A 60-W light bulb emits spherical electromagnetic waves uniformly in all directions.If 50% of the power input to such a light bulb is emitted as electromagnetic radiation,what is the maximum value of the magnetic field at a distance of 2.0 m from the light bulb?
A)71 nT
B)0.35 $\mu$T
C)0.14 $\mu$T
D)0.20 $\mu$T
E)0.10 $\mu$T

Question

Quizplus · Accepted Answer

The maximum value of the magnetic field (B) at a distance (r) from a point source emitting electromagnetic waves can be calculated using the formula derived from the Poynting vector and the relation between electric and magnetic fields in electromagnetic waves. Given that 50% of the power (P) is emitted as electromagnetic radiation, the effective power is $P_{\text{eff}} = 0.5 \times 60\, \text{W} = 30\, \text{W}$. The intensity (I) of the electromagnetic wave at a distance r is given by $I = \frac{P_{\text{eff}}}{4\pi r^2}$. The magnetic field B can be related to the intensity by $I = \frac{1}{2} c \epsilon_0 B^2$, where $c$ is the speed of light and $\epsilon_0$ is the permittivity of free space. Solving for B with the given values leads to the correct answer, which is in the order of nanoteslas (nT), making option A the correct choice.

A 60-W Light Bulb Emits Spherical Electromagnetic Waves Uniformly in All