Q. A magnetic flux through a stationary loop with a resistance R varies during the time interval τ as , φ = at (τ – t). Find the amount of heat generated in the loop during that time
(a) $ \displaystyle \frac{a\tau^2}{2R}$
(b) $\displaystyle \frac{a^2 \tau^3}{2R}$
(c) $ \displaystyle \frac{2 a^2 \tau^3}{3R}$
(d) $ \displaystyle \frac{a \tau}{3 R}$
Ans: (b)
Sol: φ = at (τ – t) = a t τ – a t2
$ \displaystyle e = -\frac{d\phi}{dt} $
$ \displaystyle e = -(a \tau – 2at ) = a (2 t – \tau) $
$ \displaystyle H = \int_{0}^{\tau}\frac{e^2}{R}dt $
$ \displaystyle = \frac{a^2}{R}\int_{0}^{\tau} (2t-\tau)^2 dt$
$ \displaystyle = \frac{a^2}{R}\int_{0}^{\tau} (4t^2 -4\tau t +\tau^2) dt$
$ \displaystyle = \frac{a^2}{R} [\frac{4t^3}{3} – \frac{4 \tau t^2}{2} + \tau^2 t]_{0}^{\tau} $
$ \displaystyle = \frac{a^2}{R} [\frac{4 \tau^3}{3} – 2 \tau^3 + \tau^3 ] $
$ \displaystyle = \frac{a^2}{R} [\frac{4 \tau^3}{3} – \tau^3 ] $
$ \displaystyle = \frac{a^2}{R} .\frac{ \tau^3}{3} $
$ \displaystyle = \frac{ a^2 \tau^3}{3 R} $