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Chapter 19: Problem 55

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?

Short Answer

Expert verified
Answer: The magnitude of the torque on the square loop of wire is 0.001094 N鈰卪, and the net magnetic force on the loop is 0.

Step by step solution

01

Calculate the area of the loop

We are given the side length of the square loop, which is 3.0 cm. To find the area, we can use the formula for the area of a square: \(A = s^2\), where \(s\) is the side length. Convert the side length to meters and then calculate the area: \(A = (0.03 \mathrm{m})^2 = 0.0009 \mathrm{m^2}\).
02

Calculate the sine of the angle

We are given the angle between the magnetic field and the plane of the loop, which is 37 degrees. Calculate the sine of this angle: \(\sin(37^{\circ}) = 0.6018\).
03

Calculate the torque

Now we have all the information needed to calculate the torque on the loop. Using the formula for the magnetic torque, we can find the magnitude of the torque: \(\tau = nIA\sin(\theta)B = (1)(3.0 \mathrm{A})(0.0009 \mathrm{m^2})(0.6018)(0.67 \mathrm{T}) = 0.001094 \mathrm{N \cdot m}\).
04

Find the net magnetic force on the loop

The net magnetic force on a closed current-carrying loop is always zero: \(F_\text{net} = 0\).
05

Write the final answers

The magnitude of the torque on the square loop of wire is \(\tau = 0.001094\,\mathrm{N\cdot m}\), and the net magnetic force on the loop is \(F_\text{net} = 0\).

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