 2r mB T = 2r I mB (b) (i) Diamagnetic substance. (ii) Paramagnetic substance. Q. 10. Write three points of differences between para-, dia- and ferro- magnetic materials, giving one example for each. [CBSE 2019 (55/1/1)] Ans. Diamagnetic Paramagnetic Ferromagnetic x&1 1 –1 # | < 0 0<|<f nr & 1 n & n0 2 0 # nr < 1 1 # nr <(1 + f) 3 n < n0 n > n0 Where e is any positive constant. Examples: Diamagnetic materials: Bi, Cu, Pb, Si, water, NaCl, Nitrogen (at STP) Paramagnetic materials: Al, Na, Ca, Oxygen (at STP), Copper chloride Ferromagnetic materials: Fe, Ni, Co, Alnico. (Any one) Q. 11. (a) State Gauss’s law for magnetism. Explain its significance. [CBSE 2019 (55/1/1)] (b) Write the four important properties of the magnetic field lines due to a bar magnet. Ans. (a) Gauss’s law for magnetism states that “The total flux of the magnetic field, through any closed surface, is always zero.” Alternatively = y B. d s = 0 s This law implies that magnetic monopoles do not exist. Also magnetic field lines form closed loops. (b) Four properties of magnetic field lines (i) Magnetic field lines always form continuous closed loops. (ii) The tangent to the magnetic field line at a given point represents the direction of the net magnetic field at that point. (iii) The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field. (iv) Magnetic field lines do not intersect. Long Answer Questions [5 marks] Q. 1. Derive an expression for magnetic field intensity due to a magnetic dipole at a point on its axial line. Ans. Consider a magnetic dipole (or a bar magnet) SN of length 2l having south pole at S and north pole at N. The strength of south and north poles are – qm and + qm respectively. Magnetic moment of magnetic dipole m = qm 2l, its direction is from S to N. Consider a point P on the axis of magnetic dipole at a distance r from mid point O of dipole. The distance of point P from N-pole, r1= (r – l) 234 Xam idea Physics–XII

The distance of point P from S-pole, r2= (r + l) Let B1 and B2 be the magnetic field intensities at point P due to north and south poles respectively. The directions of magnetic field due to north pole is away from N-pole and due to south pole is towards the S-pole. Therefore, n0 qm n0 qm B1 = 4r (r – l)2 from N to P and B2 = 4r (r + l)2 from P to S Clearly, the directions of magnetic field strengths B1 and B 2 are along the same line but opposite to each other and B1>B2. Therefore, the resultant magnetic field intensity due to bar magnet has magnitude equal to the difference of B1 and B2 and direction from N to P. n0 qm n0 qm i.e., B= B1 – B2 = 4r (r – l)2 – 4r (r + l)2 = n0 qm >(r 1 – 1 H = n0 qm > (r + l)2 – (r – l) 2 4r – l)2 + 4r (r2 – l2) (r l) 2 2 H = n0 qm > 4rl 2 H = n0 2 (qm 2l) r 4r (r2 – l2) 4r (r2 – l2)2 But qm 2l =m (magnetic dipole moment) n0 2m.r ∴ B = 4r (r2 – l2)2 …(1) If the bar magnet is very short and point P is far away from the magnet, the r >> l, therefore, equation (1) takes the form n0 2mr B = 4r r4 or B= n0 2m …(2) 4r r3 This is the expression for magnetic field intensity at axial position due to a short bar magnet. Q. 2. Derive an expression for magnetic field intensity due to a magnetic dipole at a point lies on its equatorial line. Ans. Consider a point P on equatorial position (or broad side on position) of short bar magnet of length 2l, having north pole (N) and south pole (S) of strength +qm and – qm respectively. The distance of point P from the mid point (O) of magnet is r. Let B1 and B2 be the magnetic field intensities due to north and south poles respectively. NP=SP= r2 + l2 . n0 qm B1 = 4r r2 + l2 along N to P θ θ n0 qm θθ 4r r2 + l2 B 2 = along P to S Clearly, magnitudes of B1 and B 2 are equal θ i.e., | B1 | = | B 2 | or B1 = B2 To find the resultant of B1 and B 2 , we resolve them along and perpendicular to magnetic axis SN. Components of B1 along and perpendicular θ θ to magnetic axis are B1 cosθ and B2 sinθ respectively. Components of B 2 along and perpendicular to magnetic axis are B2 cos θ and B2 sin θ respectively. Clearly, components of B1 and B 2 perpendicular to axis SN. B1 sin θ and B2 sin θ are equal in magnitude and opposite in direction and hence, cancel each other; while the components of B1 Magnetism and Matter 235

