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What is the radius of a tightly wound solenoid of circular cross-section that has 180 turns if a change in its internal magnetic field of 3.0 t/s causes a 6.0 acurrent to flow? the resistance of the circuit that contains the solenoid is 17 ω. the only emf source for the circuit is the induced emf.show all work for credit?

Respuesta :

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Current, I = 6.0 Amps, Resistance, R = 17 ohms, number of turns, N = 180, Rate of magnetic field, dB/dt = 3.0 T/s

Using electromagnetic equations:
Induced voltage, V = IR = 6*17 = 102 V

On the other hand, V = N(dФ/dt) => dФ/dt = V/N = 102/180 = 0.567 Wb/s
Additionally,
dФ/dt = (dB/dt)A, where A = area of the solenoid.
Therefore,
A = (dФ/dt)/(dB/dt) = 0.567/3.0 = 0.189 m^2

But, A = πr^2, where r = radius of the solenoid.

Then,
r = Sqrt (A/π) = Sqrt (0.189/π) = 0.245 m

The radius of the solenoid is 0.245 m