A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 sec in earth's horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be

Options

A. 1 s
B. 2 s
C. 3 s
D. 4 s

Answer

To solve the problem, we need to determine the new time period of the bar magnet when a horizontal magnetic field is applied opposite to the Earth's magnetic field. Let's break this down step by step.

### Step 1: Understand the given data
- Initial time period \( T_1 = 2 \) seconds
- Earth's horizontal magnetic field \( B_1 = 24 \) microtesla
- Opposing magnetic field \( B_2 = 18 \) microtesla

### Step 2: Calculate the resultant magnetic field
When the opposing magnetic field is applied, the resultant magnetic field \( B \) can be calculated as:
\[ B = B_1 - B_2 \]
Substituting the values:
\[ B = 24 \, \mu T - 18 \, \mu T = 6 \, \mu T \]

### Step 3: Use the formula for the time period of oscillation
The time period \( T \) of a magnet in a magnetic field is given by:
\[ T = 2\pi \sqrt{\frac{I}{B}} \]
Where \( I \) is the moment of inertia and \( B \) is the magnetic field. Since the moment of inertia \( I \) remains constant for the same magnet, we can express the relationship between the time periods and magnetic fields as:
\[ \frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}} \]

### Step 4: Substitute the known values
From the previous steps, we know:
- \( T_1 = 2 \) seconds
- \( B_1 = 24 \, \mu T \)
- \( B_2 = 6 \, \mu T \)

Now substituting these values into the equation:
\[ \frac{T_2}{2} = \sqrt{\frac{24}{6}} \]

### Step 5: Simplify the equation
Calculating the right side:
\[ \sqrt{\frac{24}{6}} = \sqrt{4} = 2 \]
Thus, we have:
\[ \frac{T_2}{2} = 2 \]

### Step 6: Solve for \( T_2 \)
Multiplying both sides by 2:
\[ T_2 = 2 \times 2 = 4 \text{ seconds} \]

### Final Answer
The new time period of the magnet when the opposing magnetic field is applied is \( T_2 = 4 \) seconds.

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