and B 2 along the axis are in the same direction and hence, add up to give to resultant magnetic field parallel to the direction NS. ∴ Resultant magnetic field intensity at P. B = B1 cos θ + Bq2mcos θ But B1 = B2 n0 r2 + l2 and cos θ = = 4r ON = l = l PN r2 + l2 (r2 + l2)1/2 ∴ B = 2B1 cos i = 2× n0 qm × (r2 l = n0 2qm l 4r (r2 + l2) + l2)1/2 4r (r2 + l2)3/2 ∴Bu t qm.2 l=m, magBn=eti4ncr0m(or2m+emnl2t)o3/f2 magnet …(1) If the magnet is very short and point P is far away, we have l<<r; so l2 may be neglected as compared to r2 and so equation (1) takes the form B = n0 m …(2) 4r r3 This is expression for magnetic field intensity at equatorial position of the magnet. Q. 3. (a) A small compass needle of magnetic moment ‘m’ is free to turn about an axis perpendicular to the direction of uniform magnetic field ‘B’. The moment of inertia of the needle about the axis is ‘I’. The needle is slightly disturbed from its stable position and then released. Prove that it executes simple harmonic motion. Hence deduce the expression for its time period. (b) A compass needle, free to turn in a vertical plane orients itself with its axis vertical at a certain place on the earth. Find out the values of (i) horizontal component of earth’s magnetic field and (ii) angle of dip at the place. [CBSE Delhi 2013] Ans. (a) If magnetic compass of dipole moment m is placed at angle θ in uniform magnetic field, and released it experiences a restoring torque. +qm –qm Restoring torque, x = – magnetic force × perpendicular distance    = – qmB . (2a sin θ), τ = – mB.sin θ, where qm = pole strength, m = qm.2a (magnetic moment) Negative sign shows that restoring torque acts in the opposite direction to that of defecting torque. In equilibrium, the equation of motion, ⇒ I d2 i =– mBi   (For small angle sin i . i) dt2 ⇒ d2 i =– mB i ⇒ d2 i = –c mB mi dt2 I dt2 I Since d2 i \\ i ⇒ d2 i =– ~2 i dt2 dt2 236 Xam idea Physics–XII

It represents the simple harmonic motion with angular frequency ~2 = mB ⇒ T = 2r = 2r I I ~ mB (b) If compass needle orients itself with its axis vertical at a place, then (i) BH = 0 because BV = |B| BV (ii) tan δ = BH = 3 Angle of dip δ = 90°, Concept: It is possible only on magnetic north or south poles. Self-Assessment Test Time allowed: 1 hour Max. marks: 30 1. Choose and write the correct option in the following questions. (3 × 1 = 3) (i) A permanent magnet (a) attracts all substances (b) attracts only ferromagnetic substances (c) attracts ferromagnetic substances and repels all others (d) attracts some substances and repels others (ii) If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is (a) repelled by the north pole and attracted by the south pole (b) attracted by the north pole and repelled by the south pole (c) attracted by both the poles (d) repelled by both the poles (iii) A bar magnet having a magnetic moment of 2 × 104 J T–1 is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 × 10–4 T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is (a) 12 J (b) 6 J (c) 2 J (d) 0.6 J 2. Fill in the blanks. (2 × 1 = 2) (i) The temperature of transition from ferromagnetic to paramagnetism is called the __________. (ii) Substances which at room temperature retain their ferromagnetic property for a long period of time are called ______________. 