Question

A long solenoid that has 1 200 turns uniformly distributed over a length of 0.420 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

The current required winding is
`2.65*10^-^2 mA`

Step-by-step explanation:

We can use the expression B=μ₀*n*I-------1 for the magnetic field that enters a coil and

n= N/L (number of turns per unit length)

Given data

The number of turns n= 1200 turns

length L= 0.42 m

magnetic field B= 1*10^-4 T

μ₀=
`4\pi*10^-^7 T.m/A`

Applying the equation B=μ₀*n*I

I= B/μ₀*n

I= B*L/μ₀*n

`I= (1*10^-^4*0.42)/(4\pi*10^-^7*1.2*10^3 )`

`I= 2.65*10^-^2 mA`