A 12 V battery connected to a 6 Ω, 10 mH coil through a switch drives a constant current in the circuit. The switch is suddenly opened. Assuming that it took 1 ms to open the switch, the average emf induced across the coil would be

The correct option is 20 V.

Let us break down the physical process and calculate the average induced electromotive force (emf) step-by-step.

Step 1: Determine the initial steady-state current in the circuit
Before the switch is opened, the battery of voltage $V=12V$ is connected across the coil. The coil has a resistance of $R=6\Omega$ and an inductance of $L=10\mathrm{mH}$.
Under constant (steady-state) current conditions, the inductor acts as a short circuit, so the current is limited only by the resistance of the coil. Using Ohm's Law, the initial current $I_i$ is given by:

$$I_i=\frac{V}{R}=\frac{12V}{6\Omega }$$

Step 2: Determine the final current
When the switch is suddenly opened, the circuit is broken, and the current drops to zero. Therefore, the final current $I_f$ is:

$$I_f=0A$$

Step 3: Calculate the change in current and the time interval
The change in current ($\Delta I$) is:

$$\Delta I=I_f-I_i=0A-2A=-2A$$

The time taken to open the switch ($\Delta t$) is given as 1 ms, which in seconds is:

$$\Delta t=1\mathrm{ms}=1\times {10}^{-3}s$$

The self-inductance of the coil is $L=10\mathrm{mH}=10\times {10}^{-3}H={10}^{-2}H$.

Step 4: Calculate the average induced electromotive force (emf)
According to Faraday's law of electromagnetic induction, the average magnitude of the induced emf ($e$) across a coil is proportional to the rate of change of current:

$$e=-L\frac{\Delta I}{\Delta t}$$

Substituting the values into the equation to find the average magnitude of the induced emf:

$$e=-\left(10\times {10}^{-3}H\right)\times \frac{-2A}{1\times {10}^{-3}s}$$

$$e=10\times 2V=20V$$

Thus, the average induced emf across the coil is 20 V.