Question

33.2 energy transport and the poynting vector
a detector on your space probe happens to measure ( b_m ) to be 0.00000239 t. knowing that the power output of the sun is ( p_s = 3.85\times10^{26}\text{ w} ), how far away is your space probe from the sun?
( 6.36\text{e}+11\text{ m} )
( 1.06\text{e}+11\text{ m} )
( 2.12\text{e}+11\text{ m} )
( 2.82\text{e}+11\text{ m} )

Explanation:

Step1: Find the rms value of the magnetic field

The relationship between the maximum value \(B_{m}\) and the rms value \(B_{rms}\) of a sinusoidal - varying magnetic field is \(B_{rms}=\frac{B_{m}}{\sqrt{2}}\). Given \(B_{m} = 2.39\times10^{-6}\text{ T}\), then \(B_{rms}=\frac{2.39\times 10^{-6}}{\sqrt{2}}\text{ T}\approx1.7\ \times10^{-6}\text{ T}\).

Step2: Relate the intensity \(I\) to the magnetic field

The intensity of an electromagnetic wave \(I = \frac{cB_{rms}^{2}}{\mu_{0}}\), where \(c = 3\times10^{8}\text{ m/s}\) and \(\mu_{0}=4\pi\times 10^{-7}\text{ T}\cdot\text{m/A}\).
Substitute the values: \(I=\frac{3\times 10^{8}\times(1.7\times 10^{-6})^{2}}{4\pi\times 10^{-7}}\).

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Step3: Use the formula for intensity in terms of power

The intensity \(I=\frac{P}{4\pi r^{2}}\), where \(P = P_{s}=3.85\times 10^{26}\text{ W}\).
Rearrange for \(r\): \(r=\sqrt{\frac{P}{4\pi I}}\).
Substitute \(P = 3.85\times 10^{26}\text{ W}\) and \(I = 69.1\text{ W/m}^2\)

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Answer:

\(1.06\text{E}+11\text{ m}\)