Electricity and Magnetism IIT-JAM Past Paper PDF

fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Electricity and Magnetism IIT-JAM 2005 Q1. A small loop of wire of area A  0.01 m 2 , N  40 turns an...

fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Electricity and Magnetism IIT-JAM 2005 Q1. A small loop of wire of area A  0.01 m 2 , N  40 turns and resistance R  20  is initially kept in a uniform magnetic field B in such a way that the field is normal to the loop. When it is pulled out of the magnetic field, a total charge of Q  2  105 C flows through the coil. The magnitude of magnetic field B is (a) 1 103 T (b) 4  103 T (c) zero (d) unobtainable, as the data is insufficient Ans.: (a) Solution: Magnetic flux through the loop   NBA d 1 d dQ 1 Induced e.m.f    and induced current i      d  dQ. dt R dt dt R  40  B  0.01  2  10 5  B  1  10 3 T. 1 Thus , 20 Q2. Two point charges  q1 and  q 2 are fixed with a finite distance d between them. It is desired to put a third charge q3 in between these two charges on the line joining them so that the charge q3 is in equilibrium. This is (a) possible only if q3 is positive (b) possible only if q3 is negative (c) possible irrespective of the sign of q3 (d) not possible at all Ans. : (c) q3 q3 Solution: If q3 is positive,  q1  q2 F2 d F1 q3 q3 If q3 is negative,  q1  q2 F1 d F2 In both case there is possibility that charge q3 may be in equilibrium. H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 1  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics IIT-JAM 2006 Q3. Two electric dipoles P1 and P2 are placed at  0, 0, 0  and 1, 0, 0  respectively with both of them pointing in the  z direction. Without changing the orientations of the dipoles P2 is moved to  0, 2, 0 . The ratio of the electrostatic potential energy of the dipoles after moving to that before moving is 1 1 1 1 (a) (b) (c) (d) 16 2 4 8 Ans. : (d) 1 U 2 r13 1 Solution: Electrostatic potential energy U  3    r U 1 r23 8 Q4. A small magnetic dipole is kept at the origin in the x-y plane. One wire L1 is located at z  a in the x - z plane with a current I flowing in the positive x direction. Another wire L2 is at z   a in y - z plane with the same current I as in L1 , flowing in the positive y -direction. The angle  made by the magnetic dipole with respect to the positive x -axis is (a) 2250 (b) 1200 (c) 450 (d) 2700 Ans.: (a) Solution: Magnetic field at z  0 due to wire at z  a is B   Byˆ. Magnetic field at z  0 due to wire at z   a is B   Bxˆ. Resultant magnetic field at z  0 makes an angle of 45 0 with  x̂ and 225 0 with x̂. H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 2  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics IIT-JAM 2007 Q5. A uniform and constant magnetic field B coming out of the plane of the paper exists in a rectangular region as shown in B the figure. A conducting rod PQ is rotated about O with a  uniform angular speed  in the plane of the paper. The emf P EPQ induced between P and Q is best represented by the O graph Q E PQ (a) (b) E PQ O t O t (c) E PQ (d) E PQ O t O t Ans.: (a) Solution: When point P is inside due to motional emf , potential PQ is positive. When point Q is inside potential QP is positive or potential PQ is negative. IIT-JAM 2008 Q6. If the electrostatic potential at a point  x, y  is given by V   2 x  4 y  volts, the electrostatic energy density at that point  in J / m3  is  0 2 x  4 y 2 1 (a) 5 0 (b) 10 0 (c) 20 0 (d) 2 Ans.: (b) H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 3  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics  Solution: E  V  2 xˆ  4 yˆ  E  20V / m 1 2 1 Electrostatic energy density   0 E   0  20 10 0 J / m 3 2 2 IIT-JAM 2009 Q7. An oscillating voltage V  t   V0 cos t is applied across a parallel V t   V0 cos  t plate capacitor having a plate separation d. The displacement current density through the capacitor is d  0V0 cos t  0  0V0 cos t (a) (b) d d  0  0V0 cos t  0V0 sin t (c)  (d)  d d Ans.: (d) E  0 V t   V sin t Solution: Displacement current density J d   0   0 0 t d t d Q8.     