Question

A toroid with a square cross section 3.0 cm ✕ 3.0 cm has an inner radius of 25.1 cm. It is wound with 600 turns of wire, and it carries a current of 3.0 A.
What is the strength of the magnetic field (in T) at the center of the square cross section?

Answer 1

Answer:

B = 1.353 x 10⁻³ T

Explanation:

The Magnetic field within a toroid is given by

B = μ₀ NI/2πr, where N is the number of turns of the wire, μ₀ is the permeability of free space, I is the current in each turn and r is the distance at which the magnetic field is to be determined from the center of the toroid.

To find r we need to add the inner radius and outer radius and divide the value by 2. Hence,

r = (a + b)/2, where a is the inner radius and b is the outer radius which can be found by adding the length of a square section to the inner radius.

b = 25.1 + 3 = 28.1 cm

a = 25.1 cm

r = (25.1 + 28.1)/2 = 26.6 cm = 0.266m

B = 4π x 10⁻⁷ x 600 x 3/2π x 0.266

B = 1.353 x 10⁻³ T

The strength of the magnetic field at the center of the square cross section is 1.3 x 10⁻³ T