Solenoids and Toroids — Question 8

Question 8

Reversed turns reduce a toroidal field twice as much as inactive turns

A toroidal winding is intended to have N=200 turns all carrying I with the same orientation. In an idealized distributed winding, k=5 turns are instead reversed while carrying the same current. Keep the geometry and current magnitude fixed; neglect leakage. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Find the field relative to its intended value and the current correction needed to restore that value.

2. Compare with five electrically bypassed, inactive turns and explain the factor-of-two difference in field loss.

Question 8 - Solution

Reversed turns reduce a toroidal field twice as much as inactive turns

(a) Method and calculation

$$NI\big|_{\mathrm{actual}}=(N-k)I-kI=(N-2k)I.$$$$\boxed{B/B_0=190/200=.950,\quad I\prime/I=200/190=20/19}.$$

(b) Interpretation and further calculation

$$\text{inactive turns:}\quad B/B_0=(N-k)/N=195/200=.975.$$$$\boxed{\Delta B_{\mathrm{reversed}}/B_0=2k/N,\quad\Delta B_{\mathrm{inactive}}/B_0=k/N}.$$

Physical check

A reversed turn removes its intended positive contribution and adds an equal negative contribution. This model concerns linked ampere-turns, not localized leakage near an individual defective turn.