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Worked Examples · Example 4.7

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

Figure 4.13
Figure 4.13
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Using Ampère's circuital law, the magnetic field inside a uniformly current-carrying wire grows linearly with distance from the axis, while outside it falls off as 1/r1/r. The results are Binside=μ0Ir2πa2B_{\text{inside}} = \frac{\mu_0 I r}{2\pi a^2} and Boutside=μ0I2πrB_{\text{outside}} = \frac{\mu_0 I}{2\pi r}.

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 μ0\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 rr from the axis. So we pick circular Amperian loops centred on the wire, and the integral becomes simply B×(2πr)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>ar > a (outside the wire) Take a circular Amperian loop of radius r>ar > a, concentric with the wire. The entire current II passes through this loop. By symmetry, BB is constant in magnitude along the loop and tangential to it. Ampère's law gives:

∮B⋅dl=B⋅(2πr)=μ0I\oint \mathbf{B} \cdot d\mathbf{l} = B \cdot (2\pi r) = \mu_0 I

So:

B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}

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.

  1. Region r<ar < a (inside the wire) Now take a loop of radius r<ar < 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:

J=Iπa2J = \frac{I}{\pi a^2}

The area enclosed by the loop is πr2\pi r^2, so the current through it is:

Ienc=J⋅πr2=Ir2a2I_{\text{enc}} = J \cdot \pi r^2 = \frac{I r^2}{a^2}

Apply Ampère's law:

B⋅(2πr)=μ0Ienc=μ0Ir2a2B \cdot (2\pi r) = \mu_0 I_{\text{enc}} = \mu_0 \frac{I r^2}{a^2}

Hence:

B=μ0Ir2πa2B = \frac{\mu_0 I r}{2\pi a^2}

Watch out

A common mistake is to use the full current II for the inside region. Remember: Ampère's law cares only about the current enclosed by the loop. For r<ar < a, that's less than II. …

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