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

Answer:

$I = 2.4 *10^{-2} A = 2.4 mA$

Explanation:

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

$B = \frac{N \mu_o I}{L}$

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

$1.00*10^{-4} = \frac{1140*4*\pi *10^{-7} I}{0.350}$

$I = 2.4 *10^{-2} A = 2.4 mA$