A very long straight solenoid with a crosssectional area of

A very long, straight solenoid with a cross-sectional area of 6.25 cm 2 is wound with 48 turns of wire per centimeter, and the windings carry a current of 0.275 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 4.75×10²s.

Part A

What is the average induced emf in the secondary coil?

Solution

Here ,

A = 6.25 cm^2

N = 4800 turns/m

I = 0.275 A

N2 = 2

part A )

average induced emf = change in flux through the secondary coil/time

average induced emf = u0 * N * I * N2 * Area/(time )

average induced emf = 4pi *10^-7 * 4800 * 2 * 6.25 *10^-4/(4.75 *10^-2)

average induced emf = 1.58 *10^-4 V

the average induced emf in the secondary is 1.58 *10^-4 V