The current density inside a long, solid cylindrical wire of radius $a$ is parallel to the central axis and varies linearly with radial distance $r$ from the axis according to $J = J_{0} \frac{r}{a}$ . If the magnetic field inside the wire at a radial distance r is $\frac{μ_{0} J_{0} r^{2}}{k a}, f i n d k .$

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answer is B.

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Detailed Solution

Using Ampere's Law,

$\oint \vec{B} . \vec{d l} = μ_{0} i_{i n} o r B \times 2 πr = μ_{0} \int_{0}^{r} J d A = μ_{0} \int_{0}^{r} (J_{0} \frac{r}{a}) \times 2 πr d r ∴ B = \frac{μ_{0} J_{0} r^{2}}{3 a}$

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