3. (i) Name the three elements of the earth’s magnetic field. (ii) Where on the surface of the earth is the vertical component of the earth’s magnetic field zero? 1 4. The susceptibility of a magnetic material is 1.9 × 10–5. Name the type of magnetic materials it represents. 1 5. Depict the behaviour of magnetic field lines in the presence of a diamagnetic material. 1 6. The given graph shows the variation of intensity of magnetisation I with strength of applied magnetic field H for two magnetic materials P and Q. Magnetism and Matter 237

(i) Identify the materials P and Q. (ii) For material P, plot the variation of intensity of magnetisation with temperature. Justify your answer. 2 7. Explain the following: (i) Why do magnetic lines of force form continuous closed loops? (ii) Why are the field lines repelled (expelled) when a diamagnetic material is placed in an external uniform magnetic field? 2 8. The relative magnetic permeability of a magnetic material is 800. Identify the nature of magnetic material and state its two properties. 2 9. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known at to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place. 2 10. A closely wound solenoid of 2000 turns and cross sectional area 1.6 ×10–4 m2 carrying a current of 4.0 A is suspended through its centre allowing it to turn in a horizontal plane. Find (i) the magnetic moment associated with the solenoid, (ii) magnitude and direction of the torque on the solenoid if a horizontal magnetic field of 7.5 × 10–2 T is set up at an angle of 30° with the axis of the solenoid. 3 11. A uniform conducting wire of length 12a and resistance R is wound up as a current carrying coil in the shape of (i) an equilateral triangle of side a; (ii) a square of sides a and, (iii) a regular hexagon of sides a. The coil is connected to a voltage source V0. Find the magnetic moment of the coils in each case. 3 12. (i) How does angle of dip change as one goes from magnetic pole to magnetic equator of the Earth? (ii) A uniform magnetic field gets modified as shown below when two specimens X and Y are placed in it. Identify whether specimens X and Y are diamagnetic, paramagnetic or ferromagnetic. Y X (iii) How is the magnetic permeability of specimen X different from that of specimen Y? 3 13. (a) Draw the magnetic field lines due to a circular loop of area A carrying current I. Show that it acts as a bar magnet of magnetic moment m = IA. (b) Derive the expression for the magnetic field due to a solenoid of length ‘2l’, radius ‘a’ having ‘n’ number of turns per unit length and carrying a steady current ‘I’ at a point on the axial line, distant ‘r’ from the centre of the solenoid. How does this expression compare with the axial magnetic field due to a bar magnet of magnetic moment ‘m’? 5 Answers 1. (i) (b) (ii) (d) (iii) (b) 2. (i) curie temperature (ii) permanent magnets 9. Be = 0.8 G 10. Magnetic moment = 1.28 A–m2, Torque = 0.048 N–m 11. (i) M1= 3 a2I (ii) M2=3a2 I (iii) M3 = 3 3 a2I zzz 238 Xam idea Physics–XII

Electromagnetic Chapter –6 Induction 1. Electromagnetic Induction The phenomenon of generation of induced emf and induced current due to change in magnetic field lines associated with a closed circuit is called electromagnetic induction. 