91Ó°ÊÓ

a) Radius of a circle, $r=5.29×{10}^{-11}\text{m}$

b) Velocity of a charge, $V=2.19×{10}^6\text{m/s}$ .

c) The magnetic field, $B=7.10\text{mT or}7.10×{10}^{-3}\text{T}$

The current flowing in a conductor is as follows:

$I=\frac{q}{T}$ …… (i)

Here,$q$is the amount of charge flow,$T$is the time taken for the flow.

The time taken by a circular body to complete a revolution is as follows:

$T=\frac{V}{2\mathrm{π}r}$ …… (ii)

Here,$V$is the velocity of the body,$2\mathrm{Ï€}r$is the radius of the circular path.

The torque acting at a point inside a magnetic field is as follows:

$\mathrm{τ}=NiABsin\mathrm{θ}$ ……. (iii)

Here, $N$is the number of turns in the coil, $i$is the current of the wire, $A$is the area of the conductor, $B$is the magnetic field, $\mathrm{θ}$is the angle made by the conductor with the magnetic field.

Using these equations (i) and (ii), the equation of torque can be given as follows:

$$\begin{gathered} \mathrm{τ}=N×\frac{q}{T}×A×B×sin\mathrm{θ} \\ =N×\frac{q}{2\mathrm{π}r}×\mathrm{π}{r}^2×B×sin\mathrm{θ}\left(\text{QArea of the circular path,}A=\mathrm{π}{r}^2\right) \\ =N×\frac{q×V×r}{2}×B×sin\mathrm{θ} \end{gathered}$$

As the torque is perpendicular to the radius and magnetic field$\mathrm{θ}=90°$

So, the value of the maximum possible torque can be given using the given data in the above equation as follows:

$$\begin{gathered} \mathrm{τ}=1×\frac{\left(1.6×{10}^{-19}\text{C}\right)×\left(2.19×{10}^6\frac{\text{m}}{\text{s}}\right)×\left(5.29×{10}^{-11}\text{m}\right)}{2}×\left(7.10×{10}^{-3}\text{T}\right)×\text{sin}\left(90°\right) \\ =6.58×{10}^{-26}\text{N}·\text{m} \end{gathered}$$

Hence, the value of the maximum possible torque is $6.58×{10}^{-26}\text{N}·\text{m}$.