Question

A solenoid of 300 turns/m is carrying a current of 5A.The length of the solenoid is 0.5m and has a radius of 1cm.Find the magnitude of magnetic field inside the solenoid.

Answer 1

The magnitude of magnetic field inside the solenoid is given as,

$$\begin{gathered} B={\mu }_{o}nI \\ where, \\ {\mu }_o=4\pi \times {10}^{-7}Tesla/Amp.m \\ n=numberofturnsperunitlength \\ I=current \end{gathered}$$
On substituting the values we get,

$$\begin{gathered} B=\left(4\pi \times {10}^{-7}Tesla/Amp.m\right)\times \left(300turns/m\right)\times \left(0.5Amp\right) \\ B=1.88\times {10}^{-4}Tesla \end{gathered}$$