Question

When the current in a toroidal solenoid is changing at a rate of 0.0210 A/s , the magnitude of the induced emf is 12.6 mV . When the current equals 1.50 A , the average flux through each turn of the solenoid is 0.00441 Wb .How many turns does the solenoid have?

Answer 1

Solution) di/dt = 0.0210 A/s

E = 12.6 mV = 12.6×10^(-3) V

i = 1.50 A

Flux , phi = 0.00441 wb

Number of turns , N = ?

E = N(d(phi)/dt)

phi = iA

A = (phi)/(i)

E = N(d(iA)/dt)

E = NA(di/dt)

E = N((phi)/(i))(di/dt)

12.6×10^(-3) = N(0.00441/1.50)(0.0210)

N = (12.6×10^(-3)×1.50)/(0.00441×0.0210)

N = 204.08

N = 204 turns