The magnetic field of a plane electromagnetic wave is given by :
$$ \overrightarrow{\mathrm{B}}=2 \times 10^{-8} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \,\hat{j} \mathrm{~T}$$.
The amplitude of the electric field would be :
Solution
<p>Speed of light $$c = {\omega \over k} = {{1.5 \times {{10}^{11}}} \over {0.5 \times {{10}^3}}} = 3 \times {10^8}$$ m/sec</p>
<p>So, ${E_0} = {B_0}c$</p>
<p>$= 2 \times {10^{ - 8}} \times 3 \times {10^8}$</p>
<p>$= 6$ V/m</p>
<p>Direction will be along z-axis.</p>
About this question
Subject: Physics · Chapter: Electromagnetic Waves · Topic: Maxwell's Equations
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