In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of $2.0\times10^{10}\ Hz$ and amplitude $48\ Vm^{-1}$. Then the amplitude of oscillating magnetic field is: (Speed of light in free space $=3\times10^8\ ms^{-1}$)

A $1.6\times10^{-6}\ T$

B $1.6\times10^{-9}\ T$

C $1.6\times10^{-8}\ T$

D $1.6\times10^{-7}\ T$

The answer is D.

Solution

$B_0=\frac{E_0}{c}=1.6\times10^{-7}\ T$.

$B=\frac{E}{C}=\frac{48}{3\times10^8}=16\times10^{-8}=1.6\times10^{-7}\ T$

Concept B = E/c

49 questions from this chapter have appeared across 24 exam years.