91Ó°ÊÓ

$\mathrm{n}=100/0.01\mathrm{m}$

$\mathrm{r}=2.30\mathrm{cm}=0.023\mathrm{m}$

$\mathrm{v}=0.0460\mathrm{c}$

The area in which the force of magnetism acts around a magnetic material or a moving electric charge is known as the magnetic field.

We can find the orbital radius of the electron by equating the magnitude of magnetic force and the centripetal force. For a long solenoid, we can find the current$\mathit{i}$ by using equation$\mathbf{29}\mathbf{-}\mathbf{23}$.

Formula:

$\mathrm{B}={\mathrm{μ}}_0\mathrm{in}$

$\mathrm{F}=\mathrm{qvB}$

$\mathrm{F}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}$

For a charged particle in equilibrium, the centripetal force must be equal to the magnetic force, i.e.

$\frac{{\mathrm{mv}}^2}{\mathrm{r}}=\mathrm{qvB}$

$\frac{\mathrm{mv}}{\mathrm{r}}=\mathrm{qB}$

$\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}$

This is the orbital radius of the electron.

But for solenoid, the magnetic field is$\mathrm{B}={\mathrm{μ}}_0\mathrm{in}$

Substituting this in the equation for radius

$\mathrm{r}=\frac{\mathrm{mv}}{{\mathrm{\pm çμ}}_0\mathrm{in}}$

Rearranging this equation for current

$\begin{array}{c}\mathrm{i}=\frac{\mathrm{mv}}{{\mathrm{\pm çμ}}_0\mathrm{rn}}\\ =\frac{9.1×{10}^{−31}\text{ k²µ}×0.0460×3×{10}^8\text{m/s}}{1.6×{10}^{−19}\text{ C}×4\mathrm{Ï€}×{10}^{−7}{\text{ N´¡}}^{\text{-2}}×0.023\text{″¾}×\left(\frac{100}{0.01}\right)\text{turns/m}}\\ =0.272\mathrm{A}\end{array}$

The current $\mathit{i}$in the solenoid is$\mathrm{i}=0.272\mathrm{A}$