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NEET PHYSICSEasy

In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0×10102.0 \times 10^{10} Hz and amplitude 4848 V m−1^{-1}. Then the amplitude of oscillating magnetic field is (Speed of light in free space = 3×1083 \times 10^8 m s−1^{-1})

A

1.6×10−91.6 \times 10^{-9} T

B

1.6×10−81.6 \times 10^{-8} T

C

1.6×10−71.6 \times 10^{-7} T

D

1.6×10−61.6 \times 10^{-6} T

Step-by-Step Solution

The relation between electric field amplitude E0E_0 and magnetic field amplitude B0B_0 is B0=E0cB_0 = \frac{E_0}{c}. Given E0=48E_0 = 48 V m−1^{-1} and c=3×108c = 3 \times 10^8 m s−1^{-1}, B0=483×108=16×10−8=1.6×10−7B_0 = \frac{48}{3 \times 10^8} = 16 \times 10^{-8} = 1.6 \times 10^{-7} T.

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