Question

An electron is shot into one end of a solenoid. As it enters the uniform magnetic field within the solenoid, its speed is $\mathbf{800}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$ and its velocity vector makes an angle of$\mathbf{30}\mathbf{°}$ with the central axis of the solenoid. The solenoid carries $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{A}$and has$\mathbf{8000}\mathbf{}\mathbf{turns}$ along its length. How many revolutions does the electron make along its helical path within the solenoid by the time it emerges from the solenoid’s opposite end? (In a real solenoid, where the field is not uniform at the two ends, the number of revolutions would be slightly less than the answer here.)

Short answer

The number of revolutions the electron makes along the helical path within the solenoid is $1.62\times {10}^6$.

Step-by-step solution

Listing the given quantities:

The velocity, $\mathrm{v}=800\mathrm{m}/\mathrm{s}$

The angle, $\mathrm{\theta }=30°$
the current$\mathrm{i}=4\mathrm{A}$

The number of turns, $\mathrm{N}=8000\text{turns}$

Understanding the concept of magnetic field and solenoid:

When an electric charge moves through a magnetic field, it experiences a force. This force is called the Lorentz force. It is given by the product of charge, speed of charge, and magnetic field. The charge moving a circular orbit will experience a centripetal force which is equal to the product of mass and square of speed divided by the radius of the orbit.

You can find the orbital radius of the electron by equating the magnitude of magnetic force and centripetal force. Then we can find the period of revolution and the total time required to travel the length of the solenoid. The number of revolutions is the ratio of the time in the solenoid to the period of the revolution.

Formula:

The equation for the magnetic field is,

$\mathrm{B}={\mathrm{\mu }}_0\mathrm{in}$ ….. (1)

Here, $\mathrm{B}$ is the magnitude of a magnetic field, ${\text{μ}}_{\text{0}}$ is the permeability constant having a value of $4\mathrm{\pi }\times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$, $\mathit{i}$ is current, and $\mathrm{n}$ is the number of turns per unit length.

The equation for the magnetic force is,

$\mathrm{F}=\mathrm{qvB}$ ….. (2)

Here,$\mathrm{F}$ is the magnetic force, $\mathrm{q}$is the charge, $\mathrm{v}$is the speed of charge, $B$ is the magnetic field.

The expression for the force is,

$\mathrm{F}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}$ ….. (3)

Here, $\mathrm{F}$ is the force,$\mathrm{v}$ is speed, and$\mathrm{r}$ is radius of the orbit.

Calculating the number of revolutions by electron:

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

$\begin{array}{c}\frac{{{\mathrm{mv}}_{\perp }}^2}{\mathrm{r}}={\mathrm{qv}}_{\perp }\mathrm{B}\\ \frac{{\mathrm{mv}}_{\perp }}{\mathrm{r}}=\mathrm{qB}\\ \mathrm{r}=\frac{{\mathrm{mv}}_{\perp }}{\mathrm{qB}}\end{array}$

Where,${\mathrm{v}}_{\perp }$is the velocity perpendicular to the magnetic field.

Write the equation for tension as below.

$\mathrm{T}=\frac{2\mathrm{\pi r}}{\mathrm{v}}$

By substituting the value of $\mathrm{r}$ in the above equation, you get

$\mathrm{T}=\frac{2\mathrm{\pi m}}{\mathrm{qB}}$ ….. (4)

Now the time required to travel the length of the solenoid is,

$\mathrm{t}=\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{v}}_{\parallel }$}\right.$

Where ${\mathrm{v}}_{\parallel }$ is the component of the velocity in the direction of the field and is equal to $\mathrm{vcos\theta }$.

As know that

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

With $\mathrm{n}=\frac{\mathrm{N}}{\mathrm{L}}$

Therefore, the magnetic field will be,

$\mathrm{B}={\mathrm{\mu }}_0\mathrm{i}\frac{\mathrm{N}}{\mathrm{L}}$

Rearrange this equation for length

$\mathrm{L}=\frac{{\mathrm{\mu }}_0\mathrm{iN}}{\mathrm{B}}$

You can find the velocity component${\mathrm{v}}_{\parallel }$as,

$\begin{array}{c}{\mathrm{v}}_{\parallel }=\mathrm{v}\text{cos}\mathrm{\theta }\\ =800\text{m}/\text{s}\times \cos 30°\\ =800\text{\hspace{0.17em}m}/\text{s}\times \frac{\sqrt{3}}{2}\\ =693\text{\hspace{0.17em}m}/\text{s}\end{array}$

Therefore, time required is

$\mathrm{t}=\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{v}}_{\parallel }$}\right.$

$\mathrm{t}=\frac{{\mathrm{\mu }}_0\mathrm{iN}}{693\mathrm{B}}$

Now the number of revolutions is given as$\raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{T}$}\right.$.

Thus, from equation(4) and (5), you have

$\begin{array}{c}\frac{\mathrm{t}}{\mathrm{T}}=\frac{{\mathrm{\mu }}_0\mathrm{iN}}{693\mathrm{B}}\times \frac{\mathrm{qB}}{2\mathrm{\pi m}}\\ =\frac{{\mathrm{\mu }}_0\mathrm{iN}}{693}\times \frac{\mathrm{q}}{2\mathrm{\pi m}}\end{array}$

$\begin{array}{c}\frac{\mathrm{t}}{\mathrm{T}}=\frac{(4\mathrm{\pi }\times {10}^{-7}\text{T}\cdot \text{m}/\text{A})\times \left(4\text{A}\right)\times \left(8000\right)}{693\text{m}/\text{s}}\times \frac{1.6\times {10}^{-19}\text{C}}{2\mathrm{\pi }\times 9.1\times {10}^{-31}\text{kg}}\\ =1.62\times {10}^6\end{array}$

The number of revolutions the electron makes along the helical path within the solenoid is $1.62\times {10}^6$.