Question

A 12 V battery connected to a 6 \(\Omega\), 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

• A. 10 V
• B. 20 V
• C. 200 V
• D. 12 V

Hint:

In a steady DC circuit, an inductor acts as a short circuit (its resistance is zero unless specified, here the coil itself has resistance). The current is simply given by \(I = V/R\). The induced EMF is only generated when the current is changing, such as when a switch is opened or closed.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves the concept of self-induction in an inductor (coil). When the current through an inductor changes, it induces an electromotive force (EMF) that opposes this change. The magnitude of this induced EMF is proportional to the rate of change of the current.

Step 2: Key Formula or Approach:
The average EMF (\(\mathcal{E}\)) induced in a coil with self-inductance \(L\) is given by Faraday's law of induction: \[ \mathcal{E}_{avg} = -L \frac{\Delta I}{\Delta t} \] where \(\Delta I\) is the change in current (\(I_{final} - I_{initial}\)) and \(\Delta t\) is the time taken for this change. We are interested in the magnitude of the EMF.

Step 3: Detailed Explanation:
Part 1: Calculate the initial steady current.
When the switch is closed for a long time, a constant (steady) current flows. In this DC steady state, the inductor behaves like a simple wire, and the current is determined by the battery voltage \(V\) and the resistance \(R\). \[ I_{initial} = \frac{V}{R} = \frac{12 \, \text{V}}{6 \, \Omega} = 2 \, \text{A} \] Part 2: Determine the change in current.
When the switch is opened, the circuit is broken, and the current drops to zero. \[ I_{final} = 0 \, \text{A} \] The change in current is: \[ \Delta I = I_{final} - I_{initial} = 0 \, \text{A} - 2 \, \text{A} = -2 \, \text{A} \] Part 3: Calculate the average induced EMF.
Given data:
Inductance, \(L = 10 \, \text{mH} = 10 \times 10^{-3} \, \text{H}\).
Time interval, \(\Delta t = 1 \, \text{ms} = 1 \times 10^{-3} \, \text{s}\).
Using the formula for the magnitude of the average induced EMF: \[ |\mathcal{E}_{avg}| = \left| -L \frac{\Delta I}{\Delta t} \right| = L \left| \frac{\Delta I}{\Delta t} \right| \] \[ |\mathcal{E}_{avg}| = (10 \times 10^{-3} \, \text{H}) \times \left| \frac{-2 \, \text{A}}{1 \times 10^{-3} \, \text{s}} \right| \] The \(10^{-3}\) terms in the numerator and denominator cancel out. \[ |\mathcal{E}_{avg}| = 10 \times 2 = 20 \, \text{V} \]

Step 4: Final Answer:
The average EMF induced across the coil would be 20 V.