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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(θ) 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 θ 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(θ) 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 (θ) is such that sin(θ) = 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 ≠ 0 and I ≠ 0. A length of zero for the wire makes no physical sense, so L ≠ 0 also. That leaves us with the possibility that sin(θ) = 0.
04

Explore when sin(θ) = 0

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

An electron is moving with a constant velocity. When it enters an electric field that is perpendicular to its velocity, the electron will follow a ________ trajectory. When the electron enters a magnetic field that is perpendicular to its velocity, it will follow a ________ trajectory.

A simple galvanometer is made from a coil that consists of \(N\) loops of wire of area \(A .\) The coil is attached to a mass, \(M\), by a light rigid rod of length \(L\). With no current in the coil, the mass hangs straight down, and the coil lies in a horizontal plane. The coil is in a uniform magnetic field of magnitude \(B\) that is oriented horizontally. Calculate the angle from the vertical of the rigid rod as a function of the current, \(i\), in the coil.

The velocity selector described in Solved Problem 27.2 is used in a variety of devices to produce a beam of charged particles of uniform velocity. Suppose the fields in such a selector are given by \(\vec{E}=\left(1.00 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\right) \hat{x}\) and \(\vec{B}=(50.0 \mathrm{mT}) \hat{y} .\) Find the velocity in the \(z\) -direction with which a charged particle can travel through the selector without being deflected.

The magnitude of the magnetic force on a particle with charge \(-2 e\) moving with speed \(v=1.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) is \(3.0 \cdot 10^{-18} \mathrm{~N}\). What is the magnitude of the magnetic field component perpendicular to the direction of motion of the particle?

An electron is moving at \(v=6.00 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's magnetic field. If the field strength is \(0.500 \cdot 10^{-4} \mathrm{~T}\), what is the radius of the electron's circular path?
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