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]

Question

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]

Smart Achievers · Accepted Answer

(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.

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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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, ∮B→·dl→=μ0×current losed by path ⇒B·2πr=μ0×Ir2a2 (∵R=a) ⇒B=μ0Ir2πa2 (for r<a) At surface r=a, so B=μ0I2πa ∴ B=μ0I2πr (for r>a) The variation of magnetic field with distance 'r' from the axis is given by the above figure.

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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]