JEE MAIN - Physics (2025 - 24th January Evening Shift - No. 9)
A long straight wire of a circular cross-section with radius ' a ' carries a steady current I . The current I is uniformly distributed across this cross-section. The plot of magnitude of magnetic field B with distance $r$ from the centre of the wire is given by
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Explanation
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$$\begin{aligned} & \mathrm{B}_{\text {in }}=\frac{\mu_0(\overline{\mathrm{~J}} \times \overline{\mathrm{r}})}{2} \\ & \mathrm{~B}_{\text {in }} \propto \mathrm{r} \\ & \mathrm{~B}_{\text {out }}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \\ & \mathrm{~B}_{\text {net }} \propto \frac{1}{\mathrm{r}} \end{aligned}$$
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