In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0×1010 Hz and amplitude 48 Vm-1. Then the amplitude of oscillating magnetic field is: ( Speed of light in free space = 3×108 ms-1 )

We know, in an EM wave EoBo=cB0=E0CgIven, E0=48Vm−1B0=483×108=16×10−8=1.6×10−7T

In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0×1010 Hz and amplitude 48 Vm-1. Then the amplitude of oscillating magnetic field is: ( Speed of light in free space = 3×108 ms-1 )

We know, in an EM wave EoBo=cB0=E0CgIven, E0=48Vm−1B0=483×108=16×10−8=1.6×10−7T

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

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

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

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

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

answer is D.

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Detailed Solution

We know, in an EM wave $\frac{E_{o}}{B_{o}} = c$
$B_{0} = \frac{E_{0}}{C}$
gIven, $E_{0} = 48 {Vm}^{- 1}$
$B_{0} = \frac{48}{3 \times 10^{8}} = 16 \times 10^{- 8} = 1.6 \times 10^{- 7} T$

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In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0×1010 Hz and amplitude 48 Vm-1. Then the amplitude of oscillating magnetic field is: ( Speed of light in free space = 3×108 ms-1 )

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