The current in ampere in an inductor is given by I=5 +16 t where t is in s. The self induced e.m.f in it is 10mv. The self inductance is

Options

A. `5.55 xx 10^(-5)` H
B. `6.25 xx 10^(-4)` H
C. `5.26 xx10^(-6)` H
D. `7.5 xx 10^(-7)` H

Explanation

To solve the problem, we need to find the self-inductance \( L \) of the inductor given the current \( I \) and the self-induced electromotive force (e.m.f).

### Step-by-Step Solution:

1. **Identify the given values:**
- The current \( I \) is given by the equation:
\[ I = 5 + 16t \]
- The self-induced e.m.f \( E \) is given as \( 10 \, \text{mV} = 10 \times 10^{-3} \, \text{V} \).

2. **Use the formula for self-induced e.m.f:**
The self-induced e.m.f in an inductor is given by:
\[ E = -L \frac{dI}{dt} \]
Here, \( L \) is the self-inductance, and \( \frac{dI}{dt} \) is the rate of change of current with respect to time.

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

4. **Substitute values into the e.m.f equation:**
Rearranging the e.m.f equation to solve for \( L \):
\[ L = -\frac{E}{\frac{dI}{dt}} \]
Substitute \( E = 10 \times 10^{-3} \, \text{V} \) and \( \frac{dI}{dt} = 16 \, \text{A/s} \):
\[ L = -\frac{10 \times 10^{-3}}{16} \]

5. **Calculate \( L \):**
\[ L = -\frac{10}{16} \times 10^{-3} = -0.625 \times 10^{-3} \, \text{H} \]
Since we are interested in the magnitude of self-inductance, we take the positive value:
\[ L = 0.625 \times 10^{-3} \, \text{H} = 6.25 \times 10^{-4} \, \text{H} \]

6. **Final Answer:**
The self-inductance \( L \) of the inductor is:
\[ L = 6.25 \times 10^{-4} \, \text{H} \]