In an inductor, the current / varies with time to `l = 5A + 16.(A//s)t`. If-induced emf in the induct is 5 mV, the self inductance of the inductor is

Options

A. `3.75 X 10^(-3)H`
B. ` 3.75 X 10^(-4)H`
C. `3.125 X 10^(-3)H`
D. `3.125 X 10^(-4)H`

Answer

To find the self-inductance of the inductor given the current variation and the induced EMF, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given information:**
- Current \( I(t) = 5 \, \text{A} + 16 \, \text{A/s} \cdot t \)
- Induced EMF \( \mathcal{E} = 5 \, \text{mV} = 5 \times 10^{-3} \, \text{V} \)

2. **Recall the formula for induced EMF in an inductor:**
\[ \mathcal{E} = -L \frac{dI}{dt} \]
where \( L \) is the self-inductance and \( \frac{dI}{dt} \) is the rate of change of current with respect to time.

3. **Differentiate the current with respect to time:**
\[ I(t) = 5 + 16t \]
Taking the derivative:
\[ \frac{dI}{dt} = 0 + 16 = 16 \, \text{A/s} \]

4. **Substitute the values into the EMF formula:**
\[ 5 \times 10^{-3} = -L \cdot 16 \]

5. **Solve for \( L \):**
Rearranging the equation gives:
\[ L = -\frac{5 \times 10^{-3}}{16} \]
Since self-inductance cannot be negative, we take the absolute value:
\[ L = \frac{5 \times 10^{-3}}{16} \]

6. **Calculate \( L \):**
\[ L = \frac{5}{16} \times 10^{-3} = 0.3125 \times 10^{-3} \, \text{H} \]

7. **Convert to standard form:**
\[ L = 3.125 \times 10^{-4} \, \text{H} \]

### Final Answer:
The self-inductance of the inductor is \( L = 3.125 \times 10^{-4} \, \text{H} \).

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