A 10-turn ideal solenoid has an inductance of 3.5 mH. When the solenoid carries a current of 2.0 A the magnetic flux through each turn is: A) 0 B) 3.5* 10^-⁴ wb C) 7.0 * 10^-⁴ wb D) 7.0 * 10^-³ wb E) 7.0 * 10^-² wb

Question

Quizplus · Accepted Answer

The inductance L of a solenoid is related to the magnetic flux Φ through each turn and the current I by the formula L = NΦ/I, where N is the number of turns. Rearranging gives Φ = LI/N. Substituting L = 3.5 mH = 3.5 x 10^-3 H, I = 2.0 A, and N = 10, we find Φ = (3.5 x 10^-3 H)(2.0 A)/10 = 7.0 x 10^-4 Wb.

A 10-Turn Ideal Solenoid Has an Inductance of 3