Question

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.

Short answer

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

Step-by-step solution

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.

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:

$$\begin{gathered} \tau =NiAB\text{sin}\theta \\ i=\frac{q}{T} \\ T=\frac{2\pi r}{v} \\ A=\pi r^2 \end{gathered}$$

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