Question

A solenoid of the length $50cm$of the inner radius of $1cm$ and is made up of $500$ turns of copper wire for a current of $5A$ in it. What will be the magnitude of the magnetic field inside the solenoid?

Answer 1

Given data

1. The total number of turns in the solenoid is $500$.
2. The length of the solenoid $50cm$

Solenoid

1. A solenoid is a coil of wire, if current flow through a solenoid it will act as an electromagnet.
2. The electric field inside a solenoid is $B={\mu }_{o}nI$, where, ${\mu }_o$ is the magnetic permeability of free space, n is the number of the turns in unit length and I is the current through the circuit.

Diagram

Finding the magnetic field

As we know the magnetic field inside a solenoid is $B={\mu }_{o}nI$.

So,

$$\begin{gathered} B={\mu }_{o}nI=\left(4\mathrm{\pi }\times {10}^{-7}\right)\times \frac{500}{0.5}\times 5 \\ orB=\left(4\mathrm{\pi }\times {10}^{-7}\right)\times 5000 \\ orB=6.28\times {10}^{-3}T. \end{gathered}$$

Therefore, the magnitude of the magnetic field inside the solenoid is $6.28\times {10}^{-3}T.$