Question

A solenoid 150 cm long has a radius of 4 cm. Approximately how many turns are needed to make the solenoid have a self-inductance of 0.004 H?

A. 9746 turns
B. 975 turns
C. 689 turns
D. 6892 turns

Answer 1

B.975 Turns

Step-by-step explanation:

Given that

l = 150 cm

r= 4 cm

L = 0.004 H

A=πr²

We know that self inductance given as

`L=(\mu_o N^2A)/(l)`

`N=\sqrt{(lL)/(\mu_o A)}`

Now by putting the values

`N=\sqrt{(0.004* 1.5)/(4* \pi * 10^(-7)* \pi * 0.04^2)}`

N=974.62

Therefore the number of turns N= 975

The answer is B.