An electric field E r   rˆ   sin  cos ˆ exists in space. What will be the total charge enclosed in a sphere of unit radius centered at the origin? (a) 4 0 (b) 4 0     (c) 4 0     (d) 4 0  Ans.: (a)      Solution: Qenc   0  E  da   0  rˆ   sin  cos ˆ  r 2 sin ddrˆ  4 0 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 4  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics IIT-JAM 2010  Q9. The magnetic field associated with the electric field vector E  E 0 sin kz  t  ˆj is given by  E  E (a) B   0 sin kz  t iˆ (b) B  0 sin kz  t iˆ c c  E  E (c) B  0 sin kz  t  ˆj (d) B  0 sin kz  t kˆ c c Ans.: (a)    k  E kzˆ  E0 sin  kz  t  ˆj kE E Solution: B     0 sin  kz  t  iˆ   0 sin  kz  t  iˆ    c Q10. Assume that z  0 plane is the interface between two linear and homogeneous dielectrics (see figure). The relative permittivities are  r  5 for z r  5 z  0 and  r  4 for z  0. The electric field in the z0   region z  0 is E 1  3iˆ  5 ˆj  4kˆ k V m. If there are r  4 no free charges on the interface, the electric field in the region z  0 is given by 3 5  (a) E 2   iˆ  ˆj  kˆ k V m 4 4    (b) E 2  3iˆ  5 ˆj  kˆ k V m   (c) E 2  3iˆ  5 ˆj  5kˆ k V m   (d) E 2  3iˆ  5 ˆj  5kˆ k V m Ans.: (d) Solution:  E1  E2  E2  3iˆ  5 ˆj and  f  0  D1  D2  E 2  1  5 2   E1   4kˆ  5kˆ 4    E 2  3iˆ  5 ˆj  5kˆ k V m H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 5  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q11. A closed Gaussian surface consisting of a hemisphere and a circular disc of radius R , is  placed in a uniform electric field E , as shown in the figure. The circular disc makes an angle   300 with the vertical. The flux of the electric field vector coming out of the curved surface of the hemisphere is 1 (a)  R2 E 2 E 3  (b)  R2 E 2 (c)  R 2 E (d) 2 R 2 E Ans.: (b) x̂ 3 1 Solution: E  E cos 30 zˆ  E sin 30 xˆ  E zˆ  E xˆ E 2 2    3  300  E   E  da     1  E zˆ  E xˆ   R 2 sin ddrˆ  ẑ S  2 2   / 2 2  3  sin dd  1 E  R        2  2 E cos E sin cos   0  0  2   / 2 2  / 2 2 ER 2   cos  sin  dd  ER 2   sin 2  cos  dd 3 1 E  2  0  0 2  0  0 3 1 3 2 E  ER 2  2   0  R E 2 2 2 OR   3 E   E  da  E cos 300   R 2   R2 E S 2 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 6  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics IIT-JAM 2011 Q12. Equipotential surface corresponding to a particular charge distribution are given by  4 x 2   y  2   z 2  Vi , where the values of Vi are constants. The electric field E at the 2 origin is     (a) E  0 (b) E  2 xˆ (c) E  4 yˆ (d) E  4 yˆ Ans.: (d)    Solution: E  V  8 xxˆ  2  y  2  yˆ  2 zzˆ  E  0, 0, 0   4 yˆ IIT-JAM 2012 Q13. A parallel plate air-gap capacitor is made up of two plates of area 10 cm 2 each kept at a distance of 0.88 mm. A sine wave of amplitude 10 V and frequency 50 Hz is applied across the capacitor as shown in the ~ figure. The amplitude of the displacement current density (in mA / m 2 ) between the plates will be closest to (a) 0.03 (b) 0.30 (c) 3.00 (d) 30.00 Ans.: (a) E  0 V t   V sin t Solution: Displacement current density, J d   0   0 0 t d t d  0V0 2 0 fV0 Amplitude of the displacement current density (in mA/m2) , J 0 d   d d fV0 1 50  10  J 0 d  4 0  5  0.03 mA / m 2 2d 9  10 2  88 10 9 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 7  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q14. A segment of a circular wire of radius R, extending from   0 to  / 2 , carries a constant linear charge density . The electric field at origin O is  y (a)  xˆ  yˆ  4 0 R   1 1  (b)  xˆ  yˆ  4 0 R  2 2    1 1  R (c)   xˆ  yˆ  4 0 R  2 2   O x (d) 0 Ans.: (a) y  Solution: E   Ex xˆ  E y yˆ where Ex   dE cos , E line y   dE sin . line dl 1  dl R and dE . 4 0 R 2 O  x 1  dl   /2 Rd  Ex   cos    cos  2 line 4 0 R 2 4 0 0 R  dE   sin  0   /2  Ex  4 0 R 4 0 R  /2 1  dl  Rd Similarly E y   sin    sin  line 4 0 R 2 4 0 0 R2     cos  0   /2  Ey  4 0 R 4 0 R   Thus E   Ex xˆ  E y yˆ    xˆ  yˆ  4 0 R H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 8  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics IIT-JAM 2014 Q15. A particle of mass m carrying charge q is moving in a circle in a magnetic field B. According to Bohr’s model, the energy of the particle in the nth level is 1  hqB   hqB   hqB   hqB  (a) 2   (b) n  (c) n  (d) n  n  m   m   2  m   4  m  Ans.: (d) q 2 B 2 rn2 mv m n n Solution: En   mvn rn  n and rn  n  rn  rn2  2m qB qB mrn qB q 2 B 2 rn2 q 2 B 2 n  qBh   En     n  2m 2m qB  4 m  Q16. A conducting slab of copper PQRS is kept on the x - y plane in a uniform magnetic field along x - axis as indicted in the Z figure. A steady current I flows S R through the cross section of the slab Q along the y - axis. The direction of P I Y the electric field inside the slab, B arising due to the applied magnetic X field is along the (a) negative Y direction (b) positive Y direction (c) negative Z direction (d) positive Z direction Ans.: (c) Q17. In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thickness 5 cm each and dielectric constants K 1  2 and K 2  4 respectively, are inserted between the plates. A potential of 100 V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is K2  4 10 cm 100 V K1  2 2000 1000 2000 (a)  0 (b)  0 (c)  250 0 (d) 0 3 3 3 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 9  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Ans.: (a)     3  Solution: V  E1d  E2 d  d  d  d d d K2  4  2 1 2 2 0 4 0 4 0  2 K1  2  1  1 V  100 volts, d  5 102 cm  4 0 4 0 4 104   V  100  0 3d 3  5 102 15      P1   0  e E1   0  K1  1 E1   1   0   2 0 2     3 P2   0  e E2   0  K 2  1 E2   2  3 0   4 0 4  3  1 4 104 2000    1   2      0   0 2 4 4 4 15 3 Q18. 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 P Q 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 2 BI 3BI BI (a) 0 (b) (c) (d) 3M 2M M Ans.: (c)    Solution: Torque   r  F  I m      d  r  dF  l  IBdl  F  I  dl  B  dF  IBdl  L IBL2   IB  ldl  0 2 ML2 Moment of inertia about point P , I m  3 IBL2 ML2 3 BI   I m      2 3 2M H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 10  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q19. A steady current in a straight conducting wire produces a surface charge on it. Let E out and Ein be the magnitudes of the electric fields just outside and just inside the wire, respectively. Which of the following statements is true for these fields? (a) E out is always greater than Ein (b) E out is always smaller than Ein (c) E out could be greater or smaller than Ein (d) E out is equal to Ein Ans.: (a) Solution: In this case Ein  0, Eout  0. So Eout  Ein Q20. A small charged spherical shell of radius 0.01 m is at a potential of 30V. The electrostatic energy of the shell is (a) 10 10 J (b) 5  10 10 J (c) 5  10 9 J (d) 10 9 J Ans.: (b) q q2 Solution: V  and W . 4 0 R 8 0 R  4 0VR  2 4 0V 2 R 900 102 Thus, W     0.5 109  5 1010 Joules 8 0 R 2 9 10  2 9 Q21. A ring of radius R carries a linear charge density . It is rotating with angular speed . The magnetic field at its center is 3 0   0   0  (a) (b) (c) (d)  0  2 2  Ans.: (b) 0 I 0 Solution: B  , where I   v   R. Thus B . 2R 2 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 11  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics IIT-JAM 2015      Q22. The electric field of a light wave is given by E  E 0 iˆ sin t  kz   ˆj sin  t  kz  .   4  The polarization state of the wave is (a) Left handed circular (b) Right handed circular (c) Left handed elliptical (d) Right handed elliptical Ans.: (c)   Solution: Ex  E0 sin t  kz  , E y  E0 sin  t  kz  .  4 Thus resultant is elliptically polarized wave.   At z  0, Ex  E0 sin t  , E y  E0 sin  t    4 E  E When t  0, Ex  0, E y   0 and when t  , Ex  0 , E y  0 2 4 2 Q23. A charge q is at the center of two concentric spheres. The outward electric flux through the inner sphere is  , while that through the outer sphere is 2. The amount of charge contained in the region between the two spheres is (a) 2q (b) q (c)  q (d)  2q Ans.: (b) q q  q Solution:   ,    2   q  q 0 0 Q24. A positively charged particle, with a charge q , enters a region in which there is a uniform   electric field E and a uniform magnetic field B , both directed parallel to the positive y -axis. At t  0 , the particle is at the origin and has a speed v0 directed along the positive x - axis. The orbit of the particle, projected on the x- z plane, is a circle. Let T be the time taken to complete one revolution of this circle. The y -coordinate of the particle at t  T is given by  2 mE 2 2 mE  2 mE v0 m 2mv0 (a) 2 (b) (c) 2  (d) 2qB qB 2 qB qB qB H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 12  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Ans.: (b) z 2 1 1 qE  2 m  2 2 mE Solution: y  u y t  a y t 2  y       2 2 m  qB  qB 2 E, B y v0 x Q25. A hollow, conducting spherical shell of inner radius R1 and outer radius R2 encloses a charge q inside, which is located at a R1 distance d  R1  from the centre of the spheres. The potential at q d the centre of the shell is R2 1 q (a) Zero (b) 4 0 d 1 q q  1 q q q  (c)    (d)     4 0  d R1  4 0  d R1 R2  Ans.: (d) Solution: Charge induced on inner surface is q and charge induced on outer surface is  q. 1 q q q  Thus, V     . 