A coil of `10^(-2)` H inductance carries a current `I = 2sin (100t)` A . When current is half of its maximum value , then at that instant the induced emf in the coil will be :

Options

A. 1V
B. `sqrt(2)`V
C. `sqrt(3)`V
D. 2V

Answer

To solve the problem, we need to find the induced emf in a coil when the current is half of its maximum value. Let's break down the solution step by step.

### Step 1: Identify the given values
- Inductance \( L = 10^{-2} \, \text{H} \)
- Current \( I(t) = 2 \sin(100t) \, \text{A} \)
- Maximum current \( I_{\text{max}} = 2 \, \text{A} \)

### Step 2: Determine the half of the maximum current
The maximum current is \( I_{\text{max}} = 2 \, \text{A} \). Therefore, half of the maximum current is:
\[ I = \frac{I_{\text{max}}}{2} = \frac{2}{2} = 1 \, \text{A} \]

### Step 3: Find the corresponding angle when the current is half of its maximum value
We set the equation for current:
\[ 1 = 2 \sin(100t) \]
Solving for \( \sin(100t) \):
\[ \sin(100t) = \frac{1}{2} \]
The angles for which \( \sin \) gives \( \frac{1}{2} \) are:
\[ 100t = \frac{\pi}{6} \quad \text{or} \quad 100t = \frac{5\pi}{6} \]

### Step 4: Calculate the induced emf
The induced emf \( \mathcal{E} \) in an inductor is given by:
\[ \mathcal{E} = -L \frac{dI}{dt} \]
First, we need to find \( \frac{dI}{dt} \):
\[ I(t) = 2 \sin(100t) \]
Differentiating with respect to \( t \):
\[ \frac{dI}{dt} = 2 \cdot 100 \cos(100t) = 200 \cos(100t) \]

### Step 5: Evaluate \( \frac{dI}{dt} \) at \( 100t = \frac{\pi}{6} \)
Now we substitute \( 100t = \frac{\pi}{6} \):
\[ \cos(100t) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \frac{dI}{dt} = 200 \cdot \frac{\sqrt{3}}{2} = 100\sqrt{3} \]

### Step 6: Calculate the induced emf
Now substituting back into the emf equation:
\[ \mathcal{E} = -L \frac{dI}{dt} = -10^{-2} \cdot 100\sqrt{3} \]
Calculating this gives:
\[ \mathcal{E} = -1\sqrt{3} \, \text{V} = -\sqrt{3} \, \text{V} \]
The negative sign indicates the direction of the induced emf according to Lenz's law.

### Final Answer
The induced emf in the coil when the current is half of its maximum value is:
\[ \mathcal{E} = \sqrt{3} \, \text{V} \]

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