Discussion
The induced surface current density \(\vec{K}\) on a conducting sphere, in a uniform background magnetic field, is \(\vec{K} = \left(\frac{3}{2}\right)\frac{(\vec{B} \times \hat{n})}{\mu_{0}}\). Where \(\hat{n}\) is the outward normal on the surface of the sphere. Now consider a current I flows in a long, straight wire placed along the z-axis. A conducting sphere of radius R is placed at a distance d from the wire (\(R \ll d\)). The induced surface current density on the sphere is proportional to ..........
Answer: C
The induced surface current density is proportional to \(I/d\).
The given result says \(K = \frac{3}{2}(\vec{B}\times\hat{n})/\mu_{0}\), so \(K\) is proportional to the field the sphere sits in. Since \(R \ll d\), the field of the wire is essentially uniform across the sphere and equal to its value at distance \(d\):
\[ \begin{aligned} B &= \frac{\mu_{0}I}{2\pi d} \\ K &\sim \frac{3}{2}\cdot\frac{I}{2\pi d} \propto \frac{I}{d} \end{aligned} \]
The radius \(R\) does not appear in the formula for \(K\), so \(I/R\) and \(I/R^{2}\) are wrong; \(1/d^{2}\) would be the fall-off of a dipole-like field, not of a long straight wire, whose field falls as \(1/d\).
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