Question

An electric bulb is rated as $200\text{W}.$ What will be the peak magnetic field at $4\text{m}$ distance produced by the radiations coming from this bulb? Consider this bulb as a point source with $3.5\%$ efficiency.

• A. $1.71\times {10}^{-8}\text{T}$
• B. $1.19\times {10}^{-8}\text{T}$
• C. $3.36\times {10}^{-8}\text{T}$
• D. $0.84\times {10}^{-8}\text{T}$

Answer 1

The correct option is A $1.71\times {10}^{-8}\text{T}$

We know that,

Intensity, $I=\frac{Power}{Area}=\frac{B^2}{2{\mu }_0}c$

$\Rightarrow 200\times \frac{1}{4\pi \times 4^2}\times \frac{3.5}{100}=\frac{B_0^2}{2{\mu }_0}c$

As, ${\mu }_0=4\pi \times {10}^{-7}\text{Tm/A}$
$c=3\times {10}^8\text{m/sec}$

$\Rightarrow 200\times \frac{1}{4\pi \times 4^2}\times \frac{3.5}{100}=\frac{B_0^2}{2\times 4\pi \times {10}^{-7}}\times 3\times {10}^8$

$\Rightarrow B_0=1.71\times {10}^{-8}\text{T}$

Hence, $\left(B\right)$ is the correct answer.