JEE Main · PhysicsHard

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

Show correct answer & step-by-step solution

Correct answer: A — $2.1 \times 1 0^{- 8} T along the z -direction$

Solution

$Given the electric field E = 6.3 \hat{j} V/m and the wave propagation direction \overset{n}{^} = \hat{i}, the magnitude of the magnetic field is B = E / c .$
$Substituting the values E = 6.3 V/m and c = 3 \times 1 0^{8} m/s yields B = 6.3/ (3 \times 1 0^{8}) = 2.1 \times 1 0^{- 8} T .$
$The direction of the magnetic field is given by the unit vector \hat{B} = \overset{n}{^} \times \hat{E} = \hat{i} \times \hat{j} = \hat{k}, which is the positive z -direction.$
$Hence the answer is (A).$