When the current changes from +2 A to -2 A in 0.05 s, an emf of 8 V is induced in a coil. The coeffi

Q: Why is the change in current $-4\text{ A}$ and not $0\text{ A}$?

The current changes from $+2\text{ A}$ to $-2\text{ A}$. The change is calculated as Final - Initial. So, $-2 - (+2) = -4\text{ A}$. It represents a total swing of $4\text{ A}$ in the negative direction.

Q: What does the negative sign in the formula $\varepsilon = -L \frac{\Delta I}{\Delta t}$ mean?

The negative sign represents Lenz's Law. It indicates that the induced EMF creates a current that opposes the change in the original current. If the original current is decreasing, the induced EMF tries to keep it flowing.

Q: Why do we use $\Delta I / \Delta t$ instead of the derivative $dI/dt$?

We use $\Delta I / \Delta t$ because we are given a finite time interval ($0.05\text{ s}$) over which the current changes uniformly. The derivative $dI/dt$ is used for instantaneous rates of change.

Q: What is the unit of self-inductance?

The SI unit of self-inductance is the Henry (H). It is equivalent to Volt-seconds per Ampere ($\text{V}\cdot\text{s}/\text{A}$).

Q: Would the induced EMF be different if the current changed from $-2\text{ A}$ to $+2\text{ A}$?

The magnitude of the induced EMF would be exactly the same ($8\text{ V}$), but its polarity (direction) would be reversed because the change in current $\Delta I$ would be $+4\text{ A}$ instead of $-4\text{ A}$.

Question

When the current changes from $+2	ext{ A}$ to $-2	ext{ A}$ in $0.05	ext{ s}$, an emf of $8	ext{ V}$ is induced in a coil. The coefficient of self-induction of the coil is

Options:

(A) $0.2	ext{ H}$

(B) $0.4	ext{ H}$

(C) $0.8	ext{ H}$

(D) $0.1	ext{ H}$

Sigma Percentile · Accepted Answer

### The Phenomenon of Self-Induction

Imagine a simple coil of wire. When you pass a current through it, a magnetic field is generated around it. But what happens when you rapidly change that current? The magnetic field changes, and according to Faraday's Law of Electromagnetic Induction, this changing magnetic flux induces an electromotive force (EMF) right back into the very same coil! This fascinating phenomenon is called **self-induction**.

In our problem, we are given a coil where the current is aggressively reversed. It goes from $+2	ext{ A}$ to $-2	ext{ A}$ in a mere $0.05	ext{ s}$. This rapid change induces an EMF of $8	ext{ V}$. Our goal is to find the coil's intrinsic property that dictates how strongly it reacts to this change: its **coefficient of self-induction**, denoted by $L$.

### Setting Up the Master Equation

The mathematical bridge connecting the induced EMF, the rate of change of current, and the self-inductance is given by the formula:

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

Here, $\varepsilon$ is the induced EMF, $\Delta I$ is the change in current, and $\Delta t$ is the time interval. The negative sign is a mathematical representation of **Lenz's Law**, indicating that the induced EMF always acts to oppose the change in current that created it.

### Executing the Calculation

First, let's carefully calculate the change in current, $\Delta I$. It's crucial to remember that change is always the final value minus the initial value.

$$\Delta I = I_{	ext{final}} - I_{	ext{initial}} = -2	ext{ A} - (+2	ext{ A}) = -4	ext{ A}$$

Now, we can substitute our known values into the master equation. We know $\varepsilon = 8	ext{ V}$ and $\Delta t = 0.05	ext{ s}$.

$$8 = -L \left( \frac{-4}{0.05} ight)$$

The negative signs beautifully cancel each other out. Let's simplify the fraction. Dividing by $0.05$ is the same as multiplying by $20$ (since $1 / 0.05 = 20$).

$$8 = L 	imes (4 	imes 20)$$

$$8 = L 	imes 80$$

Finally, isolating $L$, we get:

$$L = \frac{8}{80} = 0.1	ext{ H}$$

The coefficient of self-induction for this coil is $0.1	ext{ Henry}$. This value tells us exactly how "stubborn" the coil is when we try to change the current flowing through it!

Animated Solution for Physics - Electromagnetic Induction: When the current changes from $+ 2 A$ to $- 2 A$ in $0.05 s$ , an emf of $8 V$ is induced in a coil. The coefficient of self-induction of the coil is

Select Answer:

Visualized Solution

Visualizing the Coil

Faraday's Law of Induction

Substituting the Values

Simplifying the Equation

Calculating Self-Inductance

The Way Forward

The Sigma Insight: Self and Mutual Inductance

The Phenomenon of Self-Induction

Setting Up the Master Equation

Executing the Calculation

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