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## A toroid having a square cross section, 5.49 cm on a side, and an inner radius of 18.1 cm has 450 turns and carries a current of 0.859 A. (I

Question

A toroid having a square cross section, 5.49 cm on a side, and an inner radius of 18.1 cm has 450 turns and carries a current of 0.859 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

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Physics
3 years
2021-08-03T08:44:56+00:00
2021-08-03T08:44:56+00:00 1 Answers
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## Answers ( )

Answer:(a)

4.27 x 10^-4 Telsa(b) 3.28 x 10^-4 TelsaExplanation:side of square, a = 5.49 cm

inner radius, r = 18.1 cm = 0.181 m

number of turns,N = 450

current, i = 0.859 A

(a)

The magnetic field due to a solenoid due to inner radius isB = 4.27 x 10^-4 Telsa(b)

The outer radius is R = 18.1 + 5.49 = 23.59 cm = 0.236 mThe magnetic field due to the outer radius is

B = 3.28 x 10^-4 Tesla