The current in ampere in an inductor is given by `I=4t^2+6t` where is in s. The self induced e.m.f in it is 20 mV. The self inductance of the coil t=0

Options

A. 5.55 mH
B. `2.55 xx 10^(-3)` H
C. `3.33 mH`
D. `7.5 xx 10^(-3)` H

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Write down the expression for current
The current \( I \) in the inductor is given by the equation:
\[ I = 4t^2 + 6t \]
where \( t \) is in seconds.

### Step 2: Differentiate the current with respect to time
To find the self-induced electromotive force (e.m.f), we need to calculate the rate of change of current with respect to time. We differentiate \( I \):
\[ \frac{dI}{dt} = \frac{d}{dt}(4t^2 + 6t) = 8t + 6 \]

### Step 3: Substitute \( t = 0 \) into the derivative
Now, we substitute \( t = 0 \) to find the rate of change of current at that instant:
\[ \frac{dI}{dt} \bigg|_{t=0} = 8(0) + 6 = 6 \, \text{A/s} \]

### Step 4: Use the formula for self-induced e.m.f
The self-induced e.m.f (\( \mathcal{E} \)) in an inductor is given by the formula:
\[ \mathcal{E} = -L \frac{dI}{dt} \]
where \( L \) is the self-inductance of the coil. Given that the self-induced e.m.f is \( 20 \, \text{mV} = 20 \times 10^{-3} \, \text{V} \), we can write:
\[ 20 \times 10^{-3} = L \cdot 6 \]

### Step 5: Solve for self-inductance \( L \)
Now, we can rearrange the equation to solve for \( L \):
\[ L = \frac{20 \times 10^{-3}}{6} \]

### Step 6: Calculate \( L \)
Calculating the value:
\[ L = \frac{20 \times 10^{-3}}{6} = \frac{20}{6} \times 10^{-3} = \frac{10}{3} \times 10^{-3} \approx 3.33 \times 10^{-3} \, \text{H} \]

### Step 7: Convert to milliHenries
Converting to milliHenries:
\[ L \approx 3.33 \, \text{mH} \]

### Final Answer
The self-inductance of the coil at \( t = 0 \) is approximately:
\[ L \approx 3.33 \, \text{mH} \]

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