A coil of N turns and area A is rotated at the rate of n rotations per second in a magnetic field of intensity B, the magnitude of the maximum magnetic flux will be __

Options

A. NAB
B. nAB
C. NnAB
D. `2pinNAB`

Answer

To find the magnitude of the maximum magnetic flux through a coil of N turns and area A, rotated in a magnetic field of intensity B, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Magnetic Flux**:
The magnetic flux (Φ) through a single coil is given by the formula:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
where:
- \(B\) is the magnetic field intensity,
- \(A\) is the area of the coil,
- \(\theta\) is the angle between the magnetic field and the normal (perpendicular) to the surface of the coil.

2. **Considering Multiple Turns**:
If the coil has \(N\) turns, the total magnetic flux (Φ_total) through the coil is:
\[ \Phi_{\text{total}} = N \cdot \Phi = N \cdot (B \cdot A \cdot \cos(\theta)) \]

3. **Finding Maximum Magnetic Flux**:
The maximum value of the magnetic flux occurs when \(\cos(\theta)\) is maximized. The maximum value of \(\cos(\theta)\) is 1, which occurs when \(\theta = 0^\circ\) (i.e., the magnetic field is perpendicular to the area vector of the coil). Thus, we can write:
\[ \Phi_{\text{max}} = N \cdot (B \cdot A \cdot 1) = N \cdot B \cdot A \]

4. **Final Result**:
Therefore, the magnitude of the maximum magnetic flux through the coil is:
\[ \Phi_{\text{max}} = N \cdot B \cdot A \]

### Conclusion:
The magnitude of the maximum magnetic flux will be \(NAB\).