2. Magnetic Flux Magnetic flux through a surface of area A placed in a uniform magnetic field is zm = B. A = BA cos i, θ being angle between B and normal to A . If magnetic field is not uniform, then zm = yA B.dA, where integral extends for whole area A. The SI unit of magnetic flux is weber. Magnetic flux is a scalar quantity; because of being scalar product of two vectors B and A. 3. Faraday’s Laws of Electromagnetic Induction (i) Whenever there is a change in magnetic flux linked with a coil, an emf is induced in the coil. The induced emf is proportional to the rate of change of magnetic flux linked with the coil. i.e., f ? Tz Tt (ii) emf induced in the coil opposes the change in flux, i.e., f ? – Tz & f = –k Tz Tt Tt where k is a constant of proportionality. Negative sign represents opposition to change in flux. Tz Tt In SI system φ is in weber, t in second, f in volt, when k = 1, f = – If the coil has N-turns, then f= –N Tz Tt 4. Induced Current and Induced Charge If a coil is closed and has resistance R, then current induced in the coil, I = f = – N Tz R R Tt Induced charge, q = I Tt = – NTz = Total flux linkage R Resistance 5. Lenz’s Law It states that the direction of induced emf is such that it tends to produce a current which opposes the change in magnetic flux producing it. Electromagnetic Induction 239

6. EMF Induced in a Moving Conducting Rod EMF induced in a conducting rod of length l moving with velocity v in a magnetic field of induction B, such that B, l and v are mutually perpendicular, is given by ε = Bvl B2l2v r force required to keep the rod in constant motion is F = BIL = 7. Self Induction When the current in a coil is changed, an induced emf is produced in the same coil. This phenomenon is called self-induction. If L is self-inductance of coil, then Nz ? I or Nz = LI & L = Nz I L is also called coefficient of self induction. The graph between effective magnetic flux (Nφ) and current I is straight line of slope self inductance L. TI Also induced emf f = –L Tt The unit of self inductance is henry (H). The self induction acts as inertia in electrical circuits; so it is also called electrical inertia. The self inductance of a solenoid consisting core of relative permeability µr is L= µr µ0 n2Al where n= N is the number of turns per metre length. l 8. Mutual Induction When two coils are placed nearby and the current in one coil (often called primary coil) is changed, the magnetic flux linked with the neighbouring coil (often called secondary coil) changes; due to which an emf is induced in the neighbouring coil. This effect is called the mutual induction. If M is mutual inductance of two coils, then φ2 ∝ I1 or φ2 =MI1 Definition of mutual inductance: M= z2 . I1 The mutual inductance of two coils is defined as the magnetic flux linked with the secondary coil when the current in primary coil is 1 ampere. TI1 Also induced emf in secondary coil f2 = – M Tt &thMe e=mfTiInf1d2/Tutc.ed The mutual inductance of two coils is defined as in the secondary coil when the rate of change of current in the primary coil is 1 A /s. The SI unit of mutual inductance is also henry (H). The mutual inductance of two coils does not depend on the fact which coil carries the current and in which coil emf is induced i.e., M12 =M21 = M This is also called reciprocity theorem of mutual inductance. If L1 and L2 are self-inductances of two coils with 100% flux linkage between them, then M = L1 L2, otherwise M = k L1 L2 , where k is coefficient of flux linkage between the coils. Mutual Inductance of solenoid-coil system n0 N1 N2 A M = l where A is area of coil, l is length of solenoid, N1 is number of turns in solenoid and N2 is number of turns in coil. 9. Eddy Currents When a thick piece of a conductor is placed in a varying magnetic field the magnetic flux linked with the conductor changes, so currents are induced in the body of conductor, which causes heating of conductor. 240 Xam idea Physics–XII

