fiziksInstitute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ELECTROMAGNETIC THEORY JRF/NET–(JUNE-2011) Q1. The electrostatic potential V(x, y) in free space in a region where the charge density ρ is zero is given by V ( x, y ) = 4e 2 x + f ( x ) − 3 y 2 . Given that the x-component of the electric field Ex, and V are zero at the origin, f(x) is (a) 3x 2 − 4e 2 x + 8 x (b) 3x 2 − 4e 2 x + 16 x (c) 4e 2 x − 8 (d) 3x 2 − 4e 2 x Ans: (d) V = 4e 2 x + f ( x ) − 3 y 2 . Since ρ = 0 ⇒ ∇ 2V = 0 ⇒ 16e 2 x + f ′′( x ) − 6 = 0 . [ ] Since E x = 0 at origin ⇒ E = −∇V ⇒ E x = − 8e 2 x + f ′( x ) E x (0, 0) = −[8 + f ′(0)] = 0 ⇒ f ′(0 ) = −8 . Since V (0, 0 ) = 0 ⇒ 4 + f (0 ) = 0 ⇒ f (0 ) = −4 Solve equation 16e 2 x + f ′′( x ) − 6 = 0 ⇒ f ′′( x ) = 6 − 16e 2 x ⇒ f ′(x ) = 6 x − 8e 2 x + c1 , since f ′(0 ) = −8 ⇒ c1 = 0 . Again Integrate f ′( x ) = 6 x − 8e 2 x ⇒ f ( x ) = 3 x 2 − 4e 2 x + c 2 since f (0 ) = −4 ⇒ c 2 = 0 . Thus f ( x ) = 3x 2 − 4e 2 x Q2. For constant uniform electric and magnetic field E = E 0 and B = B0 , it is possible to choose a gauge such that the scalar potential φ and vector potential A are given by (a) φ = 0 and A = ( 1 B0 × r 2 ) (c) φ = − E 0 ⋅ r and A = 0 (b) φ = − E 0 ⋅ r and A = ( 1 B0 × r 2 ) (d) φ = 0 and A = − E 0 t Ans: (a) Let E = E 0 ( xˆ + yˆ + zˆ ) and B = B0 ( xˆ + yˆ + zˆ ) since they are constant vector. Lorentz Gauge condition is ∇ ⋅ A = − μ 0 ε 0 ∂V ∂t Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
[email protected] 1 fiziks Institute for NET/JRF, GATE, IIT‐JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES since B × r = B 0 ( z − y )xˆ − B 0 ( z − x )yˆ + B 0 ( y − x )zˆ { Q3. } (a) ∂φ = 0 and ∇ ⋅ A = 0 ∂t (b) ∂φ ≠ 0, and ∇ ⋅ A = 0 ∂t (c) ∂φ ≠ 0 and ∇ ⋅ A = 0 ∂t (d) ∂φ ≠ 0 and ∇ ⋅ A ≠ 0 ∂t A plane electromagnetic wave is propagating in a lossless dielectric. The electric field is given by [ { )}] ( E ( x, y, z , t ) = E 0 ( xˆ + Azˆ ) exp ik 0 − ct + x + 3 z , where c is the speed of light in vacuum, E0, A and k0 are constant and xˆ and zˆ are unit vectors along the x- and z-axes. The relative dielectric constant of the medium εr and the constant A are 1 3 (a) ε r = 4 and A = − (b) ε r = 4 and A = + (c) ε r = 4 and A = 3 1 3 (d) ε r = 4 and A = − 3 Ans: (a) [ { ( )}] 3 zˆ ) and ω = k c . E ( x, y, z , t ) = E 0 ( xˆ + Azˆ ) exp ik 0 − ct + x + 3 z . ( Comparing with term e i (k ⋅r −ωt ) ⇒ k = k 0 xˆ + Since v = ω k = k0c k 02 + 3k 02 = ( 0 c ⇒ Refractive index n = ε r = 2 ⇒ ε r = 4. 2 ) ( ) Since k ⋅ nˆ = 0 ⇒ k 0 xˆ + 3 zˆ ⋅ ( xˆ + Azˆ ) = 0 ⇒ k 0 1 + A 3 = 0 ⇒ A = − Q4. A static, spherically symmetric charge distribution is given by ρ (r ) = 1 3 A − Kr e where A and r K are positive constants. The electrostatic potential corresponding to this charge distribution varies with r as (a) re − Kr (b) 1 − Kr re r (c) 1 − Kr re r2 (d) ( 1 1 − e − Kr r ) Head office Branch office fiziks, H.No. 23, G.F, Jia Sarai, Anand Institute of Mathematics, Near IIT, Hauz Khas, New Delhi‐16 28‐B/6, Jia Sarai, Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas, New Delhi‐16 Website: www.physicsbyfiziks.com Email:
[email protected] 2 com 3 . x a z b y A.33 cm. Jia Sarai.4π a b where n = 1 ⇒ a = 3.0 GHz (c) 7.physicsbyfiziks. where x and y are in cm. ⎜ ⎟ + ⎜ ⎟ ⇒ f1.4π y ) . The dimensions of the waveguide are (a) a = 3.5 GHz < f < 9.30 cm (c) a = 0.33cm.1 = 2 2 ⎝ a ⎠ ⎝b⎠ 2 a b Head office Branch office fiziks. IIT‐JAM. New Delhi‐16 28‐B/6. New Delhi‐16 Website: www. 23. b = 2. G. The magnetic field of the TE11 mode of a rectangular waveguide of dimensions a × b as shown in the figure is given by H z = H 1 cos(0.50cm B.5 GH z .No. n 2 c ⎛m⎞ ⎛n⎞ c 1 1 = + 2 = 7.0 GHz (d) 7. b = 1. b = 2.0 GHz < f < 7.4πy ) ⇒ mπ nπ = 0.80
[email protected] GHz (b) 7.3π where m = 1 and = 0. Hauz Khas. Anand Institute of Mathematics.40 cm.66 cm.3π x ) cos(0. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. b = 0.5 GHz < f Ans: (d) 2 f m.com Email: fiziks. The entire range of frequencies f for which the TE11 mode will propagate is (a) 6.5 GHz < f < 12.fiziks Institute for NET/JRF. Near IIT.3πx ) cos(0. TIFR and GRE in PHYSICAL SCIENCES Ans: (b) since ∇ 2V = − ρ / ε 0 ∇ 2V must be proportional to 1 ∂ ⎛ 2 ∂V ⎞ A − kr e where ∇ 2V = 2 ⎜r ⎟. GATE. Jia Sarai.50 cm (b) a = 0.F. H.25 cm Ans: (a) Since H z = H 0 cos(0.60 cm (d) a = 1. b = 0. r r ∂r ⎝ ∂r ⎠ Q5. JEST. there is only one point where field vanishes. New Delhi‐16 28‐B/6. 2 I2 = I0 cos 2 θ . H. Jia Sarai. Near IIT. TIFR and GRE in PHYSICAL SCIENCES For propagation. The intensity of light emerging from P3 is (a) 0 (b) I0 2 (c) I0 sin 2 2θ 8 (d) I0 sin 2 2θ 4 Ans: (c) I = I 0 cos 2 θ (Malus Law) ⇒ I1 = Q7. Head office Branch office fiziks. 2 8 Four equal point charges are kept fixed at the four vertices of a square.
