Question

A 28 cm long solenoid 1.25 cm in diameter is to produce a field of 0.353 T at its center. How much current should the solenoid carry if it has 97 turns of the wire?

Answer 1

810.9 A

Step-by-step explanation:

Length of the solenoid = 28 cm =0.28 m

Magnetic field B =0.353 T

Number of turns N =97

We know that the magnetic field due to solenoid
`B=\mu _0ni` where
`n=(N)/(l)`

So
`0.353=(4\pi * 10^(-7)* 97* i)/(0.28)`

Current i =810.9 A

So the current in the solenoid will be 810.9 A