When current changes from 13 A to 7A in 0.5 s through a coil, the emf induced is `3 xx10^(-4)`V. The coefficient of self induction is :

Options

A. `25 xx 10^(-6) H`
B. `25 xx 10^(-5) H`
C. `25 xx 10^(-4) H`
D. `25 xx 10^(-3) H`

Answer

To find the coefficient of self-induction (L) for the coil, we can use the formula for induced electromotive force (emf) due to self-induction:

\[ \text{emf} = -L \frac{dI}{dt} \]

Where:
- \( \text{emf} \) is the induced electromotive force,
- \( L \) is the self-inductance,
- \( \frac{dI}{dt} \) is the rate of change of current.

### Step 1: Identify the given values
From the problem, we have:
- Initial current, \( I_1 = 13 \, \text{A} \)
- Final current, \( I_2 = 7 \, \text{A} \)
- Time interval, \( dt = 0.5 \, \text{s} \)
- Induced emf, \( \text{emf} = 3 \times 10^{-4} \, \text{V} \)

### Step 2: Calculate the change in current (\( dI \))
The change in current (\( dI \)) can be calculated as follows:

\[ dI = I_2 - I_1 = 7 \, \text{A} - 13 \, \text{A} = -6 \, \text{A} \]

### Step 3: Calculate the rate of change of current (\( \frac{dI}{dt} \))
Now, we can calculate the rate of change of current:

\[ \frac{dI}{dt} = \frac{dI}{dt} = \frac{-6 \, \text{A}}{0.5 \, \text{s}} = -12 \, \text{A/s} \]

### Step 4: Substitute values into the emf formula
Now we can substitute the values into the emf equation:

\[ 3 \times 10^{-4} = -L \cdot (-12) \]

### Step 5: Solve for \( L \)
Rearranging the equation to solve for \( L \):

\[ 3 \times 10^{-4} = 12L \]

\[ L = \frac{3 \times 10^{-4}}{12} \]

\[ L = \frac{3}{12} \times 10^{-4} = \frac{1}{4} \times 10^{-4} = 0.25 \times 10^{-4} \, \text{H} \]

### Step 6: Convert to microhenries
To express \( L \) in microhenries (µH):

\[ L = 0.25 \times 10^{-4} \, \text{H} = 25 \times 10^{-6} \, \text{H} = 25 \, \mu\text{H} \]

### Final Answer
The coefficient of self-induction \( L \) is:

\[ L = 25 \, \mu\text{H} \]