[email protected]. 2 I3 = I0 I cos 2 θ × cos 2 (90 − θ ) = 0 sin 2 2θ . G. points where the electric field vanishes) will be found inside the square? (a) 1 (b) 4 (c) 5 (d) 7 Ans: (a) Inside the square. frequency of incident wave must be greater than cutoff frequency.No. Jia Sarai. A beam of unpolarized light of intensity I0 is incident on P1 as shown. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. I0 . How many neutral points (i.com Email: fiziks. Consider three polarizer’s P1. Anand Institute of Mathematics. 23. P2 P1 P3 θ (unpolarized) I0 The pass axis of P1 and P3 are at right angles to each other while the pass axis of P2 makes an angle θ with that of P1.com 4 . IIT‐JAM. New Delhi‐16 Website: www.fiziks Institute for NET/JRF. JRF/NET-(DEC-2011) Q6. Hauz Khas.physicsbyfiziks. P2 and P3 placed along an axis as shown in the figure. GATE. Jia Sarai. New Delhi‐16 28‐B/6.physicsbyfiziks. Hauz Khas. IIT‐JAM.F. where α and R are positive constants. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. TIFR and GRE in PHYSICAL SCIENCES Q8. H. So none of the options given are correct.fiziks Institute for NET/JRF. in which a time dependent current I = I0 sinωt (where ωR/c << 1) flows. Anand Institute of Mathematics. the minimum value of d for the central fringe to be dark is (a) λD (μ − 1) (b) D +L 2 2 λD (c) (μ − 1)L λ (d) (μ − 1) λ 2(μ − 1) Ans: (d) For central fringe to be dark.com 5 . Qenc = 4παε 0 ⎜1 − ⎟ .physics@gmail. If a glass slab of refractive index μ and thickness d is placed in the path of one of the beams. GATE. (μ − 1)d = λ nλ ⇒d = 2 2(μ − 1) Q10.No. The charge contained within a sphere of radius R. The magnitude of the electric field at a perpendicular distance r < R from the axis of symmetry of the solenoid. Near IIT.com Email: fiziks. the slits are at a distance 2L from each other and the screen is at a distance D from the slits. 23. is (a) 0 (b) 1 ωμ 0 nI 0 R 2 cos ωt 2r (c) 1 ωμ 0 nI 0 r sin ωt 2 (d) 1 ωμ 0 nI 0 r cos ωt 2 Ans: (d) ∫ E ⋅ dl = −∫ (B = μ nI (t )zˆ ). Consider a solenoid of radius R with n turns per unit length. ( A static charge distribution gives rise to an electric field of the form E = α 1 − e − r / R ) rˆ r2 . ∂B ⋅ da . In a Young’s double slit interference experiment. ⎝ e⎠ Q9. New Delhi‐16 Website: www. Jia Sarai. ∂t ⇒ E × 2πr = − μ 0 n dI dt 0 r ∫ 2πr ′dr ′ = − μ 0 n × I 0ω cos ωt × r ′=0 2πr 2 2 Head office Branch office fiziks. G. JEST. centred at the origin is (a) παε 0 e R2 (b) παε 0 ( Ans: Qenc = ε 0 ∫ E ⋅ da = αε 0 ∫ 1 − e − r / R e2 R2 ) rˆ ⋅ (r r (c) 4παε 0 2 2 R e (d) παε 0 π 2π ) sin θdθdφrˆ = αε 0 × ∫ ∫ (1 − e −r / R R2 e )sin θdθdφ 0 0 ⎛ 1⎞ at r = R . Jia Sarai.com Email: fiziks. let 2 r ⎠ ⎝ xˆ F = F0 ( xˆ + yˆ + zˆ ) . is (b) − F (a) F (c) F + 30 r r4 (d) F − 30 r r4 Ans: (a) B = ∇× A = [ ( )] 1 r ⎞ ⎛ ∇ × F × r + 10⎜ ∇ × 3 ⎟ . The magnetic field corresponding to the vector potential A = 1 10 F × r + 3 r where F is 2 r a constant vector. F × r = F0 x yˆ F0 y zˆ F0 = xˆ ( z − y )F0 − yˆ ( z − x )F0 + zˆ ( y − x )F0 z Head office Branch office fiziks. Since F is a constant vector. TIFR and GRE in PHYSICAL SCIENCES 1 ⇒ E = − × ωμ 0 nI 0 r cos ωt 2 Q11. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas.
