Question

A long solenoid that has 920 turns uniformly distributed over a length of 0.380 m produces a magnetic field of magnitude 1.00 10-4 T at its center. What current is required in the windings for that to occur?

Answer 1

Answer: 0.0328 A

Step-by-step explanation:

Given

No of turns
`N=920`

Length of solenoid
`L=0.380\ m`

Magnetic field
`B=10^(-4)\ T`

the magnetic field at the center of the solenoid is

`\Rightarrow B=\mu nI=\mu (N)/(L)I`

Putting values we get

`\Rightarrow 10^(-4)=4\pi * 10^(-7)* (920)/(0.380)* I\\\\\Rightarrow I=(0.380* 1000)/(920* 4\pi)=0.0328\ A`