Q34 / 68 · NTA Abhyas (JEE Main) 2020 · Physics · Electromagnetic Induction

Magnetic flux $ϕ$ linked with a stationary loop of resistance $R$ varies with time $t$ as $ϕ = a t (T - t)$ . The amount of heat generated in the loop, during the time interval $T$ , is

A
$\frac{a T}{3 R}$
B
$\frac{a^{2} T^{2}}{3 R}$
C
$\frac{a^{2} T^{2}}{R}$
Correct answer
D
$\frac{a^{2} T^{3}}{3 R}$

Solution

| E | = |\frac{d ϕ}{d t}| = (a T - 2 a t)

H = \int \frac{E^{2}}{R} d t = \frac{a^{2}}{R} \int_{0}^{T} {(T - 2 t)}^{2} d t

= \frac{a^{2}}{R} \int_{0}^{T} (T^{2} + 4 t^{2} - 4 T t) d t

= \frac{a^{2}}{R} (T^{2} t_{0}^{T} + {\frac{4}{3} t^{3}}_{0}^{T} - {2 T t^{2}}_{0}^{T}) = \frac{a^{2} T^{3}}{3 R}

Hence, the correct answer is option: 'D'.

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