`A` bar magnet of magnetic moment `M` and moment of inertia `I` (about center and perpendicular to length) is cat into two equal pieces, perpendicular to its length. Let `T` be the period of oscillations of the original magnet about an axis through the mid-point, perpendicular to length, in a magnetic field `vec(B)`. If the similar period `T` for each piece is `(T)/(n)`, then calculate `(n)`.

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the period of oscillation for the original magnet
The period of oscillation \( T \) of a bar magnet in a magnetic field \( \vec{B} \) 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 magnetic moment, and \( B \) is the magnetic field.

### Step 2: Determine the properties of the cut pieces
When the bar magnet is cut into two equal pieces:
- Each piece will have half the mass, so the new mass \( M' = \frac{M}{2} \).
- The length of each piece will also be half, so the new length \( L' = \frac{L}{2} \).

### Step 3: Calculate the moment of inertia for each piece
The moment of inertia \( I' \) of each piece about its center and perpendicular to its length can be calculated using the formula for a rod:
\[ I' = \frac{1}{12} m L'^2 \]
Substituting \( m = \frac{M}{2} \) and \( L' = \frac{L}{2} \):
\[ I' = \frac{1}{12} \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{1}{12} \cdot \frac{M}{2} \cdot \frac{L^2}{4} = \frac{ML^2}{96} \]

### Step 4: Calculate the new period of oscillation for each piece
Now, we can find the new period of oscillation \( T' \) for each piece:
\[ T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{\frac{ML^2}{96}}{\frac{M}{2}B}} = 2\pi \sqrt{\frac{ML^2}{48MB}} = 2\pi \sqrt{\frac{L^2}{48B}} \]

### Step 5: Relate the new period to the original period
We know that the original period \( T \) was:
\[ T = 2\pi \sqrt{\frac{I}{MB}} = 2\pi \sqrt{\frac{L^2}{12B}} \]
Now, we can express \( T' \) in terms of \( T \):
\[ T' = 2\pi \sqrt{\frac{L^2}{48B}} = T \cdot \sqrt{\frac{12}{48}} = T \cdot \frac{1}{2} \]
Thus, we find:
\[ T' = \frac{T}{2} \]

### Step 6: Determine the value of \( n \)
According to the problem, if the period \( T' \) for each piece is \( \frac{T}{n} \), we have:
\[ \frac{T}{n} = \frac{T}{2} \]
This implies:
\[ n = 2 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{2} \]