When current in coil is uniformely reduced from 2A to 1A in 1 ms , the induced emf is 5V. The inductance of coil is :

Options

A. 5 H
B. 5000 H
C. 5 mH
D. 50 H

Answer

To solve the problem, we need to find the inductance of the coil using the given information about the change in current and the induced emf. Here’s the step-by-step solution:

### Step 1: Understand the relationship between induced emf, inductance, and current change
The induced emf (E) in a coil is related to the inductance (L) and the rate of change of current (di/dt) by the formula:
\[ E = -L \frac{di}{dt} \]

### Step 2: Identify the change in current (di) and the time interval (dt)
From the problem, the current changes from 2 A to 1 A:
- Initial current (I_initial) = 2 A
- Final current (I_final) = 1 A
- Change in current (di) = I_final - I_initial = 1 A - 2 A = -1 A

The time interval during which this change occurs is given as:
- dt = 1 ms = \( 1 \times 10^{-3} \) s

### Step 3: Calculate the rate of change of current (di/dt)
Now we can calculate the rate of change of current:
\[ \frac{di}{dt} = \frac{di}{dt} = \frac{-1 \, \text{A}}{1 \times 10^{-3} \, \text{s}} = -1000 \, \text{A/s} \]

### Step 4: Substitute the values into the induced emf formula
We know the induced emf (E) is given as 5 V. Substituting the values into the formula:
\[ 5 = -L \left(-1000\right) \]
This simplifies to:
\[ 5 = L \times 1000 \]

### Step 5: Solve for inductance (L)
Now, we can solve for L:
\[ L = \frac{5}{1000} = 0.005 \, \text{H} \]

### Step 6: Convert to milliHenries
To express the inductance in milliHenries:
\[ L = 0.005 \, \text{H} = 5 \, \text{mH} \]

### Final Answer
The inductance of the coil is:
\[ L = 5 \, \text{mH} \]

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