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

Explanation

To solve the problem step by step, we will use the relationship between the time period of oscillation of a magnet and the magnetic field it is subjected to.

### Step 1: Understand the relationship between time period and magnetic field
The time period \( T \) of a vibrating magnet is related to the magnetic field \( B \) by the formula:
\[ T \propto \frac{1}{\sqrt{B}} \]
This means that if the magnetic field changes, the time period will change inversely with the square root of the magnetic field.

### Step 2: Set up the initial conditions
Given:
- Initial time period \( T_1 = 2 \) seconds
- Initial magnetic field \( B_1 = 24 \) microtesla \( = 24 \times 10^{-6} \) tesla

### Step 3: Calculate the new magnetic field
When a horizontal field of \( 18 \) microtesla is applied opposite to the Earth's field, the effective magnetic field \( B_2 \) can be calculated as:
\[ B_2 = B_1 - 18 \times 10^{-6} \text{ tesla} \]
Substituting the values:
\[ B_2 = 24 \times 10^{-6} - 18 \times 10^{-6} = 6 \times 10^{-6} \text{ tesla} \]

### Step 4: Use the relationship to find the new time period
Using the relationship derived in Step 1, we can express the ratio of the time periods in terms of the magnetic fields:
\[ \frac{T_1}{T_2} = \sqrt{\frac{B_2}{B_1}} \]
Rearranging gives:
\[ T_2 = T_1 \cdot \sqrt{\frac{B_1}{B_2}} \]

### Step 5: Substitute the known values
Substituting \( T_1 = 2 \) seconds, \( B_1 = 24 \times 10^{-6} \) tesla, and \( B_2 = 6 \times 10^{-6} \) tesla:
\[ T_2 = 2 \cdot \sqrt{\frac{24 \times 10^{-6}}{6 \times 10^{-6}}} \]

### Step 6: Simplify the expression
Calculating the square root:
\[ \sqrt{\frac{24}{6}} = \sqrt{4} = 2 \]
Thus,
\[ T_2 = 2 \cdot 2 = 4 \text{ seconds} \]

### Conclusion
The new time period \( T_2 \) of the magnet when the horizontal field of \( 18 \) microtesla is applied opposite to the Earth's field is \( 4 \) seconds.