Question

A solenoid with 300 turns has a radius of 0.040 m and is 40 cm long. If this solenoid carries a current of 12 A, what is the magnitude of the magnetic field near the center of the solenoid? (μ 0 = 4π × 10-7 T · m/A)

Answer 1

The magnitude of the magnetic field near the center of the solenoid is 11.3 mT

Step-by-step explanation:

No. of turns
`N = 300`

Radius
`r = 0.040` m

Length of solenoid
`L= 40 * 10^(-2)` m

Current
`I = 12` A

For finding the magnetic field near the center of the solenoid is given by,

`B = (\mu _(o) NI )/(L)`

Where
`\mu _(o) = 4\pi * 10^(-7)`

`B = (4 \pi * 10^(-7) * 300 * 12)/(40 * 10^(-2) )`

`B = 11.3 * 10^(-3)` T

`B = 11.3` mT

Therefore, the magnitude of the magnetic field near the center of the solenoid is 11.3 mT