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JEE PHYSICS MCQ: ELECTROMAGNETIC WAVES

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Application +4 | -1
The magnetic field of a beam emerging from a filter facing a floodlight is given by

B0=12×10−8sin⁡(1.20×107z−3.60×1015t)T\mathrm{B}_0=12 \times 10^{-8} \sin ( 1.20 \times 10^7 \mathrm{z} \left.-3.60 \times 10^{15} \mathrm{t}\right) \mathrm{T}.

What is the average intensity of the beam?
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Correct option: A
1.71
Option A (1.71) is correct because it is calculated using the formula for the average intensity of an electromagnetic wave.

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