Question

An inductor ${\text{L}}$ and a resistor ${\text{R}}$ are connected in series with a direct current source of emf ${\text{E}}$. The maximum rate at which the energy is stored in the magnetic field is
(A). $\dfrac{{{E^2}}}{{4R}}$
(B). $\dfrac{{{E^2}}}{R}$
(C). $\dfrac{{4{E^2}}}{R}$
(D). $\dfrac{{2{E^2}}}{R}\;\;\;$

Answer 1

- Hint: If a circuit having some emf value is there then in the nearby area a magnetic field will set up. So if the magnetic field does not do any work directly then it will need some entity such as an electric field which can perform the work directly. So we can say that the energy will be stored in the electric and magnetic field but here we are considering the magnetic field for the above problem to perform the suitable calculations. So for the calculation of the maximum rate at which the energy is stored we need to find the maximum rate at which the energy is being stored that means we have to find the maximum value of the power.

Complete step-by-step solution -
As we know that the formula for the energy stored will be as follows
${\text{U = }}\dfrac{1}{2}L{i^2}$,
Where,
${\text{U}}$= Energy stored
${\text{L}}$= Inductance
${\text{i}}$=current.
Now we know that the equation of the current is,
${\text{i = }}{{\text{i}}_0}(1 - {e^{\dfrac{{ - t}}{\tau }}})$
Here the rate of energy is asked, which means we have to calculate the energy at which rate it is being stored. We also know that the rate of energy is power. Thus by differentiating the formula for the stored energy we will get the rate at which the energy is being stored.
Thus firstly we will write like this,
${\text{U = }}\dfrac{1}{2}L{{\text{i}}_0}^2{(1 - {e^{\dfrac{{ - t}}{\tau }}})^2}$
Now the rate at which the energy is stored i.e. power is as follows,
P=$\dfrac{{dU}}{{dt}} = \dfrac{d}{{dt}}[\dfrac{1}{2}L{{\text{i}}_0}^2{(1 - {e^{\dfrac{{ - t}}{\tau }}})^2}]$
$= \dfrac{1}{2}{\text{L}}{{\text{i}}_0}^2 \times 2(1 - {e^{\dfrac{{ - t}}{\tau }}})\left[ { - {e^{\dfrac{{ - t}}{\tau }}} \times (\dfrac{{ - 1}}{\tau })} \right]$
$\Rightarrow$P${\text{ = }}\dfrac{{\text{L}}}{\tau }{{\text{i}}_0}^2(1 - {e^{\dfrac{{ - t}}{\tau }}}){e^{\dfrac{{ - t}}{\tau }}}$------equation (1)
So this is we got power P, now for getting the maximum value we have to differentiate this power which we have got by differentiating the stored energy and then putting that value equals to zero.
So, $\dfrac{{dP}}{{dt}}{\text{ = 0}}$
$\Rightarrow$ $\dfrac{d}{{dt}}[\dfrac{{\text{L}}}{\tau }{{\text{i}}_0}^2({e^{\dfrac{{ - t}}{\tau }}} - {e^{\dfrac{{ - 2t}}{\tau }}})] = 0$
$\Rightarrow$ $\dfrac{{\text{L}}}{\tau }{{\text{i}}_0}^2[{e^{\dfrac{{ - t}}{\tau }}}(\dfrac{{ - 1}}{\tau }) - {e^{\dfrac{{ - 2t}}{\tau }}}(\dfrac{{ - 2}}{\tau })] = 0$
$\Rightarrow$ ${e^{\dfrac{{ - t}}{\tau }}}(\dfrac{{ - 1}}{\tau }) - {e^{\dfrac{{ - 2t}}{\tau }}}(\dfrac{{ - 2}}{\tau }) = 0$
\Rightarrow - \dfrac{{{e^{\dfrac{{ - t}}{\tau }}}}}{\tau } + \dfrac{{2{e^{\dfrac{{ - 2t}}{\tau }}}}}{\tau } = 0 \\\ \Rightarrow \dfrac{{{e^{\dfrac{{ - t}}{\tau }}}}}{\tau } = \dfrac{{2{e^{\dfrac{{ - 2t}}{\tau }}}}}{\tau } \\\ \Rightarrow {e^{\dfrac{{ - t}}{\tau }}} = 2{e^{\dfrac{{ - 2t}}{\tau }}} \\\ \Rightarrow \dfrac{{{e^{\dfrac{{ - 2t}}{\tau }}}}}{{{e^{\dfrac{{ - t}}{\tau }}}}} = \dfrac{1}{2} \\\ \Rightarrow {e^{\dfrac{{ - t}}{\tau }}} = \dfrac{1}{2} \\\ \
Now we will put this value in the equation of power that is equation (1)
$\ {\text{ = }}\dfrac{{\text{L}}}{\tau }{{\text{i}}_0}^2(1 - \dfrac{1}{2})\dfrac{1}{2} \\\ = \dfrac{{\text{L}}}{{4\tau }}{{\text{i}}_0}^2 \\\ = L\dfrac{{{E^2}}}{{{R^2}}}\dfrac{1}{{4L}}R \\\ = \dfrac{{{E^2}}}{{4R}} \\\ \$
So the maximum rate at which the energy is stored in the magnetic field is $\dfrac{{{E^2}}}{{4R}}$.

Hence the answer is the option (A).

Note: When a current is flowing through a conductor then a magnetic field is produced in the nearby area and this magnetic field directly will not be doing any work but it is having some storage of energy which is the ability to work.