Question

An electron moves in a circle of radius$\mathbf{r}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{29}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{\text{m}}$with speed $\mathbf{2}\mathbf{.}\mathbf{19}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\frac{\mathbf{\text{m}}}{\mathbf{\text{s}}}$. Treat the circular path as a current loop with a constant current equal to the ratio of the electron’s charge magnitude to the period of the motion. If the circle lies in a uniform magnetic field of magnitude $\mathbf{B}\mathbf{}\mathbf{=}\mathbf{}\mathbf{7}\mathbf{.}\mathbf{10}$mT, what is the maximum possible magnitude of the torque produced on the loop by the field?

Short answer

The maximum possible magnitude of the torque produced on the loop by the field is $6.58\times {10}^{-26}\text{N}·\text{m}$.

Step-by-step solution

Write the given data

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

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

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

Determine the formula for the torque and the current

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:

$T=\frac{V}{2\pi r}$ …… (ii)

Here,$V$is the velocity of the body,$2\pi r$is the radius of the circular path.

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

$\tau =NiABsin\theta$ ……. (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, $\theta$is the angle made by the conductor with the magnetic field.

Determine the maximum possible torque

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

$$\begin{gathered} \tau =N\times \frac{q}{T}\times A\times B\times sin\theta \\ =N\times \frac{q}{2\pi r}\times \pi r^2\times B\times sin\theta \left(\text{QArea of the circular path,}A=\pi r^2\right) \\ =N\times \frac{q\times V\times r}{2}\times B\times sin\theta \end{gathered}$$

As the torque is perpendicular to the radius and magnetic field$\theta =90°$

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

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

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