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

Correct answer: D. a^ 2 T^ 3 3R | E | = |(dϕ)/(dt)| = ( aT – 2at ) H = ∫ (E2)/(R)dt = (a2)/(R) ∫ _ 0 ^ T (T - 2t)^ 2 d t = (a2)/(R) ∫ _ 0 ^ T (T^ 2 + 4t^ 2 - 4Tt) d t = (a2)/(R)(T^ 2 t _ 0 ^ T + (4)/(3)t^ 3 _ 0 ^ T - 2Tt^ 2 _ 0 ^ T ) = a^ 2 T^ 3 3R

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Physics
Chapter: Electromagnetic Induction

NTA Abhyas 2020 · Physics · Electromagnetic Induction

Magnetic flux ϕ linked with a stationary loop of resistance R...

Magnetic flux

ϕ

linked with a stationary loop of resistance

R

varies with time

t

ϕ = at ( T – t )

. The amount of heat generated in the loop, during the time interval

T

, is

Options

A $\dfrac{aT}{3R}$
B $\dfrac{a^{2}T^{2}}{3R}$
C $\dfrac{a^{2}T^{2}}{R}$
D $\dfrac{a^{2}T^{3}}{3R}$

Correct answer

D. $\dfrac{a^{2}T^{3}}{3R}$

Step-by-step solution

| E | = \left|\dfrac{dϕ}{dt}\right| = ( aT – 2at )

H = ∫ \dfrac{E2}{R}dt = \dfrac{a2}{R}{∫}_{0}^{T}\left(T - 2t\right)^{2} d t

= \dfrac{a2}{R}{∫}_{0}^{T}\left(T^{2} + 4t^{2} - 4Tt\right) d t

= \dfrac{a2}{R}\left(T^{2}{t}_{0}^{T} + {\dfrac{4}{3}t^{3}}_{0}^{T} - {2Tt^{2}}_{0}^{T}\right) = \dfrac{a^{2}T^{3}}{3R}

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