A thick current carrying cable of radius carries current uniformly distributed across its cross-section. The variation of magnetic field due to the cable with the distance from the axis of the cable is represented by [2021]

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Solution

Q.
A thick current carrying cable of radius $‘ R ’$ carries current $‘ I ’$ uniformly distributed across its cross-section. The variation of magnetic field $B (r)$ due to the cable with the distance $‘ r ’$ from the axis of the cable is represented by [2021]

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Ans.
(4)

By Ampere's circuital law,

$\oint \vec{B} \cdot d \vec{l} = μ_{0} \times current losed by path$

$\Rightarrow B \cdot 2 π r = μ_{0} \times \frac{I r^{2}}{a^{2}} (∵ R = a)$

$\Rightarrow B = \frac{μ_{0} I r}{2 π a^{2}} (for r < a)$

At surface $r = a,$ so $B = \frac{μ_{0} I}{2 π a}$

$∴$ $B = \frac{μ_{0} I}{2 π r} (for r > a)$

The variation of magnetic field with distance $' r'$ from the axis is given by the above figure.

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