fiziks Institute for NET/JRF, GATE, IIT JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Electricity and Magnetism OBJECTIVE QUESTIONS IT-JAM-2005 (b)

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1 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS lectricit and Magnetism OBJTIV QUSTIONS IT-JAM-5 Q. A small loop of wire of area A=. m, N = 4 turns and resistance = Ω is initiall kept in a uniform magnetic field B in such a wa that the field is normal to the loop. When it is pulled out of the magnetic field, a total charge of Q 5 = flows through the coil. The magnetic of the field B is (a) T 4 T (c) zero Solution: Magnetic flu through the loopφ = NBA (d) unobtainable, as the data is insufficient dφ dφ Induced e.m.f ε = and induced currenti = = dt dt dq dt dφ = dq. 5 Thus ( 4 B.) = B = T. Q. Two point charges + q and + q are fied with a finite distance d between them. It is desired to put a third charge q in between these two charges on the line joining them so that the charge q is in equilibrium. This is (a) possible onl if q is positive possible onl if q is negative (c) possible irrespective of the sign of q (d) not possible at all Ans.: (c) fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 7

2 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS IIT-JAM-6 Q. Two electric dipoles P and P are placed at (,, ) and (,, ) respectivel with both of them pointing in the + z direction. Without changing the orientations of the dipoles P is moved to(,, ). The ratio of the electrostatic potential energ of the dipoles after moving to that before moving is (a) 6 Ans: (d) U Solution: lectrostatic potential energu r U r = r = (c) 4 8 (d) 8 Q4. A small magnetic dipole is kept at the origin in the - plane. One wire L is located at z = a in the -z plane with a current I flowing in the positive direction. Another wire L is at z = + a in -z plane with the same current I as in L, flowing in the positive - direction. The angle φ made b the magnetic dipole with respect to the positive -ais is (a) 5 o o (c) 45 o (d) 7 o Solution: Magnetic field at z = due to wire at z = a is B = B. Magnetic field at z = due to wire at z = + a is B = B. esultant magnetic field at z = makes an angle of 45 with and 5 with. fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 7

3 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS IIT-JAM-7 Q5. A uniform and constant magnetic field B coming out of the plane of the paper eists in a rectangular region as shown in B the figure. A conducting rod PQ is rotated about O with a ω uniform angular speed ω in the plane of the paper. The emf PQ induced between P and Q is best represented b the graph P O ( ) PQ a ( b ) PQ Q O t O t ( ) PQ c ( d ) PQ O t O t IIT-JAM-8 Q6. If the electrostatic potential at a point(, ) electrostatic energ densit at that point ( / ) is given b V ( 4) in J m is (a) 5ε ε (c) = + volts, the ε + 4 ε (d) ( ) fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 7

4 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS Solution: = V = 4 = V / m lectrostatic energ densit = ε = = ε ε J / m IIT-JAM-9 Q7. An oscillating voltage V () t = V cosωt is applied across a parallel plate capacitor having a plate separation d. The displacement current densit through the capacitor is V () t = V cosω t d ε ωv cosωt ε μωv cosωt (a) d d ε μωv cosωt (c) (d) d Ans.: (d) Solution: Displacement current densit J d = ε t Q8. An electric field ( r ) = ( αr + β sinθ cosφφ ) ( t) ε V = d t = ε ωv sinωt d ε ωv sinωt d eists in space. What will be the total charge enclosed in a sphere of unit radius centered at the origin? (a) α (α + β) (c) (α - β) (d) β Q enc = da r sin cos r sin d d r 4 Solution: ε = ε ( α + β θ φφ ) ( θ θ φ ) = παε fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 7

5 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS IIT-JAM- Q9. The magnetic field associated with the electric field vector = sin( kz t)j ω is given b (a) B = sin( kz ωt)i c B = sin( kz ωt)i c (c) B = sin( kz ωt)j (d) B = sin( kz ωt)k c c k kz ( ) sin kz ωt k Solution: B = = ( ) = sin kz ωt = sin ( kz ωt) ω ω ω c Q. Assume that z = plane is the interface between two linear and homogenous dielectrics (see figure). The relative permittivities are ε = 5 for z > and ε = 4 for z <. The electric field in the region z > is ( i j + k) kv m on the interface, the electric field in the region z < is given b z ε r r = 5 4. If there are no free charges = 5 z = r ε r = 4 (a) Ans.: (d) Solution: = i 5 j + k kv 4 4 m = ( i j + k) kv m (c) = ( i j 5k) kv m (d) = ( i j + 5k) kv m = and = 5 = i f 5 j σ D = D = = ( + k) 5k ε = ε ( i 5 j 5k ) kv m = fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 74

