The self inductance of a coil which produces 5V when the current changes from 3A to 2A in one millisecond is

Options

A. 5 mili henry
B. 50 milli henry
C. 5 henry
D. 10 henry

Explanation

To find the self-inductance \( L \) of the coil, we can use the formula for the induced electromotive force (emf) in terms of self-inductance:

\[ E = -L \frac{dI}{dt} \]

Where:
- \( E \) is the induced emf (in volts),
- \( L \) is the self-inductance (in henries),
- \( \frac{dI}{dt} \) is the rate of change of current (in amperes per second).

### Step 1: Identify the values given in the problem
From the question, we have:
- The induced voltage \( E = 5 \, \text{V} \)
- The change in current \( \Delta I = I_f - I_i = 2 \, \text{A} - 3 \, \text{A} = -1 \, \text{A} \)
- The time interval \( dt = 1 \, \text{ms} = 1 \times 10^{-3} \, \text{s} \)

### Step 2: Calculate the rate of change of current \( \frac{dI}{dt} \)
The rate of change of current can be calculated as:

\[ \frac{dI}{dt} = \frac{\Delta I}{dt} = \frac{-1 \, \text{A}}{1 \times 10^{-3} \, \text{s}} = -1000 \, \text{A/s} \]

### Step 3: Substitute the values into the formula
Now we can rearrange the formula for \( L \):

\[ L = -\frac{E}{\frac{dI}{dt}} \]

Substituting the known values:

\[ L = -\frac{5 \, \text{V}}{-1000 \, \text{A/s}} = \frac{5}{1000} = 0.005 \, \text{H} \]

### Step 4: Convert to millihenries
Since \( 1 \, \text{H} = 1000 \, \text{mH} \):

\[ L = 0.005 \, \text{H} = 5 \, \text{mH} \]

### Final Answer
The self-inductance of the coil is \( 5 \, \text{mH} \).

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