Question

An 87 mH solenoid inductor is wound on a form 0.80 m in length and 0.10 m in diameter. A coil is tightly wound around the solenoid at its center. the coil resistance is 6.1 ohms. the mutual inductance of the coil and solenoid is 30 micro H. At a given instant, the current in the solenoid is 980 mA, and is decreasing at the rate of 2.5 A/s the number of turns in the winding of the solenoid is closest to:

Answer 1

Mutual inductance L = N*phi/i

and flux phi = u0iNA/l

Now L = u0*N^2*A/l

N = sqrt(L*l/u0A)

= sqrt(87e-3*0.80/(4pi*10^-7*pi*0.05^2))

= 2655 turns answer