The electric field in a plane electromagnetic wave is given by: Ey = 69 [0.6 10^3 x - 1.8 10^11 t] V/m. The expression for magnetic field associated with this electromagnetic wave is T.

The electric field in a plane electromagnetic wave is given by: E… — Solution | JEE Physics

MCQEasyJEE 2026Characteristics of EM Waves

The electric field in a plane electromagnetic wave is given by : $E_y = 69 \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t]$ $\text{V/m}$ . The expression for magnetic field associated with this electromagnetic wave is _____ $\text{T}$ .

A
$B_z = 2.3 \times 10^{-7} \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t]$
B
$B_y = 69 \sin[0.6 \times 10^3 x + 1.8 \times 10^{11} t]$
C
$B_z = 2.3 \times 10^{-7} \sin[0.6 \times 10^3 x + 1.8 \times 10^{11} t]$
D
$B_y = 2.3 \times 10^{-7} \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t]$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The electric field is

E_y = 69 \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t] \; \text{V/m}

Find: The corresponding magnetic field expression.

For a plane electromagnetic wave in vacuum, the amplitudes satisfy

B_0 = \frac{E_0}{c}

where

E_0 = 69 \; \text{V/m}, \qquad c = 3 \times 10^8 \; \text{m/s}

Direction, amplitude and phase

The amplitude of the magnetic field is

B_0 = \frac{69}{3 \times 10^8} = 23 \times 10^{-8} \, \text{T} = 2.3 \times 10^{-7} \, \text{T}

The wave term is of the form

kx - \omega t

so the wave propagates in the $+x$ direction. The electric field is along $y$ . Using

\vec{E} \times \vec{B} \parallel \vec{k}

we need

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

Therefore the magnetic field is along the $+z$ direction.

The electric and magnetic fields are in phase, so the sinusoidal factor remains the same. Hence

B_z = 2.3 \times 10^{-7} \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t] \; \text{T}

Therefore, the correct option is A.

Quick direction-and-amplitude check

Use the cyclic relation

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

and rewrite it as

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

Since propagation is along $+x$ and the electric field is along $+y$ , the magnetic field must be along $+z$ .

For amplitude, use

B_0 = \frac{E_0}{c} = \frac{69}{3 \times 10^8} = 2.3 \times 10^{-7} \; \text{T}

Keep the same phase factor as the electric field. So the field is

B_z = 2.3 \times 10^{-7} \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t] \; \text{T}

Hence the correct option is A.

Common mistakes

Choosing $B_y$ is incorrect because in an electromagnetic wave, $\vec{E}$ , $\vec{B}$ , and the propagation direction are mutually perpendicular. Since $\vec{E}$ is along $y$ , $\vec{B}$ cannot also be along $y$ . Use $\vec{E} \times \vec{B}$ to determine the correct perpendicular direction.
Changing the sign in the sinusoidal term to $kx + \omega t$ is wrong here because $kx - \omega t$ represents propagation in the $+x$ direction. The magnetic field remains in phase with the electric field, so the same sinusoidal part must be retained.
Using $B_0 = E_0 c$ instead of $B_0 = \frac{E_0}{c}$ gives a completely incorrect magnitude. For electromagnetic waves in vacuum, divide the electric-field amplitude by the speed of light to get the magnetic-field amplitude.

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Answer

Step-by-step solution

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