From Ampere's circuital law for a long straight wire of circular...

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

Options

Aa linearly increasing function of distance upto the boundary of the wire and then linearly decreasing for the outside region.
Ba linearly increasing function of distance $r$ upto the boundary of the wire and then decreasing one with $\dfrac{1}{r}$ dependence for the outside region.
Ca linearly decreasing function of distance upto the boundary of the wire and then a linearly increasing one for the outside region
Duniform and remains constant for both the regions.

Correct answer

B. a linearly increasing function of distance $r$ upto the boundary of the wire and then decreasing one with $\dfrac{1}{r}$ dependence for the outside region.

Step-by-step solution

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

For $r < R$

$∮ \overset{\to}{B} \cdot d\overset{\to}{l} = {\mu}_{0}i$

$⇒ B2\pir = {\mu}_{0}J\pi{r}^{2}$

$⇒ B = \dfrac{{\mu}_{0}J}{2}r$

$⇒ B ∝ r$

For $r > R$

$∮ \overset{\to}{B} \cdot d\overset{\to}{l} = {\mu}_{0}\overset{\to}{i}$

$⇒ B2\pir = {\mu}_{0}\pi{R}^{2}J$

$⇒ B = \dfrac{{\mu}_{0}{R}^{2}J}{2r}$

$⇒ B ∝ \dfrac{1}{r}$

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