4 0  d R1 R2  Q26. A conducting wire is in the shape of a regular hexagon, which is inscribed inside an imaginary circle of radius R , as shown. A current I R I flows through the wire. The magnitude of the magnetic field at the C center of the circle is 3 0 I 0 I 3 0 I 3 0 I (a) (b) (c) (d) 2R 2 3R R 2R Ans.: (c) C 3 Solution: d  R cos 300  R 2 R d 0 I 600 B   sin  2  sin 1  4 d I 0 I 0 I 0 I  B1  2sin 300  2sin 300  4 d 3 2 3 R 4 R 2 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 13  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics The magnitude of the magnetic field at center of the circle is 0 I 30 I 30 I  B  6 B1  6    2 3 R 3 R R Q27. For an electromagnetic wave traveling in free space, the electric field is given  by E  100 cos10 8 t  kx  ˆj. Which of the following statements are true? V m (a) The wavelength of the wave in meter is 6 (b) The corresponding magnetic field is directed along the positive z direction (c) The Poynting vector is directed along the positive z direction (d) The wave is linearly polarized Ans.: (a) and (d)  Solution: E  100 cos 108 t  kx  ˆj V / m 2 c 2  3 108   108   108     6. Option (a) is true  108     B  kˆ  E    xˆ  yˆ    zˆ. Option (b) is wrong  S  kˆ   xˆ. Option (c) is wrong. Option (d) is true. Q28. Consider the circuit, consisting of an AC function generator V t   V0 sin 2vt with V0  5V an inductor L  8.0mH , resistor R  5 and a capacitor C  100 F. Which of the following statements are true if we vary the frequency? L R C (a) The current in the circuit would be maximum at  178Hz (b) The capacitive reactance increases with frequency (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 Ans.: (a) and (c) H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 14  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics 1 1 Solution:    178 Hz. Option (a) is true. 2 LC 2  3.14 8 10 100 10  3 6 1 XC   X C  as  . Option (b) is wrong C Option (c) is true Option (d) is wrong  Q29. A unit cube made of a dielectric material has a polarization P  3iˆ  4 ˆj units. The edges of the cube are parallel to the Cartesian axes. Which of the following statements are true? (a) The cube carries a volume bound charge of magnitude 5 units (b) There is a charge of magnitude 3 units on both the surfaces parallel to the y  z plane (c) There is a charge of magnitude 4 units on both the surfaces parallel to the x  z plane (d) There is a net non-zero induced charge on the cube Ans.: (b) and (c)    Solution:  P  3iˆ  4 ˆj  b  .P  0. Option (a) is wrong         At x  0 ,  b  P.nˆ  3iˆ  4 ˆj. iˆ  3 , At x  1 ,  b  P.nˆ  3iˆ  4 ˆj. iˆ  3 Option (b) is true         At y  0 ,  b  P.nˆ  3iˆ  4 ˆj.  ˆj  4 , At y  1 ,  b  P.nˆ  3iˆ  4 ˆj. ˆj  4 Option (c) is true. Option (d) is wrong Q30. The power radiated by sun is 3.8  10 26 W and its radius is 7  10 5 km. The magnitude of W the Poynting vector (in ) at the surface of the sun is……………… cm 2 Ans.: 6174 P 3.8  1026 Solution: I   W / cm 2  6174 W / cm 2 A 4   7 10  10 2 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 15  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q31. In an experiment on charging of an initially uncharged capacitor, an RC circuit is made with the resistance R  10k and the capacitor C  1000F along with a voltage source of 6V. The magnitude of the displacement current through the capacitor (in A ), 5 seconds after the charging has started, is………………… Ans.: 364 V  t / RC 6 6 6 6 6 Solution: I   e 5/1010 100010  4 e 5/10    364  A 3 e R 10 10 3 10 e  10 1.65 10 4 4 Q32. In a region of space, a time dependent magnetic field B t   0.4t tesla points vertically upwards. Consider a horizontal, circular loop of radius 2 cm in this region. The magnitude of the electric field (in mV / m ) induced in the loop is……………. Ans.: 4  B  r B 2 102 Solution: E  2 r     r 2  E   0.4  4 mV / m t 2 t 2 Q33. A plane electromagnetic wave of frequency 5  1014 Hz and amplitude 103 V / m traveling in a homogeneous dielectric medium of dielectric constant 1.69 is incident normally at the interface with a second dielectric medium of dielectric constant 2.25. The ratio of the amplitude of the transmitted wave to that of the incident wave is……………… Ans.: 0.93  2n1  E0T  2  r1   2 1.69  Solution: E0T    E0 I      0.93  n1  n2  E0 I   r1   r2   1.69  2.25    H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 16  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics IIT-JAM 2016 Q34. For an infinitely long wire with uniform line-charge density,  along the