Q: The instantaneous flux associated with a closed circuit of 10 ohm resistance is indicated by the following reaction φ = 6t2 – 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 $
At t = 0.25 sec
$\displaystyle e = = – (12 \times 0.25 – 5 ) $
e = 2 volt
$\displaystyle i = \frac{e}{R} = \frac{2}{10}$
i = 0.2 amp