Question

A long solenoid has $\mathbf{100}\mathbf{\text{\hspace{0.17em}turns/cm}}$and carries current $\mathbf{\text{i}}$An electron moves within the solenoid in a circle of radius $\mathbf{\text{2.30cm}}$perpendicular to the solenoid axis. The speed of the electron is $\mathbf{0}\mathbf{.0460}\mathbf{c}$(c= speed of light). Find the current$\mathit{i}$in the solenoid.

Short answer

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

Step-by-step solution

Identification of given data

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

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

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

Significance of magnetic field

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:

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

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

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

(a) Determining the current i in the solenoid  

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

Substituting this in the equation for radius

$\mathrm{r}=\frac{\mathrm{mv}}{{\mathrm{q\mu }}_0\mathrm{in}}$

Rearranging this equation for current

$\begin{array}{c}\mathrm{i}=\frac{\mathrm{mv}}{{\mathrm{q\mu }}_0\mathrm{rn}}\\ =\frac{9.1\times {10}^{-31}\text{\hspace{0.17em}kg}\times 0.0460\times 3\times {10}^8\text{m/s}}{1.6\times {10}^{-19}\text{\hspace{0.17em}C}\times 4\mathrm{\pi }\times {10}^{-7}{\text{\hspace{0.17em}NA}}^{\text{-2}}\times 0.023\text{\hspace{0.17em}m}\times \left(\frac{100}{0.01}\right)\text{turns/m}}\\ =0.272\mathrm{A}\end{array}$

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