A circular coil of n turns and area of crosssection A , carrying a current i, rests with its plane normal to an external magnetic field B. The coil is free to turn about an axis in its plane perpendicular to the field direction. If the moment of inertia of the coil about its axis of rotation is I, its frequency of oscillation about its stable equilibrium is given by

Options

A. `f=1/(2pi)((niAB)/I)^(1//2)`
B. `f=1/(2pi)((niA)/(BI))^(1//2)`
C. `f=1/(2pi)((iAB)/(nI))^(1//2)`
D. `f=1/(2pi)((nIAB)/i)^(1//2)`

Answer

To find the frequency of oscillation of a circular coil in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Parameters**:
- Number of turns in the coil: \( n \)
- Area of cross-section of the coil: \( A \)
- Current flowing through the coil: \( i \)
- External magnetic field: \( B \)
- Moment of inertia of the coil about its axis of rotation: \( I \)

2. **Understand the Torque Acting on the Coil**:
- The torque (\( \tau \)) acting on the coil in a magnetic field is given by:
\[ \tau = n \cdot i \cdot A \cdot B \cdot \sin(\theta) \]
- Here, \( \theta \) is the angle between the area vector of the coil and the magnetic field direction.

3. **Approximate for Small Angles**:
- For small angles, we can use the approximation:
\[ \sin(\theta) \approx \theta \]
- Thus, the torque can be rewritten as:
\[ \tau = n \cdot i \cdot A \cdot B \cdot \theta \]

4. **Relate Torque to Angular Acceleration**:
- The torque can also be expressed in terms of moment of inertia (\( I \)) and angular acceleration (\( \alpha \)):
\[ \tau = I \cdot \alpha \]
- Setting the two expressions for torque equal gives:
\[ I \cdot \alpha = n \cdot i \cdot A \cdot B \cdot \theta \]

5. **Substitute Angular Acceleration**:
- Angular acceleration (\( \alpha \)) can be expressed as:
\[ \alpha = \frac{d^2\theta}{dt^2} = \omega^2 \cdot \theta \]
- Substituting this into the torque equation gives:
\[ I \cdot \omega^2 \cdot \theta = n \cdot i \cdot A \cdot B \cdot \theta \]

6. **Cancel Out \( \theta \)**:
- Assuming \( \theta \neq 0 \), we can cancel \( \theta \) from both sides:
\[ I \cdot \omega^2 = n \cdot i \cdot A \cdot B \]

7. **Solve for Angular Frequency (\( \omega \))**:
- Rearranging gives:
\[ \omega^2 = \frac{n \cdot i \cdot A \cdot B}{I} \]
- Therefore, the angular frequency (\( \omega \)) is:
\[ \omega = \sqrt{\frac{n \cdot i \cdot A \cdot B}{I}} \]

8. **Relate Angular Frequency to Frequency**:
- The relationship between angular frequency (\( \omega \)) and frequency (\( f \)) is given by:
\[ \omega = 2 \pi f \]
- Substituting for \( \omega \) gives:
\[ 2 \pi f = \sqrt{\frac{n \cdot i \cdot A \cdot B}{I}} \]

9. **Solve for Frequency (\( f \))**:
- Rearranging for \( f \) gives:
\[ f = \frac{1}{2 \pi} \sqrt{\frac{n \cdot i \cdot A \cdot B}{I}} \]

### Final Answer:
The frequency of oscillation of the coil is:
\[ f = \frac{1}{2 \pi} \sqrt{\frac{n \cdot i \cdot A \cdot B}{I}} \]