JEE MAIN - Physics (2022 - 25th June Morning Shift - No. 12)
The electric field in an electromagnetic wave is given by E = 56.5 sin $$\omega$$(t $$-$$ x/c) NC$$-$$1. Find the intensity of the wave if it is propagating along x-axis in the free space.
(Given : $$\varepsilon $$0 = 8.85 $$\times$$ 10$$-$$12C2N$$-$$1m$$-$$2)
5.65 Wm$$-$$2
4.24 Wm$$-$$2
1.9 $$\times$$ 10$$-$$7 Wm$$-$$2
56.5 Wm$$-$$2
Explanation
$$I = {1 \over 2}{\varepsilon _0}E_0^2c$$
$$ = {1 \over 2} 8.5 \times {10^{ - 12}} \times {(56.5)^2} \times 3 \times {10^8}$$
$$ = 4.24$$ W/m2
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