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Marina86 [1]
3 years ago
5

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 num

ber of turns in this solenoid, assuming that it is an ideal solenoid?
Mathematics
1 answer:
mixer [17]3 years ago
4 0

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

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