Two coils have the mutual inductance of 0.05 H. The current changes in the first coil as `I=I_(0)sin omegat`, where `I_(0)=1A` and `omega=100pi"rad/s"`. The maximum emf induced in secondary coil is

Options

A. 2.5 V
B. 10 V
C. `6piV`
D. `5piV`

Explanation

To solve the problem, we need to find the maximum electromotive force (emf) induced in the secondary coil due to the changing current in the primary coil. We will use the formula for mutual inductance and the relationship between current and induced emf.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mutual inductance (M) = 0.05 H
- Current in the first coil: \( I(t) = I_0 \sin(\omega t) \)
- \( I_0 = 1 \, \text{A} \)
- \( \omega = 100\pi \, \text{rad/s} \)

2. **Find the Expression for \( \frac{dI}{dt} \):**
- The current \( I(t) = I_0 \sin(\omega t) \).
- To find \( \frac{dI}{dt} \), we differentiate \( I(t) \):
\[ \frac{dI}{dt} = I_0 \frac{d}{dt}(\sin(\omega t)) = I_0 \cdot \omega \cos(\omega t) \]
- Substituting \( I_0 \) and \( \omega \):
\[ \frac{dI}{dt} = 1 \cdot 100\pi \cos(100\pi t) = 100\pi \cos(100\pi t) \]

3. **Find the Maximum Value of \( \frac{dI}{dt} \):**
- The maximum value of \( \cos(\omega t) \) is 1.
- Therefore, the maximum value of \( \frac{dI}{dt} \) is:
\[ \left(\frac{dI}{dt}\right)_{\text{max}} = 100\pi \cdot 1 = 100\pi \]

4. **Calculate the Maximum Induced emf (E_max):**
- The formula for the induced emf in the secondary coil is given by:
\[ E = -M \frac{dI}{dt} \]
- The maximum induced emf will be:
\[ E_{\text{max}} = M \left(\frac{dI}{dt}\right)_{\text{max}} = 0.05 \cdot 100\pi \]

5. **Final Calculation:**
- Now, substituting the values:
\[ E_{\text{max}} = 0.05 \cdot 100\pi = 5\pi \, \text{V} \]

### Conclusion:
The maximum emf induced in the secondary coil is \( 5\pi \, \text{V} \).

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