[email protected] Institute for NET/JRF. New Delhi‐16 28‐B/6. Near IIT.F. G. H. The electric field at a distance r (< ct) is (c) cμ 0 I 2π c t − r 2 2 2 ( 1 ˆ ˆ i+ j 2 ) (d) − cμ 0 I 2π c t − r 2 2 2 kˆ Ans: (d) E = −∇φ − ⇒E= μ I ∂A r ∂A =− ⇒E=− 0 ∂t ∂t 2π ct + c 2 t 2 − r 2 ( − cμ 0 I 2π c t − r 2 2 2 ) ⎡1 ⎛ 2c 2 t ⎜ c + ⎢ ⎜ 2 c 2t 2 − r 2 ⎢⎣ r ⎝ ⎞⎤ ⎟⎥ . IIT‐JAM. The vector potential at a perpendicular distance r from the wire is given by A = (a) 0 ( kˆμ 0 I ⎡ 1 ln ⎢ ct + c 2 t 2 − r 2 2π ⎣r μ0 I 1 ˆ ˆ (i − j ) 2π t 2 (b) )⎤⎥⎦ . GATE.physicsbyfiziks. Hauz Khas. A constant electric current I in an infinitely long straight wire is suddenly switched on at t = 0. 23.No.com 6 . New Delhi‐16 Website: www. Jia Sarai. ⎟ ⎠⎥⎦ kˆ JRF/NET-(JUNE-2012) Q12. Anand Institute of Mathematics. G.No. Thus B = F 2 r Q13. New Delhi‐16 28‐B/6. TIFR and GRE in PHYSICAL SCIENCES xˆ yˆ zˆ ∂ ∂ ∂ ∇× F ×r = = xˆ[F0 + F0 ] − yˆ [− F0 − F0 ] + zˆ[F0 + F0 ] = 2 F0 ( xˆ + yˆ + zˆ ) ∂x ∂y ∂z (z − y )F0 (x − z )F0 ( y − x )F0 ( ⇒ ) [ ( )] 1 r ∇ × F × r = F0 ( xˆ + yˆ + zˆ ) = F . as shown in the figure. The magnetic field at a distance R from a long straight wire carrying a steady current I is proportional to (b) I /R2 (a) IR (c) I2/R2 (d) I / R Ans: (d) Q15. Jia Sarai. GATE. of the reflected wave into the water is found to remain the same for all angles of incidence. Near IIT. Which of the following questions is Lorentz invariant? (a) E × B 2 2 (b) E − B 2 2 (c) E + B 2 2 (d) E B 2 Ans: (b) Q16.com 7 . The phase of the perpendicular component of the electric field.F. New Delhi‐16 Website: www. The phase of the magnetic field H (a) does not change (b) changes by 3π/2 (c) changes by π/2 (d) changes by π Ans: (d) Q14.2Q C ˆj a a (b) + 3aQ ˆj (c) − 3aQ ˆj (d) 0 A Q a B Q iˆ Ans: (c) Head office Branch office fiziks. ∇ × 3 = 0 . Jia Sarai.physics@gmail. IIT‐JAM.fiziks Institute for NET/JRF. The dipole moment of this configuration of charges. H. Anand Institute of Mathematics.physicsbyfiziks. is (a) + 2aQ iˆ . Q and -2Q are placed on the vertices of an equilateral triangle ABC of sides of length a. 23. irrespective of the chice of origin. JEST. Charges Q. An electromagnetic wave is incident on a water-air interface. E ⊥ . Hauz Khas. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas.com Email: fiziks. TIFR and GRE in PHYSICAL SCIENCES Let coordinates of A is (l. If the electric field at the centre of the circle is zero. JEST. at the point r . m). GATE. then ⎡⎛ a ⎞ ⎛ 3a ⎞ ˆ ⎤ ⎟ j⎥ p = qi ri ′ = Q liˆ + mˆj + Q (l + a )iˆ + mˆj − 2Q ⎢⎜ l + ⎟iˆ + ⎜⎜ m + 2 ⎟⎠ ⎥⎦ ⎢⎣⎝ 2 ⎠ ⎝ [ [ ] [ ] ] [ ] [ )] ( p = Q liˆ + mˆj + Q (l + a )iˆ + mˆj − Q (2l + a )iˆ + 2m + 3a ˆj ⇒ p = − 3aQˆj Q17. The two charges Q subtend an angle 90° at the centre Q Q of the circle. Jia Sarai. is (a) 3myz r5 3mxy r5 (b) − (c) 3mxz r5 (d) 3m(z 2 − xy ) r5 Ans: (c) m = mzˆ and B = ∇ × A = B= ( ) m 1 2 cos θrˆ + sin θθˆ = 3 [3(m ⋅ rˆ )rˆ − m] 3 r r ⎤ 1 ⎡ ⎛ xxˆ + yyˆ + zzˆ ⎞ r ˆ 3 m z ⋅ ⎜ ⎟ − mzˆ ⎥ ⇒ ⎢ 3 r r ⎣ ⎝ ⎠r ⎦ Bx = 3mxz r5 JRF/NET-(DEC-2012) Q18.No. H.F. The charge q is symmetrically placed with respect to the charges Q.com Email: fiziks. New Delhi‐16 Website: www. Three charges are located on the circumference of a circle of radius R as shown in the figure below.fiziks Institute for NET/JRF. G.com 8 . Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas.physics@gmail. Jia Sarai. Hauz Khas. IIT‐JAM. r3 If m is directed along the positive z-axis. Anand Institute of Mathematics.physicsbyfiziks. Near IIT. Thus E3 = E ⇒ Q = q 2 Head office Branch office fiziks. the x-component of the magnetic field. what is the magnitude of Q? (a) q / 2 (b) 2q (c) 2q (d) 4q q Ans: (a) E1 = E 2 = Q 1 q and E3 = 2 4πε 0 R 4πε 0 R 2 1 Resultant of E1 and E 2 is E = E12 + E 22 = 2E1 . New Delhi‐16 28‐B/6. The vector potential A due to a magnetic moment m at a point r is given by A = m×r . 23. com 9 . 23. Head office Branch office fiziks. H. G. Consider a hollow charged shell of inner radius a and outer radius b.fiziks Institute for NET/JRF. IIT‐JAM. The intensity of the resulting wave is given by ε0 2 E 2 . Anand Institute of Mathematics. TIFR and GRE in PHYSICAL SCIENCES Q19.physicsbyfiziks. Hauz Khas. Jia Sarai. GATE. New Delhi‐16 Website: www.F.physics@gmail. Consider the interference of two coherent electromagnetic waves whose electric field vectors are given by E1 = iˆE 0 cos ω t and E 2 = ˆjE 0 cos(ω t + ϕ ) where ϕ is the phase difference. Near IIT. New Delhi‐16 28‐B/6. Jia Sarai. