A magnet freely suspended in a vibration magnetometer makes 10 oscillations per minute at a place A and 20 oscillations per minute at a place B. If the horizontal component of earth's magnetic field at A is `36xx10^(-6)T`, then its value at B is

Options

A. `36xx10^(-6)T`
B. `72xx10^(-6)T`
C. `144xx10^(-6)T`
D. `288xx10^(-6)T`

Explanation

To solve the problem, we need to relate the oscillation frequencies of the magnet in the vibration magnetometer at two different locations (A and B) to the horizontal component of the Earth's magnetic field at those locations.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The time period \( T \) of oscillation of a magnet in a magnetic field is given by:
\[ T = 2\pi \sqrt{\frac{I}{M B_H}} \]
where \( I \) is the moment of inertia, \( M \) is the magnetic moment, and \( B_H \) is the horizontal component of the magnetic field.

2. **Frequency and Time Period**:
The frequency \( f \) is the reciprocal of the time period:
\[ f = \frac{1}{T} \]
Thus, we can express the time period in terms of frequency:
\[ T = \frac{1}{f} \]

3. **Inversely Proportional Relationship**:
Since \( T \) is inversely proportional to \( \sqrt{B_H} \), we can write:
\[ T \propto \frac{1}{\sqrt{B_H}} \]
Therefore, the ratio of the time periods at points A and B can be expressed as:
\[ \frac{T_A}{T_B} = \sqrt{\frac{B_{H_B}}{B_{H_A}}} \]

4. **Using Given Frequencies**:
At point A, the magnet makes 10 oscillations per minute, and at point B, it makes 20 oscillations per minute. The frequencies are:
\[ f_A = 10 \text{ oscillations/min} = \frac{10}{60} \text{ Hz} = \frac{1}{6} \text{ Hz} \]
\[ f_B = 20 \text{ oscillations/min} = \frac{20}{60} \text{ Hz} = \frac{1}{3} \text{ Hz} \]

5. **Calculating Time Periods**:
The time periods at A and B are:
\[ T_A = \frac{1}{f_A} = 6 \text{ s} \]
\[ T_B = \frac{1}{f_B} = 3 \text{ s} \]

6. **Setting Up the Equation**:
Now substituting into the ratio:
\[ \frac{T_A}{T_B} = \sqrt{\frac{B_{H_B}}{B_{H_A}}} \]
This gives:
\[ \frac{6}{3} = \sqrt{\frac{B_{H_B}}{36 \times 10^{-6}}} \]
Simplifying:
\[ 2 = \sqrt{\frac{B_{H_B}}{36 \times 10^{-6}}} \]

7. **Squaring Both Sides**:
Squaring both sides results in:
\[ 4 = \frac{B_{H_B}}{36 \times 10^{-6}} \]

8. **Solving for \( B_{H_B} \)**:
Multiplying both sides by \( 36 \times 10^{-6} \):
\[ B_{H_B} = 4 \times 36 \times 10^{-6} = 144 \times 10^{-6} \text{ T} \]

### Final Answer:
The value of the horizontal component of the Earth's magnetic field at point B is:
\[ B_{H_B} = 144 \times 10^{-6} \text{ T} \]