z - axis, the electric field at a point  a, b, 0  away from the origin is ( eˆx , eˆy and eˆz are unit vectors in Cartesian – coordinate system)   (a)  eˆ  eˆy  (b)  aeˆ  beˆy  2 0 a 2  b 2 x  2 0 a 2  b 2  x   (c) eˆx (d) eˆz 2 0 a 2  b 2 2 0 a 2  b 2 Ans.: (b)      Solution: E  2 0 r rˆ  2 0 r 2 r 2 0  a 2  b 2   aeˆx  beˆy   r  a 2  b2 Q35. A 1 W point source at origin emits light uniformly in all the directions. If the units for both the axes are measured in centimeter, then the Poynting vector at the point 1,1, 0  in W is cm 2 (a) 8 1 2  eˆ x  eˆy  (b) 1 16  eˆx  eˆy  (c) 16 1 2  eˆ x  eˆy  (d) 1 4 2  eˆ x  eˆy  Ans.: (a)   P  P Solution: I  S  rˆ  A P r 2  4 r r 4 r 3 r 1 4  2 2  eˆx  eˆy   1 8 2  eˆx  eˆy   r  12  12  2 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 17  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics  Q36. A charged particle in a uniform magnetic field B  B0 eˆz starts moving from the origin  with velocity v   3eˆx  2eˆz  m / s. The trajectory of the particle and the time t at which it reaches 2 meters above the xy - plane are ( eˆx , eˆy and eˆz are unit vectors in Cartesian-coordinate system) (a) Helical path; t  1 s (b) Helical path; t  2 / 3 s (c) Circular path; t  1 s (d) Circular path; t  2 / 3 s Ans.: (a) 2m Solution: v  3 m / s and v  2 m / s , thus t   1 sec v Q37. The phase difference   between input and output voltage for the following circuits (i) C R and (ii) vi C vo vi C vo will be (i) (ii) (a) 0 and 0 (b)  / 2 and 0     / 2 respectively (c)  / 2 and  / 2 (d) 0 and 0     / 2 respectively Ans.: (d) XC v 1 Solution: (i) vo  vi  o  , phase difference   is 0. XC  XC vi 2 XC v 1 1 1 (ii) vo  vi  o    e  iCR R  XC vi 1  R / X C 1  iCR 1  CR  2 Phase difference   is 0     / 2. H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 18  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q38. In the following RC circuit, the capacitor was charged in two different ways. (i) The capacitor was first charged to 5V by moving the toggle switch to position P and then it was charged to 10V by moving the toggle switch to position Q. (ii) The capacitor was directly charged to10V , by keeping the toggle switch at position Q. Assuming the capacitor to be ideal, which one of the following statements is correct? R C P 5V 10V Q (a) The energy dissipation in cases (i) and (ii) will be equal and non-zero (b) The energy dissipation for case (i) will be more than that for case (ii) (c) The energy dissipation for case (i) will be less than that for case (ii) (d) The energy will not be dissipated in either case. Ans.: (c) 1 1 Solution: The energy dissipation in cases (i) is  C  5   C 10  5   25C 2 2 2 2 1 The energy dissipation in cases (ii) is  C 10   50 C 2 2 1 Q39. In the following RC network, for an input signal frequency f  , the voltage gain 2 RC vo and the phase angle  between vo and vi respectively are vi R C vi C R vo 1 1 1  1  (a) and 0 (b) and 0 (c) and (d) and 2 3 2 2 3 2 Ans.: (b) 1 1 Solution:  f  , then X C    jR 2 RC j 2 fC RX C  jR 2  jR  j 1  j  R ZP     and Z S  R  X C  R  jR  R 1  j  R  X C R  jR 1  j 2 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 19  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics ZP v 1 1 1 j 1  j  R vo  vi  o     ZP  ZS vi 1  Z S R 1  j  2 R 1  j  jR  R  2 R 1  j  1 1 ZP  j 1  j  R j 1  j  R 2 v j 1  j  R j 1  j  R  j  1 1  o     , and phase angle   0 vi jR  R  2 R 1  j  3 jR  3R 3  j  1 3 Q40. An arbitrarily shaped conductor encloses a charge q and is surrounded by a conducting hollow sphere as shown in the figure. q Four different regions of space 1, 2,3 and 4 are indicated in the 1 3 4 figure. Which one of the following statements is correct? 2 (a) The electric field lines in region 2 are not affected by the position of the charge q (b) The surface charge density on the inner wall of the hollow sphere is uniform (c) The surface charge density on the outer surface of the sphere is always uniform irrespective of the position of charge q in region 1 (d) The electric field in region 2 has a radial symmetry Ans.: (c) Solution: From the given statement only option (c) is correct. Q41. Consider a small bar magnet undergoing simple harmonic motion (SHM) along the x - axis. A coil whose plane is perpendicular to the x - axis 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 (b) The frequency of the induced current in the coil is half of the frequency of the SHM (c) Induced e.m.f. in