Question

An electromagnet has stored $648\;J$ of magnetic energy. When a current of $9\;A$ exists in its coils. What average emf is induced if the current is reduced to zero in $0.45\;s$?
A. $120V$
B. $32V$
C. $320V$
D. $240V$

Answer 1

Hint: The electromagnet is considered to be an inductor. Hence, the energy stored in the inductor always resides in the region of its magnetic field. The formula for the energy stored in a pure inductor is applied in-order to find the value of inductance as the only element that is present is the inductor. This value is used in the equation for the emf induced due to the current produced to determine the value of average emf.

Formula used:
General formula for magnetic energy which is,
$U = \dfrac{1}{2}L{I^2}$
where $U$ is magnetic energy stored, $L$ is inductance and I is the current and other formula what we have to use is $e = L\dfrac{{dI}}{{dt}}$ where $e$ is the emf induced and $\dfrac{{dI}}{{dt}}$ is the rate of change of current.

Complete step by step answer:
The problem revolves around the concept of the energy stored in an inductor. However in the question it is mentioned that the electromagnet stores the amount of magnetic energy in it. This is because an electromagnet is said to be an inductor with more number of wounds or more number of turns in it.

The question specifies to find the average emf induced when the current is reduced to zero because the average current over a time period, that is, over one complete cycle is proved to be zero. This is what is considered in a circuit with a pure inductor where the emf produced is sinusoidal and hence the current will also be sinusoidal. This is why the question means to find the emf when the current is reduced to zero.

The given values in the question are $U$ which is $648\;J$, change in current which is equal to $9\;A$ and change in time which is equal to $0.45\;s$. Now we have to use the formula for magnetic energy. This is given by the equation:
$U = \dfrac{1}{2}L{I^2}$ -----($1$)
Given, $U = 648\;J$, $I = 9\;A$ and $t = 0.45\;s$.
Now we will put the value of $U$ and $I$ in the above equation ($1$).
$648 = \dfrac{1}{2} \times L\left( {{9^2}} \right)$ .
By rearranging the terms to make the unknown value, that is, the inductance value to be the subject, we get:
$L = \dfrac{{648 \times 2}}{{{9^2}}}$

On solving we will get the value of inductance that is:
$L = 16\;H$
Let us now understand the concept behind the working of an inductor. When a current is produced in its coil an emf is produced due to the source supply voltage but there is also a back emf that is set-up in-order to oppose the increase in the applied emf.
$e = - L\dfrac{{dI}}{{dt}}$
Here we are neglecting the negative sign because we consider only the magnitude to be found out. The negative sign indicates the back emf produced.

Next, we will substitute the obtained value of $L$ in the formula of emf induced which is given by the equation and the other given values in the above equation omitting the negative sign. Therefore, we get:
$e = \dfrac{{16 \times 9}}{{0.45}}$
On solving this we will get the value of emf:
$\therefore e = 320$ $V$
Hence the emf induced in the electromagnet is $320V$.

Therefore, the correct answer is option C.

Note: The negative sign in the equation for the induced emf does not specify the magnitude or the value to be negative but instead refers to the back emf induced which opposes the flow of current and tries to reduce it. An ideal inductor connected to an ac source will not dissipate any power because the net power consumed in the half circle of the current cycle cancels out each other.