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. The total intensity is (a) 0 (b) ε 0 E 02 (c) ε 0 E 02 sin 2 ϕ (d) ε 0 E 02 cos 2 ϕ Ans: (a) Since waves are polarized in perpendicular direction hence there will be no interference.No.com Email: fiziks. JEST. The volume charge density is ρ (r ) = k (k is constant) in the region a < r < b. where E 2 is the time average of E2. The magnitude of the electric r2 field produced at distance r > a is (a) k (b − a ) for all r > a .da = E(4πr ) = E(4πr 2 ) = 1 1 1 k 2 Qenc = ∫ ρdV = ∫ r sin θdrdθdφ ε0 ε0 ε0 r 2 4πk r 4πk k ⎛ r−a ⎞ (r − a) ⇒ E = ⎜ rˆ ∫a dr = ε0 ε0 ε0 ⎝ r 2 ⎟⎠ For r > a E4πr 2 = 4πk b 4πk k ⎛ b −a ⎞ (b − a) ⇒ E = ⎜ rˆ ∫a dr = ε0 ε0 ε0 ⎝ r 2 ⎟⎠ Q20. ε 0r 2 (b) k (b − a ) kb for a < r < b and for r > b 2 ε 0r ε 0r 2 (c) k (r − a ) k (b − a ) for a < r < b and for r > b 2 ε 0r ε 0r 2 (d) k (r − a ) k (b − a ) for a < r < b and for r > b 2 ε 0a ε 0r 2 Ans: (c) For r < a 2 ∫ E. physics@gmail. Jia Sarai. H.fiziks Institute for NET/JRF. For what value of d will the charge remains stationary? (a) q / 4 mgπε 0 (b) q / mgπε 0 (c) There is no finite value of d (d) mgπε 0 / q Ans: (a) There is attractive force between point charge q and grounded conducting sheet that can be calculate from method of images i. At the point P which is at a distance R q −q from the centre (R >> a). TIFR and GRE in PHYSICAL SCIENCES Q21.physicsbyfiziks. Near IIT. Jia Sarai. IIT‐JAM.e. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. A point charges q of mass m is kept at a distance d below a grounded infinite conducting sheet which lies in the xy . An infinite solenoid with its axis of symmetry along the z-direction carries a steady current I. Four charges (two + q and two –q) are kept fixed at the four vertices of a square of side a as shown. Hauz Khas. Anand Institute of Mathematics.plane. the potential is proportional to (a) 1/R (b) 1/R2 (c) 1/R3 (d) 1/R4 a Ans: (c) R −q P q Given configuration is quadrupole. New Delhi‐16 28‐B/6.F.No.com 10 . New Delhi‐16 Website: www. GATE. 23. G. zˆ The vector potential A at a distance R from the axis (a) is constant inside and varies as R outside the solenoid (b) varies as R inside and is constant outside the solenoid (c) varies as R 1 inside and as R outside the solenoid R (d) varies as R inside and as 1 outside the solenoid R Ans: (d) Head office Branch office fiziks. JEST. Q22.com Email: fiziks. q2 q = mg ⇒ d = 2 4πε 0 (2d ) 4 mgπε 0 1 Q23.
[email protected] 11 . IIT‐JAM. Jia Sarai.F. 23. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. the electric field E and the Poynting vector S = 1 E × B at a larger distance r from emitter vary as μ0 1 and rn 1 respectively. TIFR and GRE in PHYSICAL SCIENCES Q24. Anand Institute of Mathematics. The vector potential at by A = (a) μ0 K 4c (ct − z )2 iˆ . Hauz Khas. H. z ) μ0 K 2c (ct − z )iˆ μ0 K (d) − 2c (ct − z ) ˆj Ans: (d) B = ∇× A = μ K ∂Ax yˆ = − 0 (ct − z ) ˆj 2c ∂z Q25. the natural frequency of oscillation of the particle is R R − + − e.fiziks Institute for NET/JRF. m 2d 2d (a) 6eR 2 πε 0 md 5 (b) 6eR πε 0 md 4 + (c) 6ed 2 πε 0 mR 5 (d) 6ed πε 0 mR 4 Head office Branch office fiziks. Assuming that the length of the dipoles is much shorter than their separation. where K is a constant. JEST. Consider an infinite conducting sheet in the xy-plane with a time dependent current density Kt iˆ . Which of the following choices for n and m are correct? rm (a) n = 1 and m = 1 (b) n = 2 and m = 2 (c) n = 1 and m = 2 (d) n = 2 and m = 4 Ans: (c) JRF/NET-(JUNE-2013) Q26. A particle of charge e and mass m is located at the midpoint of the line joining two fixed collinear dipoles with unit charges as shown in the figure. y .physicsbyfiziks. (The particle is constrained to move only along the line joining the dipoles).com Email: fiziks. Near IIT. GATE. G. New Delhi‐16 28‐B/6. The magnetic field μ 0 Kt ˆ j (b) − 2 μ 0 Kz ˆ j is given B is (c) − 2c ( x. Jia Sarai.No. When a charged particle emits electromagnetic radiation. New Delhi‐16 Website: www. F. Then the resultant electric field at point A is given by 2p ⎡ 1 1 ⎤ 6d =− x where p = 