What is the self -inductance of a coil which produce 5 V when the current changes from 3 A to 2 A in one millisecond ?

Options

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

Explanation

To find the self-inductance of a coil that produces 5 V when the current changes from 3 A to 2 A in one millisecond, we can follow these steps:

### Step 1: Understand the relationship between induced EMF and self-inductance
According to Faraday's law of electromagnetic induction, the induced electromotive force (EMF) in a coil is given by the formula:
\[ E = -L \frac{di}{dt} \]
where:
- \( E \) is the induced EMF,
- \( L \) is the self-inductance of the coil,
- \( \frac{di}{dt} \) is the rate of change of current.

### Step 2: Identify the change in current
The current changes from 3 A to 2 A, which means:
\[ \Delta I = I_f - I_i = 2 \, \text{A} - 3 \, \text{A} = -1 \, \text{A} \]

### Step 3: Determine the time interval
The time interval over which this change occurs is given as 1 millisecond:
\[ dt = 1 \, \text{ms} = 1 \times 10^{-3} \, \text{s} \]

### Step 4: Calculate the rate of change of current
Now, we can calculate the rate of change of current:
\[ \frac{di}{dt} = \frac{\Delta I}{dt} = \frac{-1 \, \text{A}}{1 \times 10^{-3} \, \text{s}} = -1000 \, \text{A/s} \]

### Step 5: Substitute values into the induced EMF equation
We know the induced EMF \( E \) is 5 V. Substituting the values into the induced EMF equation:
\[ 5 = -L \left(-1000\right) \]

### Step 6: Solve for self-inductance \( L \)
Rearranging the equation gives:
\[ L = \frac{5}{1000} = 0.005 \, \text{H} \]

### Step 7: Convert to milliHenries
Since self-inductance is often expressed in milliHenries (mH):
\[ L = 0.005 \, \text{H} = 5 \, \text{mH} \]

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

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