Chapter 4: Q48E (page 950)

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

Expert verified

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:

$B = \frac{μ_{0} N I}{2 π r}$

$Where μ_{0} = 4 π \times 10^{- 7} H / m$ 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,

$B = \frac{μ_{0} N I}{2 π r}$

Substitute the values in the modified equation we have,

$\begin{aligned} N = \frac{2 π r B}{μ_{0} I} \\ = \frac{2 π (0.140 m) (3.75 \times 10^{- 3} T)}{4 π \times 10^{- 7} T \cdot m / A \times 1.50 A} \\ = 1750 \end{aligned}$