1. Jia Sarai. of the Poynting vector S outside the beam at a radial distance r (much larger than the width of the beam) from the axis. m A − + 2d 2d Let us displace the charge particle by small amount x at A. The direction and magnitude.fiziks Institute for NET/JRF. A current I is created by a narrow beam of protons moving in vacuum with constant Then ω = velocity u . are (a) S ⊥ u and S = (c) S || u and S = I2 (b) S || (− u ) and S = 4π 2 ε 0 u r 2 I2 (d) S || u and S = 4π 2 ε 0 u r 2 I2 4π 2 ε 0 u r 4 I2 4π 2 ε 0 u r 4 Ans: (c) Let charge per unit length be λ . New Delhi‐16 Website: www.com Email: fiziks. H. Jia Sarai. TIFR and GRE in PHYSICAL SCIENCES R Ans: (d) R x − + e. GATE. JEST.No. The magnetic field at a distance r is B = The electric field at a distance r is E = μ0 I ˆ φ. 2πr I λ rˆ = rˆ . Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. 2πε 0 r 2πε 0 ur Head office Branch office fiziks. hence I = λ u in z-direction. IIT‐JAM.physicsbyfiziks. respectively. Near IIT. − ⎢ 3 3 ⎥ 4πε 0 ⎣ (R + x ) (R − x ) ⎦ πε 0 R 4 6ed F = eE = − x πε 0 R 4 E= k 6ed = m πε 0 mR 4 Q27. Anand Institute of Mathematics. New Delhi‐16 28‐B/6. G.physics@gmail. Hauz Khas.com 12 .2d = 2d . 23. z ) (c) (0. A2 . Anand Institute of Mathematics. yzt . 23. zxt . 2 ⎛r ⎞ ρ = ρ 0 ⎜ 0 ⎟ e − r / r0 cos 2 ϕ ⎝r⎠ The radial component of the dipole moment due to this charge distribution is (a) 2πρ 0 r04 (b) πρ 0 r04 (c) ρ 0 r04 (d) πρ 0 r04 / 2 Ans: (a) 2 ⎛r ⎞ p = ∫ r ′ρ (r ′)dτ ′ = ∫ ∫ ∫ r ′ × ρ 0 ⎜ 0 ⎟ e − r ′ / r0 cos 2 ϕ × r ′ 2 sin θdr ′dθdϕ ⎝ r′ ⎠ V p=ρ r 2 0 0 r0 π 2π 0 0 −r / r 2 4 ∫ r ′e 0 dr ′∫ sin θdθ ∫ cos ϕdϕ = 2πρ 0 r0 r ′=0 ′ Q30. New Delhi‐16 Website: www. If the electric and magnetic fields are unchanged when the potential A changes (in suitable units) according to A → A + rˆ.F. Jia Sarai. Az ) ⇒ φ = −kxyz. GATE. xy ) (b) k ( x.com 13 . Az = kxyt Since ∇φ = k (− yzxˆ − xzyˆ − xyzˆ ) and ∂A = k ( yzxˆ + xzyˆ + xyzˆ ) ∂t Head office Branch office fiziks. Hauz Khas. zx. then the scalar potential Φ must simultaneously change to (a) Φ − r (b) Φ + r (c) Φ − ∂ r / ∂ t (d) Φ + ∂ r / ∂ t Ans: (c) A′ = A + ∇λ = A + rˆ ⇒ ∂ λ / ∂ r = 1 ⇒ λ = r + C V ′ =V − ∂λ / ∂t =V − ∂ r / ∂t Q29. Ax = kyzt . JEST.com Email: fiziks. The three components of the electric field are (a) k ( yz. H. 0. The components of a vector potential Aμ ≡ ( A0 . Ay = kzxt . G. 0 ) (d) k ( xt . zt ) Ans: (c) Aμ = (φ . xyt ) where k is a constant.fiziks Institute for NET/JRF.physicsbyfiziks. yt . Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. IIT‐JAM. Ay . A3 ) are given by Aμ = k (− xyz. where r = r (t )rˆ. A1 .No. Consider an axially symmetric static charge distribution of the form. Jia Sarai. y. TIFR and GRE in PHYSICAL SCIENCES Hence Poynting vector S = E×B μ0 = I2 4π 2 ε 0 ur 2 zˆ Q28. Ax .physics@gmail. Near IIT. New Delhi‐16 28‐B/6. The force between two long and parallel wires carrying currents I 1 and I 2 and separated by a distance D is proportional to (1) I 1 I 2 / D (2) (I 1 + I 2 ) / D (3) (I 1 I 2 / D ) 2 (4) I 1 I 2 / D 2 Ans: (1) Head office Branch office fiziks. Near IIT.com 14 . Anand Institute of Mathematics.0 cm2 kept in the xy -plane.com Email: fiziks. Jia Sarai. The rate of total energy radiated per unit area from the surfaces of the metal sheet at a distance of 100 m is ( ) ω / (12πε c ) ( ) / (24πε c ) (a) I 0ω / 12πε 0 c 3 (b) I 02ω 2 / 12πε 0 c 3 (c) I 02 (d) I 0ω 2 3 0 3 0 Ans: (b) JRF/NET-(DEC-2013) Q32.No. A horizontal metal disc rotates about the vertical axis in a uniform magnetic field pointing up as shown in the figure. TIFR and GRE in PHYSICAL SCIENCES ∂A E = −∇φ − = k ( yzxˆ + xzyˆ + xyzˆ ) − k ( yzxˆ + xzyˆ + xyzˆ ) = 0 ⇒ E = (0.0 ) ∂t Q31.physicsbyfiziks. GATE. G. JEST. The circuit that flows through the resistor is B B A S (1) zero (2) DC from A to B (3) DC from B to A (4) AC Ans: (2) Q33.0. 23. Jia Sarai. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas.fiziks Institute for NET/JRF. Hauz Khas. A circuit is made by connecting one end A of a resistor to the centre of the disc and the other end B to its edge through a sliding contact.physics@gmail. New Delhi‐16 Website: www. H.F. IIT‐JAM. An oscillating current I (t ) = I 0 exp(− iωt ) flows in the direction of the y-axis through a thin metal sheet of area 1. New Delhi‐16 28‐B/6.
