A long straight wire of circular cross-section (radius a) carries a steady current I. The current is uniformly distributed across this cross-section. The magnitude of the magnetic field produced at a point at a distance $(a/2)$ from the axis of the wire will be

This tests Ampere's circuital law applied inside a solid conductor with uniform current density.

Concept

• Current density $J = I/(\pi a^2)$. The current enclosed within radius r (< a) is $I_{enc} = J\cdot\pi r^2 = I r^2/a^2$.
• Ampere's law $B\cdot(2\pi r) = \mu_0 I_{enc}$ gives $B = \mu_0 I r/(2\pi a^2)$, so inside the wire B grows linearly with r.
• Substituting $r = a/2$: $B = \mu_0 I(a/2)/(2\pi a^2) = \mu_0 I/(4\pi a)$.

Why the distractors are wrong

• A (Zero): the field is zero only on the axis ($r = 0$), not at $r = a/2$.
• B ($\mu_0 I/2\pi a$): this is the field at the surface $r = a$; it is the maximum value, twice the correct answer here.
• D ($\mu_0 I/6\pi a$): corresponds to $r = a/3$, not $a/2$ — an arithmetic slip.

Exam trap

• Students reflexively use $B = \mu_0 I/(2\pi r)$ (valid only OUTSIDE the wire). Inside, the enclosed current is reduced by the factor $r^2/a^2$, changing the dependence to $B \propto r$.