12.5 Ampère’s Law

A fundamental property of a static magnetic field is that, unlike an electrostatic field, it is not conservative. A conservative vector field is one whose line integral between two end points is the same regardless of the path chosen. Magnetic fields do not have such a property. Instead, there is a relationship between the magnetic field and its source, electric current. It is expressed in terms of the line integral of $\vec{B}$ and is known as Ampère’s law. This law can also be derived directly from the Biot-Savart law. We now consider that derivation for the special case of an infinite, straight wire.

Figure 12.14 shows an arbitrary plane perpendicular to an infinite, straight wire whose current I is directed out of the page. The magnetic field lines are circles directed counterclockwise and centered on the wire. To begin, let’s consider $∮ B → times d l →$ over the closed paths M and N. Notice that one path (M) encloses the wire, whereas the other (N) does not. Since the field lines are circular, $\vec{B} \cdot d \vec{l}$ is the product of B and the projection of dl onto the circle passing through $d \vec{l} .$ If the radius of this particular circle is r, the projection is $r d θ,$ and

$\vec{B} \cdot d \vec{l} = B r d θ .$

Figures A and B show an arbitrary plane perpendicular to an infinite, straight wire whose current I is directed out of the page. The magnetic field lines are circles directed counterclockwise and centered on the wire. Ampere path M demonstrated in the Figure A encloses the wire. Ampere path N demonstrated in the Figure B does not enclose the wire. — **Figure 12.14** The current I of a long, straight wire is directed out of the page. The integral $∮ d θ$ equals $2 π$ and 0, respectively, for paths M and N.

With $\vec{B}$ given by Equation 12.9,

$\oint \vec{B} \cdot d \vec{l} = \oint (\frac{μ_{0} I}{2 π r}) r d θ = \frac{μ_{0} I}{2 π} \oint d θ .$

For path M, which circulates around the wire, $∮ sub M d θ equals 2 π$ and

$\oint_{M} \vec{B} \cdot d \vec{l} = μ_{0} I .$

Path N, on the other hand, circulates through both positive (counterclockwise) and negative (clockwise) $d θ$ (see Figure 12.14), and since it is closed, $∮ sub N d θ equals 0.$ Thus for path N,

$\oint_{N} \vec{B} \cdot d \vec{l} = 0 .$

The extension of this result to the general case is Ampère’s law.

To determine whether a specific current I is positive or negative, curl the fingers of your right hand in the direction of the path of integration, as shown in Figure 12.14. If I passes through S in the same direction as your extended thumb, I is positive; if I passes through S in the direction opposite to your extended thumb, it is negative.

Example 1 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.

Figure 12.15 The possible components of the magnetic field B due to a current I, which is directed out of the page. The field has no radial component, so there is no radial contribution to the integral.

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, $B_{r}$ and $B_{θ},$ are shown at arbitrary points on a circle of radius r centered on the wire. Since the field is cylindrically symmetric, neither $B_{r}$ nor $B_{θ}$ 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 would mean, however, that there must be a net magnetic flux across an arbitrary cylinder concentric with the wire. However, because magnetic field lines are continuous, forming closed loops without a beginning or end (see the discussion in Magnetic Fields and Lines) the net magnetic flux through any closed test surface must be zero. We can therefore conclude that the radial component of the magnetic field must be zero. Therefore, we can apply Ampère’s law to the circular path as shown.

Solution Over this path $\vec{B}$ is constant and parallel to $d \vec{l},$ so

$\oint \vec{B} \cdot d \vec{l} = B_{θ} \oint d l = B_{θ} (2 π r) .$

Thus Ampère’s law reduces to

$B sub θ open parenthesis 2 π r close parenthesis equals μ sub 0 I.$

Finally, since $B_{θ}$ is the only component of $\vec{B},$ we can drop the subscript and write

$B = \frac{μ_{0} I}{2 π r} .$

This agrees with the Biot-Savart calculation above.

Significance Ampère’s law works well if you have a path to integrate over which $\vec{B} \cdot d \vec{l}$ 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 2 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 $I sub 0$ that is distributed uniformly over its cross-section. Find the magnetic field both inside and outside the wire.

Figure 12.16 (a) A model of a current-carrying wire of radius a and current

I sub 0.

(b) A cross-section of the same wire showing the radius a and the Ampère’s loop of radius r.

Strategy This problem has the same geometry as Example 1, 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 For any circular path of radius r that is centered on the wire,

$\oint \vec{B} \cdot d \vec{l} = \oint B d l = B \oint d l = B (2 π r) .$

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 $(r \leq a)$ 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 $I sub 0 / π a squared.$ Therefore the current I passing through the area enclosed by the path is

$I = \frac{π r^{2}}{π a^{2}} I_{0} = \frac{r^{2}}{a^{2}} I_{0} .$

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

$B (2 π r) = μ_{0} (\frac{r^{2}}{a^{2}}) I_{0},$

and the magnetic field inside the wire is

$B = \frac{μ_{0} I_{0}}{2 π} \frac{r}{a^{2}} (r \leq a) .$

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

$B = \frac{μ_{0} I_{0}}{2 π r} (r \geq a) .$

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

Figure 12.17 Variation of the magnetic field produced by a current

I sub 0

in a long, straight wire of radius a.

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 3 Using Ampère’s Law with Arbitrary Paths

Use Ampère’s law to evaluate $∮ B → times d l →$ for the current configurations and paths in Figure 12.18.

Figure 12.18 Current configurations and paths for Example 3.

