The time period of a bar magnet oscillating in a uniform magnetic field is 3 seconds. if the magnet is cut into two equal parts along the equatorial line of the magnet and one part is made to vibrate in the same field, what is the time period ?

Answer

To solve the problem, we need to understand how the time period of oscillation of a bar magnet is affected when it is cut into two equal parts. The time period \( T \) of a bar magnet oscillating in a magnetic field is given by the formula:

\[ T = 2\pi \sqrt{\frac{I}{mB}} \]

where:
- \( I \) is the moment of inertia of the magnet,
- \( m \) is the mass of the magnet,
- \( B \) is the magnetic field strength.

### Step-by-Step Solution:

1. **Identify the initial time period**:
The time period of the original bar magnet is given as \( T = 3 \) seconds.

2. **Understand the effect of cutting the magnet**:
When the bar magnet is cut into two equal parts, each part will have half the mass of the original magnet. Therefore, if the mass of the original magnet is \( m \), the mass of each half will be \( \frac{m}{2} \).

3. **Calculate the new moment of inertia**:
The moment of inertia \( I \) for a bar magnet about its end is given by \( I = \frac{1}{3} m L^2 \) (where \( L \) is the length of the magnet). When the magnet is cut in half, the moment of inertia for one half becomes:

\[ I' = \frac{1}{3} \left(\frac{m}{2}\right) L^2 = \frac{1}{6} m L^2 \]

4. **Substituting into the time period formula**:
The new time period \( T' \) for one half of the magnet can be calculated using the modified moment of inertia:

\[ T' = 2\pi \sqrt{\frac{I'}{mB}} = 2\pi \sqrt{\frac{\frac{1}{6} m L^2}{\frac{m}{2} B}} = 2\pi \sqrt{\frac{L^2}{3B}} \]

5. **Relate the new time period to the original time period**:
Since we know the original time period \( T = 2\pi \sqrt{\frac{L^2}{mB}} \), we can express the new time period \( T' \) in terms of \( T \):

\[ T' = 2\pi \sqrt{\frac{L^2}{3B}} = \sqrt{\frac{1}{3}} \cdot 2\pi \sqrt{\frac{L^2}{mB}} = \sqrt{\frac{1}{3}} \cdot T \]

6. **Calculate the new time period**:
Since \( T = 3 \) seconds, we find:

\[ T' = \sqrt{\frac{1}{3}} \cdot 3 = \sqrt{3} \text{ seconds} \approx 1.732 \text{ seconds} \]

### Final Answer:
The time period of the half magnet oscillating in the same magnetic field is approximately \( 1.732 \) seconds.