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 t^{2}

$ \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} $

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