Answer:
This is the complete question,
A long solenoid that has 1 040 turns uniformly distributed over a length of 0.390 m produces a magnetic field of magnitude 1.00 x [tex]10^{-4}[/tex] T at its center. What current is required in the windings for that to occur ? mA
The current required would be 29.84 mA
Explanation:
Given that
The number of turns N = 1040 turns
length of the solenoid L= 0.390 m
the magnetic field B = 1.00 x [tex]10^{-4}[/tex] T
The magnetic field at the center of a solenoid can be obtained with the expression bellow;
B = μ[tex]_{0}[/tex] n l ............1
Where μ[tex]_{0}[/tex] is the permeability of free space = 4π x [tex]10^{-7}[/tex] T.m/A
I is the current on the solenoid
n is the turns per unit length = N/L
Substituting the new value of n into equation 1 we have;
B = μ[tex]_{0}[/tex] N l / L
making the current (I) the subject formula, it would be;
I = BL /μ[tex]_{0}[/tex] N
I = (1.00 x [tex]10^{-4}[/tex] T x 0.390 m) / ( 4π x [tex]10^{-7}[/tex] T.m/A x 1040)
I = 0.000039 / 0.0013069
I = 0.029841
I = 29.84 x [tex]10^{-3}[/tex] A
I = 29.84 mA
Therefore the current required would be 29.84 mA
