Q.Current in a circuit falls from $5.0\ \text{A}$ to $0.0\ \text{A}$ in $0.1\ \text{s}$. If an average emf of $200\ \text{V}$ induced, give an estimate of the self-inductance of the circuit.

The key idea is self-inductance: the induced emf opposes the change in current, given by $\mathcal{E} = -L \frac{\Delta I}{\Delta t}$.

1. The magnitude of the average induced emf is $|\mathcal{E}| = L \frac{|\Delta I|}{\Delta t}$.
2. Here, $|\Delta I| = 5.0\ \text{A} - 0.0\ \text{A} = 5.0\ \text{A}$, $\Delta t = 0.1\ \text{s}$, and $|\mathcal{E}| = 200\ \text{V}$. …

The key idea here is self-inductance — a circuit’s property that opposes a change in current by inducing an emf. When the current changes, the magnetic flux through the circuit itself changes, and that induces an emf (back emf) given by:

The negative sign is Lenz’s law: the induced emf opposes the change. But for magnitude, we drop the sign and use the average values.

Since the current falls uniformly from $5.0\ \text{A}$ to $0.0\ \text{A}$ in $0.1\ \text{s}$, the average rate of change is:

The magnitude of this rate is $50\ \text{A/s}$.

The average induced emf is given as $200\ \text{V}$. Using the magnitude form of Faraday’s law:

So:

Therefore:

$L = \frac{200}{50} = 4.0\ \text{H}$ …

Method: Faraday's Law of Self-Induction (using average emf)

This problem uses the average emf form of Faraday's law for self-inductance.

Steps

1. Recall the formula for average induced emf due to self-inductance

   The average emf induced in a circuit due to a change in its own current is:

where:

• $\mathcal{E}$ = average induced emf (in volts)
• $L$ = self-inductance (in henries)
• $\Delta I$ = change in current (in amperes)
• $\Delta t$ = time interval (in seconds)

The negative sign indicates Lenz's law (opposition to change). For magnitude, we take the absolute value.

2. Identify the given values

   • Initial current, $I_i = 5.0\ \text{A}$
   • Final current, $I_f = 0.0\ \text{A}$
   • Time interval, $\Delta t = 0.1\ \text{s}$
   • Average induced emf (magnitude), $|\mathcal{E}| = 200\ \text{V}$

3. Calculate the change in current

The magnitude of change is $|\Delta I| = 5.0\ \text{A}$.

4. Rearrange the formula to solve for $L$

   Using magnitudes: …

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting the Negative Sign in Faraday's Law

The mistake: Students often write:

and plug in values without the sign, getting confused about the answer.

Why it's wrong: The correct relation is:

The negative sign indicates Lenz's law — the induced emf opposes the change in current. When current decreases ($\frac{dI}{dt}$ is negative), the induced emf is positive (it tries to keep current flowing).

How to avoid: Always write the full equation with the sign. Then, when using magnitudes, take absolute values:

Mistake 2: Using $\Delta I = 5.0\ \text{A}$ Instead of the Change

The mistake: Some students take $\Delta I = 5.0\ \text{A}$ (the final value) or get confused about the direction of change.

Why it's wrong: The change in current is:

The magnitude of change is $|\Delta I| = 5.0\ \text{A}$.

How to avoid: Always compute $\Delta I = I_f - I_i$ explicitly. For magnitude problems, use $|\Delta I|$.

Mistake 3: Confusing $\Delta t$ with Time Constant or Period

The mistake: Students think $0.1\ \text{s}$ is the time constant ($\tau = L/R$) or the period of oscillation.

Why it's wrong: Here, $0.1\ \text{s}$ is simply the time interval over which the current changes. It has nothing to do with circuit time constants.

How to avoid: Read the problem carefully. The phrase "falls from ... to ... in 0.1 s" clearly indicates a time interval $\Delta t$, not a time constant.

Mistake 4: Incorrect Unit Handling

The mistake: Mixing up units — writing $L = \frac{200}{5/0.1}$ without tracking units.

Why it's wrong: This leads to errors in the final unit (should be henry, not ohm or volt-second).

How to avoid: Write the calculation with units:

Remember: $1\ \text{H} = 1\ \text{V} \cdot \text{s} / \text{A}$.

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