Solenoids and Toroids — Question 6

Question 6

An ideal toroid has a stronger field at its inner edge than at its outer edge

An ideal tightly wound air-core toroid has N=500 turns, current I=.400 A, inner radius a=.0800 m and outer radius b=.120 m. Approximate the windings by a continuous symmetric distribution. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Use Ampère law to find the field within the winding region and calculate its inner and outer edge magnitudes.

2. Find the fields in the central hole and outside the toroid under this idealization, explaining the enclosed currents.

Question 6 - Solution

An ideal toroid has a stronger field at its inner edge than at its outer edge

(a) Method and calculation

$$2\pi rB(r)=\mu_0NI\Rightarrow\boxed{B(r)=\frac{\mu_0NI}{2\pi r}}\quad(a<r<b).$$$$\boxed{B(a)=.500\,\mathrm{mT},\quad B(b)=.333333\,\mathrm{mT}}.$$

(b) Interpretation and further calculation

$$r<a:\ I_{\mathrm{enc}}=0;\quad r>b:\ I_{\mathrm{enc}}=NI-NI=0.$$$$\boxed{B=0\text{ in the hole and exterior}}\quad\text{by the ideal winding symmetry}.$$

Physical check

Zero circulation alone would not establish a zero local field; the ideal toroidal symmetry is an essential part of the argument.