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 step by step, we need to find the induced EMF in a coil with a given inductance when the current is half of its maximum value.

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

### Step 2: Determine the maximum current
The maximum current \( I_0 \) can be identified from the equation:
\[ I_0 = 2 \, \text{A} \]

### Step 3: Calculate the current when it is half of its maximum value
Half of the maximum current is:
\[ I = \frac{I_0}{2} = \frac{2}{2} = 1 \, \text{A} \]

### Step 4: Set up the equation for current
We need to find the time \( t_1 \) when the current is \( 1 \, \text{A} \):
\[ I(t_1) = 2 \sin(100 t_1) \]
Setting this equal to \( 1 \, \text{A} \):
\[ 1 = 2 \sin(100 t_1) \]
Dividing both sides by 2:
\[ \sin(100 t_1) = \frac{1}{2} \]

### Step 5: Solve for \( t_1 \)
The sine function equals \( \frac{1}{2} \) at \( 30^\circ \) (or \( \frac{\pi}{6} \) radians):
\[ 100 t_1 = 30^\circ \]
Converting degrees to radians:
\[ 100 t_1 = \frac{\pi}{6} \]
Thus:
\[ t_1 = \frac{\pi}{600} \, \text{s} \]

### Step 6: Find the induced EMF
The formula for the induced EMF \( \mathcal{E} \) in an inductor is given by:
\[ \mathcal{E} = -L \frac{dI}{dt} \]

### Step 7: Calculate the derivative of the current
First, we need to differentiate the current \( I(t) \):
\[ I(t) = 2 \sin(100t) \]
Thus:
\[ \frac{dI}{dt} = 2 \cdot 100 \cos(100t) = 200 \cos(100t) \]

### Step 8: Substitute \( t_1 \) into the derivative
Now, we substitute \( t_1 = \frac{\pi}{600} \) into the derivative:
\[ \frac{dI}{dt} \bigg|_{t = t_1} = 200 \cos(100 \cdot \frac{\pi}{600}) \]
Calculating \( 100 \cdot \frac{\pi}{600} = \frac{\pi}{6} \):
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
Thus:
\[ \frac{dI}{dt} \bigg|_{t = t_1} = 200 \cdot \frac{\sqrt{3}}{2} = 100\sqrt{3} \]

### Step 9: Calculate the induced EMF
Now substituting back into the EMF equation:
\[ \mathcal{E} = -L \frac{dI}{dt} \]
Substituting \( L = 10^{-2} \, \text{H} \):
\[ \mathcal{E} = -10^{-2} \cdot 100\sqrt{3} \]
\[ \mathcal{E} = -\sqrt{3} \, \text{V} \]

### 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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