JEE Main · 2025 · 23 Jan 2025 · Shift 2 · EM Waves

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A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is

E_{y} = 9.3 Vm^{- 1}

. Then, the magnetic field vector of the wave at that point is-

(A) $B_{z} = 9.3 \times 1 0^{- 8} T$
(B) $B_{z} = 1.55 \times 1 0^{- 8} T$
(C) $B_{z} = 6.2 \times 1 0^{- 8} T$
(D) $B_{z} = 3.1 \times 1 0^{- 8} T$ ✓ Correct

Answer: (D)

Solution

For a plane electromagnetic wave travelling in free space, the electric and magnetic field magnitudes are related by

E = c B

B = \frac{E}{c} = \frac{9.3}{3 \times 1 0 ^{8}} = 3.1 \times 1 0^{- 8} T

Since the wave travels along

+ x

with

E

along

+ y

, the direction of propagation fixes

B

along

+ z

B_{z} = 3.1 \times 1 0^{- 8} T

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option

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