If `phi = 0.02 cos 100` at weber/turns and number of turns is 50 in the coil , the maximum induced emf is equal to -

Options

A. 314 volt
B. 100 volt
C. 31.4 volt
D. 6.28 volt

Explanation

To find the maximum induced emf in the coil, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- Magnetic flux per turn, \( \phi(t) = 0.02 \cos(100t) \) Weber/turn.
- Number of turns in the coil, \( N = 50 \).

2. **Understand the formula for induced emf**:
The induced emf (\( \mathcal{E} \)) in a coil is given by Faraday's law of electromagnetic induction:
\[ \mathcal{E} = -N \frac{d\phi}{dt} \]
where \( N \) is the number of turns and \( \phi \) is the magnetic flux.

3. **Differentiate the magnetic flux**:
We need to differentiate \( \phi(t) \) with respect to time \( t \):
\[ \phi(t) = 0.02 \cos(100t) \]
The derivative is:
\[ \frac{d\phi}{dt} = 0.02 \cdot \frac{d}{dt}(\cos(100t)) = 0.02 \cdot (-100 \sin(100t)) = -2 \sin(100t) \]

4. **Substitute the derivative into the induced emf formula**:
Now substitute \( \frac{d\phi}{dt} \) into the induced emf formula:
\[ \mathcal{E} = -N \frac{d\phi}{dt} = -50 \cdot (-2 \sin(100t)) = 100 \sin(100t) \]

5. **Find the maximum induced emf**:
The maximum value of \( \sin(100t) \) is 1. Therefore, the maximum induced emf is:
\[ \mathcal{E}_{\text{max}} = 100 \cdot 1 = 100 \, \text{V} \]

### Final Answer:
The maximum induced emf is \( 100 \, \text{V} \).