A 10 V battery connected to `5omega` resistance coil having inductance 10 H through a switch drives a constant current in the circuit. The switch is suddenly opened and the time taken to open it is 2ms. The average emf induced across the coil is

Options

A. `4xx10^(4)V`
B. `2xx10^(4)V`
C. `2xx10^(2)V`
D. `1xx10^(4)V`

Explanation

To find the average emf induced across the coil when the switch is opened, we can follow these steps:

### Step 1: Calculate the Initial Current (I1)
The initial current in the circuit can be calculated using Ohm's law:
\[ I_1 = \frac{E}{R} \]
Where:
- \( E = 10 \, \text{V} \) (emf of the battery)
- \( R = 5 \, \Omega \) (resistance)

Substituting the values:
\[ I_1 = \frac{10 \, \text{V}}{5 \, \Omega} = 2 \, \text{A} \]

### Step 2: Determine the Final Current (I2)
When the switch is opened, the current drops to zero:
\[ I_2 = 0 \, \text{A} \]

### Step 3: Calculate the Change in Current (di)
The change in current (\( di \)) is:
\[ di = I_2 - I_1 = 0 - 2 = -2 \, \text{A} \]

### Step 4: Calculate the Time Interval (dt)
The time taken to open the switch is given as:
\[ dt = 2 \, \text{ms} = 2 \times 10^{-3} \, \text{s} \]

### Step 5: Calculate the Rate of Change of Current (di/dt)
The rate of change of current (\( \frac{di}{dt} \)) can be calculated as:
\[ \frac{di}{dt} = \frac{di}{dt} = \frac{-2 \, \text{A}}{2 \times 10^{-3} \, \text{s}} = -1000 \, \text{A/s} \]

### Step 6: Calculate the Induced EMF (E)
Using the formula for induced emf in an inductor:
\[ E = -L \frac{di}{dt} \]
Where:
- \( L = 10 \, \text{H} \) (inductance)

Substituting the values:
\[ E = -10 \, \text{H} \times (-1000 \, \text{A/s}) = 10000 \, \text{V} = 10^4 \, \text{V} \]

### Final Answer
The average emf induced across the coil is:
\[ \text{Average EMF} = 10^4 \, \text{V} \]

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