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 10-4 T at its center. What current is required in the windings for that to occur

Answer 1

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
`10^(-4)` T at its center. What current is required in the windings for that to occur ? mA

The current required would be 29.84 mA

Step-by-step 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
`10^(-4)` T

The magnetic field at the center of a solenoid can be obtained with the expression bellow;

B = μ
`_(0)` n l ............1

Where μ
`_(0)` is the permeability of free space = 4π x
`10^(-7)` 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
`_(0)` N l / L

making the current (I) the subject formula, it would be;

I = BL /μ
`_(0)` N

I = (1.00 x
`10^(-4)` T x 0.390 m) / ( 4π x
`10^(-7)` T.m/A x 1040)

I = 0.000039 / 0.0013069

I = 0.029841

I = 29.84 x
`10^(-3)` A

I = 29.84 mA

Therefore the current required would be 29.84 mA