8.Electromagnetic waves
easy

The magnetic field of a beam emerging from a filter facing a floodlight is given by B${B_0} = 12 \times {10^{ - 8}}\,\sin \,(1.20 \times {10^7}\,z - 3.60 \times {10^{15}}t)T$. What is the average intensity of the beam ?

Option A
Option B
Option C
Option D

Solution

Comparing $\mathrm{B}=12 \times 10^{-8} \sin \left(1.20 \times 10^{7} z-3.60 \times 10^{15} t\right)$ with equation $\mathrm{B}=\mathrm{B}_{0} \sin \omega t$ $\mathrm{B}_{0}=12 \times 10^{-8} \mathrm{~T}$

Average intensity of beam,

$\mathrm{I}_{\text {average }}=\frac{\mathrm{B}_{0}^{2}}{2 \mu_{0}} \mathrm{C}$

$=\frac{1}{2} \times \frac{\left(12 \times 10^{-8}\right)^{2} \times 3 \times 10^{8}}{4 \times 3.14 \times 10^{-7}}$

$\therefore I_{\text {average }} =1.71 \mathrm{~W} / \mathrm{m}^{2}$

Standard 12
Physics

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