m58465 Ampère’s Law By the end of this section, you will be able to: Explain how Ampère’s law relates the magnetic field produced by a current to the value of the current Calculate the magnetic field from a long straight wire, either thin or thick, by Ampère’s law cc30d0fc-3dfa-411b-aa04-2b98c0ef20cf 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 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. 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→·dl→ 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, B→·dl→ is the product of B and the projection of dl onto the circle passing through dl→. If the radius of this particular circle is r, the projection is rdθ, and B→·dl→=Brdθ.

The current I of a long, straight wire is directed out of the page. The integral ∮dθ equals 2π and 0, respectively, for paths M and N.

With B→ given by , ∮B→·dl→=∮(μ0I2πr)rdθ=μ0I2π∮dθ. For path M, which circulates around the wire, ∮Mdθ=2π and ∮MB→·dl→=μ0I. Path N, on the other hand, circulates through both positive (counterclockwise) and negative (clockwise) dθ (see ), and since it is closed, ∮Ndθ=0. Thus for path N, ∮NB→·dl→=0. The extension of this result to the general case is Ampère’s law. Over an arbitrary closed path, ∮B→·dl→=μ0I 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. 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 . 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. To calculate the magnetic field created from current in wire(s), use the following steps: Identify the symmetry of the current in the wire(s). If there is no symmetry, use the Biot-Savart law to determine the magnetic field. Determine the direction of the magnetic field created by the wire(s) by right-hand rule 2. Choose a path loop where the magnetic field is either constant or zero. Calculate the current inside the loop. Calculate the line integral ∮B→·dl→ around the closed loop. Equate ∮B→·dl→ with μ0Ienc and solve for B→. 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 .

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.

Consider an arbitrary plane perpendicular to the wire, with the current directed out of the page. The possible magnetic field components in this plane, Br and Bθ, are shown at arbitrary points on a circle of radius r centered on the wire. Since the field is cylindrically symmetric, neither Br nor Bθ 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. Over this path B→ is constant and parallel to dl→, so ∮B→·dl→=Bθ∮dl=Bθ(2πr). Thus Ampère’s law reduces to Bθ(2πr)=μ0I. Finally, since Bθ is the only component of B→, we can drop the subscript and write B=μ0I2πr. This agrees with the Biot-Savart calculation above. Ampère’s law works well if you have a path to integrate over which B→·dl→ 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. The radius of the long, straight wire of is a, and the wire carries a current I0 that is distributed uniformly over its cross-section. Find the magnetic field both inside and outside the wire.

(a) A model of a current-carrying wire of radius a and current I0. (b) A cross-section of the same wire showing the radius a and the Ampère’s loop of radius r.

This problem has the same geometry as , 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 ). For any circular path of radius r that is centered on the wire, ∮B→·dl→=∮Bdl=B∮dl=B(2πr). 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≤a) such as that shown in part (a) of . 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 I0/πa2. Therefore the current I passing through the area enclosed by the path is I=πr2πa2I0=r2a2I0. 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(2πr)=μ0(r2a2)I0, and the magnetic field inside the wire is B=μ0I02πra2(r≤a). Outside the wire, the situation is identical to that of the infinite thin wire of the previous example; that is, B=μ0I02πr(r≥a). The variation of B with r is shown in .

Variation of the magnetic field produced by a current I0 in a long, straight wire of radius a.

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. Use Ampère’s law to evaluate ∮B→·dl→ for the current configurations and paths in .

Current configurations and paths for .

Ampère’s law states that ∮B→·dl→=μ0I where I is the total current passing through the enclosed loop. The quickest way to evaluate the integral is to calculate μ0I 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. (a) The current going downward through the loop equals the current going out of the loop, so the net current is zero. Thus, ∮B→·dl→=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 ∮B→·dl→=μ0(2A)=2.51×10−6T⋅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 ∮B→·dl→=μ0(9A)=1.13×10–5T⋅m. 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. Consider using Ampère’s law to calculate the magnetic fields of a finite straight wire and of a circular loop of wire. Why is it not useful for these calculations? In these cases the integrals around the Ampèrian loop are very difficult because there is no symmetry, so this method would not be useful.

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.

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.

A current I flows around the rectangular loop shown in the accompanying figure. Evaluate ∮B→·dl→ for the paths A, B, C, and D. a. μ0I; b. 0; c. μ0I; d. 0 Evaluate ∮B→·dl→ for each of the cases shown in the accompanying figure. 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→·dl→ for the paths indicated. a. 3μ0I; b. 0; c. 7μ0I; d. −2μ0I 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 r1=3.0 cm and outer radius r2=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=4.0cm,andr=6.0 cm. 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. 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≤r1, (b) r2≥r≥r1, (c) r3≥r≥r2, and (d) r≥r3. Assume that the current is distributed uniformly over the cross sections of the two parts of the cable.

Ampère’s law physical law that states that the line integral of the magnetic field around an electric current is proportional to the current