6 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS Q. A closed Gaussian surface consisting of a hemisphere and a circular disc of radius, is placed in a uniform electric field, as shown in the figure. The circular disc makes an angle θ = with the vertical. The flu of the electric field vector coming out of the curved surface of the hemisphere is (a) π π (c) π θ (d) π Ans.: Solution: = cos z + sin = z + φ = da = z + r S ( sinθdθdφ) ẑ φ = π / π θ = φ = cosθ + sinθ cosφ ( sinθdθdφ ) π / π = ( cos sin ) π / π φ θ θ dθdφ + ( ) sin θ cosφ dθdφ θ = φ = θ = φ = φ = π + = π O φ π = da = cos π = S fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 75

7 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS IIT-JAM- Q. quipotential surface corresponding to a particular charge distribution are given b ( ) 4 z Vi + + = where the values of V i are constants. The electric field at the origin is (a) = = Ans.: (d) = V = 8+ + zz, = 4 Solution: ( ) ( ) (c) = 4 (d) = 4 IIT-JAM- Q. A parallel plate air-gap capacitor is made up of two plates of area cm each kept at a distance of.88 mm. A sine wave of amplitude V and frequenc 5 Hz is applied across the capacitor as shown in the figure. The amplitude of the displacement current densit (in ma/m ) between the plates will be closest to ~ (a).. (c). (d). Solution: Displacement current densit J d = ε t ( t) ε V = d t = ε ωv sinωt Amplitude of the displacement current densit (in ma/m ε ωv πεfv ) Jd = = d d fv 5 Jd = = =. ma/ m 9 5 d 9 88 d fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 76

8 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS Q4. A segment of a circular wire of radius, etending from θ = to π / linear charge densit λ. The electric field at origin O is λ (a) ( ) πε 4, carries a constant λ πε 4 (c) (d) λ πε 4 O θ Solution: = where = dcosθ, = dsinθ. line line dl λdl and d =. π / λdl λ dθ = cosθ = cosθ line λ π [ sin ] / λ = θ = d O θ θ Similarl π / λdl λ dθ = sinθ = sinθ line λ π [ cos ] / λ = θ = λ Thus = = ( ) fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 77

9 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS IIT-JAM-4 Q5. A particle of mass m carring charge q is moving in a circle in a magnetic field B. According to Bohr s model, the energ of the particle in the n th level is (a) n hqb π m n hqb m (c) hqb π n π m (d) hqb n 4π m Ans.: (d) Solution: n qbr = m n mv m n n mv r = n and r = r = r = qb qb mr qb n n n n n n n qbrn qb n qbh n = = = n m m qb 4π m Q6. A conducting slab of copper PQS is kept on the plane in a uniform magnetic field along -ais as indicted in the figure. A stead current I flows through the cross section of the slab along the -ais. The direction of the electric field inside the slab, arising due to the applied magnetic field is along X Z P S B Q I Y the (a) negative Y direction positive Y direction (c) negative Z direction (d) positive Z direction Ans.: (c) fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 78

10 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS Q7. In a parallel plate capacitor the distance between the plates is cm. Two dielectric slabs of thickness 5 cm each and dielectric constants K = and K = 4 respectivel, are inserted between the plates. A potential of V is applied across the capacitor as shown in the figure. The value of the net bound surface charge densit at the interface of the two dielectrics is K = 4 cm V K = (a) ε ε (c) 5ε (d) ε σ σ σ σ σ σ Solution: V = d + d = d + d = d + d = d K +σ ε ε ε 4ε 4ε = 4 σ K +σ = V = volts, d = 5 cm + σ 4 4ε 4ε 4 σ = V = = ε d 5 5 σ σ P = εχ e = ε( K ) σ = ε = ε σ σ P = εχ e = ε( K ) σ = ε = 4ε 4 4 σ σ σ 4 σ = σ σ = = = ε = ε Q8. A rigid uniform horizontal wire PQ of mass M, pivoted at P, carries a constant current I. It rotates with a constant angular speed in a uniform vertical magnetic field B. If the current were switched off, the angular acceleration of the wire, in terms of B, M and I would be P Q (a) BI BI BI (c) (d) M M M Ans.: (c) fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 79

