Understanding Induced EMF and Self-Inductance Calculation

This question tests the understanding of electromagnetic induction, specifically the concept of self-inductance ($L$) and its relationship with the average induced electromotive force (EMF) ($\epsilon_{avg}$) in a coil. The formula relating these quantities is derived from Faraday's Law:

$ \epsilon_{avg} = -L \frac{\Delta I}{\Delta t} $

Where:

• $\epsilon_{avg}$ is the average induced EMF.
• $L$ is the self-inductance of the coil.
• $\Delta I$ is the change in current through the coil.
• $\Delta t$ is the time interval over which the current changes.

We are often interested in the magnitude of the induced EMF, so the formula can be written as:

$ |\epsilon_{avg}| = L \left| \frac{\Delta I}{\Delta t} \right| $

To verify the statements (A) and (B), we rearrange the formula to calculate the self-inductance $L$ based on the given values and compare it with the provided inductance for each case.

$ L = \frac{|\epsilon_{avg}|}{\left| \frac{\Delta I}{\Delta t} \right|} $

Analyzing Statement (A): Self-Inductance Calculation

Details for Statement (A):

• Average EMF, $|\epsilon_{avg}| = 2$ V
• Time Interval, $\Delta t = 0.4$ s
• Current Change: From $5$ A to $2$ A.
• Given Self-Inductance, $L_{given} = 0.266$ mH

Step-by-Step Calculation:

1. Calculate the change in current ($\Delta I$): $ \Delta I = (\text{Final Current}) - (\text{Initial Current}) = 2 \text{ A} - 5 \text{ A} = -3 \text{ A} $
2. Calculate the rate of change of current ($\frac{\Delta I}{\Delta t}$): $ \frac{\Delta I}{\Delta t} = \frac{-3 \text{ A}}{0.4 \text{ s}} = -7.5 \text{ A/s} $
3. Calculate the self-inductance ($L$) using the formula: $ L = \frac{|\epsilon_{avg}|}{\left| \frac{\Delta I}{\Delta t} \right|} = \frac{2 \text{ V}}{| -7.5 \text{ A/s} |} = \frac{2 \text{ V}}{7.5 \text{ A/s}} $ $ L \approx 0.26666... \text{ H} $
4. Convert the calculated inductance from Henry (H) to millihenry (mH): $ L \approx 0.26666... \text{ H} \times 1000 \text{ mH/H} \approx 266.67 \text{ mH} $

Verification of Statement (A):

The calculated self-inductance is approximately $266.67$ mH. The value provided in statement (A) is $0.266$ mH. Since $266.67 \neq 0.266$, statement (A) is incorrect.

Analyzing Statement (B): Self-Inductance Calculation

Details for Statement (B):

• Average EMF, $|\epsilon_{avg}| = 2$ V
• Time Interval, $\Delta t = 0.4$ s
• Current Change: From $4$ A to $4$ A in the opposite direction. This means the change is from $+4$ A to $-4$ A (or $-4$ A to $+4$ A).
• Given Self-Inductance, $L_{given} = 0.10$ mH

Step-by-Step Calculation:

1. Calculate the change in current ($\Delta I$): If the current changes from $4$ A to $-4$ A, then: $ \Delta I = (-4 \text{ A}) - (4 \text{ A}) = -8 \text{ A} $ The magnitude of the change in current is $|\Delta I| = 8$ A.
2. Calculate the rate of change of current ($\frac{\Delta I}{\Delta t}$): $ \frac{\Delta I}{\Delta t} = \frac{-8 \text{ A}}{0.4 \text{ s}} = -20 \text{ A/s} $ The magnitude of the rate of change is $|\frac{\Delta I}{\Delta t}| = 20$ A/s.
3. Calculate the self-inductance ($L$) using the formula: $ L = \frac{|\epsilon_{avg}|}{\left| \frac{\Delta I}{\Delta t} \right|} = \frac{2 \text{ V}}{| -20 \text{ A/s} |} = \frac{2 \text{ V}}{20 \text{ A/s}} $ $ L = 0.1 \text{ H} $
4. Convert the calculated inductance from Henry (H) to millihenry (mH): $ L = 0.1 \text{ H} \times 1000 \text{ mH/H} = 100 \text{ mH} $

Verification of Statement (B):

The calculated self-inductance is $100$ mH. The value provided in statement (B) is $0.10$ mH. Since $100 \neq 0.10$, statement (B) is also incorrect.

Conclusion on Correctness

After analyzing both statements based on the principles of electromagnetic induction:

• Statement (A) claims the self-inductance is $0.266$ mH, but our calculation shows it should be approximately $266.67$ mH for the given conditions. Thus, (A) is incorrect.
• Statement (B) claims the self-inductance is $0.10$ mH, but our calculation shows it should be $100$ mH for the given conditions. Thus, (B) is incorrect.

Since both statements (A) and (B) are incorrect, the correct option is the one stating that both are incorrect.