Question

A long solenoid that has 1,140 turns uniformly distributed over a length of 0.350 m produces a magnetic field of magnitude 1.00 X 10^-4 T at its center. What current is required in the windings for that to occur? Apply the expression for the magnetic field along the axis of a solenoid to find the current.

Answer 1

`I = 2.4 *10^(-2) A = 2.4 mA`

Step-by-step explanation:

The magnetic field B inside long solenoid with current I is given as

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

where

N is number of turn of solenoids = 1140 turns

`\mu_0 = 4*\pi *10^(-7) T.m`

I is current that passes through solenoid

L is length along which current pass = 0.350 m

plugging value to get required value of current

`1.00*10^(-4) = (1140*4*\pi *10^(-7) I)/(0.350)`

`I = 2.4 *10^(-2) A = 2.4 mA`