Ncert Exercises & Solutions

The magnetic field of a beam emerging from filter facing a floodlight given by B₀ = 12 × 10^-8 sin (1.20 × 10⁷ z - 3.60 × 10¹⁵ t)T.
What is the average intensity of the beam?

Hint: The average intensity of the beam depends on the magnitude of the magnetic field.

Step 1: Find the magnitude of the magnetic field.

Magnetic field, B=

B_{0} sinωt

Given, equation B =

12 \times 10^{- 8} \sin (1.20 \times 10^{7} z - 3.60 \times 10^{15} t) T

On comparing this equation with the standard equation, we get;

B_{0} = 12 \times 10^{- 8} T

Step 2: Use the formula of average intensity.

The average intensity of the beam;

I_{av} = \frac{1}{2} \frac{B_{0}^{2}}{μ_{0}} . c = \frac{1}{2} \times \frac{{(12 \times 10^{- 8})}^{2} \times 3 \times 10^{8}}{4 π \times 10^{- 7}}

= 1.71 W / m^{2}

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