11 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS Q9. A stead current in a straight conducting wire produces a surface charge on it. Let out and in be the magnitudes of the electric fields just outside and just inside the wire, respectivel. Which of the following statements is true for these fields? (a) out is alwas greater than in (c) out is alwas smaller than in out could be greater or smaller than in (d) out is equal to in Q. A small charged spherical shell of radius. m is at a potential of V. The electrostatic energ of the shell is (a) Ans.: J 5 J 9 (c) 5 J (d) 9 J Solution: V = q and W = q 8πε. ( ) V V 9 9 Thus W = = = =.5 = 5 Joules 9 8πε 9 Q. A ring of radius carries a linear charge densit λ. It is rotating with angular speed ω. The magnetic field at its center is μ (a) λω μ λω μ (c) λω (d) μ λω π Ans.: μi Solution: B = where I = λv= λω. Thus B = μ λω. fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 8

12 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS IIT-JAM-5 π Q. The electric field of a light wave is given b isin( ωt kz) sin j ωt kz = + 4. The polarization state of the wave is (a) Left handed circular ight handed circular (c) Left handed elliptical (d) ight handed elliptical Ans.: (c) π Solution: = sin ( ωt kz), = sin ωt kz 4. Thus resultant is ellipticall polarized wave. π At z =, = sin ( ωt), = sin ωt 4 When ω t =, π =, = and when ω t =, =, 4 = Q. A charge q is at the center of two concentric spheres. The outward electric flu through the inner sphere is φ while that through the outer sphere is φ. The amount of charge contained in the region between the two spheres is (a) q q (c) q (d) q Ans.: q Solution: φ =, ε q+ q φ = φ = q = q ε Q4. A positivel charged particle, with a charge q, enters a region in which there is a uniform electric field and a uniform magnetic field B, both directed parallel to the positive -ais. Att =, the particle is at the origin and has a speed v directed along the positive - ais. The orbit of the particle, projected on the - z plane, is a circle. Let T be the time taken to complete one revolution of this circle. The -coordinate of the particle at π m (a) qb t = T is given b π m qb (c) π m qb v π m qb + (d) πmv qb fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 8

13 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS Ans.: z Solution: ut at q πm π m = + = = m qb qb Q5. A hollow, conducting spherical shell of inner radius and v, B outer radius encloses a charge q inside, which is located at a distance d ( < ) from the centre of the spheres. The potential at the centre of the shell is (a) Zero q q (c) d Ans.: (d) q d q q q (d) + d Solution: charge induced on inner surface is q and charge induced on outer surface is + q. q d q q q ThusV = +. d Q6. A conducting wire is in the shape of a regular heagon, which is inscribed inside an imaginar circle of radius, as shown. A current I flows through the wire The magnitude of the magnetic field at the center of the circle is I (a) Ans.: (c) μ I μ I π π Solution: d = cos = μi B = sinθ sinθ 4π d ( ) fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / (c) μi μi μi B = sin = sin = 4π d π 4π μ I π d 6 8 B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 8 I (d) μ I π

14 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS μi μi μi B= 6B = 6 = = π π π STION B: MSQ Q7. For an electromagnetic wave traveling in free space, the electric field is given = cos 8 V b ( ) m t + k (a) The wavelength of the wave in meter is 6 π j. Which of the following statements are true? The corresponding magnetic field is directed along the positive z direction (c) The Ponting vector is directed along the positive z direction (d) The wave is linearl polarized and (d) 8 = cos t+ k j V / m Solution: ( ) 8 8 πc 8 π ω = = λ = = 6π. Option (a) is true 8 λ B k z. Option is wrong ( ) ( ) S k. Option (c) is wrong. Option (d) is true. = with Q8. onsider the circuit, consisting of an A function generator V ( t) V sin πvt V = 5V an inductor L = 8. mh, resistor = 5Ω and a capacitor = μf. Which of the following statements are true if we var the frequenc? L (a) The current in the circuit would be maimum at ν = 78Hz The capacitive reactance increases with frequenc (c) At resonance, the impedance of the circuit is equal to the resistance in the circuit (d) At resonance, the current in the circuit is out of phase with the source voltage fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 8

