Question

Choose the correct answer from the alternatives given.
Assume a bulb of efficiency 2.5% as a point source. The peak values of electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m is respectively

• A. $2.5V{m}^1,3.6\times {10}^{-8}T$
• B. $4.2V{m}^1,2.8\times {10}^{-8}T$
• C. $4.08V{m}^1,1.36\times {10}^{-8}T$
• D. $3.6V{m}^1,3.6\times {10}^{-8}T$

Answer 1

Answer: (C)

$4.08V{m}^1,1.36\times {10}^{-8}T$

Here intensity, $I=\frac{power}{area}$

= $\frac{100\times 2.5}{4\pi (3{)}^2\times 100}=\frac{2.5}{36\pi }W{m}^{-2}$

Half of this intensity belongs to electronic field and half of that to magnetic field.

$\therefore \frac{I}{2}=\frac{1}{4}{\epsilon }_0{E}_0^{2}c$

or

$E_o=\sqrt{\frac{2I}{{ϵ}_{0}c}}\Rightarrow \sqrt{\frac{2\times \frac{2.5}{36\pi }}{\frac{1}{4\pi \times 9\times {10}^9}}\times 3\times {10}^8}\Rightarrow 4.08V{m}^{-i}$

$\therefore B_o=\frac{E_0}{c}=\frac{4.08}{3\times {10}^8}=1.36\times {10}^{-8}T$