The oscillating magnetic field in a plane electromagnetic wave is given by $\mathrm{B}_{\mathrm{y}}=5 \times 10^{-6} \sin 1000 \pi\left(5 \mathrm{x}-4 \times 10^{8} \mathrm{t}\right)$ T. The amplitude of electric field will be:

(D) $4 \times 10^{2} \mathrm{Vm}^{-1}$

(A) $15 \times 10^{2} \mathrm{Vm}^{-1}$

(B) $5 \times 10^{-6} \mathrm{Vm}^{-1}$

(C) $16 \times 10^{12} \mathrm{Vm}^{-1}$

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · EM Waves

The oscillating magnetic field in a plane electromagnetic wave is given by

B_{y} = 5 \times 1 0^{- 6} sin 1000 π (5 x - 4 \times 1 0^{8} t)

T. The amplitude of electric field will be:

(A) $15 \times 1 0^{2} Vm^{- 1}$
(B) $5 \times 1 0^{- 6} Vm^{- 1}$
(C) $16 \times 1 0^{12} Vm^{- 1}$
(D) $4 \times 1 0^{2} Vm^{- 1}$ ✓ Correct

Answer: (D)

Solution

From the given equation, the amplitude of magnetic field is

B_{0} = 5 \times 1 0^{- 6}

T. The wave speed is

v = \frac{ω}{k} = \frac{4 \times 1 0 ^{8} \times 1000 π}{5 \times 1000 π} = 8 \times 1 0^{7}

m/s. For an EM wave,

E_{0} = v B_{0} = 8 \times 1 0^{7} \times 5 \times 1 0^{- 6} = 4 \times 1 0^{2}

V/m.

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option

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