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Question 7.34: A fat wire, radius a, carries a constant current I, uniforml...

A fat wire, radius a, carries a constant current I, uniformly distributed over its cross section. A narrow gap in the wire, of width w \ll a, forms a parallel-plate capacitor, as shown in Fig. 7.45. Find the magnetic field in the gap, at a distance s < a from the axis.

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The displacement current density (Sect. 7.3.2) is  J _{d}=\epsilon_{0} \frac{\partial E }{\partial t}=\frac{I}{A}=\frac{I}{\pi a^{2}} \hat{ z} . Drawing an “amperian loop” at radius s, 

\oint B \cdot d l=B \cdot 2 \pi s=\mu_{0} I_{ d _{ enc }}=\mu_{0} \frac{I}{\pi a^{2}} \cdot \pi s^{2}=\mu_{0} I \frac{s^{2}}{a^{2}} \Rightarrow B=\frac{\mu_{0} I s^{2}}{2 \pi s a^{2}} ; \quad B =\frac{\mu_{0} I s}{2 \pi a^{2}} \hat{\phi} .

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