The currents induced in the conductor are called the eddy currents. In varying magnetic field, the free electrons of conductor experience Lorentz force and traverse closed paths; which are equivalent to small current loops. These currents are the eddy currents; they cause heating effect and sometimes the conductor becomes red-hot. Eddy current losses may be reduced by using laminated soft iron cores in galvanometers, transformers, etc., and making holes in the core. Few of the application of eddy currents is in induction furnace, induction motor and many more. Selected NCERT Textbook Questions Induced emf Q. 1. A 1.0 m metallic rod is rotated with an angular velocity of 400 rad/s about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. Ans. EMF developed between the centre of ring and the point on the ring. f = 1 B ~l2 2 Given B = 0.5 T, ω=400 rad/s, l=1.0 m. ∴ f = 1 # 0.5 # 400 # (1.0) 2 = 100 volt 2 Q. 2. A rectangular wire loop of sides 8 cm × 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed if the velocity of the loop is 1 cms–1 in a direction normal to the (i) longer side (ii) shorter side of the loop? For how long does the induced voltage last in each case? Ans. Given l=8 cm =8 ×10–2 m, b = 2 cm = 2 × 10–2 m v = 1 cm s–1 = 1 × 10–2 m/s, B = 0.3 T (i) When velocity is normal to the longer side Induced emf, ε = Bvl = 0.3 × 1×10–2 × 8 ×10–2 = 24 × 10–5 V emf will last only so long as the loop is in the magnetic field. Time taken = distance = b = 2 # 10–2 = 2s velocity v 1 # 10–2 (ii) When velocity is normal to the shorter side ε2 = Bvb = 0.3 × 1 × 10–2 × 2 × 10–2 = 6 × 10–5 V Time taken = l = 8 # 10–2 = 8s v 1 # 10–2 Q. 3. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 ms–1 at right angles to the horizontal component of earth’s magnetic field equal to 0.30 ×10–4 Wbm–2. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of emf? (c) Which emf of the wire is at the higher electrical potential ? Electromagnetic Induction 241

An s. (a) IGnivsteann,taHn=eo0u.3s0e×mf1, 0f–4=T,Bvn=vl5=.0 Hvl l =10 m N (j) (–vk) ms–1, z H ∴ ε = 0.30 × 10–4 × 5.0 × 10 = 1.5×10–3 V=1.5 mV W e– +E u (b) By Fleming’s right hand rule, the direction of induced current induced emf in wire is from west to east, therefore, direction of emf is from west to east. (c) The direction of electron flow according to relation F m = qv×B = –e (–vkt) × (Btj) = –evBit S i.e., along negative x-axis, i.e., from east to west. The induced emf will oppose the flow of electrons from east to west, so eastern end will be at higher potential. Q. 4. A jet plane is travelling westward at a speed of 1800 km/h. What is the potential difference developed between the ends of a wing 25 m long? Its earth’s magnetic field at the location has a magnitude of 5.0 ×10–4 T and the dip angle is 30°. [CBSE (AI) 2009] Ans. The wing of horizontal travelling plane will cut the vertical component of earth’s magnetic field, so emf is induced across the wing. The vertical component of earth’s field is given by V = Be sin θ; where Be is earth’s magnetic field and θ is angle of dip Induced emf of wing ε = V v l = (Be sin θ) v l Given Be= 5.0 ×10–4 T, l=25 m, θ = 30°, v = 1800 km/h = 1800 # 5 m/s = 500 m/s 18 ∴ ε = (5.0 × 10–4 ×sin 30°) × 500 × 25 = (5.0 ×10–4 ×0.5) × 500 × 25 = 3.1 V Induced emf and Power Q. 5. A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad/s in a uniform horizontal magnetic field of magnitude 3.0 × 10–2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to joule heating. Where does the power come from? Ans. Magnetic flux linked with the coil, z = \" : \" B A = NBA cosθ = NBA cos ωt (where θ = ωt) EMF induced in the coil f = –N dz dt = –N d (BA cos ~t) = NBA ~ sin ~t dt Maximum emf induced εmax =NBAω = NB (πr2) ω Given N = 20, r = 8.0 cm = 8.0 × 10–2 m, B=3.0 × 10–2 T, ω = 50 rad/s ∴ εmax = 20×3.0×10–2 ×3.14× (8.0×10–2)2 ×50 = 0.603 volt Average emf = NBA ω (sin ωt)av = 0 (Since average value of sin ωt over a complete cycle is zero.) Maximum current induced, fmax Imax = R = 0.603 = 0.0603 A 10 Average power loss due to joule heating 242 Xam idea Physics–XII

