Question

A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm. It carries a current of 1.50 A. How many equally spaced turns must it have so that it will produce a magnetic field of 3.75 MT at points within the coils 14.0 cm from its centre?

Short answer

The number of equally spaced turns is 1750. It must have produced a magnetic field of 3.75 MT at points within the coils 14.0 cm from its center.

Step-by-step solution

Concept of the magnetic due to the toroidal solenoid

The magnetic due to the toroidal solenoid is given by:

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{N}\mathbf{I}}{\mathbf{2}\mathbf{\pi }\mathbf{r}}$

$\mathbf{\text{Where}}{\mathit{\mu }}_{\mathbf{0}}\mathbf{=}\mathbf{4}\mathit{\pi }\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{H}\mathbf{/}\mathbf{m}$is the permeability of free space, N is the number of turns, I is the current passing through the coil, and r is the distance from the center of the torus to the point

Find the number of equally spaced turns

Given data:

Consider a toroidal solenoid with an inner radius of 12.0 cm and an outer radius of 15.0 cm. It carries a current of I 1.50 A.

Within the space enclosed by the windings, at a distance r from the symmetry axis, the magnetic field due to a toroidal solenoid is given by,

$B=\frac{{\mu }_{0}NI}{2\pi r}$

Substitute the values in the modified equation we have,

$\begin{array}{r}N=\frac{2\pi rB}{{\mu }_{0}I}\\ =\frac{2\pi \left(0.140\mathrm{m}\right)\left(3.75\times {10}^{-3}\mathrm{T}\right)}{4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\times 1.50\mathrm{A}}\\ =1750\end{array}$

Therefore, the number of equally spaced turns is 1750.