15 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS and (c) Solution: ν = = = 78Hz. Option (a) is true. π L ( )( ) ω X = X as ω. Option is wrong Option (c) is true Option (d) is wrong Q9. A unit cube made of a dielectric material has a polarization P = i + 4 j units. The edges of the cube are parallel to the artesian aes. Which of the following statements are true? (a) The cube carries a volume bound charge of magnitude 5 units There is a charge of magnitude units on both the surfaces parallel to the (c) There is a charge of magnitude 4 units on both the surfaces parallel to the (d) There is a net non-zero induced charge on the cube Ans.: and (c) Solution: P= i+ 4j ρ b =. P =. Option (a) is wrong At σ. 4. b = Pn= i+ j i =, At σ. 4. b = Pn= i+ j i = =, ( ) ( ) Option is true At σ. 4. b = Pn= i+ j j = 4, At =, ( ) ( ) Option (c) is true. Option (d) is wrong =, ( )( ) σ =. = + 4. = 4 =, ( )( b Pn i j j) z plane z plane Q. The power radiated b sun is W and its radius is 7 km. The magnitude of W the Ponting vector (in cm Ans.: 674 ) at the surface of the sun is Solution: 6 P.8 I = = W / cm = 674 W / cm A 4π 7 ( ) fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 84

16 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS Q. In an eperiment on charging of an initiall uncharged capacitor, an circuit is made with the resistance = kω and the capacitor = μf along with a voltage source of 6 V. The magnitude of the displacement current through the capacitor (in μ A ), Ans.: 64 Solution: 5 seconds after the charging has started, is V 6 / 6 t 5/ 6 5/ 6 6 I = e = e = e = = = 64μ A e.65 Q. In a region of space, a time dependent magnetic field B( t). 4t Ans.: 4 = tesla points verticall upwards. onsider a horizontal, circular loop of radius cm in this region. The magnitude of the electric field (in mv / m ) induced in the loop is. Solution: B r B πr = πr = =.4= 4 mv / m t t 4 Q. A plane electromagnetic wave of frequenc5 Hz and amplitude Ans.:.9 Vm / traveling in a homogeneous dielectric medium of dielectric constant.69 is incident normall at the interface with a second dielectric medium of dielectric constant.5. The ratio of the amplitude of the transmitted wave to that of the incident wave is Solution: n ε.69 T r T = I = = = n+ n I εr + ε r fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 85

17 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS IIT-JAM-6 Q4. For an infinitel long wire with uniform line-charge densit, λ along the z -ais, the electric field at a point ( ab,,) awa from the origin is ( e, e and e z are unit vectors in artesian coordinate sstem) (a) ( e + e ) (c) πε λ a λ + b πε a + b e (d) πε λ ( ) ( ae ) + be a + b λ πε a + b e z Ans.: Solution: λ λ λ = r= r = ae + be πε r πε r πε a b ( ) ( ) + r = a + b Q5. A W point source at origin emits light uniforml in all the directions. If the units for both the aes are measured in centimeter, then the Ponting vector at the point (,, ) in W cm is (a) ( e + e ) ( e + e ) 8π 6π (c) ( e + e ) (d) ( e + e ) 6π 4π P P r P A πr r πr π π Solution: I =< S >= r= = r = ( + ) = ( + ) r = + = fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 86

18 velocit = ( + ) fiziks Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS Q6. A charged particle in a uniform magnetic field B = Be z starts moving from the origin with v e e m/ s. The trajector of the particle and the time t at which it reaches meters above the - plane are z ( e, e and e z are unit vectors in artesian-coordinate sstem) (a) Helical path; t = s Helical path; t = /s (c) ircular path; t = s (d) ircular path; t = /s Solution: v = m/ s and v = m/ s m, thus t = = sec v Q7. The phase difference ( ) and (ii) δ between input and output voltage for the following circuits (i) v i vo v i vo Ans.: (d) (i) (ii) will be (a) and π / and < δ π / respectivel (c) π / and π / (d) and < δ π / respectivel v v o o X = X + X X = X v i v v (i) o i =, phase difference ( ) vi δ is. vo = = = vi + / X + i + ω + ω ( ) Phase difference( δ ) is < δ π /. (ii) e iω fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 87

