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Physics Question on Electromagnetic Spectrum

The oscillating magnetic field in a plane electromagnetic wave is given by By = 5×10–65 × 10^{–6} sin  1000  π(5x–4×108  t)Tsin\;1000\;π(5x – 4 × 108\;t)T. The amplitude of electric field will be:

A

15×10215 × 10^2 Vm–1Vm^{–1}

B

5×10–65 × 10^{–6} Vm–1Vm^{–1}

C

16×1012  Vm–116 × 10^{12} \;Vm^{–1}

D

4×1024 × 10^2 Vm–1Vm^{–1}

Answer

4×1024 × 10^2 Vm–1Vm^{–1}

Explanation

Solution

We know that the Speed of light : c=ωkc = \frac{ω }{ k}

= 4×1085=0.8×108\frac{4 × 10^8}{5} = 0.8 × 10^8 m/secm/sec

Therefore, E0=cB0E_0 = cB_0

=0.8×108×5×10–6 0.8 × 10^8 × 5 × 10^{–6}

= 400400 V/mV/m

= 4×10–24 × 10^{–2}

Hence, the correct option is (D): 4×10–2  Vm−14 × 10^{–2}\; Vm^{-1}