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A particle of charge \(q\) moves in a circle of radius \(r\) with speed \(\underline{v}\) Treating the circular path as a current loop with an average current, find the maximum torque exerted on the loop by a uniform field of magnitude \(B\).

Short Answer

Expert verified
The maximum torque is \(\frac{qvr}{2} B\).

Step by step solution

01

Determine the current

To treat the circular path as a current loop, we need to calculate the current. A charge completing a full circular revolution in one period is equivalent to a current. The period \(T\) is found by using the relation between speed \(v\) and radius \(r\): \(T = \frac{2\pi r}{v}\). The current \(I\) is given by \(I = \frac{q}{T} = \frac{qv}{2\pi r}\).
02

Calculate the magnetic moment

The magnetic moment \(\mu\) of the loop is given by \(\mu = I \, A\), where \(A\) is the area of the loop. The area of a circle is \(A = \pi r^2\). Therefore, \(\mu = \frac{qv}{2\pi r} \times \pi r^2 = \frac{qvr}{2}\).
03

Find the maximum torque

The torque \(\tau\) in a magnetic field \(B\) is \(\tau = \mu B \sin \theta\), where \(\theta\) is the angle between the magnetic moment and the field. The maximum torque occurs when \(\sin \theta = 1\), so \(\tau_{\text{max}} = \mu B = \frac{qvr}{2} B\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Current Loop
When discussing magnetic fields and their interactions with electrical currents, the concept of a current loop is essential. Imagine a charge moving in a circular path. Here, this moving charge can be thought of as generating an electrical current around that loop. By definition, a current loop is when current flows continuously around a closed path, creating a magnetic field in its surroundings.
To find the current in a loop like the one described, you can use the formula for current, which is the charge transferred per unit of time, represented as:
  • current, \(I = \frac{q}{T}\)
  • where \(q\) is the charge and \(T\) is the period of one complete revolution.
In this scenario, the charge moves in a circle of radius \(r\) at a speed \(v\), so the period \(T\) is calculated by \(T = \frac{2\pi r}{v}\). This leads to the conclusion that the current is \(I = \frac{qv}{2\pi r}\). Understanding this concept of the current loop helps in finding other important values like magnetic moment and torque.
Magnetic Moment
The magnetic moment is a crucial concept when analyzing how current-carrying loops interact with magnetic fields. It's essentially a measure of the strength and direction of a magnetic field produced by the current loop.
To determine the magnetic moment of a loop, you use the formula:
  • magnetic moment, \(\mu = I \cdot A\)
  • where \(I\) is the current and \(A\) is the area of the loop.
For a circular loop, the area is given by \(A = \pi r^2\). Inserting this into the equation and using the previously determined current, we get:
  • magnetic moment, \(\mu = \frac{qv}{2\pi r} \times \pi r^2 = \frac{qvr}{2}\).
This tells us the amount of torque that can be exerted by the field when subjected to a magnetic field. It's key in understanding how energy is generated or lost in magnetic systems.
Uniform Magnetic Field
A uniform magnetic field is one in which the magnetic field lines have the same strength and direction at every point. Imagine a field where each point experiences the identical magnetic force and directionality. Such fields are crucial in experiments and theories since they provide consistent conditions to observe effects like torque on a current loop.
The torque experienced by a current-carrying loop in a magnetic field is given by the formula:
  • torque, \(\tau = \mu B \sin \theta\)
  • where \(\mu\) is the magnetic moment, \(B\) is the magnetic field's strength, and \(\theta\) is the angle between the magnetic moment and the magnetic field lines.
For maximum torque, the angle should be \(90^\circ\) (since \(\sin 90^\circ = 1\)), meaning the magnetic moment and magnetic field are perpendicular. Thus, the maximum torque is \(\tau_{\text{max}} = \mu B\). In our case, it calculates to: \(\tau_{\text{max}} = \frac{qvr}{2} B\). Comprehending how torque operates in a uniform magnetic field aids in applications within electric motors and generators where such fields are common.

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Most popular questions from this chapter

A circular loop of wire having a radius of \(8.0 \mathrm{~cm}\) carries a current of \(0.20\) A. A vector of unit length and parallel to the dipole moment \(\vec{\mu}\) of the loop is given by \(0.60 \hat{\mathrm{i}}-0.80 \mathrm{j}\). (This unit vector gives the orientation of the magnetic dipole moment vector.) If the loop is located in a uniform magnetic field given by \(\vec{B}=\) \((0.25 \mathrm{~T}) \hat{\mathrm{i}}+(0.30 \mathrm{~T}) \hat{\mathrm{k}}\), find \((\mathrm{a})\) the torque on the loop (in unit-vec- tor notation) and (b) the orientation energy of the loop.

An electron has an initial velocity of \((12.0 \hat{\mathrm{j}}+15.0 \mathrm{k}) \mathrm{km} / \mathrm{s}\) and a constant acceleration of \(\left(2.00 \times 10^{12} \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}\) in a region in which uniform electric and magnetic fields are present. If \(\vec{B}=(400 \mu \mathrm{T}) \hat{\mathrm{i}}\), find the electric field \(\vec{E}\).

What uniform magnetic field, applied perpendicular to a beam of electrons moving at \(1.30 \times 10^{6} \mathrm{~m} / \mathrm{s}\), is required to make the electrons travel in a circular arc of radius \(0.350 \mathrm{~m}\) ?

A proton circulates in a cyclotron, beginning approximately at rest at the center. Whenever it passes through the gap between dees, the electric potential difference between the dees is \(200 \mathrm{~V}\). (a) By how much does its kinetic energy increase with each passage through the gap? (b) What is its kinetic energy as it completes 100 passes through the gap? Let \(r_{100}\) be the radius of the proton's circular path as it completes those 100 passes and enters a dee, and let \(r_{101}\) be its next radius, as it enters a dee the next time. (c) By what percentage does the radius increase when it changes from \(r_{100}\) to \(r_{101} ?\) That is, what is $$ \text { percentage increase }=\frac{r_{101}-r_{100}}{r_{100}} 100 \% ? $$

An electron follows a helical path in a uniform magnetic field of magnitude \(0.300 \mathrm{~T}\). The pitch of the path is \(6.00 \mu \mathrm{m}\), and the magnitude of the magnetic force on the electron is \(2.00 \times 10^{-15} \mathrm{~N}\). What is the electron's speed?

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