Consider an ideal toroid with an average radius 16.0 cm with 240 turns. A current of 10 A is maintained in it. The magnitude of the magnetic field inside the toroid is [(\(\frac{μ_0}{4π}\)) = 10-7 Tm/A:

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AAI ATC Junior Executive 27 July 2022 Shift 2 Official Paper

Answer 1

Concept:

If a solenoid is bent in a circular shape and the ends are joined, we get a toroid.
The magnetic field inside a toroid at point P is, \(B = \frac {μ _0 Ni}{2\pi r}\)
Where μ₀ = permeability of vacuum, N = total number of turns, i = current, r = average radius.

F3 Vilas Engineering 30.12.2022 D1

Calculation:

Given, average radius, r = 16.0 cm = 0.16 m, N = 240, i = 10 A, \(\frac{μ_0}{4π}\) = 10^-7 Tm/A

The magnetic field inside a toroid at point P is, \(B = \frac {μ _0 Ni}{2\pi r}\)

\(\Rightarrow B = \frac {2 × 10^{-7}× 240 × 10}{0.16}\)

= 3 × 10^-3 T = 3.0 mT

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