With a standard rectangular bar magnet the time period of a vibration magnetometer is 4 s. The bar magnet is cut parallel to its length into four equal pieces. The time period of vibration magnetometer when one piece is used (in second) (bar magnet breadth is small) is

Options

A. 16
B. 8
C. 4
D. 2

Answer

To solve the problem, we need to determine the new time period of the vibration magnetometer when one piece of the bar magnet is used after it has been cut into four equal pieces.

### Step-by-step Solution:

1. **Understand the Initial Conditions**:
- The time period \( T \) of the vibration magnetometer with the whole bar magnet is given as \( 4 \) seconds.

2. **Cutting the Magnet**:
- The bar magnet is cut parallel to its length into 4 equal pieces.
- When the bar magnet is cut, the length of each piece remains unchanged, but the pole strength changes.

3. **Determine the Pole Strength**:
- The original pole strength of the bar magnet is denoted as \( m \).
- When the magnet is cut into 4 pieces, the pole strength of each piece becomes \( \frac{m}{4} \).

4. **Moment of Inertia**:
- The moment of inertia \( I \) of the magnet about its center of mass is given by \( I = \frac{m l^2}{3} \), where \( l \) is the length of the magnet.
- Since the width is small, we can assume that the moment of inertia remains unchanged when we use one piece of the magnet.

5. **Magnetic Moment**:
- The magnetic moment \( M \) of the magnet is given by \( M = m \cdot l \).
- For one piece, the magnetic moment becomes \( M' = \frac{m}{4} \cdot l \).

6. **Time Period Relation**:
- The time period \( T \) of the vibration magnetometer is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{M}} \]
- Since \( I \) remains constant and \( M \) changes, we can express the new time period \( T' \) when using one piece:
\[ T' = 2\pi \sqrt{\frac{I}{M'}} \]

7. **Finding the New Time Period**:
- The ratio of the new time period \( T' \) to the original time period \( T \) can be expressed as:
\[ \frac{T'}{T} = \sqrt{\frac{M}{M'}} \]
- Substituting \( M' = \frac{m}{4} \):
\[ \frac{T'}{T} = \sqrt{\frac{M}{\frac{m}{4}}} = \sqrt{4} = 2 \]
- Therefore, we have:
\[ T' = 2T \]
- Given \( T = 4 \) seconds, we find:
\[ T' = 2 \times 4 = 8 \text{ seconds} \]

### Final Answer:
The time period of the vibration magnetometer when one piece is used is **8 seconds**.