Pmax = (I 2) av R = (f2) av R 1 1 Since average value of sin2 ωt for a complete cycle is 2 , i.e., (sin2 ωt) av = 2 ∴ Pmax = 1 N2 B2 A2 ~2 2 R = 1 (NBA~)e NBA~ o = 1 fmax Imax 2 R 2 = 1 # 0.603 # 0.0603 = 0.018 W 2 The current induced causes a torque which opposes the rotation of the coil. An external agency (rotor) must supply torque to counter this torque in order to keep the coil rotating uniformly. The source of power dissipated as heat is the rotor. Q. 6. A rectangular loop of sides 8 cm × 2 cm with a small cut is stationary in a uniform magnetic field produced by an electromagnet. If the current feeding the electromagnet is gradually reduced so that the magnetic field decreases from its initial value of 0·3 T at the rate of 0·02 Ts–1. If the cut is joined and the loop has a resistance of 1·6 Ω, how much power is dissipated by the loop as heat ? What is the source of this power? Ans. Area of loop, A = 8 cm × 2 cm = 16 cm2=16×10–4 m2 Induced emf, f = – Tz = – T (BA) = – A TB Tt Tt Tt TB Here, Tt = –0.02 Ts–1 ∴ Induced emf, ε=–(16 × 10–4) × (–0.02) = 3.2 × 10–5 V Induced current, I= f = 3.2×10–5 = 2×10–5 A R 1.6 Power dissipated, P=I2R=(2×10–5)2 × 1.6 = 6.4 × 10–10 W The source of the power is the external source feeding the electromagnet Self Inductance and Mutual Inductance Q. 7. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, calculate the self-induction of the circuit. [CBSE (F) 2011] Ans. Induced emf E = –L TI …(i) Tt Here, E= 200 V, TI I2 – I1 0.0 – 5.0 Tt = Tt = 0.1 = – 50 A/s ∴ Substituting these values in (i), we get L = E = 200 = 4H (–TI/Tt) 50 Q. 8. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside normal to the axis of the solenoid. The current carried by the solenoid changes steadily from 2 A to 4 A in 0.1 s, what is the induced emf in the loop while the current is changing? [CBSE (F) 2016] Ans. Mutual inductance of solenoid coil system M = n0 N1 N2 A2 Here l N1 = 15, N2 = 1, l = 1 cm = 10–2 m, A2 = 2.0 cm2 = 2.0 # 10–4 m2 ∴ M = 4r×10–7 # 15 # 1 # 2.0 # 10–4 10–2 Electromagnetic Induction 243

= 120 π × 10–9 H Induced emf, in the loop f2 = M DI1 (numerically) Dt   = 120 r×10–9 (4 – 2) 0.1 2   = 120 # 3.14×10–9 # 0.1 = 7.5 # 10–6 V = 7.5 nV Q. 9. An air cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 500 carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10–3 s. How much is the average back emf induced across the ends of the open switch in the circuit ? Ignore the variation in magnetic field near the ends of the solenoid. Ans. Induced emf in a solenoid, f = –L TI …(i) Tt Inductance of solenoid L = n0 N2 A …(ii) l N2 ∴ Induced emf f = –e n0 l A o TI Tt Here N = 500, A = 25 cm2, = 25 × 10–4 m2, l=30 cm =0.30 m and TI = I2 – I1 = 0 – 2.5 = – 2.5 # 103 A/s Tt t 10-3 ∴ f = – 4r×10–7 # (500) 2 # 25 # 10–4 # (–2.5 # 103) 0.30 3.14 # 25 2.5 = 3 # # 10–1 = 6.5 V Q. 10. (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side ‘a’ as shown in fig. (b) Evaluate the induced emf in the loop if the wire carries a current x v of 50 A and the loop has an instantaneous velocity v =10 ms–1 at the location x = 0.2 m as shown. Take a = 0.1 m and assume that the loop has a large resistance. Ans. (a) Suppose the loop is formed of a number of small elements parallel to the length of wire. Consider an element of width dr at a distance r from the wire. The magnetic field at the vicinity of n0 I wire, B = 2rr downward perpendicular to the plane of paper. The magnetic flux linked with this element z2 = \"\" n0 I B . d A2 2rr = B dA2 cos r = (a dr) = n0 Ia dr 2r r n0 Ia Total magnetic flux linked with the loop, z2 = 2r yxx+a dr r     = n0 Ia 8loge rBxx + a = n0 Ia loge d x + a n 2r 2r x ∴ Mutual inductance, M= z2 = n0 a loge d1 + a n I 2r x 244 Xam idea Physics–XII

