Question

long solenoid that has 1 080 turns uniformly distributed over a length of 0.390 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

28.72 mA

Explanation:

The computation is shown below:-

The expression for the magnetic field at the center

`B = \mu_0\ nI`

Where I indicate current in the solenoid

n indicates the number of turns per unit length of the solenoid

indicates permeability of free space

Now,

The expression for the number of turns per unit length

`n = (N)/(l)`

where

n indicates the number of turns

and l indicates the length of the solenoid

now we will combine n and b to reach the value of I

`B = \mu_o (N)/(l) I`

So,

`I = (Bl)/(\mu_0N)`

`= ((1.0 * 10^(-4)T) (0.390\ m))/(4\pi * 10^(-7)(T.m)/(A) (1,080))`

`= 28.72 * 10^-3\ A`

= 28.72 mA