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?

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

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

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