Question

The inductance of a coil is$L=10H$ and resistance $R=5\Omega$. If the applied voltage of the battery is $10V$ and it switches off in$1ms,$ find induced emf of the inductor.

• A. $2\times {10}^{4}V$
• B. $1.2\times {10}^{4}V$
• C. $2\times {10}^{-4}V$
• D. None of these

Answer 1

The correct option is A

$2\times {10}^{4}V$

Step 1: Given data:

The inductance of a coil$\left(L\right)=10H$

The resistance $\left(R\right)=5\Omega$

The battery voltage$\left(V\right)=10V$

Time (t) = $1ms$ = $1\times {10}^{-3}sec$

Step 2: Calculating current

The formula,

$\varphi =LI$, where $\varphi$ is the flux, $L$ is the inductance and $I$ is the current.

Now from the Ohm's law,

$V=IR$

Substitute the value of $V$and $R$in the Ohm's law.

$$\begin{gathered} I=\frac{V}{R} \\ I=\frac{10}{5} \\ I=2A \end{gathered}$$

Therefore the current through the battery will be $2A$.

Step 3: Formula used the Faraday's law of induction

According to Faraday's law of induction, the electromotive force (EMF) generated by a coil is proportional to the rate at which its flux changes.

$\epsilon =\frac{-d\left(LI\right)}{dt}$ , $\epsilon$are the flux and $L$ inductance.

Differentiate the function and substitute the given values,

$$\begin{gathered} \epsilon =-L\left(\frac{dI}{dt}\right) \\ \epsilon =-10\times \left(\frac{2}{1\times {10}^{-3}}\right) \\ \epsilon =-2\times {10}^{4}V \end{gathered}$$

Therefore the negative sign shows the direction as it opposes the change in flux or the flow of current. therefore $\epsilon =2\times {10}^{4}V$.

Hence, the correct option is A.