The magnetic field of a plane electromagnetic wave is given by $\overrightarrow{\mathrm{B}}=2 \times 10^{-8} \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{j}} \mathrm{T}$ The amplitude of the electric field would be

(C) $6 \mathrm{Vm}^{-1}$ along z-axis

(A) $6 \mathrm{Vm}^{-1}$ along x-axis

(B) $3 \mathrm{Vm}^{-1}$ along z-axis

(D) $2 \times 10^{-8} \mathrm{Vm}^{-1}$ along z-axis

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · EM Waves

The magnetic field of a plane electromagnetic wave is given by

B = 2 \times 1 0^{- 8} sin (0.5 \times 1 0^{3} x + 1.5 \times 1 0^{11} t) \hat{j} T

The amplitude of the electric field would be

(A) $6 Vm^{- 1}$ along x-axis
(B) $3 Vm^{- 1}$ along z-axis
(C) $6 Vm^{- 1}$ along z-axis✓ Correct
(D) $2 \times 1 0^{- 8} Vm^{- 1}$ along z-axis

Answer: (C)

Solution

For an EM wave,

E_{0} = c B_{0}

.

Given

B_{0} = 2 \times 1 0^{- 8} T

E_{0} = 3 \times 1 0^{8} \times 2 \times 1 0^{- 8} = 6 V/m

Since

B

is along

\hat{j}

and the wave propagates along negative x-axis (from the phase term), by

E \times B ∥ v

E

must be along the z-axis.

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option

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