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Chapter 27: Problem 14

A current-carrying wire is positioned within a large, uniform magnetic field, \(\vec{B}\). However, the wire experiences no force. Explain how this might be possible.

Short Answer

Expert verified
In order for a current-carrying wire within a uniform magnetic field to experience no magnetic force, the angle between the magnetic field's direction and the current's direction must be either 0 degrees (aligned in the same direction) or 180 degrees (aligned in opposite directions). It is under these conditions that the sine of the angle becomes zero, resulting in a magnetic force of zero as well.

Step by step solution

01

State the formula for magnetic force on a current-carrying wire

The magnetic force on a current-carrying wire can be calculated using the formula: F = ILBsin(胃) where F is the magnetic force, I is the current flowing through the wire, L is the wire's length, B is the magnetic field's strength, and 胃 is the angle between the magnetic field and the current's direction.
02

Analyze the terms contributing to magnetic force

For the wire to experience no magnetic force, the formula F = ILBsin(胃) must be equal to zero. This can happen if any of the following conditions are met: 1. The strength of the magnetic field (B) is zero. 2. No current flows through the wire (I = 0). 3. The length of the wire (L) is zero. 4. The angle between the magnetic field and the current's direction (胃) is such that sin(胃) = 0.
03

Determine the possible scenario for no magnetic force

Since the exercise states that the magnetic field is uniform and the wire is carrying a current, that means B 鈮� 0 and I 鈮� 0. A length of zero for the wire makes no physical sense, so L 鈮� 0 also. That leaves us with the possibility that sin(胃) = 0.
04

Explore when sin(胃) = 0

The sine function equals zero at two possible angles: 胃 = 0 degrees and 胃 = 180 degrees. If 胃 = 0 degrees, the magnetic field and the current flow are aligned in the same direction. If 胃 = 180 degrees, the magnetic field and current flow are in opposite directions.
05

Explain the scenario for no magnetic force

The wire experiences no magnetic force if the angle between the direction of the magnetic field and the direction of current flow is either 0 degrees (aligned in the same direction) or 180 degrees (aligned in opposite directions). In both cases, the sine of the angle is zero, which results in a magnetic force of zero.

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Most popular questions from this chapter

A proton moving at speed \(v=1.00 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters a region in space where a magnetic field given by \(\vec{B}=\) \((-0.500 \mathrm{~T}) \hat{z}\) exists. The velocity vector of the proton is at an angle \(\theta=60.0^{\circ}\) with respect to the positive \(z\) -axis. a) Analyze the motion of the proton and describe its trajectory (in qualitative terms only). b) Calculate the radius, \(r\), of the trajectory projected onto a plane perpendicular to the magnetic field (in the \(x y\) -plane). c) Calculate the period, \(T,\) and frequency, \(f\), of the motion in that plane. d) Calculate the pitch of the motion (the distance traveled by the proton in the direction of the magnetic field in 1 period).

The work done by the magnetic field on a charged particle in motion in a cyclotron is zero. How, then, can a cyclotron be used as a particle accelerator, and what essential feature of the particle's motion makes it possible? A blue problem number indicates a worked-out solution is available in the Student Solutions Manual. One \(\bullet\) and two \(\bullet\) indicate increasing level of problem difficulty.

Show that the magnetic dipole moment of an electron orbiting in a hydrogen atom is proportional to its angular momentum, \(L: \mu=-e L / 2 m,\) where \(-e\) is the charge of the electron and \(m\) is its mass.

An electron in a magnetic field moves counterclockwise on a circle in the \(x y\) -plane, with a cyclotron frequency of \(\omega=1.2 \cdot 10^{12} \mathrm{~Hz}\). What is the magnetic field, \(\vec{B}\) ?

A circular coil with a radius of \(10.0 \mathrm{~cm}\) has 100 turns of wire and carries a current, \(i=100 . \mathrm{mA} .\) It is free to rotate in a region with a constant horizontal magnetic field given by \(\vec{B}=(0.0100 \mathrm{~T}) \hat{x}\). If the unit normal vector to the plane of the coil makes an angle of \(30.0^{\circ}\) with the horizontal, what is the magnitude of the net magnetic torque acting on the coil? 27.61 At \(t=0\) an electron crosses the positive \(y\) -axis (so \(x=0\) ) at \(60.0 \mathrm{~cm}\) from the origin with velocity \(2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. It is in a uniform magnetic field. a) Find the magnitude and the direction of the magnetic field that will cause the electron to cross the \(x\) -axis at \(x=60.0 \mathrm{~cm}\). b) What work is done on the electron during this motion? c) How long will the trip take from \(y\) -axis to \(x\) -axis?
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