Electromagnetic Induction

JEE Main level, numerical answer, medium

104 of 153

Question 104 of 153

In the given figure, an inductor and resistor are connected in series with a battery of emfemf EE volt. Ea2bJs−1\dfrac{{E}^{a}}{2b}J{s}^{-1} represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of ba\dfrac{b}{a} will be _____.

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Answer: 25

The formula to calculate the current flowing through an L-R circuit is given by
i=ER[1−e−RtL]...(1)i=\dfrac{E}{R}\left[1-{e}^{\dfrac{-Rt}{L}}\right]...\left(1\right)
The energy stored within the inductor can be calculated as follows:
Einduction=12Li2=12LE2R2[1−e−RtL]2...(2)\begin{matrix}{E}_{\text{induction}} & = & \dfrac{1}{2}L{i}^{2} \\ & = & \dfrac{1}{2}L\dfrac{{E}^{2}}{{R}^{2}}{\left[1-{e}^{\dfrac{-Rt}{L}}\right]}^{2}...\left(2\right)\end{matrix}
The rate of change of energy stored within the inductor is given by
ddt(Einduction)=2[1−e−RtL]⋅e−RtL⋅RL⋅E2L2R2=E2R[e−RtL−e−2RtL]...(3)\begin{matrix}\dfrac{d}{dt}\left({E}_{\text{induction}}\right) & = & 2\left[1-{e}^{\dfrac{-Rt}{L}}\right]\cdot {e}^{\dfrac{-Rt}{L}}\cdot \dfrac{R}{L}\cdot \dfrac{{E}^{2}L}{2{R}^{2}} \\ & = & \dfrac{{E}^{2}}{R}\left[{e}^{\dfrac{-Rt}{L}}-{e}^{\dfrac{-2Rt}{L}}\right]...\left(3\right)\end{matrix}
For maximum rate,
d2Edt2=0...(4)\dfrac{{d}^{2}E}{d{t}^{2}}=0...\left(4\right)
From equation (3) and (4), it can be written that
ddt[E2R[e−RtL−e−2RtL]]=0⇒e−RtL=2e−2RtL⇒eRtL=2...(5)\begin{matrix}\dfrac{d}{dt}\left[\dfrac{{E}^{2}}{R}\left[{e}^{\dfrac{-Rt}{L}}-{e}^{\dfrac{-2Rt}{L}}\right]\right] & = & 0 \\ & \Rightarrow & {e}^{\dfrac{-Rt}{L}}=2{e}^{\dfrac{-2Rt}{L}} \\ & \Rightarrow & {e}^{\dfrac{Rt}{L}}=2...\left(5\right)\end{matrix}
From equation (3) and (5), it can be written, under maximum energy stored condition for the circuit, that
ddt(Einduction)=E2R[2−1−2−2]=E24R\begin{matrix}\dfrac{d}{dt}\left({E}_{\text{induction}}\right) & = & \dfrac{{E}^{2}}{R}\left[{2}^{-1}-{2}^{-2}\right] \\ & = & \dfrac{{E}^{2}}{4R}\end{matrix}
Substitute the values of the resistance and compare with the given expression to obtain the values of the parameters.
a=2b=2R=50\begin{matrix}a & = & 2 \\ b & = & 2R \\ & = & 50\end{matrix}
Hence, the required ratio is given by
ba=25\dfrac{b}{a}=25

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