Question

a 60 - w light bulb radiates electromagnetic waves uniformly in all directions. at a distance of 1.0 m from the bulb, the light intensity is ( i_0 ), the average energy density of the waves is ( u_0 ), and the rms electric and magnetic field values are ( e_0 ) and ( b_0 ), respectively.
part c
at 2.0 m from the bulb, what is the rms magnetic field value?
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( b_0 )
( sqrt{\frac{1}{2}}b_0 )
( \frac{1}{2}b_0 )
( \frac{1}{4}b_0 )
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although the average energy density has dropped by a factor of 4, the rms magnetic field has not decreased by the same amount

Explanation:

Step1: Recall the formula for intensity

The intensity \(I=\frac{P}{4\pi r^{2}}\), where \(P\) is the power of the source and \(r\) is the distance from the source.

Step2: Relate intensity and magnetic field

The intensity \(I = \frac{E_{rms}^{2}}{\mu_{0}c}=\frac{(cB_{rms})^{2}}{\mu_{0}c}=\frac{cB_{rms}^{2}}{\mu_{0}}\) (since \(E = cB\)). Also, \(I\propto\frac{1}{r^{2}}\) and \(B_{rms}\propto\frac{1}{r}\) (because \(I\propto B_{rms}^{2}\) and \(I\propto\frac{1}{r^{2}}\)).

Step3: Calculate the ratio of magnetic fields

Let \(B_{1}\) be the magnetic field at \(r_{1} = 1m\) and \(B_{2}\) be the magnetic field at \(r_{2}=2m\). Using \(B\propto\frac{1}{r}\), we have \(\frac{B_{2}}{B_{1}}=\frac{r_{1}}{r_{2}}\). Substituting \(r_{1} = 1m\) and \(r_{2}=2m\), we get \(B_{2}=\frac{1}{2}B_{1}\).

Answer:

\(\frac{1}{2}B_{0}\)