12.5 Ampère’s Law

Ampère’s law

Over an arbitrary closed path,

where I is the total current passing through any open surface S whose perimeter is the path of integration. Only currents inside the path of integration need be considered.

Example

Using Ampère’s Law to Calculate the Magnetic Field Due to a Wire

Use Ampère’s law to calculate the magnetic field due to a steady current I in an infinitely long, thin, straight wire as shown in Figure 12.15.

Strategy

Consider an arbitrary plane perpendicular to the wire, with the current directed out of the page. The possible magnetic field components in this plane, [latex]{B}_{r}[/latex] and [latex]{B}_{\theta },[/latex] are shown at arbitrary points on a circle of radius r centered on the wire. Since the field is cylindrically symmetric, neither [latex]{B}_{r}[/latex] nor [latex]{B}_{\theta }[/latex] varies with the position on this circle. Also from symmetry, the radial lines, if they exist, must be directed either all inward or all outward from the wire. This means, however, that there must be a net magnetic flux across an arbitrary cylinder concentric with the wire. The radial component of the magnetic field must be zero because [latex]{\stackrel{\to }{\textbf{B}}}_{r}\cdot d\stackrel{\to }{\textbf{l}}=0.[/latex] Therefore, we can apply Ampère’s law to the circular path as shown.

Solution

Significance

Ampère’s law works well if you have a path to integrate over which [latex]\stackrel{\to }{\textbf{B}}·d\stackrel{\to }{\textbf{l}}[/latex] has results that are easy to simplify. For the infinite wire, this works easily with a path that is circular around the wire so that the magnetic field factors out of the integration. If the path dependence looks complicated, you can always go back to the Biot-Savart law and use that to find the magnetic field.

Example

Calculating the Magnetic Field of a Thick Wire with Ampère’s Law

The radius of the long, straight wire of Figure 12.16 is a, and the wire carries a current [latex]{I}_{0}[/latex] that is distributed uniformly over its cross-section. Find the magnetic field both inside and outside the wire.

Strategy

This problem has the same geometry as Example 12.6, but the enclosed current changes as we move the integration path from outside the wire to inside the wire, where it doesn’t capture the entire current enclosed (see Figure 12.16).

Solution

Show Answer

For any circular path of radius r that is centered on the wire,

[latex]\oint \stackrel{\to }{\textbf{B}}·d\stackrel{\to }{\textbf{l}}=\oint Bdl=B\oint dl=B\left(2\pi r\right).[/latex]

From Ampère’s law, this equals the total current passing through any surface bounded by the path of integration.

Consider first a circular path that is inside the wire [latex]\left(r\le a\right)[/latex] such as that shown in part (a) of Figure 12.16. We need the current I passing through the area enclosed by the path. It’s equal to the current density J times the area enclosed. Since the current is uniform, the current density inside the path equals the current density in the whole wire, which is [latex]{I}_{0}/\pi {a}^{2}.[/latex] Therefore the current I passing through the area enclosed by the path is

[latex]I=\frac{\pi {r}^{2}}{\pi {a}^{2}}{I}_{0}=\frac{{r}^{2}}{{a}^{2}}{I}_{0}.[/latex]

We can consider this ratio because the current density J is constant over the area of the wire. Therefore, the current density of a part of the wire is equal to the current density in the whole area. Using Ampère’s law, we obtain

[latex]B\left(2\pi r\right)={\mu }_{0}\left(\frac{{r}^{2}}{{a}^{2}}\right){I}_{0},[/latex]

and the magnetic field inside the wire is

[latex]B=\frac{{\mu }_{0}{I}_{0}}{2\pi }\phantom{\rule{0.2em}{0ex}}\frac{r}{{a}^{2}}\phantom{\rule{0.2em}{0ex}}\left(r\le a\right).[/latex]

Outside the wire, the situation is identical to that of the infinite thin wire of the previous example; that is,

[latex]B=\frac{{\mu }_{0}{I}_{0}}{2\pi r}\phantom{\rule{0.2em}{0ex}}\left(r\ge a\right).[/latex]

The variation of B with r is shown in Figure 12.17.

Significance

The results show that as the radial distance increases inside the thick wire, the magnetic field increases from zero to a familiar value of the magnetic field of a thin wire. Outside the wire, the field drops off regardless of whether it was a thick or thin wire.

This result is similar to how Gauss’s law for electrical charges behaves inside a uniform charge distribution, except that Gauss’s law for electrical charges has a uniform volume distribution of charge, whereas Ampère’s law here has a uniform area of current distribution. Also, the drop-off outside the thick wire is similar to how an electric field drops off outside of a linear charge distribution, since the two cases have the same geometry and neither case depends on the configuration of charges or currents once the loop is outside the distribution.

Example

Using Ampère’s Law with Arbitrary Paths

Use Ampère’s law to evaluate [latex]\oint \stackrel{\to }{\textbf{B}}·d\stackrel{\to }{\textbf{l}}[/latex] for the current configurations and paths in Figure 12.18.

Strategy

Ampère’s law states that [latex]\oint \stackrel{\to }{\textbf{B}}·d\stackrel{\to }{\textbf{l}}={\mu }_{0}I[/latex] where I is the total current passing through the enclosed loop. The quickest way to evaluate the integral is to calculate [latex]{\mu }_{0}I[/latex] by finding the net current through the loop. Positive currents flow with your right-hand thumb if your fingers wrap around in the direction of the loop. This will tell us the sign of the answer.

Solution

Significance

If the currents all wrapped around so that the same current went into the loop and out of the loop, the net current would be zero and no magnetic field would be present. This is why wires are very close to each other in an electrical cord. The currents flowing toward a device and away from a device in a wire equal zero total current flow through an Ampère loop around these wires. Therefore, no stray magnetic fields can be present from cords carrying current.