In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of Hz and amplitude 48 . Then the amplitude of oscillating magnetic field is (Speed of light in free space = )

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Solution

Q.
In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of $2.0 \times 10^{10}$ Hz and amplitude 48 ${Vm}^{- 1}$ . Then the amplitude of oscillating magnetic field is
(Speed of light in free space = $3 \times 10^{8}$ $m$ $s^{- 1}$ ) [2023]

1 $1.6 \times 10^{- 7} T$

2 $1.6 \times 10^{- 6} T$

3 $1.6 \times 10^{- 9} T$

4 $1.6 \times 10^{- 8} T$

Ans.
(1)

Frequency of E.M. wave, $υ = 2.0 \times 10^{10} Hz$

Electric field amplitude, $E_{0} = 48$ $V$ $m^{- 1}$

Speed of light, $c = 3 \times 10^{8} m/s$

Magnetic field strength is given as:

$B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}} = 1.6 \times 10^{- 7} T$

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