Question

The average emf induced in a coil is 2 V when current is changed in 0.4 s
(A) from 5 A to 2 A and the self-inductance of the coil is 0.266 mH
(B) from 4 A to 4 A in the opposite direction, the self-inductance of the coil is 0.10 mH

• A. (A) is correct, (B) is incorrect
• B. Both (A) and (B) are correct
• C. (A) is incorrect, (B) is correct
• D. Both (A) and (B) are incorrect

Hint:

Be extremely careful with SI prefixes like 'milli' (m, \(10^{-3}\)). A common mistake in exams is to misread or miscalculate units. Always perform calculations in base SI units (Volts, Amperes, Seconds, Henries) and then convert to the required prefix at the end to avoid errors.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question deals with self-inductance, which is the property of a coil to induce an EMF in itself when the current flowing through it changes. 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 due to self-inductance (\(L\)) is given by: \[ \mathcal{E} = -L \frac{\Delta I}{\Delta t} \] where \(\Delta I\) is the change in current and \(\Delta t\) is the time interval over which the change occurs. We are interested in the magnitude, so we use: \[ |\mathcal{E}| = L \left| \frac{\Delta I}{\Delta t} \right| \implies L = \frac{|\mathcal{E}| \cdot \Delta t}{|\Delta I|} \]

Step 3: Detailed Explanation:
Given common data:
Average induced EMF, \(|\mathcal{E}| = 2 \, \text{V}\).
Time interval, \(\Delta t = 0.4 \, \text{s}\).
Analysis of Statement (A):
The current changes from \(I_{initial} = 5 \, \text{A}\) to \(I_{final} = 2 \, \text{A}\). The change in current is \(\Delta I = I_{final} - I_{initial} = 2 - 5 = -3 \, \text{A}\). The magnitude of the change is \(|\Delta I| = 3 \, \text{A}\). Now, let's calculate the self-inductance \(L\): \[ L = \frac{2 \, \text{V} \times 0.4 \, \text{s}}{3 \, \text{A}} = \frac{0.8}{3} \, \text{H} \approx 0.267 \, \text{H} \] Converting to millihenries (mH): \(0.267 \, \text{H} = 267 \, \text{mH}\). Statement (A) says the self-inductance is 0.266 mH. Our calculated value is approximately 267 mH. There is a factor of 1000 difference. Thus, statement (A) is incorrect.
Analysis of Statement (B):
The current changes from \(I_{initial} = 4 \, \text{A}\) to \(I_{final} = -4 \, \text{A}\) (4 A in the opposite direction). The change in current is \(\Delta I = I_{final} - I_{initial} = -4 - 4 = -8 \, \text{A}\). The magnitude of the change is \(|\Delta I| = 8 \, \text{A}\). Now, let's calculate the self-inductance \(L\): \[ L = \frac{2 \, \text{V} \times 0.4 \, \text{s}}{8 \, \text{A}} = \frac{0.8}{8} \, \text{H} = 0.1 \, \text{H} \] Converting to millihenries (mH): \(0.1 \, \text{H} = 100 \, \text{mH}\). Statement (B) says the self-inductance is 0.10 mH. Our calculated value is 100 mH. Again, there is a factor of 1000 difference. Thus, statement (B) is incorrect.

Step 4: Final Answer:
Since both statements (A) and (B) are incorrect, the correct option is (D).