Question

A magnetic flux through a stationary loop with a resistance R varies during the time interval $\tau\varphi = at(\tau - t)$ What is the amount of heat generated in the loop during that time ?

• A. $\frac{a^{2}\tau^{3}}{4R}$
• B. $\frac{a^{2}\tau^{3}}{3R}$
• C. $\frac{a^{2}\tau^{3}}{6R}$
• D. $\frac{a^{2}\tau^{3}}{2R}$

Answer 1

Answer: (B)

$\frac{a^{2}\tau^{3}}{3R}$

Answer 2

The flux through the stationary loop is

$\varphi = at(\tau - t)$

Induced emf,

$\varepsilon = \frac{d\varphi}{dt} = - \frac{d}{dt}\lbrack at(\tau - t) = \lbrack a\tau - 2at\rbrack = (2at - a\tau)$

The amount of heat generated in the loop during a small time interval dt is

$dQ = \frac{\varepsilon^{2}}{R}dt = \frac{(2at - a\tau)^{2}}{R}dt$

Hence the total amount of heat generated is

$Q = \int_{0}^{\tau}\frac{(2at - a\tau)^{2}}{dt} = \frac{1}{R}\int_{0}^{\tau}{(4a^{2}t^{2} + a^{2}\tau^{2} - 4a^{2}t)dt}$

$= \frac{1}{R}\left\lbrack \frac{4}{3}a^{3}t^{3} + a^{2}\tau^{2}t - \frac{4}{2}a^{2}\tau t^{2} \right\rbrack_{0}^{\tau} = \frac{1}{3}\frac{a^{2}\tau^{3}}{R}$