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Expert-verified Found in: Page 835 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A particle of charge q moves in a circle of radius r with speed 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.

The maximum torque exerted on the loop by a magnetic field is ${\tau }_{max}=\frac{1}{2}qvrB$

See the step by step solution

## Step 1: Given

Radius of a circle is r.

The charge on particle q.

The speed of particle is v.

The magnitude of magnetic field is B.

## Step 2: Understanding the concept

Here, we need to use the equations of torque exerted on a current loop by a magnetic field. For the maximum torque, we can consider .

Formulae:

$\tau =NiAB\text{sin}\theta \phantom{\rule{0ex}{0ex}}i=\frac{q}{T}\phantom{\rule{0ex}{0ex}}T=\frac{2\pi r}{v}\phantom{\rule{0ex}{0ex}}A=\pi {r}^{2}$

## Step 3: Calculate the maximum torque exerted on the loop by a magnetic field.

We have the equation for torque exerted on the current loop as

$\tau =NiAB\text{sin}\theta$

For the maximum torque, $sin\theta$ should be maximum, so consider . Here we are also considering the current loop of moving charge, so N=1.

${\tau }_{max}=iAB$

Now, the current due to the moving charge can be expressed as

$i=\frac{q}{T}$

But we have the equation of period in terms of velocity as

$T=\frac{2\pi r}{v}$

So, the equation for current will become

$i=\frac{qv}{2\pi r}$

Also, the area of cross-section is

$A=\pi {r}^{2}$

Substituting the equation of current and area in the equation of maximum torque, we get,

$\begin{array}{c}{\tau }_{max}=\frac{qv}{2\pi r}×\pi {r}^{2}\\ {\tau }_{max}=\frac{1}{2}qvrB\end{array}$

Hence, the maximum torque exerted on the loop by a magnetic field is ${\tau }_{max}=\frac{1}{2}qvrB$

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