(b) The square loop is moving in non-uniform magnetic field. The magnetic flux linked with the loop at any instant is n0 Ia z = 2r loge b1 + a l x Induced emf set up in the loop, f =– dz =– dz . dx =– v dz dt dx dt dx =– v d = n0 Ia loge b1 + a lG dx 2r x =– v. n0 Ia . loge 1 a .f– a p = n0 . a2 v .I 2r b1 + x x2 2r x (x + a) l = 4r #10–7 # (0.1)2 #10 # 50 2r 0.2 (0.2 + 0.1) = 1.67 × 10–5 V b 1.7 × 10–5 V. Q. 11. Two concentric circular coils, one of small radius r2 and the other of large radius r1, such that r2<<r1 are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. [CBSE Chennai 2015] Ans. Mutual Inductance of two plane coils: Consider two concentric circular C1 plane coils C1 and C2 placed very near to each other. The number of turns in the primary coil is N1 and radius is r1 while the number of turns r1 C2 N1 in the secondary coil is N2 and its radius is r2. If I1 is the current in the primary coil, then magnetic field produced at its centre, O r2 N2 B1 = n0 N1 I1 ... (i) 2r1 If we suppose this magnetic field to be uniform over the entire plane of secondary coil, then total effective magnetic flux linkage with secondary coil z2 = N2 B1 A2 = N2e n0 N1 I1 oA2 = n0 N1 N2 A2 I1 2r1 2r1 By definition, Mutual Inductance, M = z2 = n0 N1 N2 A2 I1 2r1 But A2 = r r22 ` M = n0 N1 N2rr22 2r1 n0 rr22 Special case: If both coils have one turn each; then N1 = N2 = 1, so mutual inductance M = 2r1 Multiple Choice Questions [1 mark] Choose and write the correct option(s) in the following questions. 1. Whenever the flux linked with a circuit changes, there is an induced emf in the circuit. This emf in the circuit lasts (a) for a very short duration (b) for a long duration (c) forever (d) as long as the magnetic flux in the circuit changes. 2. Theareaofasquareshapedcoilis10–2m2.Itsplaneisperpendiculartoamagneticfieldofstrength 10–3 T. The magnetic flux linked with the coil is (a) 10 Wb (b) 10–5 Wb (c) 105 Wb (d) 100 Wb Electromagnetic Induction 245

3. An area A = 0.5 m2 shown in the figure is situated in a uniform magnetic field B = 4.0 Wb/m2 and its normal makes an angle of 60° with the field. The magnetic flux passing through the area A would be equal to (a) 2.0 weber (b) 1.0 weber (c) 3 weber (d) 0.5 weber 4. A square of side L meters lies in the X-Y plane in a region, where the magnetic field is given by B = Bo (2it + 3tj + 4kt) T, where Bo is constant. The magnitude of flux passing through the square is [NCERT Exemplar] (a) 2 Bo L2 Wb (b) 3 Bo L2 Wb (c) 4 Bo L2 Wb (d) 29 Bo L2 Wb 5. A loop, made of straight edges has six corners at A(0, 0, 0), B(L, O, 0), C(L, L, 0), D(0, L, 0) E(0, L, L) and F(0, 0, L). A magnetic field B = Bo (it + kt) T is present in the region. The flux passing through the loop ABCDEFA (in that order) is [NCERT Exemplar] (a) Bo L2 Wb (b) 2 Bo L2 Wb (c) 2 Bo L2 Wb (d) 4 Bo L2 Wb 6. An emf is produced in a coil, which is not connected to an external voltage source. This can be due to [NCERT Exemplar] (a) the coil being in a time varying magnetic field. (b) the coil moving in a time varying magnetic field. (c) the coil moving in a constant magnetic field. (d) the coil is stationary in external spatially varying magnetic field, which does not change with time. 7. A magnet is dropped with its north pole towards a closed circular coil placed on a table then (a) looking from above, the induced current in the coil will be anti-clockwise. (b) the magnet will fall with uniform acceleration. (c) as the