When the current changes from `+2A` to `-2A` in `0.05` second, an e.m.f. of `8V` is induced in a coil. The coefficient of self-induction of the coil is

Options

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

Explanation

To find the coefficient of self-induction (L) of the coil, we can use the formula for induced electromotive force (e.m.f.) in terms of self-inductance:

\[ \text{e.m.f.} = -L \frac{dI}{dt} \]

Where:
- e.m.f. is the induced electromotive force (in volts),
- L is the coefficient of self-induction (in henries),
- \( \frac{dI}{dt} \) is the rate of change of current (in amperes per second).

### Step 1: Calculate the change in current (dI)
The current changes from +2 A to -2 A. Therefore, the change in current (dI) is:

\[ dI = I_{\text{final}} - I_{\text{initial}} = -2 A - (+2 A) = -4 A \]

### Step 2: Calculate the rate of change of current (dI/dt)
Given that this change occurs over a time interval of 0.05 seconds, we can calculate the rate of change of current:

\[ \frac{dI}{dt} = \frac{dI}{\Delta t} = \frac{-4 A}{0.05 s} = -80 A/s \]

### Step 3: Substitute values into the e.m.f. equation
We know the induced e.m.f. is 8 V. Using the formula for e.m.f., we can rearrange it to solve for L:

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

This simplifies to:

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

### Step 4: Solve for L
Now, we can solve for L:

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

### Conclusion
The coefficient of self-induction of the coil is:

\[ L = 0.1 \, \text{H} \, \text{(Henries)} \]

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