Question

A 32-cm-long solenoid, 1.8 cm in diameter, is to produce a 0.30-T magnetic field at its center. If the maximum current is 4.5 A, how many turns must the solenoid have?

Answer 1

Answer:

16,931 turns

Explanation:

The magnetic field produced is expressed using the formula

$B = \frac{\mu_0NI}{L}$

B is the magnetic field = 0.30T

I is the current produced in the coil = 4.5A

$\mu_0$ is the magnetic permittivity in vacuum = 1.26*10^-6Tm/A

L is the length of the solenoid = 32 cm = 0.32 m

N is the number of turns in the solenoid.

Making N the subject of the formula from the equation above;

$B = \frac{\mu_0NI}{L}\\\\BL = \mu_0NI\\\\Dividing\ both\ sides \ by \ \mu_0I\\\\\frac{BL}{\mu_0I} =\frac{\mu_oNI}{\mu_0I}$

$N = \frac{BL}{\mu_0I}$

Substituting the give values to get N;

$N = \frac{0.3*0.32}{1.26*10^{-6} * 4.5}\\\\N = \frac{0.096}{0.00000567} \\\\N = 16,931.21$

The number of turns the solenoid must have is approximately 16,931 turns