JEE MAIN - Physics (2022 - 26th July Morning Shift - No. 4)

The magnetic field of a plane electromagnetic wave is given by :

$$ \overrightarrow{\mathrm{B}}=2 \times 10^{-8} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \,\hat{j} \mathrm{~T}$$.

The amplitude of the electric field would be :

$$6\, \mathrm{Vm}^{-1}$$ along $$x$$-axis

$$3\, \mathrm{Vm}^{-1}$$ along $$z$$-axis

$$6\, \mathrm{Vm}^{-1}$$ along $$z$$-axis

$$2 \times 10^{-8} \,\mathrm{Vm}^{-1}$$ along $$z$$-axis

Speed of light $$c = {\omega \over k} = {{1.5 \times {{10}^{11}}} \over {0.5 \times {{10}^3}}} = 3 \times {10^8}$$ m/sec

So, $${E_0} = {B_0}c$$

$$ = 2 \times {10^{ - 8}} \times 3 \times {10^8}$$

$$ = 6$$ V/m

Direction will be along z-axis.