Question

The magnetic field B inside a long solenoid, carrying a current of 5.00 A, is 3.14 × 10⁻² T. Find the number of turns per unit length of the solenoid.

Answer 1

Given:
Magnitude of current, i = 5 A
Magnetic field intensity, B = 3.14 × 10⁻² T
We know that the magnetic field inside a long solenoid having n turns per unit length is given by
$$\begin{gathered} B={\mathrm{\mu }}_{0}ni \\ 3.14\times {10}^{-2}=4\mathrm{\pi }\times {10}^{-7}\times \mathrm{n}\times 5 \\ \Rightarrow n=\frac{{10}^{-2}}{20\times {10}^{-7}} \\ =5\times {10}^3=5000\mathrm{turns}/\mathrm{m} \end{gathered}$$