Q: The instantaneous flux associated with a closed circuit of 10 ohm resistance is indicated by the following reaction φ = 6t^{2} – 5t + 1 , then the value in amperes of the induced current at t = 0.25 sec will be:

(A) 1.2

(B) 0.8

(C) 6

(D) 0.2

Solution : $\displaystyle \phi = 6 t^2 – 5 t + 1$

$\displaystyle e = – \frac{d\phi}{dt} = -12 t – 5 $