The magnetic field of an E.M. wave is given by $\overrightarrow{\mathrm{B}}=\left(\frac{\sqrt{3}}{2} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}\right) 30 \sin \left[\omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\right]$ (S.I. Units) The corresponding electric field in S.I. units is :

(A) $\overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)\right]$

(B) $\vec{E}=\left(\frac{3}{4} \hat{i}+\frac{1}{4} \hat{j}\right) 30 c \cos \left[\omega\left(t-\frac{z}{c}\right)\right]$

(C) $\overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right) 30 \mathrm{csin}\left[\omega\left(\mathrm{t}+\frac{\mathrm{z}}{\mathrm{c}}\right)\right]$

(D) $\overrightarrow{\mathrm{E}}=\left(\frac{\sqrt{3}}{2} \hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}+\frac{\mathrm{z}}{\mathrm{c}}\right)\right]$

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · EM Waves

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The magnetic field of an E.M. wave is given by

B = (\frac{3}{2} \hat{i} + \frac{1}{2} \hat{j}) 30 sin [ω (t - \frac{z}{c})]

(S.I. Units)
The corresponding electric field in S.I. units is :

(A) $E = (\frac{1}{2} \hat{i} - \frac{3}{2} \hat{j}) 30 c sin [ω (t - \frac{z}{c})]$ ✓ Correct
(B) $E = (\frac{3}{4} \hat{i} + \frac{1}{4} \hat{j}) 30 c cos [ω (t - \frac{z}{c})]$
(C) $E = (\frac{1}{2} \hat{i} + \frac{3}{2} \hat{j}) 30 csin [ω (t + \frac{z}{c})]$
(D) $E = (\frac{3}{2} \hat{i} - \frac{1}{2} \hat{j}) 30 c sin [ω (t + \frac{z}{c})]$

Answer: (A)

Solution

For a plane electromagnetic wave, the electric field, the magnetic field and the direction of propagation

\hat{k}

form a right-handed mutually perpendicular set, with

E = c (B \times \hat{k})

and

E_{0} = c B_{0}

.

The given field is

B = (\frac{3}{2} \hat{i} + \frac{1}{2} \hat{j}) 30 sin [ω (t - \frac{z}{c})]

The phase depends on

t - z / c

, so the wave travels along

+ \hat{k}

.

Using

\hat{i} \times \hat{k} = - \hat{j}

and

\hat{j} \times \hat{k} = \hat{i}

B \times \hat{k} = 30 [\frac{3}{2} (- \hat{j}) + \frac{1}{2} (\hat{i})] sin [ω (t - \frac{z}{c})] = (\frac{1}{2} \hat{i} - \frac{3}{2} \hat{j}) 30 sin [ω (t - \frac{z}{c})]

Multiplying by

c

E = (\frac{1}{2} \hat{i} - \frac{3}{2} \hat{j}) 30 c sin [ω (t - \frac{z}{c})]

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 28 Jan 2025 · Shift 2 · EM Waves

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