Suggested languages for you:

Americas

Europe

90P

Expert-verifiedFound in: Page 835

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 ${\tau}_{max}=\frac{1}{2}qvrB$

Radius of a circle is r.

The charge on particle q.

The speed of particle is v.

The magnitude of magnetic field is B.

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:

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

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,

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

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

94% of StudySmarter users get better grades.

Sign up for free