JEE Main · PhysicsHard

$A plane electromagnetic wave of frequency f = 50 MHz travels in free space along the positive x -direction. At a particular point and time, the electric field is E = 6.3 \hat{j} V/m . The magnetic field B at this point and time is:$

Show correct answer & step-by-step solution

Correct answer: A — $2.1 \times 1 0^{- 8} T \hat{k}$

Solution

$The magnitude of the magnetic field is given by B = \frac{E}{c} .$
$Substituting E = 6.3 V/m and c = 3 \times 1 0^{8} m/s, we get B = \frac{6.3}{3 \times 1 0 ^{8}} = 2.1 \times 1 0^{- 8} T .$
$The direction of propagation is \hat{i}, and the electric field is \hat{j} .$
$Since B = \frac{1}{c} (\hat{k} \times E), the direction is \hat{i} \times \hat{j} = \hat{k} .$
$Hence the answer is (A).$