In a coil of self-inuctance `0.5` henry, the current varies at a constant rate from zero to `10` amperes in `2` seconds. The e.m.f. generated in the coil is

Options

A. 10 V
B. 5 V
C. 2.5 V
D. 1.25 V

Explanation

To solve the problem, we need to find the electromotive force (e.m.f.) generated in a coil due to self-induction when the current through it changes over time. Here’s a step-by-step solution:

### Step 1: Understand the formula for e.m.f. due to self-induction
The e.m.f. (ε) generated in a coil due to self-induction is given by the formula:
\[ \epsilon = -L \frac{di}{dt} \]
where:
- \(L\) is the self-inductance of the coil (in henries),
- \(di\) is the change in current (in amperes),
- \(dt\) is the change in time (in seconds).

### Step 2: Identify the given values
From the problem, we have:
- Self-inductance, \(L = 0.5 \, \text{H}\)
- Initial current, \(I_i = 0 \, \text{A}\)
- Final current, \(I_f = 10 \, \text{A}\)
- Time interval, \(dt = 2 \, \text{s}\)

### Step 3: Calculate the change in current (di)
The change in current (\(di\)) can be calculated as:
\[ di = I_f - I_i = 10 \, \text{A} - 0 \, \text{A} = 10 \, \text{A} \]

### Step 4: Calculate the rate of change of current (di/dt)
Now, we can find the rate of change of current:
\[ \frac{di}{dt} = \frac{10 \, \text{A}}{2 \, \text{s}} = 5 \, \text{A/s} \]

### Step 5: Substitute values into the e.m.f. formula
Now, substituting the values into the e.m.f. formula:
\[ \epsilon = -L \frac{di}{dt} = -0.5 \, \text{H} \times 5 \, \text{A/s} \]
\[ \epsilon = -2.5 \, \text{V} \]

### Step 6: Determine the magnitude of the e.m.f.
Since the question asks for the magnitude of the e.m.f., we take the absolute value:
\[ |\epsilon| = 2.5 \, \text{V} \]

### Conclusion
The magnitude of the e.m.f. generated in the coil is \(2.5 \, \text{V}\).

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