[email protected] y − 1000t )](3iˆ + 4 ˆj ) (2) 10 −4 E 0 cos[π (0.3 x + 0. IIT‐JAM. Anand Institute of Mathematics. The net force on the charge (in units of 1 / 4πε 0 ) is ( ) ( ) (1) q2 2 2 − 1 away from the corner 8d 2 (2) q2 2 2 − 1 towards the corner 8d 2 (3) (4) q2 2 2d 2 q d d towards the corner 3q 2 away from the corner 8d 2 −q d d F1 Ans: (2) F1 = F 2 = k q q and F 3 = k 2 2 4d 8d 2 F3 q F2 d 2 Resultant of F 1 .com Email: fiziks.4 yˆ ).4 ˆj ) (4) 10 E cos[π (0.com 15 .3x + 0. JEST.3x + 0. Jia Sarai. ω = 1000t B= k×E ω = 1 ω π (0.3iˆ + 0. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. F 2 is F12 = F12 + F22 = 2 2k q2 .F.No.4 y − 1000t )]kˆ .3 xˆ + 0. A point charge q is placed symmetrically at a distance d from two perpendicularly placed grounded conducting infinite plates as shown in the figure. H.4 y − 1000t )] 4iˆ − 3 ˆj ) Q35.4 y − 1000t )](0. TIFR and GRE in PHYSICAL SCIENCES Q34. New Delhi‐16 Website: www. Near IIT. GATE. G.4 y − 1000t )]kˆ ( ⇒ B = 10 −4 E 0 cos[π (0. 8d 2 +q d −q 2d Head office Branch office fiziks. Jia Sarai.3 x + 0. The electric field of an electromagnetic wave is given by E = E 0 cos[π (0.4 y − 1000t )] 4iˆ − 3 ˆj 0 2 0 Ans: (2) k = π (0.3 xˆ + 0. New Delhi‐16 28‐B/6.3 x + 0.fiziks Institute for NET/JRF. The associated magnetic field B is (1) 10 −3 E 0 cos[π (0.3 x + 0.3x + 0.4 yˆ ) × E 0 cos[π (0. Hauz Khas.4 y − 1000t )]kˆ ( ) (3) E cos[π (0. Jia Sarai. Anand Institute of Mathematics. TIFR and GRE in PHYSICAL SCIENCES 2 q Net force F = k 2 2 2 − 1 (towards the corner) 8d ( ) Q36. IIT‐JAM. New Delhi‐16 28‐B/6.fiziks Institute for NET/JRF. Let four point charges q. 23. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. q −q/2 −q q −q/2 Let V (r ) be the electrostatic potential at a point P at a distance r >> a from the centre of the square. G. New Delhi‐16 Website: www. H.F. q and − q / 2 be placed at the vertices of a square of side a .No. − q / 2. Then V (2r ) / V (r ) is Head office Branch office fiziks.com Email: fiziks.physics@gmail. JEST. Hauz Khas. If the electrostatic potential V (r . GATE. Near IIT. then the functional form of f (r ) (in the following a and b are constants) is: (1) ar 2 + b r (2) ar + b r2 (3) ar + ⎛r⎞ (4) a ln⎜ ⎟ ⎝b⎠ b r Ans: (2) ∇ 2V = ∂ ⎛ ∂V ⎞ 1 ∂ ⎛ 2 ∂V ⎞ 1 1 ⎜r ⎟+ 2 ⎜ sin θ ⎟+ 2 2 2 r ∂r ⎝ ∂r ⎠ r sin θ ∂θ ⎝ ∂θ ⎠ r sin θ ⇒ 1 ∂ ⎛ 2 ∂f 1 ∂ ⎞ cos θ ⎟ + 2 ( sin θ f × − sin θ ) = 0 ⎜r 2 r ∂r ⎝ ∂r ⎠ r sin θ ∂θ ⇒ cos θ r2 ⇒ r2 ⎛ ∂ 2V ⎜ 2 ⎝ ∂φ ⎞ ⎟=0 ⎠ ⎛ 2 ∂2 f ∂f ⎞ f ( 2sin θ cos θ ) = 0 ⎜ r 2 + 2r ⎟ − 2 ∂r ⎠ r sin θ ⎝ ∂r ∂2 f ∂f + 2r − 2 f ( r ) = 0 2 ∂r ∂r f (r ) = ar + b satisfy the above equation. θ . r2 Q37.physicsbyfiziks. φ ) in a charge free region has the form V (r ⋅ θ ⋅ φ ) = f (r ) cos θ . Jia Sarai. Let another point charge − q be placed at the cnetre of the square (see the figure).com 16 . New Delhi‐16 Website: www. TIFR and GRE in PHYSICAL SCIENCES 1 1 1 (3) (4) (1) 1 (2) 2 4 8 Ans: (4) According to multipole expansion Qmono = − q (axˆ + ayˆ ) − q(axˆ − ayˆ ) + q(− axˆ − ayˆ ) − q (− axˆ + ayˆ ) + 0 = 0 2 2 p = q(axˆ + ayˆ ) − Thus V ∝ q q +q− +q−q =0 2 2 1 V (2r ) 1 ⇒ = . A time-dependent current I (t ) = Ktzˆ (where K is a constant) is switched on at t = 0 in an infinite current-carrying wire. G. New Delhi‐16 28‐B/6. 23. and ψ a scalar function. A′ denote two sets of scalar and vector potentials. Hauz Khas.F. Near IIT. Let V . Anand Institute of Mathematics. GATE. Jia Sarai. IIT‐JAM. H. JEST.com 17 . 