Light wave traveling in air along x -direction is given by $\mathrm{E}_{\mathrm{y}}=540 \sin \pi \times 10^{4}(\mathrm{x}-\mathrm{ct}) \mathrm{Vm}^{-1}$. Then, the peak value of magnetic field of wave will be (Given $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ )

(A) $18 \times 10^{-7} \mathrm{~T}$

(B) $54 \times 10^{-7} \mathrm{~T}$

(C) $54 \times 10^{-8} \mathrm{~T}$

(D) $18 \times 10^{-8} \mathrm{~T}$

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · EM Waves

Light wave traveling in air along x -direction is given by

E_{y} = 540 sin π \times 1 0^{4} (x - ct) Vm^{- 1}

. Then, the peak value of magnetic field of wave will be (Given

c = 3 \times 1 0^{8} ms^{- 1}

)

(A) $18 \times 1 0^{- 7} T$ ✓ Correct
(B) $54 \times 1 0^{- 7} T$
(C) $54 \times 1 0^{- 8} T$
(D) $18 \times 1 0^{- 8} T$

Answer: (A)

Solution

For an electromagnetic wave, the relationship between the peak electric field

E_{0}

and peak magnetic field

B_{0}

is:

B_{0} = \frac{E _{0}}{c}

From the given equation,

E_{0} = 540

V/m.

Therefore:

B_{0} = \frac{540}{3 \times 1 0 ^{8}} = 18 \times 1 0^{- 7} T

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option

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