NEET 2023 — Physics solution

NEET2023Physics

In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of $2.0 \times 1 0^{10}$ Hz and amplitude $48$ Vm $^{- 1}$ . Then the amplitude of oscillating magnetic field is: (Speed of light in free space $= 3 \times 1 0^{8}$ m s $^{- 1}$ )

A
$1.6 \times 1 0^{- 8}$ T
B
$1.6 \times 1 0^{- 7}$ T
C
$1.6 \times 1 0^{- 6}$ T
D
$1.6 \times 1 0^{- 9}$ T

Solution

For an EM wave in vacuum, $\frac{E _{0}}{B _{0}} = c$ , so $B_{0} = \frac{E _{0}}{c}$ .

$B_{0} = \frac{48}{3 \times 1 0 ^{8}} = 1.6 \times 1 0^{- 7}$ T.

(Frequency is irrelevant for this ratio.)

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