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  • A magnetic flux through a stationary loop with a resistance R varies during the time interval as <img alt="\mathrm{\tau \text { as } \phi=\text { at }(\tau-t)}" src="https://entrancecorner.onc

A magnetic flux through a stationary loop with a resistance R varies during the time interval as \mathrm{\tau \text { as } \phi=\text { at }(\tau-t)} . What is the amount of heat generated in the loop during that time?

Option: 1

\mathrm{\frac{a^2 \tau^3}{4 R}}


Option: 2

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


Option: 3

\mathrm{\frac{a^2 \tau^3}{6 R}}


Option: 4

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


Answers (1)

best_answer

The flux through the stationary loop is \mathrm{\phi=a t(\tau-t)}

Induced emf,

\mathrm{\varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\mathrm{d}}{\mathrm{dt}}[\mathrm{at}(\tau-\mathrm{t})]=-[\mathrm{a} \tau-2 \mathrm{at}]=(2 \mathrm{at}-\mathrm{a} \tau)}

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

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

Hence, the total amount of heat generated is

\mathrm{\begin{aligned} & Q=\int_0^\tau \frac{(2 a t-a \tau)^2}{R} d t=\frac{1}{R} \int_0^\tau\left(4 a^2 t^2+a^2 \tau^2-4 a^2 \tau t\right) d t \\ & =\frac{1}{R}\left[\frac{4}{3} a^2 t^3+a^2 \tau^2 t-\frac{4}{2} a^2 \tau t^2\right]_0^\tau=\frac{1}{3} \frac{a^2 \tau^3}{R} \end{aligned}}

 

Posted by

HARSH KANKARIA

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