Q131 / 200 · NEET 2022 (Phase 1) · Physics

From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is

A
a linearly increasing function of distance upto the boundary of the wire and then linearly decreasing for the outside region.
Correct answer
B
a linearly increasing function of distance $r$ upto the boundary of the wire and then decreasing one with $\frac{1}{r}$ dependence for the outside region.
C
a linearly decreasing function of distance upto the boundary of the wire and then a linearly increasing one for the outside region
D
uniform and remains constant for both the regions.

Solution

If the current density in the wire is $J$ then

For $r < R$

$\oint \vec{B} \cdot d \vec{l} = μ_{0} i$

$\Rightarrow B 2 π r = μ_{0} J π r^{2}$

$\Rightarrow B = \frac{μ_{0} J}{2} r$

$\Rightarrow B \propto r$

For $r > R$

$\oint \vec{B} \cdot d \vec{l} = μ_{0} \vec{i}$

$\Rightarrow B 2 π r = μ_{0} π R^{2} J$

$\Rightarrow B = \frac{μ_{0} R^{2} J}{2 r}$

$\Rightarrow B \propto \frac{1}{r}$

Hence, the correct answer is option: 'B'.