19 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS Q8. In the following circuit, the capacitor was charged in two different was. (i) The capacitor was first charged to 5V b moving the toggle switch to position P and then it was charged to V b moving the toggle switch to positionq. (ii) The capacitor was directl charged tov, b keeping the toggle switch at positionq. Assuming the capacitor to be ideal, which one of the following statements is correct? P 5V V Q (a) The energ dissipation in cases (i) and (ii) will be equal and non-zero The energ dissipation for case (i) will be more than that for case (ii) (c) The energ dissipation for case (i) will be less than that for case (ii) (d) The energ will not be dissipated in either case. Ans.: (c) Solution: The energ dissipation in cases (i) is = ( 5) + ( 5) = 5 = = 5 The energ dissipation in cases (ii) is ( ) Q9. In the following network, for an input signal frequenc f v v o i and the phase angle φ between v o and v i respectivel are =, the voltage gain π v i v o (a) and and (c) and π (d) and π Ans.: fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 88

20 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS Solution: f = then X = = j π jπ f X ( j j j + j) ZP = = = = and ZS = + X = j= ( j) + X j j ZP v j( + j) o vo = vi = = = = ZP + ZS v ZS ( j) ( j i + ) j j + ZP j( + j) j( + j) v j( + j) j( + j) ( j o ) = = = = vi j ( j) j ( j ) Q4. An arbitraril shaped conductor encloses a charge q and is ( ) surrounded b a conducting hollow sphere as shown in the figure. Four different regions of space,,, and 4 are indicated in the figure. Which one of the following statements is correct? (a) The electric field lines in region are not affected b the position of the charge q q 4 The surface charge densit on the inner wall of the hollow sphere is uniform (c) The surface charge densit on the outer surface of the sphere is alwas uniform irrespective of the position of charge q in region (d) The electric field in region has a radial smmetr Ans.: (c) Q4. onsider a small bar magnet undergoing simple harmonic motion (SHM) along the - ais. A coil whose plane is perpendicular to the - ais is placed such that the magnet passes in and out of it during its motion. Which one of the following statements is correct? Neglect damping effects. (a) Induced e.m.f. is minimum when the center of the bar magnet crosses the coil The frequenc of the induced current in the coil is half of the frequenc of the SHM (c) Induced e.m.f. in the coil will not change with the velocit of the magnet (d) The sign of the e.m.f. depends on the pole (N or S) face of the magnet which enters into the coil fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 89

21 fiziks Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS Q4. onsider a spherical dielectric material of radius a centered at origin. If the polarization vector, P= Pe, where P!is a constant of appropriate dimensions, then ( e, e, and e z are unit vectors in artesian- coordinate sstem) (a) the bound volume charge densit is zero. the bound surface charge densit is zero at (,, a ). (c) the electric field is zero inside the dielectric (d) the sign of the surface charge densit changes over the surface.,, (d) Solution: ρ b =. P = σ. ( ). b = Pn= P r= Psinθcosφ = at (,, a ) θ =. Q4. For an electric dipole with moment P= pe z placed at the origin, ( p is a constant of appropriate dimensions and e, e and e z are unit vectors in artesian coordinate sstem) (a) potential falls as, where r is the distance from origin r a spherical surface centered at origin is an equipotential surface (c) electric flu through a spherical surface enclosing the origin is zero (d) radial component of is zero on the - plane., (c), (d) V r, rp. pcosθ = =. r r Solution: dip ( θ ) dip r r r o p (, θ ) = ( cosθ + sinθθ ) o. fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 9

22 Institute for NT/JF, GAT, IIT JAM, JST, TIF and G in PHYSIAL SINS Q44. Three infinitel-long conductors carring currents I, I and I lie perpendicular to the plane of the paper as shown in the figure. If the value of the integral B. dl for the loops, and I I I are,4 μ μ and μ in the units of N A respectivel, then (a) I = A into the paper I = 5A out of the paper (c) I =. (d) I = A out of the paper, Solution: B dl = μi. enc I + I =, I + I = 4, I+ I + I = I = A, I = 5 A and I = A. Q45. The shape of a dielectric lamina is defined b the two curves = and =. If the charge densit of the laminaσ = 5 / m, then the total charge on the lamina is... Ans.: 8 Solution: Total charge on the lamina is 5 Q = σ da = 5dd = ( ) d S ( ) Q= d + = Q = = = = Q= = 8 5 fiziks, H.No. 4 D, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi 6 Phone: / B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi 6 mail: fiziks.phsics@gmail.com 9