JEE Main 2025 Jan 23 Shift 2, Physics Q17: Electromagnetic Waves

JEE Main2025Jan 23, Shift 2Physics

A plane electromagnetic wave of frequency $20$ MHz travels in free space along the $+ x$ direction. At a particular point in space and time, the electric field vector of the wave is $E_{y} = 9.3 V m^{- 1}$ . Then, the magnetic field vector of the wave at that point is :

A
$B_{z} = 9.3 \times 1 0^{- 8} T$
B
$B_{z} = 1.55 \times 1 0^{- 8} T$
C
$B_{z} = 6.2 \times 1 0^{- 8} T$
D
$B_{z} = 3.1 \times 1 0^{- 8} T$

Solution

For a plane EM wave in vacuum, $E$ and $B$ are in phase with $E = c B$ , and $\hat{E} \times \hat{B}$ points along the propagation direction.

Propagation along $+ \overset{x}{^}$ with $E$ along $+ \overset{y}{^}$ requires $B$ along $+ \overset{z}{^}$ (so $\overset{y}{^} \times \overset{z}{^} = \overset{x}{^}$ ).

$B_{z} = \frac{E _{y}}{c} = \frac{9.3}{3 \times 1 0 ^{8}} = 3.1 \times 1 0^{- 8} T .$

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JEE Main 2025 Jan 23 Shift 2, Physics Q17: Electromagnetic Waves

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