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kodGreya [7K]
1 year ago
5

A toroid with a square cross section 3.0 cm × 3.0

Physics
1 answer:
earnstyle [38]1 year ago
4 0

The magnetic field is 1.353 x 10⁻³ T.

<h3>What is the magnetic field at the center of the square cross section?</h3>

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:brainly.com/question/14848188

#SPJ1

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Answer:

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Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is L=I\omega. If we call 1 the situation when the student is at the rim and 2 the situation when the student is at r_2=1.39m from the center, then we have:

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\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s

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