A magnet of magnetic moment M and having moment of inertia I , is suspended to orient freely in the horizontal plane under the influence of earth's magnetic field . If it is slightly disturbed , the time period is given by

Options

A. `2pi sqrt((I)/(MB_(H)))`
B. `2 pi sqrt((MB_(H))/(I))`
C. `(1)/(2pi) sqrt((I)/(MB_(H)))`
D. `(1)/(2pi)sqrt((MB_(H))/(I))`

Answer

To find the time period of a magnet suspended to orient freely in the horizontal plane under the influence of Earth's magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a magnet with a magnetic moment \( M \) and moment of inertia \( I \).
- The magnet is suspended in a way that it can rotate freely in the horizontal plane.

2. **Torque Due to Magnetic Moment**:
- When the magnet is disturbed from its equilibrium position, it experiences a torque due to the Earth's magnetic field \( B \).
- The torque \( \tau \) acting on the magnet can be expressed as:
\[ \tau = M \times B \sin(\theta) \]
- For small angles \( \theta \), we can approximate \( \sin(\theta) \approx \theta \). Therefore, the torque becomes:
\[ \tau \approx M B \theta \]

3. **Relating Torque to Angular Acceleration**:
- According to Newton's second law for rotation, the torque is also related to angular acceleration \( \alpha \) by:
\[ \tau = I \alpha \]
- Substituting the expression for torque, we have:
\[ M B \theta = I \alpha \]

4. **Expressing Angular Acceleration**:
- The angular acceleration \( \alpha \) can be expressed in terms of the angle \( \theta \):
\[ \alpha = \frac{d^2\theta}{dt^2} \]
- Thus, we can rewrite the equation as:
\[ M B \theta = I \frac{d^2\theta}{dt^2} \]

5. **Formulating the Differential Equation**:
- Rearranging gives us:
\[ \frac{d^2\theta}{dt^2} + \frac{M B}{I} \theta = 0 \]
- This is a standard form of the simple harmonic motion equation.

6. **Identifying Angular Frequency**:
- Comparing with the standard form \( \frac{d^2\theta}{dt^2} + \omega^2 \theta = 0 \), we can identify:
\[ \omega^2 = \frac{M B}{I} \]
- Therefore, the angular frequency \( \omega \) is:
\[ \omega = \sqrt{\frac{M B}{I}} \]

7. **Finding the Time Period**:
- The time period \( T \) of the oscillation is related to the angular frequency by:
\[ T = \frac{2\pi}{\omega} \]
- Substituting for \( \omega \):
\[ T = 2\pi \sqrt{\frac{I}{M B}} \]

8. **Final Result**:
- Thus, the time period of the magnet is given by:
\[ T = 2\pi \sqrt{\frac{I}{M B}} \]