Answer

Verified

54k+ views

**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.

Recently Updated Pages

Why do we see inverted images in a spoon class 10 physics JEE_Main

A farsighted man who has lost his spectacles reads class 10 physics JEE_Main

State whether true or false The outermost layer of class 10 physics JEE_Main

Which is not the correct advantage of parallel combination class 10 physics JEE_Main

State two factors upon which the heat absorbed by a class 10 physics JEE_Main

What will be the halflife of a first order reaction class 12 chemistry JEE_Main

Other Pages

Dissolving 120g of urea molwt60 in 1000g of water gave class 11 chemistry JEE_Main

The nitride ion in lithium nitride is composed of A class 11 chemistry JEE_Main

when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Which of the following distance time graph is representing class 11 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main