magnet falls, its acceleration will be reduced. (d) no current will be induced in the coil. 8. A cylindrical bar magnet is rotated about its axis (Figure given alongside). A A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then [NCERT Exemplar] (a) a direct current flows in the ammeter A. (b) no current flows through the ammeter A. (c) an alternating sinusoidal current flows through the ammeter A with a time period T=2π/ω. (d) a time varying non-sinusoidal current flows through the ammeter A. 9. A copper ring is held horizontally and a magnet is dropped through the ring with its length along the axis of the ring. The acceleration of the falling magnet is (a) equal to that due to gravity (b) less than that due to gravity (c) more than that due to gravity (d) depends on the diameter of the ring and the length of the magnet 10. There are two coils A and B as shown in the figure. A current starts flowing in B as shown, when A is moved towards B and stops when A stops moving. The current in A is counter clockwise. B is kept stationary when A moves. We can infer that          [NCERT Exemplar] (a) there is a constant current in the clockwise direction in A. (b) there is a varying current in A. 246 Xam idea Physics–XII

(c) there is no current in A. (d) there is a constant current in the counterclockwise direction in A. 11. Same as the above problem except the coil A is made to rotate about a vertical axis refer to the figure. No current flows in B if A is at rest. The current in coil A, when the current in B (at t = 0) is counterclockwise and the coil A is as shown at this instant, t = 0, is [NCERT Exemplar] (a) constant current clockwise. (b) varying current clockwise. (c) varying current counterclockwise. (d) constant current counterclockwise. 12. Lenz’s law is essential for (b) conservation of mass (a) conservation of energy (c) conservation of momentum (d) conservation of charge 13. The self inductance L of a solenoid of length l and area of crosssection A, with a fixed number of turns N increases as [NCERT Exemplar] (a) l and A increase. (b) l decreases and A increases. (c) l increases and A decreases. (d) both l and A decrease. 14. A thin circular ring of area A is held perpendicular to a uniform magnetic field of induction B. A small cut is made in the ring and a galvanometer is connected across its ends in such a way that the total resistance of the circuit is R. When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer is B2 A (a) BR (b) AB (c) ABR (d) R2 A R 15. A conducting square loop of side L and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A magnetic induction B constant in time and space, pointing perpendicular and into the plane of the loop exists everywhere as in given figure. The current induced in the loop is (a) Blv/R clockwise (b) Blv/R anticlockwise (c) 2 Blv/R anticlockwise (d) zero. 16. Inductance plays the role of (a) inertia (b) friction (c) source of emf (d) force 17. A circular coil expands radially in a region of magnetic field and no electromotive force is produced in the coil. This can be because [NCERT Exemplar] (a) the magnetic field is constant. (b) the magnetic field is in the same plane as the circular coil and it may or may not vary. (c) the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably. (d) there is a constant magnetic field in the perpendicular (to the plane of the coil) direction. 18. When the current in a coil changes from 8A to 2A in 3 × 10–2 second, the emf induced in the coil is 2 volt. The self-inductance of the coil, in millihenry, is (a) 1 (b) 5 (c) 20 (d) 10 19. A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from (a) the current source (b) the magnetic field Electromagnetic Induction 247