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Question

(B) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around a closed path is equal to $\mu_{0}$ times the current

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$B \propto r$

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(B) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around a closed path is equal to $\mu_{0}$ times the current enclosed by the path: $\oint \vec{B} \cdot d\vec{l} = \mu_{0} I_{	ext{enclosed}}$.For a point at a distance $r The current density $J$ is uniform,so $J = \frac{I}{\pi R^{2}}$.The current enclosed by the loop of radius $r$ is $I_{	ext{enclosed}} = J \cdot (\pi r^{2}) = \frac{I}{\pi R^{2}} \cdot \pi r^{2} = I \frac{r^{2}}{R^{2}}$.Applying Ampere's law: $B(2\pi r) = \mu_{0} \left( I \frac{r^{2}}{R^{2}} ight)$.Solving for $B$,we get $B = \frac{\mu_{0} I r}{2\pi R^{2}}$.Since $\mu_{0}$,$I$,and $R$ are constants,we have $B \propto r$.

$A$ long straight wire with a circular cross-section having radius $R$ is carrying a steady current $I$. The current $I$ is uniformly distributed across this cross-section. Then the variation of magnetic field due to current $I$ with distance $r$ $(r < R)$ from its centre will be:

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