The time period of oscillation of a freely suspended bar magnet with usual notations is given by

Options

A. `T=2pisqrt((I)/(MB_(H)))`
B. `T=2pisqrt((MB_(H))/(I))`
C. `T=sqrt((I)/(MB_(H)))`
D. `T=2pisqrt((B_(H))/(MI))`

Answer

To find the time period of oscillation of a freely suspended bar magnet, we can follow these steps:

### Step 1: Understand the Time Period of a Pendulum
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where:
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

### Step 2: Modify the Formula for a Bar Magnet
In the case of a freely suspended bar magnet, we need to make some adjustments to the formula. The gravitational acceleration \( g \) is replaced by the torque due to the magnetic field. The effective force acting on the bar magnet can be expressed in terms of its magnetic moment \( M \) and the magnetic field \( B \):
\[ g \rightarrow M \cdot B \]
where:
- \( M \) is the magnetic moment of the bar magnet,
- \( B \) is the magnetic field strength.

### Step 3: Replace Length with Moment of Inertia
For a bar magnet, the length \( L \) in the pendulum formula is replaced by the moment of inertia \( I \) of the magnet about its axis of rotation:
\[ L \rightarrow I \]

### Step 4: Combine the Adjusted Values
Substituting these values into the original pendulum formula, we get:
\[ T = 2\pi \sqrt{\frac{I}{M \cdot B}} \]

### Step 5: Final Expression
Thus, the time period of oscillation of a freely suspended bar magnet is given by:
\[ T = 2\pi \sqrt{\frac{I}{M \cdot B}} \]

### Conclusion
The correct answer is that the time period of oscillation of a freely suspended bar magnet is:
\[ T = 2\pi \sqrt{\frac{I}{M \cdot B}} \]