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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⋅m, 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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Most popular questions from this chapter

The magnetic field in a cyclotron is \(0.50 \mathrm{T}\). Find the magnitude of the magnetic force on a proton with speed $1.0 \times 10^{7} \mathrm{m} / \mathrm{s}$ moving in a plane perpendicular to the field.
Find the magnetic force exerted on a proton moving east at a speed of $6.0 \times 10^{6} \mathrm{m} / \mathrm{s}\( by a horizontal magnetic ficld of \)2.50 \mathrm{T}$ directed north.
A positron \((q=+e)\) moves at \(5.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) in a magnetic field of magnitude 0.47 T. The magnetic force on the positron has magnitude \(2.3 \times 10^{-12} \mathrm{N} .\) (a) What is the component of the positron's velocity perpendicular to the magnetic field? (b) What is the component of the positron's velocity parallel to the magnetic field? (c) What is the angle between the velocity and the field?
Prove that the time for one revolution of a charged particle moving perpendicular to a uniform magnetic field is independent of its speed. (This is the principle on which the cyclotron operates.) In doing so, write an expression that gives the period \(T\) (the time for one revolution) in terms of the mass of the particle, the charge of the particle, and the magnetic field strength.
A uniform magnetic field of \(0.50 \mathrm{T}\) is directed to the north. At some instant, a particle with charge \(+0.020 \mu \mathrm{C}\) is moving with velocity \(2.0 \mathrm{m} / \mathrm{s}\) in a direction \(30^{\circ}\) north of east. (a) What is the magnitude of the magnetic force on the charged particle? (b) What is the direction of the magnetic force?
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