Question

A solenoid of length \(\pi\) m and 5 cm in diameter has a winding of 1000 turns and carries a current of 5A. Calculate the magnetic field at its centre along the axis.

Hint:

The key to solenoid problems is calculating \(n\), the turn density (turns per meter). Once you have \(n\), the formula \(B = \mu_0 n I\) is straightforward. Don't get distracted by the diameter unless you need to check if the solenoid is 'long' (\(L \gg D\)).

Answer 1

For a long solenoid, the magnetic field (\(B\)) at its centre is uniform and given by the formula: \[ B = \mu_0 n I \] where \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, T \cdot m/A\)), \(I\) is the current, and \(n\) is the number of turns per unit length.
Given:
Length, \(L = \pi \, m\)
Number of turns, \(N = 1000\)
Current, \(I = 5 \, A\)
First, calculate the number of turns per unit length (\(n\)): \[ n = \frac{N}{L} = \frac{1000}{\pi} \, \text{turns/m} \] Now, calculate the magnetic field \(B\): \[ B = (4\pi \times 10^{-7}) \times \left(\frac{1000}{\pi}\right) \times 5 \] The \(\pi\) terms cancel out: \[ B = 4 \times 10^{-7} \times 1000 \times 5 \] \[ B = 20 \times 10^{-4} \, T = 2 \times 10^{-3} \, T \] The magnetic field at the centre of the solenoid is \(2 \times 10^{-3}\) Tesla. (Note: The diameter of the solenoid is not needed for this calculation, assuming it is a long solenoid where end effects are negligible).