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

volts times amps 6 x 0.5 3w