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 \( L \) of the inductor, we can follow these steps:

### Step 1: Understand the given information
We are given the current \( I(t) \) in the inductor as:
\[ I(t) = 5 + 16t \quad \text{(in Amperes)} \]
We are also given the induced EMF \( \mathcal{E} \) as:
\[ \mathcal{E} = 5 \, \text{mV} = 5 \times 10^{-3} \, \text{V} \]

### Step 2: Use the formula for induced EMF
The induced EMF 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.

### Step 3: Calculate \( \frac{dI}{dt} \)
From the expression for \( I(t) \):
\[ I(t) = 5 + 16t \]
We can differentiate this with respect to \( t \):
\[ \frac{dI}{dt} = \frac{d}{dt}(5 + 16t) = 16 \, \text{A/s} \]

### Step 4: Substitute values into the EMF equation
Now we can substitute the values of \( \mathcal{E} \) and \( \frac{dI}{dt} \) into the EMF equation:
\[ 5 \times 10^{-3} = -L \cdot 16 \]

### Step 5: Solve for \( L \)
Rearranging the equation to solve for \( L \):
\[ L = -\frac{5 \times 10^{-3}}{16} \]
Since self-inductance is a positive quantity, we take the absolute value:
\[ L = \frac{5 \times 10^{-3}}{16} \]

### Step 6: Calculate the numerical value
Calculating \( L \):
\[ L = \frac{5}{16} \times 10^{-3} \, \text{H} \]
Calculating \( \frac{5}{16} \):
\[ \frac{5}{16} = 0.3125 \]
Thus,
\[ L = 0.3125 \times 10^{-3} \, \text{H} = 3.125 \times 10^{-4} \, \text{H} \]

### Final Answer
The self-inductance of the inductor is:
\[ L = 3.125 \, \text{mH} \]

---