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Chapter 19: Problem 49

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?

Short Answer

Expert verified
Answer: The magnetic force on each side parallel to the magnetic field is 0 N, the magnetic force on each side perpendicular to the magnetic field is 0.75 N, and the net magnetic force on the loop is 1.5 N.

Step by step solution

01

Setting up the problem

Determine the lengths of the sides of the rectangular loop. Using the given dimensions, we know that the loop has two sides with lengths of 20.0 cm and two sides with lengths of 30.0 cm.
02

Calculate the magnetic force on the sides parallel to the magnetic field

Using the formula \(F = I\mathcal{L} \times B\), we calculate the magnetic force on the two sides with lengths of 20.0 cm: \(F_{1} = I\mathcal{L}_{1} \times B = (1.0\,\mathrm{A})(0.20\,\mathrm{m})(2.5\,\mathrm{T}) = 0.5\,\mathrm{N}\). For these sides, since the current is parallel to the magnetic field, there is no magnetic force acting on them, so \(F_{1} = 0\).
03

Calculate the magnetic force on the sides perpendicular to the magnetic field

Now, we calculate the magnetic force on the two sides with lengths of 30.0 cm: \(F_{2} = I\mathcal{L}_{2} \times B = (1.0\,\mathrm{A})(0.30\,\mathrm{m})(2.5\,\mathrm{T}) = 0.75\,\mathrm{N}\). For these sides, the current is perpendicular to the magnetic field, so the magnetic force is maximized.
04

Calculate the net magnetic force

We now have the magnetic forces for each of the sides: \(F_{1} = 0\) and \(F_{2} = 0.75\,\mathrm{N}\). Since the forces on the two sides parallel to the magnetic field are zero, the net magnetic force is the sum of the forces on the other two sides: \(F_{\mathrm{net}} = 2F_{2} = 2(0.75\,\mathrm{N}) = 1.5\,\mathrm{N}\). So, the magnetic force on each side parallel to the magnetic field is 0 N, the magnetic force on each side perpendicular to the magnetic field is 0.75 N, and the net magnetic force on the loop is 1.5 N.

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Most popular questions from this chapter

At a certain point on Earth's surface in the southern hemisphere, the magnetic field has a magnitude of \(5.0 \times 10^{-5} \mathrm{T}\) and points upward and toward the north at an angle of \(55^{\circ}\) above the horizontal. A cosmic ray muon with the same charge as an electron and a mass of $1.9 \times 10^{-28} \mathrm{kg}$ is moving directly down toward Earth's surface with a speed of \(4.5 \times 10^{7} \mathrm{m} / \mathrm{s} .\) What is the magnitude and direction of the force on the muon?
A uniform magnetic field points vertically upward; its magnitude is $0.800 \mathrm{T}\(. An electron with kinetic energy \)7.2 \times 10^{-18} \mathrm{J}$ is moving horizontally eastward in this field. What is the magnetic force acting on it?
A charged particle is accelerated from rest through a potential difference \(\Delta V\). The particle then passes straight through a velocity selector (field magnitudes \(E\) and \(B\) ). Derive an expression for the charge-to-mass ratio \((q / m)\) of the particle in terms of \(\Delta V, E,\) and \(B\)
The intrinsic magnetic dipole moment of the electron has magnitude $9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ In other words, the electron acts as though it were a tiny current loop with $N I A=9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ What is the maximum torque on an electron due to its intrinsic dipole moment in a \(1.0-T\) magnetic field?
A solenoid has 4850 turns per meter and radius \(3.3 \mathrm{cm}\) The magnetic field inside has magnitude 0.24 T. What is the current in the solenoid?
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