the coil will not change with the velocity 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 Ans.: (a) Solution: From the given statement only option (a) is correct. H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 20  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q42. Consider a spherical dielectric material of radius ‘ a ’ centered at origin. If the  polarization vector, P  P0 eˆx , where P0 is a constant of appropriate dimensions, then ( eˆx , eˆy , and eˆz are unit vectors in Cartesian- coordinate system) (a) the bound volume charge density is zero. (b) the bound surface charge density is zero at  0, 0, a . (c) the electric field is zero inside the dielectric (d) the sign of the surface charge density changes over the surface. Ans.: (a), (b), (d)   Solution: b  .P  0   b  P.nˆ   P0 eˆx .rˆ  P0 sin  cos   0 at  0, 0, a    0.  Q43. For an electric dipole with momentum P  p0 eˆz placed at the origin, ( p0 is a constant of appropriate dimensions and eˆx , eˆy and eˆz are unit vectors in Cartesian coordinate system) 1 (a) potential falls as , where r is the distance from origin r2 (b) a spherical surface centered at origin is an equipotential surface (c) electric flux through a spherical surface enclosing the origin is zero  (d) radial component of E is zero on the xy - plane. Ans.: (a), (c), (d)  rˆ. p p cos  Solution: Vdip  r ,    . 4 o r 2 4 o r 2  E dip  r ,   p 4 0 r 3  2 cos rˆ  sin ˆ . H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 21  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q44. Three infinitely-long conductors carrying currents I1 , I 2 and I 3 C3 lie perpendicular to the plane of the paper as shown in the figure. I 3 C2   I2 If the value of the integral  B.dl for the loops C1 , C2 and C I1 C1 N C3 are 20 , 40 and 0 in the units of respectively, then A (a) I1  3 A into the paper (b) I 2  5 A out of the paper (c) I 3  0. (d) I 3  1A out of the paper Ans.: (a), (b)   Solution:   B.dl  0 I enc C  I1  I 2  2 , I 2  I 3  4 , I1  I 2  I 3  1  I1  3 A , I 2  5 A and I 3  1 A. Q45. The shape of a dielectric lamina is defined by the two curves y  0 and y  1  x 2. If the charge density of the lamina   15 y C / m 2 , then the total charge on the lamina is…………….. C. Ans.: 8 Solution: Total charge on the lamina is y 1 1   1 x 2 15 Q    da    2 15 ydxdy  1  x2 dx S 1 0 2 1 1 15  x3  1 x5 15 2 1 2   Q   1  x  2 x dx   x   2  4 2  5 3  1  x 1 0 1 15  1 2  1 2   15  2 4 Q 1     1       2    2  5 3  5 3  2  5 3 15 16 Q  8 C 2 15 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 22  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics IIT-JAM 2017 Q46. A current I  10 A flows in an infinitely long wire along the axis of    hemisphere (see figure). The value of    B  ds over the   hemispherical surface as shown in the figure is: I  10 A (a) 10 0 (b) 5 0 (c) 0 (d) 7.5 0 Ans. : (a)      I Solution:     B   ds   B.dl  B  2 r  2 r  2 r   I  10 0 0 0 Q47. Consider two, single turn, co-planar, concentric coils of radii R1 and R2 R2 with R1  R2. The mutual inductance between the two coils is R1 proportional to R1 R2 R22 R12 (a) (b) (c) (d) R2 R1 R1 R2 Ans. : (c) 0 I1   R22 2 B1   R 2 2 R1 R22 Solution: 2  M 21 I1  M 21   2   I1 I1 I1 R1 Q48. Consider a thin long insulator coated conducting wire carrying current I. It is now wound once around an insulating thin disc of radius R to bring I the wire back on the same side, as shown in the figure. R I The magnetic field at the centre of the disc is equal to: 0 I 0 I  2 0 I 2 0 I 1 (a) (b)  3  (c) 1  (d) 1  2R 4R    4R    2R    Ans. : (d) 0 I 0 I 0 I  1 Solution: From R.H.R. magnetic field is pointing inwards, B  2    1    4 R 2 R 2 R H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 23  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q49. The electric field of an electromagnetic wave is given by    E  2kˆ  3 ˆj  103 sin 107  x  2 y  3 z   t  . The value of  is ( c is the speed of light): (a) 14 c (b) 12 c (c) 10 c (d) 7c Ans. : (a)    Solution: E  2kˆ  3 ˆj  103 sin 107  x  2 y  3 z   t      107  k  107  xˆ  2 yˆ  3 zˆ   k  107 14,   107  , c    7    14c k 10 14 Q50. A rectangular loop of dimension L and width w moves with a constant velocity v away from an infinitely long straight wire carrying a current I in the plane of the loop as shown in the figure below. Let R be the resistance of the loop. What is the current in the loop at the instant the near –side is at a distance