Question

A solenoid of length $1.0m$, radius of $1cm$ and total turns $1000$ wound on it, carries a current of $5A$. Calculate the magnitude of the axial magnetic field inside the solenoid. If an element was to move with a speed of $104m/s$ along the axis of this current carrying solenoid, what would be the force experienced by this electron?

Answer 1

Use the formula of solenoid to find the magnetic field inside a solenoid and then apply the Lorentz’s force formula for the electron to find the force acting on it.

Complete step by step answer:
A solenoid may be a coil of wire during a corkscrew shape wrapped around a piston, often made from iron. The device creates a magnetic flux from current and uses the magnetic flux to produce and sustain linear motion. As altogether electromagnets, a magnetic flux is made when an electrical current passes through the wire. In a solenoid, the electromagnetic field causes the piston to either move backward or forward, and that is how motion is made by a solenoid coil.

In our case, let the solenoid have a length $l = 1m$ and let the radius of the solenoid be $r$ where $r = 1cm$and the total turns be n=1000. The current in the solenoid is $I = 5A$ and the speed of the element is given as $v = 104\dfrac{m}{s}$.
We know that the electric field of the solenoid is given by:

$$B = {\mu _0}nI \\\ \Rightarrow B = {\mu _0}\dfrac{N}{l}I \\\$$

Where, ${\mu _0}$is the coefficient of permeability, B is the magnetic field, $I$ is the current in the solenoid, n is the total turns per length of the solenoid, N is the total number of turns and$l$is the length of the solenoid.
Now, magnetic field here will be:

Now the force experienced by the electron would be obtained by the Lorentz’s formula which is as below:
$F = qvB$
Where F is the force experienced by the electron in the magnetic field, q is the charge of the electron, v is the velocity of the electron and B is the magnetic field.

$$F=1.6\times {{10}^{(-19)}}\centerdot 104\centerdot 6.28\times {{10}^{(-3)}} \\\ \therefore F=1.044\times {{10}^{(-19)}}N \\\$$

Hence, the force experienced by this electron is $1.044\times {{10}^{(-19)}}N$.

Note: The actual Lorentz’s force formula has a vector form where the velocity and magnetic field have a cross product between them, while in this case; they both are perpendicular to each other and hence can be directly multiplied.