Question

Flux in a closed circuit varies with time t according to the equation v is equal to 2 t squared minus 5 t + 8 bieber the magnitude of average induced emf between t is equal to zero to 28 is equal to point 5 second

Answer 1

Answer: 4V.

Explanation:

Since, e.m.f. induced is equal to negative of the rate of change of magnetic flux, and here, the flux is given as:

$\phi = 2t^2 - 5t + 8 W$,

The rate of change of magnetic flux is:

$\frac{d\phi}{dt} = 4t - 5 W/s$

Which, will give the magnitude of the emf induced.

Now, the average of a function of time, f(t), with respect to time, is given as:

$< f(t) > = \frac{\int\limits^a_b {f(t)} \, dt }{\int\limits^a_b {t} \, dt}$

So, the average e.m.f. induced is:

$< E(t) > = \frac{\int\limits^a_b {-\frac{d\phi}{dt}} \, dt }{\int\limits^a_b {t} \, dt}$

$=> < E(t) > = \frac{-(\phi(b) - \phi(a))}{b - a}$

And, given, b = 0.5s, and a = 0s.

Substituting, we get:

$< E(t) > = \frac{-(6 - 8)}{0.5 - 0} V$

Or, $< E(t) > = \frac{2}{0.5} V = 4V$.