3 V (r ) 8 r ( ) ( ) Q38.physics@gmail. The magnetic vector potential at a perpendicular distance a from the wire is given (for time t > a / c ) by μ K c t −a ct − a 2 + z 2 (a) zˆ 0 dz 4π c − c ∫t − a (a 2 + z 2 )1 / 2 (b) zˆ μ 0 K ct ct − a 2 + z 2 (c) zˆ dz 4π c −∫ct (a 2 + z 2 )1 / 2 μ K (d) zˆ 0 4π 2 2 2 2 2 2 μ0 K 4π ct ∫ dz (a − ct t 2 + z2 ) c 2t 2 − a 2 ∫ − c 2t 2 − a 2 dz (a 1/ 2 t 2 + z2 ) 1/ 2 Head office Branch office fiziks. Which of the following transformations leave the electric and magnetic fields (and hence Maxwell’s equations) unchanged? (1) A′ = A + ∇ψ and V ′ = V − ∂ψ ∂t (2) A′ = A − ∇ψ and V ′ = V + 2 (3) A′ = A + ∇ψ and V ′ = V + ∂ψ ∂t (4) A′ = A − ∇ψ and V ′ = V − ∂ψ ∂t ∂ψ ∂t Ans: (1) JRF/NET–(JUNE-2014) Q39.No.physicsbyfiziks. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas.fiziks Institute for NET/JRF. Jia Sarai.com Email: fiziks. A and V ′. com 18 .F. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. Jia Sarai.com Email: fiziks. TIFR and GRE in PHYSICAL SCIENCES Ans: (a) μ A = zˆ 0 4π ⇒ A = zˆ ∞ ∫ −∞ I ( tr ) μ dz = zˆ 0 R 4π μ0 K 4π c − c 2t 2 − a 2 ∫ ∞ ∫ K (t − R / c ) R −∞ c 2t 2 − a 2 ct − a + z (a 2 R z 2 dz I dz dz + z2 2 ) a 1/2 •P Q40. i p (t ) 1 3 t 2 Which of the following graphs represents the current i S in the secondary coil? (a) (b) is ( t ) is ( t ) 1 (c) is ( t ) (d) 1 2 3 t 3 t 2 1 2 is ( t ) 3t 1 2 3t Head office Branch office fiziks. New Delhi‐16 28‐B/6. H. The graph of i p (t ) as a function of time t is shown in the figure below.fiziks Institute for NET/JRF. 23. Hauz Khas. A current i p flows through the primary coil of a transformer.No. G. GATE. Near IIT. Jia Sarai. Anand Institute of Mathematics.physicsbyfiziks. IIT‐JAM. JEST. New Delhi‐16 Website: www.physics@gmail. H. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. then the charge density at a distance r = r0 will be (a) ε 0ϕ 0 (b) er 2 0 Ans: (a) ∵ ∇ 2φ = − eε 0ϕ 0 2r02 (c) − ε 0ϕ 0 (d) − er 2 0 2eε 0ϕ 0 r02 ρ ⇒ ρ = −ε 0 ( ∇ 2φ ) ε0 ⎤ φ0 − r / r0 ⎞ 1 ∂ ⎛ 2 ∂φ ⎞ 1 ∂ ⎛ 2 1 φ0 ∂ 2 − r / r0 1 φ ⎡ 1 r ×e = − 2 0 ⎢ r 2 × − e − r / r0 + 2re − r / r0 ⎥ ⎜r ×− e ⎟=− 2 ⎜r ⎟= 2 2 r ∂r ⎝ ∂r ⎠ r ∂r ⎝ r0 r r0 ∂r r r0 ⎣ r0 ⎠ ⎦ ⎤ φ ⎡ 1 2 ⇒ ∇ 2φ = − 0 ⎢ − e − r / r0 + e − r / r0 ⎥ r0 ⎣ r0 r ⎦ ( ∇ 2φ = At a distance r = r0 . Near IIT.com Email: fiziks. ∇ 2φ = − ) φ0 ⎡ 1 ⎛ φ0 ⎞ φ0ε 0 φ0 2 −1 ⎤ −1 ⎢ e + e ⎥ = − 2 ⇒ ρ = −ε 0 ⎜ − 2 ⎟ = 2 r0 ⎣ r0 r0 ⎦ r0 e ⎝ r0 e ⎠ r0 e Q42. G. Jia Sarai. GATE. Jia Sarai. then the value of the integral (a) π (b) π 2 iˆ ∂ Ans: (d) ∇ × A = ∂x yz (c) π 4 ∫ C A ⋅ d is (d) 0 kˆ ∂ = iˆ ( x − x ) − ˆj ( y − y ) + kˆ ( z − z ) = 0 ∂z xy ˆj ∂ ∂y zx ( ) Since ∫ A ⋅ d = ∫ ∇ × A ⋅ d a = 0 C S Q43. Consider an electromagnetic wave at the interface between two homogenous dielectric media of dielectric constants ε 1 and ε 2 .physicsbyfiziks. New Delhi‐16 Website: www. 23.physics@gmail. Hauz Khas. TIFR and GRE in PHYSICAL SCIENCES di Ans: (c) is ∝ − p dt Q41. New Delhi‐16 28‐B/6. with the centre on the z .axis.F. If A = yziˆ + zxˆj + xykˆ and C is the circle of unit radius in the plane defined by z = 1 . IIT‐JAM. Anand Institute of Mathematics. If the electrostatic potential in spherical polar coordinates is ϕ (r ) = ϕ 0 e − r / r 0 where ϕ 0 and r0 are constants. Assuming ε 2 > ε 1 and no charges on the Head office Branch office fiziks. JEST.No.com 19 .fiziks Institute for NET/JRF. Anand Institute of Mathematics.No. H.0 ) . TIFR and GRE in PHYSICAL SCIENCES surface. 0) πε 0 d 2 (d) − (b) +e E+ = 1 e 4πε 0 ( 3d / 2 ) 10e (1. 0) πε 0 d 2 E− d E+ 2 P d −e 0 x 4e 1 e 1 4e and E− = = 2 2 4πε 0 9d 4πε 0 ( d / 2 ) 4πε 0 d 2 1 Thus resultant electric field at point P is E = E+ + E− = 4e 1 4e 1 40e 1 10e 1 10e + = = ⇒E= xˆ 2 2 2 2 4πε 0 9d 4πε 0 d 4πε 0 9d 9πε 0 d 9πε 0 d 2 1 Q45.com 20 . Jia Sarai. Near IIT. Hauz Khas. 0. 0. G. then the phase change experienced by the light upon reflection is (a) tan (2 / ρ ) (b) tan −1 (1 / ρ ) (c) tan −1 (2 / ρ ) (d) tan −1 (2 ρ ) Head office Branch office fiziks. A beam of light of frequency ω is reflected from a dielectric-metal interface at normal incidence.0 ⎟ is ⎝2 ⎠ (a) − (c) Ans: 10e (1. JEST. IIT‐JAM. The refractive index of the dielectric medium is n and that of the metal is n 2 = n(1 + iρ ) . New Delhi‐16 28‐B/6. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas.com Email: fiziks.fiziks Institute for NET/
