Question

A 15.0-cm-long solenoid with radius 0.750 cm is closely wound with 600 turns of wire. The current in the windings is 8.00 A. Compute the magnetic field at a point near the centre of the solenoid.

Short answer

The magnetic field at the center of the solenoid is 0.0402T

Step-by-step solution

Concept of the magnetic field at the center of the solenoid

The magnetic field at the center of the solenoid is given by,${\mathbf{B}}_{\mathbf{\text{solenoid}}}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}\mathbf{nl}$$\mathbf{\text{where}}\mathit{n}\mathbf{=}\frac{\mathbf{N}}{\mathbf{L}}\mathbf{\text{thus,}}{\mathit{B}}_{\mathbf{\text{solenoid}}}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}\frac{\mathbf{N}}{\mathbf{L}}\mathbf{\text{I were where,}}{\mathit{\mu }}_{\mathbf{0}}\mathbf{=}\mathbf{4}\mathit{\pi }\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{H}\mathbf{/}\mathbf{m}$is permeability of free space, N is the number of turns, I is the current passing through the coil

Find the magnetic field at the center of the solenoid

Consider a solenoid which is L= 15.0 cm long and contains N =600 circular coils R =0.750 cm in radius, it carries a current of I= 8.00 A. The magnetic field at the center of the solenoid is given by,${\mathrm{B}}_{\text{solenoid}}={\mu }_0\frac{\mathrm{N}}{\mathrm{L}}$ Substitute the values in the given equation we have,

$\begin{array}{r}{\mathrm{B}}_{\text{solenoid}}=\left(4\pi \times {10}^{-7}\mathrm{Tm}/\mathrm{A}\right)\left(\frac{600}{0.15}\right)\left(8.00\mathrm{A}\right)\\ =0.0402\mathrm{T}\end{array}$

Therefore, the magnetic field at the center of the solenoid is 0.0402T