Electromagnetic Waves

JEE Main level, single correct, medium

15 of 165

Question 15 of 165

A laser beam has intensity of 4.0×1014 W/m24.0 \times 10^{14} \mathrm{~W} / \mathrm{m}^{2}. The amplitude of magnetic field associated with beam is ____\_\_\_\_ T.
(Take ϵo=8.85×10−12C2/Nm2\epsilon_{\mathrm{o}}=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2} and c=3×108 m/s\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s})

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Answer: (B) 1.83

The intensity of an electromagnetic wave is
I=12ϵ0cE02I = \frac{1}{2} \epsilon_0 c E_0^2.
From the relation E0=cB0E_0 = c B_0, we get I=12ϵ0c3B02I = \frac{1}{2} \epsilon_0 c^3 B_0^2.
Solving for magnetic field amplitude:
B0=2Iϵ0c3=2×4.0×10148.85×10−12×(3×108)3=8.0×10142.39×1014B_0 = \sqrt{\frac{2I}{\epsilon_0 c^3}} = \sqrt{\frac{2 \times 4.0 \times 10^{14}}{8.85 \times 10^{-12} \times (3 \times 10^8)^3}} = \sqrt{\frac{8.0 \times 10^{14}}{2.39 \times 10^{14}}}
=3.347=1.83= \sqrt{3.347} = 1.83 T.

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