Q 12-08-121JEE MainJEE Main 2022 (24 Jun, Shift 2)Medium
An electric bulb is rated as $200\ \text{W}$. What will be the peak magnetic field at $4\ \text{m}$ distance produced by the radiation coming from this bulb? Consider this bulb as a point source with $3.5\%$ efficiency.
Answer: (B) $1.71\times10^{-8}\ \text{T}$
Radiated power $= 0.035\times200 = 7\ \text{W}$. Intensity at $4\ \text{m}$:
$$I = \frac{7}{4\pi(4)^2} = 0.0348\ \text{W m}^{-2}$$
Using $I = \dfrac{cB_0^2}{2\mu_0}$:
$$B_0 = \sqrt{\frac{2\mu_0 I}{c}} = \sqrt{\frac{2\times4\pi\times10^{-7}\times0.0348}{3\times10^8}} \approx 1.71\times10^{-8}\ \text{T}$$
Solution by Sreeraj P, M.Sc Physics