a. Calculate the selfinductance of a 2300turn coil that is 0.02 m in diameter. A

Question

a. Calculate the selfinductance of a 2300turn coil that is 0.02 m in diameter. Also, please calculate the selfinductance of a 3200turn solenoid that is 0.02 m in diameter and 0.04 m long.

b. A large coil of 200 turns, measuring 0.105 m in radius, is used as a primary coil to induce an Emf in a 3200turn secondary coil that measures 0.02 m in radius. The primary and secondary coils are coaxial. If the current in the primary changes at a rate of 0.1 A/s, what is the magnitude of the Emf induced in the secondary?

Explanation / Answer

(A) L = u0 N^2 A / l

= (4pi x 10^-7) (2300^2) (pi x 0.01^2) / 0.04

= 0.052 H

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L = (4pi x 10^-7) (3200^2) (pi x 0.01^2) / 0.04

= 0.101 H

(B) e = u0 N1 N2 A dI/dt

= (4pi x 10^-7) (200 x 3200) (pi x 0.02^2) (0.1)

= 1.01 x 10^-4 Volt

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