Question

A solenoid has a magnetic field of magnitude 0.50T when carrying a current of 400 A. The solenoid is 8.0 m long. What is the number of turns in this solenoid, assuming that it is an ideal solenoid?

Answer 1

The correct answer is "995.22 turns".

Explanation:

The given values are:

Magnetic field,

B = 0.50 T

Current,

i = 400 A

Length of solenoid,

L = 8.0 m

As we know,

⇒
`B=\mu_0ni`

then,

⇒
`n=(B)/(\mu_0i)`

On substituting the given values, we get

⇒
`=(0.50)/(4 \pi* 10^(-7)* 400)`

⇒
`=(0.50)/(5.024* 10^(-4))`

⇒
`=995.22 \ turns`