r from the wire? v R w L r I 0 IL wv 0 IL wv (a) (b) 2 R r  r  2 w 2 R  2r  w 0 IL wv 0 IL wv (c) (d) 2 R r  r  w 2 R 2r  r  w Ans. : (c)   r   I  IL  r  w  Solution: B   B.d a   0 Ldr  0 ln   S r 2 r 2  r  1 dB  0 IL  1 1  dr 0 ILwv I       R dt 2 R  r  w r  dt 2 Rr  r  w  H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 24  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics  Q51. For a point dipole of dipole moment p  pzˆ located at the origin, which of the following is (are) correct? (a) The electric field at  0, b, 0  is zero qp (b) The work done in moving a charge q from  0, b, 0  to  0, 0,b  is 4 0 b 2 (c) The electrostatic potential at  b, 0, 0  is zero qp (d) If a charge q is kept at  0, 0,b  it will exert a force of magnitude on the 4 0 b3 dipole. Ans. : (b) and (c) p cos   Solution: V  4 0 r 2 and E p 4 0 r 3  2 cos  rˆ  sin ˆ    (a) At  0, b, 0  ;   E0 2 (b) The work done in moving a charge q from  0, b, 0  to  0, 0,b   p  qp W  q V  0, 0, b   V  0, b, 0    q   0    4 0 b  4 0 b 2 2 (c) The electrostatic potential at  b, 0, 0  is V  b, 0, 0   0  2p (d) At  0, 0,b  ;  0 E  rˆ 4 0 b3 2qp If a charge q is kept at  0, 0,b  it will exert a force of magnitude. 4 0 b3  Q52. A dielectric sphere of radius R has constant polarization P  P0 zˆ , so that the field inside  P the sphere is Ein   0 zˆ. Then, which of the following is (are) correct? 3 0 (a) The bound surface charge density is P0 cos  1 (b) The electric field at a distance r on the z - axis varies as for r  R r2 PR (c) The electric potential at a distance 2R on the z - axis is 0 12 0 (d) The electric field outside is equivalent to that of a dipole at the origin Ans. : (a), (c) and (d) H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 25  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics  Solution:  b  P.nˆ   P0 zˆ .rˆ  P0 cos    1 p.rˆ 1 4 R 3 P.rˆ 1 4 R 3  P0 zˆ .zˆ PR Vdip     0 4 0 r 2 4 0 3 r 2 4 0 3  2 R  2 12 0 Q53. Consider a circular parallel plate capacitor of radius R with separation d between the plates  d  R . The plates are placed symmetrically about the origin. If a sinusoidal voltage V  V0 sin t is applied between the plates, which of the following statement(s) is (are) true? V02 0 R (a) The maximum value of the Poynting vector at r  R is 4d 2 (b) The average energy per cycle flowing out of the capacitor is zero (c) The magnetic field inside the capacitor is constant (d) The magnetic field lines inside the capacitor are circular with the curl being independent of r. Ans. : (a), (b) and (d) V V0 sin t I  E   R V0 cos t Solution: E   and B  0 d  0  0   R2  0 0 d d 2 R 2 R t 2 d 1  0 R V0 cos t V0 sin t  0 RV02 sin t cos t  0 RV02 S EB     sin 2t 0 2 d d 2d 2 4d 2  0 RV02 0 I d r S max  ;  S  0 , B  , inside 4d 2 2 R 2 Q54. In a coaxial cable, the radius of the inner conductor is 2 mm and that of the outer one is 5 mm. The inner conductor is at a potential of 10 V , while the outer conductor is grounded. The value of the potential at a distance of 3.5 mm from the axis is………… (Specify your answer in volts to two digits after the decimal 10V point) Ans. : 3.8 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 26  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Solution:   2V  0 1   V  In Cylindrical coordinate system, r   0  V  A ln r  B r r  r  Thus 10  A ln 2  B and 0  A ln 5  B 10 10 ln 5  10  A ln 2  A ln 5  A    10.86 and  B   17.39 ln  5 / 2  ln  5 / 2   V  r  3.5   A ln 3.5  B  3.8 V Q55. The wave number of an electromagnetic wave incident on a metal surface is  20  750i  m1 inside the metal, where i  1. The skin depth of the wave in the metal is………(Specify your answer in mm to two digits after the decimal point). Ans. : 1.33 Solution: k  k  i   20  750 i  m 1 1 1 1000 Skin depth, d   m mm  1.33 mm  750 750 Q56. A sphere of radius R has a uniform charge density . A sphere of R /2 smaller radius R / 2 is cut out from the original sphere, as shown in the figure below. The center of the cut out sphere lies at z  R / 2. After the R smaller sphere has been cut out, the magnitude of the electric field at z   R / 2 is  R / n 0. The value of the integer n is…………… Ans. : 8   r Solution: Electric field inside a uniformly charge solid sphere of radius R is E  rˆ 3 0   R3 Electric field outside a uniformly charge solid sphere of radius R is E  rˆ 3 0 r 2  R / 2   R / 2 3 R R Electric field at z   is E    n8 2 3 0 3 0 R 2 8 0 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 