Strategy Ampère’s law states that $\oint \vec{B} \cdot d \vec{l} = μ_{0} I$ where I is the total current passing through the enclosed loop. The quickest way to evaluate the integral is to calculate $μ sub 0 I$ 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 (a) The current going downward through the loop equals the current going out of the loop, so the net current is zero. Thus, $\oint \vec{B} \cdot d \vec{l} = 0.$

(b) The only current to consider in this problem is 2A because it is the only current inside the loop. The right-hand rule shows us the current going downward through the loop is in the positive direction. Therefore, the answer is $\oint \vec{B} \cdot d \vec{l} = μ_{0} (2 A) = 2.51 \times 10^{−6} T \cdot m .$

(c) The right-hand rule shows us the current going downward through the loop is in the positive direction. There are $7A + 5A = 12A$ of current going downward and –3 A going upward. Therefore, the total current is 9 A and $\oint \vec{B} \cdot d \vec{l} = μ_{0} (9 A) = 1.13 \times 10^{–5} T \cdot m .$

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.

Summary

The magnetic field created by current following any path is the sum (or integral) of the fields due to segments along the path (magnitude and direction as for a straight wire), resulting in a general relationship between current and field known as Ampère’s law.
Ampère’s law can be used to determine the magnetic field from a thin wire or thick wire by a geometrically convenient path of integration. The results are consistent with the Biot-Savart law.

Conceptual Questions

Is Ampère’s law valid for all closed paths? Why isn’t it normally useful for calculating a magnetic field?

Ampère’s law is valid for all closed paths, but it is not useful for calculating fields when the magnetic field produced lacks symmetry that can be exploited by a suitable choice of path.

Problems

A current I flows around the rectangular loop shown in the accompanying figure. Evaluate $∮ B → times d l →$ for the paths A, B, C, and D.

Figure shows rectangular loop carrying current I. Paths A and C intersect with the short sides of the loop. Path B intersects with the two long sides of the loop. Path D intersects both with the short and the long sides of the loop.

a. $μ sub 0 I;$ b. 0; c. $μ sub 0 I;$ d. 0

Evaluate $∮ B → times d l →$ for each of the cases shown in the accompanying figure.

Figure A shows a wire inside the loop that carries current of two Amperes upward through the loop. Figure B shows three wires inside the loop that carry current of five Amperes, two Amperes, and six Amperes. First and third wires carry current upward through the loop. Second wire carries current downward through the loop. Figure C shows two wires outside the loop that carry current of three Amperes and two Amperes upward through the loop. Figure D shows three wires carrying current of three Amperes, two Amperes, and four Amperes. First wire is outside the loop, second and third wires are inside the loop. First and third wires carry current downward through the loop. Second wire carries current upward through the loop. Figure D shows four wires carrying currents of four Amperes, three Amperes, two Amperes, and two Amperes. First and fourth wires are outside the loop. Second and third wires are inside the loop. First, second, and third wires carry current upward through the loop. Fourth wire carries current downward through the loop.

The coil whose lengthwise cross section is shown in the accompanying figure carries a current I and has N evenly spaced turns distributed along the length l. Evaluate $∮ B → times d l →$ for the paths indicated.

Figure shows the lengthwise cross section of a coil. Path A, running counterclockwise, intersects three coils carrying current from the plane of the paper. Path B, running clockwise, intersects four coils with two carrying current from the plane of the paper and two carrying current into the plane of the paper. Path C, running clockwise, intersects seven coils carrying current into the plane of the paper. Path D, running counterclockwise, intersects two coils carrying current into the plane of the paper.

a. $3 μ sub 0 I;$ b. 0; c. $7 μ sub 0 I;$ d. $−2 μ sub 0 I$

A superconducting wire of diameter 0.25 cm carries a current of 1000 A. What is the magnetic field just outside the wire?

A long, straight wire of radius R carries a current I that is distributed uniformly over the cross-section of the wire. At what distance from the axis of the wire is the magnitude of the magnetic field a maximum?

at the radius R

The accompanying figure shows a cross-section of a long, hollow, cylindrical conductor of inner radius $r sub 1 equals 3.0 cm$ and outer radius $r sub 2 equals 5.0 cm.$ A 50-A current distributed uniformly over the cross-section flows into the page. Calculate the magnetic field at $r = 2.0 cm, r = 4.0 cm, and r = 6.0 cm .$

Figure shows a cross-section of a long, hollow, cylindrical conductor with an inner radius of three centimeters and an outer radius of five centimeters.

A long, solid, cylindrical conductor of radius 3.0 cm carries a current of 50 A distributed uniformly over its cross-section. Plot the magnetic field as a function of the radial distance r from the center of the conductor.

Graph shows the variation of B with r. B linearly increases with r until the point a. Then it starts to decreases proportionally to the inverse of r.

A portion of a long, cylindrical coaxial cable is shown in the accompanying figure. A current I flows down the center conductor, and this current is returned in the outer conductor. Determine the magnetic field in the regions (a) $r less than or equal to r sub 1,$ (b) $r sub 2 greater than or equal to r greater than or equal to r sub 1,$ (c) $r sub 3 greater than or equal to r greater than or equal to r sub 2,$ and (d) $r greater than or equal to r sub 3.$ Assume that the current is distributed uniformly over the cross sections of the two parts of the cable.

Figure shows a long, cylindrical coaxial cable. Radius of the inner center conductor is r1. Distance from the center to the inner side of the shield is r2. Distance from the center to the outer side of the shield is r3.