91Ó°ÊÓ

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:

$$\begin{gathered} \mathrm{τ}=NiAB\text{sin}\mathrm{θ} \\ i=\frac{q}{T} \\ T=\frac{2\mathrm{π}r}{v} \\ A=\mathrm{π}{r}^2 \end{gathered}$$

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

$\mathrm{τ}=NiAB\text{sin}\mathrm{θ}$

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

${\mathrm{Ï„}}_{max}=iAB$

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

$T=\frac{2\mathrm{Ï€}r}{v}$

So, the equation for current will become

$i=\frac{qv}{2\mathrm{Ï€}r}$

Also, the area of cross-section is

$A=\mathrm{Ï€}{r}^2$

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

$\begin{array}{c}{\mathrm{τ}}_{max}=\frac{qv}{2\mathrm{π}r}×\mathrm{π}{r}^2\\ {\mathrm{τ}}_{max}=\frac{1}{2}qvrB\end{array}$

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