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

Answer:

The correct answer is "995.22 turns".

Step-by-step 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,

⇒ $n=\frac{B}{\mu_0i}$

On substituting the given values, we get

⇒    $=\frac{0.50}{4 \pi\times 10^{-7}\times 400}$

⇒    $=\frac{0.50}{5.024\times 10^{-4}}$

⇒    $=995.22 \ turns$