Question

The self induced emf of a cell is 25 volts, when the current in it is changed at uniform rate from 10 A to 25 A is 1 s. Then the change in the energy of the inductance is
A. 437.5 J
B. 637.5J
C. 740 J
D. 540 J

Answer 1

Hint : Find the inductance (L) of the coil by using the formula for magnitude of induced emf when there is changing current in the coil, i.e. $emf=L\dfrac{di}{dt}$. Then use the formula for energy stored $E=\dfrac{1}{2}L{{i}^{2}}$, to find the change in energy.
Formula used:
$emf=L\dfrac{di}{dt}$
$E=\dfrac{1}{2}L{{i}^{2}}$

Complete step by step solution :
We know that when a current passes through a coil, it produces a magnetic field inside the coil. When this current is constant and is changing with time, the flux through the coil also changes. Due to the change in the flux through the coil, an emf is induced in the coil.
The magnitude of the induced emf is given by $emf=L\dfrac{di}{dt}$ ….. (i).
Here, emf is the emf induced, L is the inductance of the coil and $\dfrac{di}{dt}$ is the rate of change of current with time.
However, the emf is induced in such a way that it resists the change in the current. In other words, it acts as an opposing and as a result, an energy is stored in the form of magnetic field inside the coil. If the current in the coil is i, then the energy stored is given by
$E=\dfrac{1}{2}L{{i}^{2}}$ …… (ii).
Therefore, when the current in the coil changes from ${{i}_{1}}$ to ${{i}_{2}}$, then the change in the energy of the inductance will be $\Delta E={{E}_{2}}-{{E}_{1}}=\dfrac{1}{2}Li_{2}^{2}-\dfrac{1}{2}Li_{1}^{2}=\dfrac{1}{2}L\left( i_{2}^{2}-i_{1}^{2} \right)$.
In the given case, ${{i}_{1}}$=10A and ${{i}_{2}}$=25A.
This means that $\Delta E=\dfrac{1}{2}L\left( {{25}^{2}}-{{10}^{2}} \right)=\dfrac{1}{2}L\left( 625-100 \right)=\dfrac{1}{2}L(525)$
$\Rightarrow \Delta E=262.5L$ …. (iii).
Now, if we find the value of inductance L, then we will find the exact value of change in energy. For this, we will use equation (i).
It is given that the emf induced is 25 volts. Hence, emf=25V.
It is also said that the rate of change of current is uniform and the current changes from 10A to 25A in 1 s. This means that $\dfrac{di}{dt}=\dfrac{\Delta i}{\Delta t}=\dfrac{25-10}{1}=15A{{s}^{-1}}$.
Substitute the values of emf and $\dfrac{di}{dt}$ in equation (i).
$\Rightarrow 25=15L$
$\Rightarrow L=\dfrac{25}{15}=\dfrac{5}{3}H$
Substitute the value of L in equation (iii).
$\Rightarrow \Delta E=262.5\times \dfrac{5}{3}=437.5J$
Hence, the correct option is A.

Note : It is said that the current is changed at a uniform rate. This means that $\dfrac{di}{dt}=k$…. (1), where k is a constant.
$\Rightarrow di=kdt$
Integrate both the sides.
$\Rightarrow \Delta i=k\Delta t$
$\Rightarrow \dfrac{\Delta i}{\Delta t}=k$ …. (2).
From (1) and (2) we gwt,
$\dfrac{di}{dt}=\dfrac{\Delta i}{\Delta t}$
Hence, $\dfrac{di}{dt}=\dfrac{25-10}{1}=15A{{s}^{-1}}$