For a vibration magnetometer, the time period of suspended bar magnet can be reduced by -

Options

A. Moving it towards south pole
B. Moving it towards north pole
C. Moving it towards equator
D. Anyone of them

Explanation

To solve the problem regarding the reduction of the time period of a suspended bar magnet in a vibration magnetometer, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Vibration Magnetometer**:
- A vibration magnetometer is used to measure the horizontal component of the Earth's magnetic field. It consists of a bar magnet suspended freely so that it can oscillate.

2. **Time Period Formula**:
- The time period \( T \) of the oscillation of the bar magnet is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mH}} \]
where:
- \( I \) = moment of inertia of the magnet,
- \( m \) = mass of the magnet,
- \( H \) = horizontal component of the Earth's magnetic field.

3. **Horizontal Component of Magnetic Field**:
- The horizontal component \( H \) of the Earth's magnetic field can be expressed as:
\[ H = B \cos \theta \]
where \( B \) is the total magnetic field and \( \theta \) is the dip angle.

4. **Analyzing the Relationship**:
- From the formula for the time period, we can see that \( T \) is inversely proportional to \( H \):
\[ T \propto \frac{1}{H} \]
- This means that as \( H \) increases, \( T \) decreases.

5. **Effect of Dip Angle**:
- Since \( H = B \cos \theta \), if \( \theta \) decreases (moving towards the equator where \( \theta = 0 \)), \( \cos \theta \) increases.
- At the equator, \( \cos(0) = 1 \), which maximizes \( H \) and minimizes \( T \).

6. **Conclusion**:
- Therefore, the time period \( T \) of the suspended bar magnet can be reduced by moving towards the equator, where the dip angle \( \theta \) is zero.

### Final Answer:
The time period of a suspended bar magnet can be reduced by moving towards the equator.

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