JEE Main · 2019 · 12 Apr 2019 · Shift 2 · EM Waves

A plane electromagnetic wave having a frequency

n = 23.9 GHz

propagates along the positive z-direction in free space. The peak value of the electric field is

60 V / m

. Which among the following is the acceptable magnetic field component in the electromagnetic wave?

(A) $B = 2 \times 1 0^{- 7} sin (1.5 \times 1 0^{2} x + 0.5 \times 1 0^{11} t) \hat{j}$
(B) $B = 60 sin (0.5 \times 1 0^{3} x + 1.5 \times 1 0^{11} t) \hat{k}$
(C) $B = 2 \times 1 0^{- 7} sin (0.5 \times 1 0^{2} z - 1.5 \times 1 0^{11} t) \hat{i}$ ✓ Correct
(D) $B = 2 \times 1 0^{7} sin (0.5 \times 1 0^{3} z + 1.5 \times 1 0^{11} t) \hat{i}$

Answer: (C)

Solution

For a wave travelling along

+ z

the magnetic field must be transverse and vary as

sin (k z - ω t)

. Its amplitude is fixed by

B_{0} = \frac{E _{0}}{c} = \frac{60}{3 \times 1 0 ^{8}} = 2 \times 1 0^{- 7}

T, and

ω = 2 π f = 2 π \times 23.9 \times 1 0^{9} \approx 1.5 \times 1 0^{11}

rad/s, with

k = \frac{ω}{c} \approx 500

^{- 1}

. The only field with the correct amplitude

2 \times 1 0^{- 7}

T, a

(k z - ω t)

argument for

+ z

travel, and

B

perpendicular to the direction of travel is

B = 2 \times 1 0^{- 7} sin (0.5 \times 1 0^{2} z - 1.5 \times 1 0^{11} t) \hat{i}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option