If the period of oscillaion of freely suspended bar magnet in earth's horizontal field H is 4 sec. When another magnet is brought near it, the period of oscillation is reduced to 2s. The magnetic field of second bar magnet is

Options

A. 4H
B. 3H
C. 2H
D. `sqrt3 H`

Answer

To solve the problem step by step, we will use the relationship between the period of oscillation of a bar magnet and the magnetic fields involved.

### Step 1: Understand the relationship between period and magnetic field
The period of oscillation \( T \) of a freely suspended bar magnet in a magnetic field \( B \) is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{M \cdot B}} \]
where:
- \( I \) is the moment of inertia of the bar magnet,
- \( M \) is the magnetic moment of the bar magnet,
- \( B \) is the magnetic field in which the magnet is oscillating.

### Step 2: Set up the equations for the two scenarios
1. When the bar magnet is oscillating in Earth's magnetic field \( H \) (which we denote as \( B_H \)), the period \( T_1 \) is given as 4 seconds:
\[ T_1 = 2\pi \sqrt{\frac{I}{M \cdot H}} \]

2. When the second magnet is brought near, the period \( T_2 \) is reduced to 2 seconds:
\[ T_2 = 2\pi \sqrt{\frac{I}{M \cdot B_{net}}} \]
where \( B_{net} \) is the net magnetic field when the second magnet is present.

### Step 3: Relate the two periods
From the equations for \( T_1 \) and \( T_2 \), we can write:
\[ \frac{T_1}{T_2} = \sqrt{\frac{B_{net}}{H}} \]
Substituting the values of \( T_1 \) and \( T_2 \):
\[ \frac{4}{2} = \sqrt{\frac{B_{net}}{H}} \]
This simplifies to:
\[ 2 = \sqrt{\frac{B_{net}}{H}} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 4 = \frac{B_{net}}{H} \]
Thus, we can express \( B_{net} \) as:
\[ B_{net} = 4H \]

### Step 5: Determine the external magnetic field
The net magnetic field \( B_{net} \) is the sum of the Earth's magnetic field \( H \) and the magnetic field \( B_{ext} \) of the second magnet:
\[ B_{net} = H + B_{ext} \]
Substituting the expression for \( B_{net} \):
\[ 4H = H + B_{ext} \]
Rearranging gives:
\[ B_{ext} = 4H - H = 3H \]

### Conclusion
The magnetic field of the second bar magnet is \( 3H \).