Question

A toroid having a square cross section, 5.83 cm on a side, and an inner radius of 19.5 cm has 590 turns and carries a current of 0.739 A. (It is made up of a square solenoid instead of round one bent into a doughnut shape.) What is the magnitude of the magnetic field inside the toroid at (a) the inner radius and (b) the outer radius?

Answer 1

a) The magnitude of the magnetic field inside the toroid at the inner radius is

0.000447 T = 0.447 mT

b) The magnitude of the magnetic field inside the toroid at the outer radius is

0.000344 T = 0.344 mT

Step-by-step explanation:

With a logical assumption that magnetic permeability of vacuum would be used,

the magnetic field at a distance r, from the centre of the loop is given as

B = μ₀I (N/2πr)

B = ?

μ₀ = (4π × 10⁻⁷) H/m

I = 0.739 A

N = 590 turns

For the inner radius,

r = 19.5 cm = 0.195 m

a) B = μ₀I (N/2πr)

B = (4π × 10⁻⁷ × 0.739 × 590) ÷ (2π × 0.195)

B = 0.0004471897 T = 0.000447 T = 0.447 mT

b) Magnetic field at the outer radius

r(outer) = r(inner) + length of the square

r(outer) = 19.5 cm + 5.83 cm = 25.33 cm = 0.2533 m

B = μ₀I (N/2πr)

B = (4π × 10⁻⁷ × 0.739 × 590) ÷ (2π × 0.2533)

B = 0.0003442637 T = 0.000344 T = 0.344 mT

Hope this Helps!!!