A moving coil galvanometer consists of a coil of N turns are area A suspended by a thin phosphor bronze strip in radial magnetic field B. The moment of inertia of the coil about the axis of rotation is l and c is the torsional constant of the phosphor bronze strip. When a current i is passed through the coil, it deffects through an angle `theta`
When a charge Q is almost instantly through the coil the angular speed `omega` acquired by the coil is

Options

A. a.`(NAB)/(QI)`
B. b.`(BAQ)/(NI)`
C. c.`(NABQ)/(I)`
D. d.`(NAQI)/(B)`

Answer

To find the angular speed \( \omega \) acquired by the coil when a charge \( Q \) is passed through it, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Magnetic Moment**:
The magnetic moment \( m \) of the coil is given by the formula:
\[ m = N \cdot i \cdot A \]
where \( N \) is the number of turns, \( i \) is the current, and \( A \) is the area of the coil.

2. **Calculating the Torque**:
The torque \( \tau \) acting on the coil in a magnetic field \( B \) is given by:
\[ \tau = m \cdot B \cdot \sin(\theta) \]
Since the coil is in a radial magnetic field and the angle \( \theta \) is \( 90^\circ \), we have:
\[ \sin(90^\circ) = 1 \]
Thus, the torque simplifies to:
\[ \tau = m \cdot B = N \cdot i \cdot A \cdot B \]

3. **Angular Impulse**:
The angular impulse is defined as the integral of torque over time:
\[ \text{Angular Impulse} = \int \tau \, dt \]
Substituting the expression for torque, we get:
\[ \text{Angular Impulse} = \int N \cdot A \cdot B \cdot i \, dt \]
Since \( N \), \( A \), and \( B \) are constants, we can factor them out:
\[ \text{Angular Impulse} = N \cdot A \cdot B \cdot \int i \, dt \]

4. **Relating Charge to Current**:
The integral \( \int i \, dt \) represents the total charge \( Q \) that has passed through the coil:
\[ \int i \, dt = Q \]
Therefore, we can rewrite the angular impulse as:
\[ \text{Angular Impulse} = N \cdot A \cdot B \cdot Q \]

5. **Using Moment of Inertia**:
The angular impulse is also equal to the change in angular momentum, which can be expressed as:
\[ \text{Angular Impulse} = I \cdot \omega \]
where \( I \) is the moment of inertia of the coil and \( \omega \) is the angular speed.

6. **Setting the Equations Equal**:
Setting the two expressions for angular impulse equal gives us:
\[ N \cdot A \cdot B \cdot Q = I \cdot \omega \]

7. **Solving for Angular Speed**:
Rearranging the equation to solve for \( \omega \):
\[ \omega = \frac{N \cdot A \cdot B \cdot Q}{I} \]

### Final Answer:
Thus, the angular speed \( \omega \) acquired by the coil is:
\[ \omega = \frac{N \cdot A \cdot B \cdot Q}{I} \]

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