12.5 Ampère’s law  (Page 2/3)

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

Strategy

Solution

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 $(r\le a)$ 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 $I_0/\pi a^2.$ Therefore the current I passing through the area enclosed by the path is

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

and the magnetic field inside the wire is

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

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

Significance

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 ${\displaystyle \oint \overrightarrow{B}}·d\overrightarrow{l}$ for the current configurations and paths in [link] .

Strategy

Ampère’s law states that ${\displaystyle \oint \overrightarrow{B}}·d\overrightarrow{l}={\mu }_{0}I$ where I is the total current passing through the enclosed loop. The quickest way to evaluate the integral is to calculate ${\mu }_{0}I$ 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, ${\displaystyle \oint \overrightarrow{B}}·d\overrightarrow{l}=0.$

(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 ${\displaystyle \oint \overrightarrow{B}}·d\overrightarrow{l}={\mu }_0(2\text{A})=2.51\times {10}^{\text{−6}}\text{T}\cdot \text{m/A}.$

(c) The right-hand rule shows us the current going downward through the loop is in the positive direction. There are $\text{7A}+\text{5A}=\text{12A}$ of current going downward and –3 A going upward. Therefore, the total current is 9 A and ${\displaystyle \oint \overrightarrow{B}}·d\overrightarrow{l}={\mu }_0(9\text{A})=5.65\times {10}^{\text{−6}}\text{T}\cdot \text{m/A}.$

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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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

Problems

A current I flows around the rectangular loop shown in the accompanying figure. Evaluate ${\displaystyle \oint \overrightarrow{B}}·d\overrightarrow{l}$ for the paths A , B , C , and D .

a. ${\mu }_{0}I;$ b. 0; c. ${\mu }_{0}I;$ d. 0

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Evaluate ${\displaystyle \oint \overrightarrow{B}}·d\overrightarrow{l}$ for each of the cases shown in the accompanying figure.

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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 ${\displaystyle \oint \overrightarrow{B}}·d\overrightarrow{l}$ for the paths indicated.

a. $3{\mu }_{0}I;$ b. 0; c. $7{\mu }_{0}I;$ d. $\text{−2}{\mu }_{0}I$

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The accompanying figure shows a cross-section of a long, hollow, cylindrical conductor of inner radius $r_1=\text{3.0 cm}$ and outer radius $r_2=\text{5.0 cm}.$ A 50-A current distributed uniformly over the cross-section flows into the page. Calculate the magnetic field at $r=\text{2.0 cm},r=4.0\text{cm},\text{and}r=\text{6.0 cm}.$

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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) $r\le r_1,$ (b) $r_2\ge r\ge r_1,$ (c) $r_3\ge r\ge r_2,$ and (d) $r\ge r_3.$ Assume that the current is distributed uniformly over the cross sections of the two parts of the cable.

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