When the current in a toroidal solenoid is changing at a rate of 0.0270 A/s, the magnitude of the induced emf is 12.5 mV. When the current equals 1.30 A, the average flux through each turn of the solenoid is 0.00245 Wb. How many turns does the solenoid have?

no of turns is 246

Explanation:

given data

current changing rate dI/dt = 0.0270 A/s

induced emf = 12.5 mV

current I = 1.30 A

average flux ∅ = 0.00245 Wb

to find out

no of turns

solution

we know from faradays law

emf = d∅/dt

and emf = L * di/dt

and L = emf / (di/dt) ... 1

so emf = N * d (∅) / dt

here we know Li = N∅

N = Li / ∅

so by equation 1

N = (emf * i) / (di∅/dt)

put here value

N = (0.00245 * 1.3) / (0.0270 * 0.00245)

N = 245.65

so no of turns is 246