Discussion
An oscillating magnetic dipole is formed by making sinusoidal current flow through a circular loop. Consider this dipole as a perfect dipole. If peak value of current flowing is doubled, then time averaged power radiated by this dipole becomes ............. times the initial power.
Answer: A
The radiated power becomes 4 times the initial power.
The time-averaged power of an oscillating magnetic dipole of moment amplitude \(m_{0}\) at angular frequency \(\omega\) is
\[ \begin{aligned} \langle P \rangle &= \frac{\mu_{0}m_{0}^{2}\omega^{4}}{12\pi c^{3}} \\ m_{0} &= I_{0}\pi b^{2} \end{aligned} \]
The moment is proportional to the current amplitude, so the power goes as \(I_{0}^{2}\). Doubling \(I_{0}\) at the same frequency and loop size multiplies the power by \(2^{2} = 4\).
16 is what one gets by confusing the current with the frequency, which enters as \(\omega^{4}\); the fractions 1/4 and 1/16 would mean the power falls when the current rises.
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