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jonny [76]
3 years ago
10

A flat, 101 turn current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 5.61×10−4 m2, and the an

gle between its magnetic dipole moment and the field is 48.9∘. Find the strength of the magnetic field that causes a torque of 2.75×10−5 N⋅m to act on the loop when a current of 0.00339 A flows in it.
Physics
1 answer:
asambeis [7]3 years ago
3 0

Answer:

0.2 tesla

Explanation:

Number of turn, N = 101

Area, A = 5.61 x 10^-4 m^2

angle between the magnetic moment and magnetic field, θ = 48.9°

Torque, τ = 2.75 x 10^-5 Nm

Current, i = 0.00339 A

Let the magnetic field strength is B.

The formula for the torque is given by

\tau = M \times B \times Sin\theta

where, M is the magnetic moment

M = N x i x A = 101 x 0.00339 x 5.61 x 10^-4 = 1.92 x 10^-4 Am^2

By substitute the values

2.75 x 10^-5 = 1.92 x 10^-4 x B x Sin 48.9°

B = 0.189 Tesla

B  = 0.2 Tesla

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The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

To find the answer, we need to know about the thermodynamic processes.

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<h3>How much work is done?</h3>
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Where, R is the universal gas constant.

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Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

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