The induced emf in a coil of 10 H inductance in which current varies from 9 A to 4 A in 0.2 s is

Options

A. 200 V
B. 250 V
C. 300 V
D. 350 V

Explanation

To solve the problem of finding the induced emf in a coil with a given inductance and changing current, we can follow these steps:

### Step 1: Identify the given values
- Inductance (L) = 10 H (Henries)
- Initial current (I_initial) = 9 A (Amperes)
- Final current (I_final) = 4 A (Amperes)
- Time interval (Δt) = 0.2 s (seconds)

### Step 2: Calculate the change in current (ΔI)
The change in current (ΔI) can be calculated as:
\[ \Delta I = I_{\text{final}} - I_{\text{initial}} = 4\, \text{A} - 9\, \text{A} = -5\, \text{A} \]

### Step 3: Calculate the rate of change of current (di/dt)
The rate of change of current (di/dt) is given by:
\[ \frac{di}{dt} = \frac{\Delta I}{\Delta t} = \frac{-5\, \text{A}}{0.2\, \text{s}} = -25\, \text{A/s} \]

### Step 4: Use the formula for induced emf
The induced emf (E) in the coil can be calculated using the formula:
\[ E = -L \frac{di}{dt} \]
Taking the magnitude, we have:
\[ |E| = L \left| \frac{di}{dt} \right| = 10\, \text{H} \times 25\, \text{A/s} = 250\, \text{V} \]

### Step 5: State the final answer
The induced emf in the coil is:
\[ E = 250\, \text{V} \]