A magnetic needle is made to vibrate in uniform field `H`, then its time period is `T`. If it vibrates in the field of intensity `4H`, its time period will be

Options

A. `2T`
B. `T/2`
C. `T/4`
D. T

Answer

To solve the problem, we need to determine the time period of a magnetic needle vibrating in a magnetic field of intensity \(4H\) given that its time period in a magnetic field of intensity \(H\) is \(T\).

### Step-by-Step Solution:

1. **Understand the formula for the time period**:
The time period \(T\) of a magnetic needle in a magnetic field \(H\) is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mH}} \]
where:
- \(I\) is the moment of inertia of the needle,
- \(m\) is the mass of the needle,
- \(H\) is the magnetic field intensity.

2. **Write the time period for the magnetic field \(H\)**:
For the magnetic field \(H\), we have:
\[ T = 2\pi \sqrt{\frac{I}{mH}} \tag{1} \]

3. **Write the time period for the magnetic field \(4H\)**:
Now, for the magnetic field \(4H\), the time period \(T'\) can be expressed as:
\[ T' = 2\pi \sqrt{\frac{I}{m(4H)}} \tag{2} \]

4. **Simplify the expression for \(T'\)**:
We can simplify equation (2):
\[ T' = 2\pi \sqrt{\frac{I}{4mH}} = 2\pi \sqrt{\frac{I}{mH}} \cdot \frac{1}{\sqrt{4}} = \frac{T}{2} \tag{3} \]
Here, we used the fact that \(\sqrt{4} = 2\).

5. **Conclusion**:
From equation (3), we find that:
\[ T' = \frac{T}{2} \]

Thus, the time period of the magnetic needle in the magnetic field of intensity \(4H\) is \(\frac{T}{2}\).

### Final Answer:
The time period \(T'\) in the field of intensity \(4H\) is \(\frac{T}{2}\).