A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2s in the earth's horizontal magnetic field of `24 mu T`. When a horizontal field of `18 mu T` is produced opposite to the earth's field by placing a current carrying wire, the new time period of the magnet will be

Answer

To find the new time period of the magnet when a horizontal magnetic field of 18 µT is applied opposite to the Earth's magnetic field, we can follow these steps:

### Step 1: Understand the relationship between time period and magnetic field
The time period \( T \) of a vibrating magnet is given by the formula:
\[ T \propto \frac{1}{\sqrt{B}} \]
where \( B \) is the magnetic field strength. This means that the time period is inversely proportional to the square root of the magnetic field.

### Step 2: Write down the initial conditions
From the question, we know:
- Initial time period \( T_1 = 2 \) s
- Initial magnetic field \( B_1 = 24 \) µT

### Step 3: Calculate the new effective magnetic field
When a horizontal field of 18 µT is applied opposite to the Earth's field, the effective magnetic field \( B_2 \) can be calculated as:
\[ B_2 = B_1 - 18 \, \mu T = 24 \, \mu T - 18 \, \mu T = 6 \, \mu T \]

### Step 4: Set up the ratio of the time periods
Using the relationship between time period and magnetic field:
\[ \frac{T_1}{T_2} = \sqrt{\frac{B_2}{B_1}} \]
Substituting the known values:
\[ \frac{2}{T_2} = \sqrt{\frac{6 \, \mu T}{24 \, \mu T}} \]

### Step 5: Simplify the ratio
Calculating the right side:
\[ \frac{2}{T_2} = \sqrt{\frac{6}{24}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Step 6: Solve for the new time period \( T_2 \)
Cross-multiplying gives:
\[ 2 = \frac{T_2}{2} \]
Thus,
\[ T_2 = 4 \, \text{s} \]

### Final Answer
The new time period of the magnet when the horizontal field of 18 µT is applied opposite to the Earth's field is \( T_2 = 4 \) s.

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