The value of the self inductance of a coil is 5 H. If the current in the coil changes steadily from 1A to 2A in 0.5 seconds, then the magnitude of induced emf is __

Options

A. 1V
B. 10V
C. 100V
D. 0.1V

Answer

To solve the problem, we need to calculate the induced electromotive force (emf) in a coil using the formula for self-inductance. The formula we will use is:

\[ \text{Induced EMF} (E) = L \frac{di}{dt} \]

Where:
- \(L\) is the self-inductance of the coil,
- \(\frac{di}{dt}\) is the rate of change of current.

### Step 1: Identify the given values
- Self-inductance \(L = 5 \, \text{H}\)
- Initial current \(I_1 = 1 \, \text{A}\)
- Final current \(I_2 = 2 \, \text{A}\)
- Time interval \(\Delta t = 0.5 \, \text{s}\)

### Step 2: Calculate the change in current (\(di\))
\[ di = I_2 - I_1 = 2 \, \text{A} - 1 \, \text{A} = 1 \, \text{A} \]

### Step 3: Calculate the rate of change of current (\(\frac{di}{dt}\))
\[ \frac{di}{dt} = \frac{di}{\Delta t} = \frac{1 \, \text{A}}{0.5 \, \text{s}} = 2 \, \text{A/s} \]

### Step 4: Substitute the values into the induced emf formula
\[ E = L \frac{di}{dt} = 5 \, \text{H} \times 2 \, \text{A/s} = 10 \, \text{V} \]

### Conclusion
The magnitude of the induced emf is \(10 \, \text{V}\).

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