Question

A coil of inductance 1 henry and resistance $10 \Omega$ is connected to an ideal battery of emf $50 \mathrm{V}$ at time $\mathrm{t}=0 .$ Calculate the ratio of the rate at which magnetic energy is stored in the coil to the rate at which energy is supplied by the battery at $\mathrm{t}=0.1$
sec.
A $e$
B.$\dfrac{1}{e-1}$
C $1-\dfrac{1}{\text{e}}$
D.$\dfrac{1}{e}$

Answer 1

Hint: A magnetic field surrounding the conductor is produced when a conductor carries a current. The magnetic flux resulting from this is proportional to the current. Because nature abhors rapid change, the voltage produced in the conductor (electromotive force, EMF) opposes the current change, which is also proportional to the change in magnetic flux.

Formula used:
$\mathrm{K}=\dfrac{\mathrm{dE}}{\mathrm{dt}}=\dfrac{\mathrm{d}\left(\dfrac{\mathrm{I}}{2} \mathrm{LI}^{2}\right)}{\mathrm{dt}}$

Complete answer:
The change in magnetic flux is proportional to the time-rate of the change in the current by a factor called inductance (L) if the current changes.
Given
According to question
$\begin{array}{l}
\mathrm{L}=1 \mathrm{H} \\
\mathrm{R}=10 \Omega \\
\mathrm{V}_{0}=50 \mathrm{V} \\
\mathrm{t}=0.1 \mathrm{sec}
\end{array}$
The current in $\mathrm{L}-\mathrm{R}$ circuit.
$I=1-\dfrac{1}{e}$ ------ (1)
where, $\tau=\dfrac{\mathrm{L}}{\mathrm{R}}$
$\Rightarrow \tau =\dfrac{1}{10}=0.1\text{sec}$
The rate at which magnetic energy is stored in the coil,
Differentiate potential energy of the inductor(E) to the time (t)
$\mathrm{K}=\dfrac{\mathrm{dE}}{\mathrm{dt}}=\dfrac{\mathrm{d}\left(\dfrac{1}{2} \mathrm{LI}^{2}\right)}{\mathrm{dt}}$
Here,
\[\text{(E)=}\dfrac{\text{1}}{\text{2}}\text{L}{{\text{I}}^{\text{2}}}\] and $\text{I}={{\text{I}}_{0}}\left( 1-{{\text{e}}^{-\text{t}/\tau }} \right)$
At $\text{t}=0.1\text{sec}$and $\tau =0.1\text{sec}$
$I=1-\dfrac{1}{e}$
On putting the values
We get
$\mathrm{K}=\dfrac{\mathrm{dE}}{\mathrm{dt}}=250\left(\dfrac{\mathrm{e}-1}{\mathrm{e}^{2}}\right)$ ----(2)
The energy supplied by battery, $\mathrm{E}^{\prime}=\mathrm{V}_{0} \mathrm{It}$
$\mathrm{E}^{\prime}=250\left(\dfrac{\mathrm{e}-1}{\mathrm{e}}\right)$
By putting the value of $\mathrm{I}$ at $\text{t}=0.1\text{sec}$ ----(3)
Dividing equation (2) by (1), we get
$\dfrac{\mathrm{K}}{\mathrm{E}^{\prime}}=\dfrac{1}{\mathrm{e}}$
\[\therefore \] The ratio of the rate at which magnetic energy is stored in the coil to the rate at which energy is $\dfrac{\mathrm{K}}{\mathrm{E}^{\prime}}=\dfrac{1}{\mathrm{e}}$

The correct option is (D).

Note:
The energy stored in a magnetic field is equal to the work necessary through the inductor to produce a current. By producing a voltage that in turn generates a current to oppose the change in magnetic flux, inductors oppose changes in current; the voltage is proportional to the change in current. The energy required to drive the original current must have an outlet due to energy conservation. This outlet is the magnetic field for an inductor; the energy stored by an inductor is equal to the work required to generate a current through the inductor.