Question

At the instant when the current in an inductor is increasing at a rate of 0.0740 A/s, the magnitude of the self-induced emf is 0.0130 V.
a) What is the inductance of the inductor?
b) If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is 0.720 A?

Answer 1

Answer:

(a) Inductance will be 0.1756 Henry

(b) Flux through each turn will be $3.16\times 10^{-4}Wb$        

Explanation:

We have given the rate of change of current $\frac{di}{dt}=0.0740A/sec$

Magnitude of self induced emf e = 0.0130 volt

(a) We know that in inductor self induced emf is given by

$e=L\frac{di}{dt}$

So $0.0130=L\times 0.0740$

L = 0.1756 Henry

Number of turns N = 400

Now magnetic flux through each turns is given by $\Phi =\frac{Li}{N}=\frac{0.1756\times 0.720}{400}=3.16\times 10^{-4}Wb$