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Question: Magnetic flux linked with a stationary loop resistance \( R \) varies with time \( t \) as \( \phi =...

Magnetic flux linked with a stationary loop resistance RR varies with time tt as ϕ=at(Tt)\phi = at\left( {T - t} \right) . Amount of heat generated in loop during time interval TT is
(A) aT3R\left( A \right){\text{ }}\dfrac{{aT}}{{3R}}
(B) a2T23R\left( B \right){\text{ }}\dfrac{{{a^2}{T^2}}}{{3R}}
(C) a2T2R\left( C \right){\text{ }}\dfrac{{{a^2}{T^2}}}{R}
(D) a2T33R\left( D \right){\text{ }}\dfrac{{{a^2}{T^3}}}{{3R}}

Explanation

Solution

Since in this question we have to find the amount of heat generated in the required time interval so for this we will use the flux equation which is given by at(Tt)at\left( {T - t} \right) . And as we know that the induced emf is the negative derivative of flux. So substituting these values in the Heat equation we will get to the solution.

Formula used:
Flux is given by
X=at(Tt)X = at\left( {T - t} \right)
Here, XX will be the flux produced.
aa , will be the acceleration
tt , will be the time.

Complete step by step solution:
Since from the formula of flux, we have the equation given as, X=at(Tt)X = at\left( {T - t} \right)
And the induced emf will be given by e=dXdte = - \dfrac{{dX}}{{dt}}
And on substituting the above values and differentiating it, the emf will be equal to
e=2atat\Rightarrow e = 2at - at
Since thee heat produced during this time period is given by
H=0Te2Rdt\Rightarrow H = \int\limits_0^T {\dfrac{{{e^2}}}{R}} dt
Now on substituting the values, we get the equation as
H=0T(2atat)2Rdt\Rightarrow H = \int\limits_0^T {\dfrac{{{{\left( {2at - at} \right)}^2}}}{R}} dt
Here, the above numerator can be written by using this formula (ab)2=a2+b22ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab , so the equation will become
H=0T4a2t2+a2T24a2TtRdt\Rightarrow H = \int\limits_0^T {\dfrac{{4{a^2}{t^2} + {a^2}{T^2} - 4{a^2}Tt}}{R}} dt
So by solving the integration, the above integration will become
H=13R(4a2T3+3a2T36a2T3)\Rightarrow H = \dfrac{1}{{3R}}\left( {4{a^2}{T^3} + 3{a^2}{T^3} - 6{a^2}{T^3}} \right)
Further solving more, the above equation will become
H=a2T33R\Rightarrow H = \dfrac{{{a^2}{T^3}}}{{3R}}
Hence, the amount of heat generated in the loop during the time interval TT is a2T33R\dfrac{{{a^2}{T^3}}}{{3R}} .
Therefore, the option (D)\left( D \right) is correct.

Note:
There are mainly two factors on which the amount of heat produced in the wire depends upon and those factors are the current flowing in the wire and the resistance of the wire. Also while solving this type of question we have to be clear about the terms in the formula as they are being restructured and then used as we have used in the above part.