A cylindrical, straight long wire having radius 'a', is carrying current 'I' uniformly distributed inside the wire. As per Ampere's circuital law, the magnetic field inside the wire at any distance r, (r < a) is given by :

$B =μ_0Ir/2π а^2$

The correct answer is Option (3) → $B =μ_0Ir/2π а^2$

For a long straight cylindrical wire carrying current I uniformly, Ampere's law gives:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$

Inside the wire ($r < a$), current enclosed by radius $r$:

$I_{\text{enclosed}} = I \frac{\pi r^2}{\pi a^2} = I \frac{r^2}{a^2}$

Magnetic field along circular path of radius $r$:

$B (2 \pi r) = \mu_0 I_{\text{enclosed}} = \mu_0 I \frac{r^2}{a^2}$

So,

$B = \frac{\mu_0 I r}{2 \pi a^2}$

Magnetic field inside the wire: $B = \frac{\mu_0 I r}{2 \pi a^2}$