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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
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Hint: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.