Question

A square loop of wire of side $3.0 \mathrm{~cm}$ carries $3.0 \mathrm{~A}$ of current. A uniform magnetic field of magnitude $0.67 \mathrm{~T}$ makes an angle of $37^{\circ}$ with the plane of the loop. (a) What is the magnitude of the torque on the loop? (b) What is the net magnetic force on the loop?

          A square loop of wire of side $3.0 \mathrm{~cm}$ carries $3.0 \mathrm{~A}$ of current. A uniform magnetic field of magnitude $0.67 \mathrm{~T}$ makes an angle of $37^{\circ}$ with the plane of the loop.
(a) What is the magnitude of the torque on the loop?
(b) What is the net magnetic force on the loop?

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Added by Harold A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Narayan Hari

08/13/2023

Step 1

The magnetic moment is given by the product of the current, the area of the loop, and the number of turns (which is 1 in this case). The area of the square loop is $A = (3.0 \times 10^{-2} \mathrm{m})^2$. So, the magnetic moment is: $\mu = IA = (3.0 Show more…

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A square loop of wire of side $3.0 \mathrm{~cm}$ carries $3.0 \mathrm{~A}$ of current. A uniform magnetic field of magnitude $0.67 \mathrm{~T}$ makes an angle of $37^{\circ}$ with the plane of the loop. (a) What is the magnitude of the torque on the loop? (b) What is the net magnetic force on the loop?

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