A long straight wire of a circular cross-section of radius 'a' is carrying steady current I. The current I is uniformly distributed across this cross-section. What will be the magnetic field in the region $r < a$ and $r > a$?

$B∝r (r< a)$ and $B ∝ r^{-1} (r > a)$

The correct answer is Option (1) → $B∝r (r< a)$ and $B ∝ r^{-1} (r > a)$

Given:

Long straight wire of radius $a$, carrying steady current $I$, uniformly distributed.

Using Ampère’s circuital law:

$\oint \mathbf{B} \cdot d\mathbf{l} = μ_0 I_{enc}$

Case 1: For r < a (inside the wire)

Current enclosed, $I_{enc} = I \cdot \frac{πr^2}{πa^2} = I \cdot \frac{r^2}{a^2}$

$B(2πr) = μ_0 I_{enc}$

$B = \frac{μ_0 I r}{2π a^2}$

Magnetic field inside the wire: $B = \frac{μ_0 I r}{2π a^2}$

Case 2: For r > a (outside the wire)

Current enclosed, $I_{enc} = I$

$B(2πr) = μ_0 I$

$B = \frac{μ_0 I}{2π r}$

Magnetic field outside the wire: $B = \frac{μ_0 I}{2π r}$

Final Answer:

For $r < a$: $B = \frac{μ_0 I r}{2π a^2}$

For $r > a$: $B = \frac{μ_0 I}{2π r}$