When the current changes from `+2A` to `-2A` in `0.05s`, and emf of `8V` is induced in a coil. The coefficient of self-induction of the coil is

Options

A. 0.2 H
B. 0.4 H
C. 0.8 H
D. 0.1 H

Answer

To find the coefficient of self-induction (L) of the coil, we can use the formula for the induced electromotive force (emf) in an inductor:

\[ \text{emf} (E) = -L \frac{di}{dt} \]

Where:
- \(E\) is the induced emf (in volts),
- \(L\) is the self-inductance (in henries),
- \(\frac{di}{dt}\) is the rate of change of current (in amperes per second).

### Step 1: Determine the change in current (\(di\))
The current changes from \(+2A\) to \(-2A\). Thus, the total change in current (\(di\)) is:

\[ di = -2A - (+2A) = -4A \]

### Step 2: Determine the time interval (\(dt\))
The time interval during which this change occurs is given as \(0.05s\).

### Step 3: Calculate the rate of change of current (\(\frac{di}{dt}\))
Now, we can calculate the rate of change of current:

\[ \frac{di}{dt} = \frac{di}{dt} = \frac{-4A}{0.05s} = -80A/s \]

### Step 4: Substitute values into the emf equation
We know the induced emf \(E = 8V\). Substituting the values into the emf equation:

\[ 8V = -L \cdot (-80A/s) \]

This simplifies to:

\[ 8V = L \cdot 80A/s \]

### Step 5: Solve for \(L\)
Now, we can solve for \(L\):

\[ L = \frac{8V}{80A/s} = \frac{8}{80} = 0.1H \]

Thus, the coefficient of self-induction of the coil is:

\[ L = 0.1 \, \text{H} \]

### Final Answer
The coefficient of self-induction of the coil is \(0.1 \, \text{H}\).

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