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A toroid with a square cross section 3.0 cm × 3.0
cm has an inner radius of 25.0 cm. It is wound with 500
turns of wire, and it carries a current of 2.0 A. What is the
strength of the magnetic field at the center of the square
cross section?

Respuesta :

pstnonsonjoku

The magnetic field is 1.353 x 10⁻³ T.

What is the magnetic field at the center of the square cross section?

Now we know that the magnetic field is given by the formula;

B = μ₀ NI/2πr

N = number of turns of the wire

μ₀ =  permeability of free space

I = current in each turn

r = distance at which the magnetic field

B = magnetic field

Given that;

r = (a + b)/2, where a = inner radius and b= outer 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

Hence the magnetic field is 1.353 x 10⁻³ T.

Learn ore about magnetic field:https://brainly.com/question/14848188

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