Question

A very long, straight solenoid with a cross-sectional area of 5.85 cm^2 is wound with 32 turns of wire per centimeter, and the windings carry a current of 0.245 A . A secondary winding of 2 turns encircles the solenoid at its center. When the primary circuit is opened, the magnetic field of the solenoid becomes zero in 4.75×10^−2 s .

What is the average induced emf in the secondary coil?

Answer 1

Solution:

Cross sectional area = A= 5.85 cm^2 = 5.85 x 10^-4 m^2

No. of turns = Np = 32 /cm = 3200 /m

Current thru the windings = 0.245 A

NUmber of turns of Secondary windings = Ns = 2

Time for the B field to become 0 = 4.75 e-2 s

The peak voltages and peak currents are related to the number of turns of primary and secondary as folows:

V₁/V₂ = I₂/I₁ = N₁/N₂ where the subscrits 1 correspond to primary and 2 to secondary.

First let us find the Magnetic field thru th eprimary :

B1= o N₁ I = (4 x10^-7) ( 3200)( 0.245) = 1909 x10^-7 T = 9.85 x10^-4 T

Flux thru the secondary = magnetic field in the coil x area of the coil = (9.85 x 10^-4 ) ( 5.85 x 10^-4 m^2)

                                                                                                             = 5.76 x 10^-7 Tm^2

Induced EF thru the secondary = V2 = flux X no. of turn in the secondary / time [ Faraday's law ]

                                                          = (5.76 x 10^-7) ( 2) /(0.0475) = 242.5 x 10^-7 Volts

                                                                                                          = 24.25 micro volts

                                                                                                          = 24 V