A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-direction. At a particular point in space and time, B = 8.0 × 10 - 8 z T. The value of electric field at this point is: (speed of light = 3 × 108 ms ⁻¹) x y z are unit vectors along x, y and z directions. (a) 2.6 x V/m (b) - 24 x V/m (c) 24 x V/m (d) - 2.6 y V/m

Answer: (b). Magnitude: E = cB = 3× 10⁸× 8.0× 10⁻⁸ = 24 V/m. Direction: E = y. Since x = -y, E = -x. So the answer is (b) -24x V/m.

A plane electromagnetic wave of frequency 500 MHz is travelling in… | JEE Main 2021 (16 Mar, Shift 1)

Q44JEE Main 2021 · 16 MAR · SHIFT 1Easy

Nature of EM Waves & Wave Equation

A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-direction. At a particular point in space and time,

$\vec{B}$ = 8.0 $\times$ 10 $-$ 8 $\hat{z}$ T. The value of electric field at this point is:

(speed of light = 3 $\times$ 108 ms ${}^{-1}$ )

$\hat{x}\ \hat{y}\ \hat{z}$ are unit vectors along x, y and z directions.

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Answer: (b)

Magnitude: $E = cB = 3\times 10^{8}\times 8.0\times 10^{-8} = 24\ \text{V/m}$ .
Direction: $\hat{E}\times\hat{z} = \hat{y}$ . Since $\hat{x}\times\hat{z} = -\hat{y}$ , $\hat{E} = -\hat{x}$ .
So the answer is (b) $-24\hat{x}$ V/m.