A bar magnet of magnetic moment M and moment of inertia I (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let T be the period of oscillation of the original magnet about an axis through the mid point, perpendicular to length, in a magnetic field `vecB`. What would be the similar period `T^'` for each piece?

Options

A. `(T)/(2)`
B. `(3T)/(4)`
C. `(5T)/(2)`
D. `T`

Explanation

To solve the problem, we need to analyze the changes in the properties of the bar magnet when it is cut into two equal pieces. We will derive the new period of oscillation for each piece step by step.

### Step 1: Understand the properties of the original bar magnet
- The original bar magnet has a magnetic moment \( M \) and a moment of inertia \( I \).
- The time period of oscillation \( T \) of the original magnet in a magnetic field \( \vec{B} \) is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{MB}} \]

### Step 2: Cut the bar magnet into two equal pieces
- When the magnet is cut into two equal pieces, the length of each piece becomes half, and the mass of each piece also becomes half.
- The pole strength remains constant.

### Step 3: Calculate the new moment of inertia for each piece
- The moment of inertia \( I \) of a bar magnet about its center is given by:
\[ I = \frac{ML^2}{12} \]
- After cutting, the new moment of inertia \( I' \) for each piece (with mass \( \frac{M}{2} \) and length \( \frac{L}{2} \)) is:
\[ I' = \frac{\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2}{12} = \frac{M}{2} \cdot \frac{L^2}{4} \cdot \frac{1}{12} = \frac{ML^2}{96} \]
- Since \( I = \frac{ML^2}{12} \), we can express \( I' \) in terms of \( I \):
\[ I' = \frac{I}{8} \]

### Step 4: Calculate the new magnetic moment for each piece
- The magnetic moment \( M' \) of each piece is half of the original magnetic moment \( M \):
\[ M' = \frac{M}{2} \]

### Step 5: Write the formula for the new period of oscillation \( T' \)
- The new time period \( T' \) for each piece can be expressed as:
\[ T' = 2\pi \sqrt{\frac{I'}{M'B}} \]
- Substituting \( I' \) and \( M' \):
\[ T' = 2\pi \sqrt{\frac{\frac{I}{8}}{\frac{M}{2}B}} = 2\pi \sqrt{\frac{I}{8} \cdot \frac{2}{MB}} = 2\pi \sqrt{\frac{I}{4MB}} = 2\pi \cdot \frac{1}{2} \sqrt{\frac{I}{MB}} = \frac{T}{2} \]

### Conclusion
The period of oscillation \( T' \) for each piece is:
\[ T' = \frac{T}{2} \]