Question

The magnetic field of a plane electromagnetic wave is given by
$\vec{ B }=2 \times 10^{-8} \sin \left(0.5 \times 10^3 x +1.5 \times 10^{11} t \right) \hat{ j } T$
The amplitude of the electric field would be

• A. $6\, Vm ^{-1}$ along $x$-axis
• B. $3\, Vm ^{-1}$ along z-axis
• C. $6\, Vm ^{-1}$ along z-axis
• D. $2 \times 10^{-8} Vm ^{-1}$ along z-axis

Answer 1

Answer: (C)

$c =\frac{ E _0}{ B _0} \Rightarrow E _0= cB _0 $
$ E _0=\left(3 \times 10^8\right)\left(2 \times 10^{-8}\right) $
$ E _0=6 \,Vm ^{-1}$
As, $\vec{ B }=$ along y-axis $\vec{ v }=$ along negative $x$-axis hence $ \vec{ E }_0=$ along z-axis