The current in ampere through an inductor is
i=(10+20t)
Here t is in second. The induced emf in the inductor 4V.
The self inductance of the indicator is, L…..H,

Options

A. `0.2`
B. `0.4`
C. `0.1`
D. `1.0`

Answer

To solve the problem, we need to find the self-inductance \( L \) of an inductor given the current \( i(t) = 10 + 20t \) and the induced EMF \( \mathcal{E} = 4 \, \text{V} \).

### Step-by-Step Solution:

1. **Understand the relationship between EMF, inductance, and current**:
The induced EMF \( \mathcal{E} \) in an inductor is given by the formula:
\[ \mathcal{E} = -L \frac{di}{dt} \]
where \( L \) is the self-inductance and \( \frac{di}{dt} \) is the rate of change of current with respect to time.

2. **Differentiate the current function**:
Given the current \( i(t) = 10 + 20t \), we need to find \( \frac{di}{dt} \):
\[ \frac{di}{dt} = \frac{d}{dt}(10 + 20t) = 0 + 20 = 20 \, \text{A/s} \]

3. **Substitute values into the EMF equation**:
Now we can substitute \( \mathcal{E} = 4 \, \text{V} \) and \( \frac{di}{dt} = 20 \, \text{A/s} \) into the EMF equation:
\[ 4 = -L \cdot 20 \]

4. **Solve for self-inductance \( L \)**:
Rearranging the equation to solve for \( L \):
\[ L = -\frac{4}{20} = -0.2 \, \text{H} \]
Since inductance cannot be negative, we take the absolute value:
\[ L = 0.2 \, \text{H} \]

### Final Answer:
The self-inductance of the inductor is \( L = 0.2 \, \text{H} \).

---