1 markMCQSection A

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

Q: 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

**Correct option: C — $\mu_0 I/(4\pi a)$.** - Inside a uniformly current-carrying wire, Ampere's law gives $B = \mu_0 Ir/(2\pi a^2)$ for $r < a$. - At $r = a/2$: $B = \mu_0 I(a/2)/(2\pi a^2)$ = **$\mu_0 I/(4\pi a)$**.

Q: Why isn't the field $\mu_0 I/2\pi r$ at the point $r = a/2$?

That formula applies only outside the wire where all the current I is enclosed. Inside, only the fraction of current within radius r contributes, namely $I_{enc} = I r^2/a^2$, which changes the field to $B = \mu_0 I r/(2\pi a^2)$.

Q: Where is the magnetic field maximum for a solid current-carrying wire?

At the surface, $r = a$, where $B = \mu_0 I/(2\pi a)$. It increases linearly from zero on the axis to this maximum, then decreases as $1/r$ outside the wire.

Moving Charges and Magnetism

Magnetic field inside a current-carrying wire (Ampere's law)

Options

(A)Zero

(B)

\frac{\mu_0 I}{2\pi a}

(C)

\frac{\mu_0 I}{4\pi a}

(D)

\frac{\mu_0 I}{6\pi a}

Official Answer

Correct option: C — $\mu_0 I/(4\pi a)$ .

Inside a uniformly current-carrying wire, Ampere's law gives $B = \mu_0 Ir/(2\pi a^2)$ for $r < a$ .
At $r = a/2$ : $B = \mu_0 I(a/2)/(2\pi a^2)$ = $\mu_0 I/(4\pi a)$ .

Ampere's circuital lawmagnetic field inside wireuniform current densityenclosed currentB proportional to r insidesolid conductor fieldfield at half radius

Marking Scheme

11 mark: correct option C ( $\mu_0 I/4\pi a$ ).
2Reasoning credited: use of enclosed current $I r^2/a^2$ and Ampere's law giving $B \propto r$ inside the wire.

Hint

The point is inside the wire ( $a/2 < a$ ). Only the current within radius $a/2$ is enclosed; use $B = \mu_0 I_{enc}/(2\pi r)$ .

Quick Oral Answer

Inside a uniform wire only the current within radius r is enclosed, giving $B = \mu_0 Ir/2\pi a^2$ ; at $r = a/2$ this is $\mu_0 I/4\pi a$ , exactly half the surface field.

Analysis & Explanation

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 $B \propto r$ .

Common Mistakes

1Using the external formula $B = \mu_0 I/(2\pi r)$ and forgetting that only part of the current is enclosed inside the wire.
2Assuming the field is zero everywhere inside the wire (it is zero only on the axis).
3Substituting $r = a$ (surface) instead of $r = a/2$ .

Interesting Facts

Inside a uniform wire the field rises linearly to a maximum of $\mu_0 I/(2\pi a)$ at the surface, then falls as $1/r$ outside — a continuous, tent-shaped profile.

This linear interior field is the magnetic analogue of the electric field inside a uniformly charged solid sphere, both a direct consequence of the enclosed-source law.

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Frequently Asked Questions

Why isn't the field $\mu_0 I/2\pi r$ at the point $r = a/2$ ?

That formula applies only outside the wire where all the current I is enclosed. Inside, only the fraction of current within radius r contributes, namely $I_{enc} = I r^2/a^2$ , which changes the field to $B = \mu_0 I r/(2\pi a^2)$ .

Where is the magnetic field maximum for a solid current-carrying wire?

At the surface, $r = a$ , where $B = \mu_0 I/(2\pi a)$ . It increases linearly from zero on the axis to this maximum, then decreases as $1/r$ outside the wire.