Question

A solenoid contains 300 coils of very thin wire evenly wrapped over a length of 20 cm. Each coil is 0.80 cm in diameter. If this solenoid carries a current of 4.0 A, what is the magnetic field at its center?

Answer 1

The magnetic field at the center of the solenoid is 7.53 mT.

Step-by-step explanation:

Given that,

Number of turns in the solenoid, N = 300

Length of the wire, L = 20 cm = 0.2 m

Diameter of each coil, d = 0.8 cm

Radius, r = 0.4 cm

Current in the solenoid, I = 4 A

We need to find the magnetic field at the center of solenoid. We know that the magnetic field is given by the below formula as :

`B=\mu_o nI`

n is number of turns per unit length

`B=(\mu_o NI)/(L)`

So,

`B=(4\pi * 10^(-7)* 300* 4)/(0.2)\\\\B=7.53* 10^(-3)\ T\\\\B=7.53\ mT`

So, the magnetic field at the center of the solenoid is 7.53 mT.