27  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics IIT-JAM 2018 Q57. A current I is flowing through the sides of an equilateral triangle of side a. The magnitude of the magnetic field at the centroid of the triangle is 9 0 I 0 I 30 I 30 I (a) (b) (c) (d) 2 a a 2 a a Ans.: (a) 3 RS 3 Solution: RS  a 2  a 2 / 4  a and OS   a R 2 3 6 For segment PQ a I 0 I 3 I O BPQ   2sin 60  0  BQR  BRP 0  3  2 a 4  a P S Q  6  9 0 I B  3BPQ  2 a Q58. Three infinite plane sheets carrying uniform charge densities  , 2 ,3 are parallel to the x  z plane at y  a,3a, 4a , respectively. The electric field at the point  0, 2a, 0  is 4 3 2  (a) ĵ (b)  ĵ (c)  ĵ (d) ĵ 0 0 0 0 Ans.: (b) z Solution: The electric field at the point P  0, 2a, 0  is  2 3 P    2 3  ˆ 3 y E     j  0   ĵ  2 0 2 0 2 0  a 3a 4a x H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 28  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q59. A rectangular loop of dimensions l and w moves with a constant v speed of v through a region containing a uniform magnetic field B directed into the paper and extending a distance of 4w. Which of l B  the following figures correctly represents the variation of emf    w with the position  x  of the front end of the loop? 0 4w x   (a) (b)  Bwv  Bwv 4w 0 0 w x 0 x 4w 0 w  Bwv  Bwv    Blv (c)  Blv (d) 0 w 4w 0 x 0 x 4w 0 w  Blv  Blv Ans.: (c) Solution: B B l   l v v dx Case-I Case-II Case-I: at x  0, 1  Blw and at, x  dx , 2  Bl  w  dx  d    Bldx      Blv dt Case-II:   Blv and direction will be opposite. When loop is inside there is no flux change so,   0. H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 29  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q60. A long solenoid is carrying a time dependent current such that the  k̂ magnetic field inside has the form B  t   B0t 2 kˆ , where k̂ is along the axis of the solenoid. The displacement current at the point P on a circle of radius r in a plane perpendicular to the axis r p (a) is inversely proportional to r and radially outward (b) is inversely proportional to r and tangential (c) increases linearly with time and is tangential. (d) is inversely proportional to r 2 and tangential Ans.: (b)    dB  Solution:   E  dl     dl dt  B0tR 2  E  2 r  2 B0t   R 2  E  r E  B R 2 1  Jd  0  Jd  0 0  Jd  t r r kr 2 , r  R Q61. Given a spherically symmetric charge density   r    ( k being a constant),  0, r  R the electric field for r  R is (take the total charge as Q ) Qr 3 3Qr 2 5Qr 3 Q (a) rˆ (b) rˆ (c) rˆ (d) rˆ 4 0 R 5 4 0 R 4 8 0 R 5 4 0 R 5 Ans.: (a)   Q  1 r  Solution:   E.d a  enc  E  4 r 2    kr 2  4 r 2 dr  S 0 0  0   1 r5  kr 3  E  4 r  2  4 k  E  E0 5 5 0 R R5 5Q  5Q r3 Qr 3  Q   kr  4 r dr  4 k 2 2 k   E    0 5 4 R 5 4 R 5 5 0 4 0 R 5 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 30  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q62. An infinitely long solenoid, with its axis along k̂ , carries a current I. In addition there is  a uniform line charge density  along thee axis. If S is the energy flux, in cylindrical   coordinates ˆ , ˆ, kˆ , then  (a) S is along ̂  (b) S is along k̂  (c) S has non zero components along ̂ and k̂  (d) S is along ˆ  kˆ  Ans. : (d)  Solution: E  E ˆ   B E ˆ B  Bkˆ    S  EB  S  ˆ  kˆ  Let the electric field in some region R be given by E  e  y iˆ  e  x ˆj. From this we may 2 2 Q63. conclude that (a) R has a non-uniform charge distribution (b) R has no charge distribution (c) R has a time dependent magnetic field. (d) The energy flux in R is zero everywhere. Ans.: (b), (c)     Solution:    E  0 and   E  0 , Thus R has no charge distribution and R has a time dependent magnetic field. H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 31  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q64. In presence of a magnetic field Bjˆ and an electric field   E  kˆ , a particle moves undeflected. Which of the following statements is (are) correct? Eˆ (a) The particle has positive charge, velocity   i B Eˆ (b) The particle has positive charge, velocity  i B Eˆ (c) The particle has negative charge, velocity   i B Eˆ (d) The particle has negative charge, velocity  i B Ans.: (b), (d)       Solution:  F  q  E  v  B   0    E v  B   E For  ve charge: a kˆ  v  xˆ B   E For ve charge: a kˆ  v  xˆ B      2  Q65.  Consider an electromagnetic plane wave E  E0 iˆ  bjˆ cos    ct  x  3 y  , where   is the wavelength, c is the speed of light and b is a constant. The