In the above problem. The plane of the coil is initially kept parallel to `B`. The coil is rotated by an angle `theta` about the diameter perpendicular to `B` and charge of amount `Q` flows through it. Choose the correct alternatives.

Options

A. (a) `theta = 90^(@), Q = (Ban//R)`
B. (b) `theta = 180^(@), Q = (Ban//R)`
C. ( c) `theta = 180^(@), Q = 0`
D. ( d) `theta = 360^(@), Q = 0`

Explanation

To solve the problem, we need to determine the amount of charge \( Q \) that flows through the coil when it is rotated by an angle \( \theta \) about a diameter perpendicular to the magnetic field \( B \). We will follow these steps:

### Step 1: Understand the Initial Conditions
Initially, the plane of the coil is parallel to the magnetic field \( B \). The area vector \( \vec{A} \) of the coil is also parallel to \( B \). Therefore, the initial angle \( \theta_i = 0^\circ \).

### Step 2: Determine the Magnetic Flux
The magnetic flux \( \Phi \) through the coil is given by:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
Where:
- \( B \) is the magnetic field strength,
- \( A \) is the area of the coil,
- \( \theta \) is the angle between the area vector and the magnetic field.

Initially, when the coil is parallel to \( B \):
\[ \Phi_i = B \cdot A \cdot \cos(0) = B \cdot A \]

### Step 3: Calculate the Final Flux After Rotation
After rotating the coil by an angle \( \theta \), the new angle \( \theta_f \) becomes \( \theta \). Thus, the final magnetic flux is:
\[ \Phi_f = B \cdot A \cdot \cos(\theta) \]

### Step 4: Calculate the Change in Flux
The change in magnetic flux \( \Delta \Phi \) is given by:
\[ \Delta \Phi = \Phi_f - \Phi_i = B \cdot A \cdot \cos(\theta) - B \cdot A = B \cdot A (\cos(\theta) - 1) \]

### Step 5: Calculate the Induced EMF
According to Faraday's law of electromagnetic induction, the induced EMF \( \mathcal{E} \) in the coil is given by:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
For a finite change, we can express it as:
\[ \mathcal{E} = -n \frac{\Delta \Phi}{\Delta t} \]
Where \( n \) is the number of turns in the coil.

### Step 6: Relate EMF to Current
Using Ohm's law, the current \( I \) flowing through the coil can be expressed as:
\[ I = \frac{\mathcal{E}}{R} \]
Where \( R \) is the resistance of the coil.

### Step 7: Calculate the Charge Flowing Through the Coil
The charge \( Q \) that flows through the coil can be calculated using the relationship:
\[ Q = I \cdot t \]
Substituting for \( I \):
\[ Q = \frac{\mathcal{E}}{R} \cdot t \]
Substituting for \( \mathcal{E} \):
\[ Q = -\frac{n}{R} \cdot \Delta \Phi \]
Substituting for \( \Delta \Phi \):
\[ Q = -\frac{n}{R} \cdot B \cdot A (\cos(\theta) - 1) \cdot t \]

### Step 8: Analyze Specific Cases
1. **When \( \theta = 0^\circ \)**:
\[ Q = 0 \quad (\text{since } \cos(0) - 1 = 0) \]

2. **When \( \theta = 90^\circ \)**:
\[ Q = -\frac{n \cdot B \cdot A}{R} \cdot (-1) \cdot t = \frac{n \cdot B \cdot A \cdot t}{R} \]

3. **When \( \theta = 180^\circ \)**:
\[ Q = -\frac{n \cdot B \cdot A}{R} \cdot (1 - (-1)) \cdot t = 0 \quad (\text{since } \cos(180) - 1 = -2) \]

4. **When \( \theta = 360^\circ \)**:
\[ Q = 0 \quad (\text{since it returns to the initial position}) \]

### Conclusion
The correct alternatives are:
- When \( \theta = 90^\circ \), \( Q \) is maximum.
- When \( \theta = 0^\circ \) or \( 180^\circ \) or \( 360^\circ \), \( Q = 0 \).