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# The self-induced emf of a coil is $20volts$, when the current in it is changed at a uniform rate from $10A$ to $25A$ in$2s$. The change in the energy of the inductance is____________. (in $J$)

Last updated date: 21st Jun 2024
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Hint: Recall that the change in energy is related to the self inductance of the coil and the change in current. First, we need to find the value of the self inductance of the coil, for that, remember the inductance is related to the rate of change of current with respect to the time and the induced emf. Further calculation should be done with the required unit conversions.

Complete step by step solution:
It is given the question that the self-induced emf of a coil is $20volts$.
Change in current is from $10A$ to $25A$.
Time taken for the uniform change of the current is $2s$.
We need to find the value of the self-inductance during this time.
For that we take the formula which relates the voltage and the self-inductance.
It is known that, $v = L\dfrac{{dI}}{{dt}}$
Where, $v$ is the self-induced emf of a coil
$L$ is self-inductance.
$dI$ is the change in the current at uniform rate
$dt$ is the time taken for the uniform change of the current.
Applying the values of the known values in the above equation we get,
$v = L\dfrac{{dI}}{{dt}}$
$\Rightarrow 20 = L\dfrac{{(25 - 10)}}{2}$
$L = \dfrac{{40}}{{15}} = \dfrac{8}{3}H$
Here, in the question we are asked to find the value of the change in the energy.
Change in energy, $\Delta E = \dfrac{1}{2}L{I^2}$
Here, the value of ${I^2}$ is given by the difference between the square of the final current and the square of the initial current.
That is, ${I^2} = ({I_2}^2 - {I_1}^2)$
We know The final current,${I_2}$ = $25A$
The initial current ${I_1}$=$10A$
Applying these values to the equation for the change in energy,
$\Delta E = \dfrac{1}{2}L{I^2}$
$\Rightarrow \Delta E = \dfrac{1}{2} \times \dfrac{8}{3}\left( {{{25}^2} - {{10}^2}} \right)$
$\Rightarrow \Delta E = \dfrac{1}{2} \times \dfrac{8}{3}\left( {625 - 100} \right)$
$\Rightarrow \Delta E = \dfrac{8}{6} \times 525$
$\therefore \Delta E = 700J$

That is the change in energy when the current is changed at a uniform rate from $10A$ to $25A$ in $2s$ will be equal to, $\Delta E = 700J$.

Note: This energy is actually stored in the magnetic field generated by the current flowing through the inductor. In a pure inductor, the energy is stored without loss, and is returned to the rest of the circuit when the current through the inductor is ramped down, and its associated magnetic field collapses.