[email protected]. 0) 9πε 0 d 2 2 = d e (1. 23. The region x ≤ 0 ⎛d ⎞ is filled uniformly with a metal.physicsbyfiziks. The electric field at the point ⎜ .F. 0) 9πε 0 d 2 (b) e (1.0. 0. GATE. where d > 0 . Jia Sarai. A charge (− e ) is placed in vacuum at the point (d . New Delhi‐16 Website: www. If the beam is polarised parallel to the interface. 0. the electric field vector E and the displacement vector D in the two media satisfy the following inequalities (a) E 2 > E1 and D2 > D1 (b) E 2 < E1 and D2 < D1 (c) E 2 < E1 and D2 > D1 (d) E 2 > E1 and D2 < D1 Q44. Near IIT.physics@gmail. Ay . ⎟ 2 ⎠ ⎛2⎞ ⇒ φ = tan −1 ⎜ ⎟ ρ ⎝ρ⎠ 2 Q46. infinitely long solenoid placed along the z .23 mm (d) 0. H. Jia Sarai.com 21 . GATE.axis contains a magnetic flux φ . Which of the following vector potentials corresponds to the magnetic field at an arbitrary point (x. 0 2 2 2 2 π π 2 2 x y x y + + ⎝ ⎠ ⎛ φ ⎞ φ y x (b) (Ax .46 mm (b) Head office Branch office fiziks. .60 mm Ans: (b) 2.0 ⎟⎟ 2 2 2 2 2 2 ⎝ 2π x + y + z 2π x + y + z ⎠ ⎛ φ x+ y φ x+ y ⎞ (c) (Ax . . JEST. IIT‐JAM.fiziks Institute for NET/JRF.3 ) (a) 4. New Delhi‐16 Website: www.F. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. Ay . Jia Sarai. Ay . Ay . Hauz Khas. .0 ⎟⎟ 2 2 2 2 ⎝ 2π x + y 2π x + y ⎠ ⎛ φ ⎞ φ x y (d) (Ax . Az ) = ⎜⎜ − . New Delhi‐16 28‐B/6. An electromagnetically-shielded room is designed so that at a frequency ω = 10 7 rad/s the intensity of the external radiation that penetrates the room is 1% of the incident radiation. . z ) ? ⎛ φ ⎞ φ y x ⎟⎟ (a) (Ax . If σ = 1 −1 × 10 6 (Ωm ) is the conductivity of the shielding material.30 mm (c) 0.No. A thin. Az ) = ⎜⎜ − . TIFR and GRE in PHYSICAL SCIENCES Ans: (c) ⎛ 1− β ⎞ Since E0 R = ⎜ ⎟ E0 I ⎝ 1+ β ⎠ where β = ⎛ ρ e−iπ /2 ⎛ −i ρ ⎞ E ⇒ E0 R = ⎜ = ⎟ 0 I ⎜⎜ 2 iθ ⎝ 2 + iρ ⎠ ⎝ 4+ ρ e v1 c/n = = 1 + iρ v2 c / n (1 + i ρ ) ⎞ ⎛ ρ ⎟ E0 I = ⎜ ⎟ ⎜ 4 + ρ2 ⎠ ⎝ Thus phase change φ = − (π / 2 + θ ) ⇒ tan φ = cot θ = ⎞ −i(π /2+θ ) ρ ⎟e E0 I where tan θ = . y. G.physicsbyfiziks.com Email: fiziks. Az ) = ⎜⎜ − . its 2π minimum thickness should be (given that ln 10 = 2. 23. Anand Institute of Mathematics. Az ) = ⎜⎜ − .0 ⎟⎟ 2 2 2 2 ⎝ 2π x + y 2π x + y ⎠ Ans: (a) B = ∇ × A = 0 Q47. Hauz Khas.No. At what distance from the plane will the particle reach the speed 2u ? (a) d / 6 Ans: (b) d / 3 (c) d / 4 (d) d / 5 x (b) F = ma = m d 2x 1 q2 d 2x A q2 ⇒ = − where A = − = . H.com Email: fiziks. Jia Sarai.physics@gmail. Jia Sarai.com 22 . Near IIT. JEST. G.F. dt 2 4πε 0 4d 2 dt 2 x2 16π mε 0 dv A dv dv A dx 1 d 2 d ⎛ A⎞ ⇒ =− 2 v ⇒v =− 2 ⇒ v = ⎜ ⎟ dt x dt dt x dt 2 dt dt ⎝ x ⎠ ( ) A v2 A ⎛1 1⎞ ⇒ = + C at ⇒ x = d . 2 x d ⎝x d⎠ 1⎞ ⎛ 1 − ⎟= Thus u = 2 A ⎜ ⎝d /2 d ⎠ 2A d ⎛1 1⎞ then 2u = 2 A ⎜ − ⎟ ⇒ x = d 5 ⎝x d⎠ P +q d 0 d −q Head office Branch office fiziks. New Delhi‐16 28‐B/6. A charged particle is at a distance d from an infinite conducting plane maintained at zero potential.30 mm 2 × 103 Q48. κ = = × × 106 × 4π × 10−7 × 107 = 103 I 2 2 2π ⇒z= 1 ln (100 ) = 2. the particle reaches a speed u at a distance d / 2 from the plane. IIT‐JAM. New Delhi‐16 Website: www. TIFR and GRE in PHYSICAL SCIENCES 1 ⎛ I0 ⎞ I = I 0 e −2κ z ⇒ z = ln 2κ ⎜⎝ I ⎟⎠ where I0 σμω 1 1 = 100.physicsbyfiziks. When released from rest. GATE.fiziks Institute for NET/JRF. 23. Near IIT Phone: 011‐26865455/+91‐9871145498 Hauz Khas. v = 0 ⇒ C = − ⇒ v = 2 A ⎜ − ⎟ . Anand Institute of Mathematics.