Two coils have a mutual inductance 0.005 H, The current changes in the first coil according to the equation `i= i_(m) sin omega t` where `i_(m) = 10 A` and `omega= 100pi rad s^(-1)`. The maximum value of the emf induced in the second coil is

Options

A. `2pi`
B. `5pi`
C. `pi`
D. `4pi`

Explanation

To solve the problem, we need to find the maximum value of the induced emf in the second coil due to the changing current in the first coil. Here are the steps to arrive at the solution:

### Step 1: Understand the given information
We have:
- Mutual inductance \( M = 0.005 \, \text{H} \)
- Current in the first coil given by \( i(t) = I_m \sin(\omega t) \)
- Maximum current \( I_m = 10 \, \text{A} \)
- Angular frequency \( \omega = 100\pi \, \text{rad/s} \)

### Step 2: Find the expression for induced emf
According to Faraday's law of electromagnetic induction, the induced emf \( \mathcal{E} \) in the second coil is given by:
\[ \mathcal{E} = -M \frac{di}{dt} \]
where \( \frac{di}{dt} \) is the rate of change of current in the first coil.

### Step 3: Differentiate the current equation
The current in the first coil is:
\[ i(t) = I_m \sin(\omega t) \]
Differentiating with respect to time \( t \):
\[ \frac{di}{dt} = I_m \omega \cos(\omega t) \]

### Step 4: Substitute the values
Substituting \( I_m \) and \( \omega \):
\[ \frac{di}{dt} = 10 \cdot (100\pi) \cos(100\pi t) = 1000\pi \cos(100\pi t) \]

### Step 5: Calculate the induced emf
Now substitute \( \frac{di}{dt} \) into the induced emf equation:
\[ \mathcal{E} = -M \cdot 1000\pi \cos(100\pi t) \]
Substituting \( M = 0.005 \):
\[ \mathcal{E} = -0.005 \cdot 1000\pi \cos(100\pi t) \]
\[ \mathcal{E} = -5\pi \cos(100\pi t) \]

### Step 6: Find the maximum value of induced emf
The maximum value of \( \cos(100\pi t) \) is 1, so the maximum induced emf is:
\[ \mathcal{E}_{\text{max}} = 5\pi \, \text{V} \]
Calculating the numerical value:
\[ \mathcal{E}_{\text{max}} \approx 5 \cdot 3.14 = 15.7 \, \text{V} \]

### Final Answer
The maximum value of the induced emf in the second coil is approximately \( 15.7 \, \text{V} \).

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