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  5. Ampère’s law

12.5 Ampère’s law  (Page 2/3)

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Calculating the magnetic field of a thick wire with ampère’s law

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

Figure A shows a long, straight wire of radius a that carries current I. Figure B shows a cross-section of the same wire with the Ampère’s loop of radius r.
(a) A model of a current-carrying wire of radius a and current I 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 [link] , 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 [link] ).

Solution

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

∮ B → · d l → = ∮ B d l = B ∮ 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 ≤ a ) such as that shown in part (a) of [link] . 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 0 / π a 2 . Therefore the current I passing through the area enclosed by the path is

I = π r 2 π a 2 I 0 = 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 ( r 2 a 2 ) I 0 ,

and the magnetic field inside the wire is

B = μ 0 I 0 2 π r a 2 ( r ≤ a ) .

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

B = μ 0 I 0 2 π r ( r ≥ a ) .

The variation of B with r is shown in [link] .

Graph shows the variation of B with r. It linearly increases with the r until the point a. Then it starts to decrease proportionally to the inverse of r.
Variation of the magnetic field produced by a current I 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.

Using ampère’s law with arbitrary paths

Use Ampère’s law to evaluate ∮ B → · d l → for the current configurations and paths in [link] .

Figure A shows four wires carrying currents of two Amperes, five Amperes, three Amperes, and four Amperes. All four wires are inside the loop. First and second wires carry current downward through the loop. Third and fourth wires carry current upward through the loop. Figure B shows three wires carrying currents of five Amperes, two Amperes, and three Amperes. First and third wires are outside the loop, second wire is inside the loop. First wire carries current upward through the loop. Second and third wires carry current downward through the loop. Figure C shows three wires carrying currents of seven Amperes, five Amperes, and three Amperes. All three wires are inside the loop. First and second wires carry current downward through the loop. Third wire carries current upward through the loop.
Current configurations and paths for [link] .

Strategy

Ampère’s law states that ∮ B → · d 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 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, ∮ B → · d 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 ∮ B → · d l → = μ 0 ( 2 A ) = 2.51 × 10 −6 T ⋅ m/A .

(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 → · d l → = μ 0 ( 9 A ) = 5.65 × 10 −6 T ⋅ m/A .

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.

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Check Your Understanding 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.

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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.

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Problems

A current I flows around the rectangular loop shown in the accompanying figure. Evaluate ∮ B → · 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 0 I ; b. 0; c. μ 0 I ; d. 0

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Evaluate ∮ B → · 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.
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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 → · d l → for the paths indicated.

Figure shows the lengthwise cross section of a coil. Path A intersects three coils carrying current from the plane of the paper. Path B 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 intersects seven coils carrying current into the plane of the paper. Path D intersects two coils carrying current into the plane of the paper.

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

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

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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

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The accompanying figure shows a cross-section of a long, hollow, cylindrical conductor of inner radius r 1 = 3.0 cm and outer radius r 2 = 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.
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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.

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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 ≤ r 1 , (b) r 2 ≥ r ≥ r 1 , (c) r 3 ≥ r ≥ r 2 , and (d) r ≥ r 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.
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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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