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4 years ago

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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. (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

1 answer:

joja [24]4 years ago

7 0

Answer:

(a) 4.27 x 10^-4 Telsa

(b) 3.28 x 10^-4 Telsa

Explanation:

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 is

$B = \frac{\mu_{0}Ni}{2\pi r_{inner}}$

$B = \frac{4\pi\times 10^{-7}\times 450\times 0.859}{2\pi \times 0.181}$

B = 4.27 x 10^-4 Telsa

(b)

The outer radius is R = 18.1 + 5.49 = 23.59 cm = 0.236 m

The magnetic field due to the outer radius is

$B = \frac{\mu_{0}Ni}{2\pi r_{outer}}$

$B = \frac{4\pi\times 10^{-7}\times 450\times 0.859}{2\pi \times 0.236}$

B = 3.28 x 10^-4 Tesla

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3 years ago

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Answer:

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