Question

A long solenoid that has 1 170 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

I = 28.6mA.

The magnetic field in the center of a solenoid is given by:

B = μ₀NI/L

Clear I from the equation above, we obtain:

I = BL/μ₀N

With B = 1.00 x 10⁻⁴T, L = 0.42m, μ₀ = 4π x 10⁻⁷T.m/A  and N = 1170turns

I = [(1.00 x 10⁻⁴T)(0.42m)]/[(4π x 10⁻⁷T.m/A)(1170turns)]

I = 0.0286A

I = 28.6mA