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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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Most popular questions from this chapter

The torque on a loop of wire (a magnetic dipole) in a uniform magnetic field is \(\tau=N I A B\) sin \(\theta,\) where \(\theta\) is the angle between \(\overline{\mathbf{B}}\) and a line perpendicular to the loop of wire. Suppose an electric dipole, consisting of two charges \(\pm q\) a fixed distance \(d\) apart is in a uniform electric field \(\overrightarrow{\mathbf{E}}\). (a) Show that the net electric force on the dipole is zero. (b) Let \(\theta\) be the angle between \(\overrightarrow{\mathbf{E}}\) and a line running from the negative to the positive charge. Show that the torque on the electric dipole is \(\tau=q d E\) sin \(\theta\) for all angles $-180^{\circ} \leq \theta \leq 180^{\circ} .$ (Thus, for both electric and magnetic dipoles, the torque is the product of the dipole moment times the field strength times $\sin \theta .\( The quantity \)q d\( is the electric dipole moment; the quantity \)N I A$ is the magnetic dipole moment.)
(a) A proton moves with uniform circular motion in a magnetic field of magnitude 0.80 T. At what frequency \(f\) does it circulate? (b) Repeat for an electron.
You want to build a cyclotron to accelerate protons to a speed of $3.0 \times 10^{7} \mathrm{m} / \mathrm{s} .$ The largest magnetic field strength you can attain is 1.5 T. What must be the minimum radius of the dees in your cyclotron? Show how your answer comes from Newton's second law.
A \(20.0 \mathrm{cm} \times 30.0 \mathrm{cm}\) rectangular loop of wire carries \(1.0 \mathrm{A}\) of current clockwise around the loop. (a) Find the magnetic force on each side of the loop if the magnetic ficld is 2.5 T out of the page. (b) What is the net magnetic force on the loop?
An electromagnet is made by inserting a soft iron core into a solenoid. The solenoid has 1800 turns, radius \(2.0 \mathrm{cm},\) and length $15 \mathrm{cm} .\( When \)2.0 \mathrm{A}$ of current flows through the solenoid, the magnetic field inside the iron core has magnitude 0.42 T. What is the relative permeability \(\overline{K_{\mathrm{B}}}\) of the iron core? (See Section 19.10 for the definition of \(\boldsymbol{K}_{\mathbf{B}} .\) )
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