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 :

Answer:

[tex]I = 2.4 *10^{-2} A = 2.4 mA[/tex]

Explanation:

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

[tex]B = \frac{N \mu_o I}{L}[/tex]

where

N is number of turn of solenoids = 1140 turns

[tex]\mu_0 = 4*\pi *10^{-7} T.m[/tex]

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

[tex]1.00*10^{-4} = \frac{1140*4*\pi *10^{-7} I}{0.350}[/tex]

[tex]I = 2.4 *10^{-2} A = 2.4 mA[/tex]

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