Question 30.6: The Magnetic Field Created by a Toroid A device called a tor......

Conceptualize Study Figure 30.15 carefully to understand how the wire is wrapped around the torus. The torus could be a solid material or it could be air, with a stiff wire wrapped into the shape shown in Figure 30.15 to form an empty toroid.

Categorize Because the toroid has a high degree of symmetry, we categorize this example as an Ampère’s law problem.

Analyze Consider the circular amperian loop (loop 1) of radius r in the plane of Figure 30.15. By symmetry, the magnitude of the field is constant on this circle and tangent to it, so \overrightarrow{\mathbf{B}} \cdot d \overrightarrow{\mathbf{s}}=B  d s. Furthermore, the wire passes through the loop N times, so the total current through the loop is N I.

Apply Ampère’s law to loop 1:

\oint \overrightarrow{\mathbf{B}} \cdot d \overrightarrow{\mathbf{s}}=B \oint d s=B(2 \pi r)=\mu_{0} N I

Solve for B :

B=\frac{\mu_{0} N I}{2 \pi r} \qquad (30.16)

Finalize This result shows that B varies as 1 / r and hence is nonuniform in the region occupied by the torus. If, however, r is very large compared with the cross-sectional radius a of the torus, the field is approximately uniform inside the torus.

For an ideal toroid, in which the turns are closely spaced, the external magnetic field is close to zero, but it is not exactly zero. In Figure 30.15, imagine the radius r of the amperian loop to be either smaller than b or larger than c. In either case, the loop encloses zero net current, so \oint \overrightarrow{\mathbf{B}} \cdot d \overrightarrow{\mathbf{s}}=0. You might think this result proves that \overrightarrow{\mathbf{B}}=0, but it does not. Consider the amperian loop (loop 2) on the right side of the toroid in Figure 30.15. The plane of this loop is perpendicular to the page, and the toroid passes through the loop. As charges enter the toroid as indicated by the current directions in Figure 30.15, they work their way counterclockwise around the toroid. Therefore, a current passes through the perpendicular amperian loop! This current is small, but not zero. As a result, the toroid acts as a current loop and produces a weak external field of the form shown in Figure 30.6. The reason \oint \overrightarrow{\mathbf{B}} \cdot d \overrightarrow{\mathbf{s}}=0 for the amperian loops of radius r<b and r>c in the plane of the page is that the field lines are perpendicular to d \overrightarrow{\mathbf{s}}, not because \overrightarrow{\mathbf{B}}=0.