A solenoid 3.0 cm long consists of 5748 loops of wire. If the magnetic field inside the solenoid is 1.0T. what is the magnitude of the current that flows through it? A) 52 A B) 0.24 A C) 4.2 A D) 3.0 A

C. I = 4.2A.

The magnetic field inside a solenoid is given by the equation:

B = μ₀NI/L

Clearing I for the equation above.

I = BL/μ₀N

With B = 1.0T, L = 3 x 10⁻²m, μ₀ = 4π x 10⁻⁷T. m/A and N = 5748turns

I = [ (1.0T) (3 x 10⁻²m) ]/[ (4π x 10⁻⁷T. m/A) (5748turns) ]

I = 4.15 ≅ 4.2A