Q.Figure 4.13 shows a long straight wire of a circular cross-section (radius $a$) carrying steady current $I$. The current $I$ is uniformly distributed across this cross-section. Calculate the magnetic field in the region $r < a$ and $r > a$.

Concept: Magnetic Field of a Straight Wire -- using Ampère's circuital law, the line integral of $\vec{B}$ around a closed loop equals $\mu_0$ times the current enclosed.

Step 1: Region $r > a$ (outside the wire)

Take a circular Amperian loop of radius $r$ concentric with the wire. The total current enclosed is $I$. By symmetry, $B$ is tangential and constant on the loop:

Thus,

Step 2: Region $r < a$ (inside the wire)

Current density $J = \frac{I}{\pi a^2}$ (uniform). For a loop of radius $r$, the enclosed current is: …

Why Ampère's Circuital Law?

Ampere's law is the cleanest tool here. It says: for any closed loop, the line integral of the magnetic field around it equals $\mu_0$ times the current passing through the loop. The symmetry of a long straight wire -- cylindrical, infinite -- tells us the field must be azimuthal (circles around the wire) and depend only on the radial distance $r$ from the axis. So we pick circular Amperian loops centred on the wire, and the integral becomes simply $B \times (2\pi r)$. The only trick is: how much current actually pierces the loop? That depends on whether the loop lies inside or outside the wire.

1. Region $r > a$ (outside the wire) Take a circular Amperian loop of radius $r > a$, concentric with the wire. The entire current $I$ passes through this loop. By symmetry, $B$ is constant in magnitude along the loop and tangential to it. Ampère's law gives:

So:

This is exactly the field of a thin wire -- as if all current were concentrated at the axis. Outside the wire, the finite radius doesn't matter.

2. Region $r < a$ (inside the wire) Now take a loop of radius $r < a$. Only a fraction of the total current passes through it. Since the current is uniformly distributed over the cross-section, the current density is:

The area enclosed by the loop is $\pi r^2$, so the current through it is:

Apply Ampère's law:

Hence:

Method: Ampère’s Circuital Law

This is the standard method for finding the magnetic field due to symmetric current distributions. It uses the fact that for a steady current, the line integral of the magnetic field around a closed loop equals $\mu_0$ times the current enclosed.

Steps

Step 1: Identify symmetry and choose an Ampèrian loop

• The wire is long, straight, and has cylindrical symmetry.
• The magnetic field $\mathbf{B}$ will be tangential (azimuthal direction) and depend only on the radial distance $r$ from the centre.
• Choose a circular loop of radius $r$, concentric with the wire, as the Ampèrian path.

Step 2: Write Ampère’s law

For a circular loop of radius $r$, the left side becomes:

Step 3: Find the enclosed current $I_{\text{enc}}$ for each region

• Region $r > a$ (outside the wire): The entire current $I$ passes through the loop.

• Region $r < a$ (inside the wire): Current is uniformly distributed over the cross-section. Current density:

Enclosed current for radius $r$:

Step 4: Apply Ampère’s law to each region

• For $r > a$: …

Common Mistakes: Magnetic Field Inside & Outside a Current-Carrying Wire

Mistake 1: Forgetting the Current Enclosed Changes with $r$

The error: Students use the same total current $I$ for both regions. Inside the wire ($r < a$), only a fraction of the total current is enclosed by the Amperian loop.

How to avoid:

• Always ask: "How much current actually passes through my Amperian loop?"
• For $r < a$: Current density $J = \frac{I}{\pi a^2}$, so enclosed current is:

• For $r > a$: The entire current $I$ is enclosed.

Mistake 2: Using the Wrong Amperian Loop Radius

The error: Students set the loop radius equal to $a$ (the wire's radius) instead of the variable $r$ (distance from centre where field is being calculated).

How to avoid:

• The Amperian loop is your choice — draw it at distance $r$ from the centre.
• $r$ is the variable in your answer; $a$ is a constant (wire radius).
• For $r < a$: loop is inside the wire.
• For $r > a$: loop is outside the wire.

Mistake 3: Misapplying Ampere's Circuital Law

The error: Writing $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$ without considering symmetry or direction.

How to avoid:

• By symmetry, $\vec{B}$ is tangential and constant in magnitude on a circular Amperian loop.
• So:

• Then equate: $B \cdot 2\pi r = \mu_0 I_{\text{enc}}$

Mistake 4: Getting the Final Expression Wrong for $r < a$

The error: Writing $B = \frac{\mu_0 I}{2\pi r}$ for inside the wire (which is actually the outside formula).

How to avoid:

• Inside ($r < a$): Substitute $I_{\text{enc}} = \frac{I r^2}{a^2}$:

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