A small bar magnet `A` oscillates in a horizontal plane with a period `T` at a place where the angle of dip is `60^(@)`. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic merdian, its period will be

Options

A. `(T)/(sqrt2)`
B. T
C. `sqrt2T`
D. `2T`

Answer

To solve the problem, we need to find the period of oscillation of a small bar magnet when it oscillates in a vertical plane coinciding with the magnetic meridian, given that its period in a horizontal plane is \( T \) and the angle of dip is \( 60^\circ \).

### Step-by-Step Solution:

1. **Understand the Period of Oscillation in a Horizontal Plane**:
The period of oscillation \( T \) for a bar magnet in a horizontal plane is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{M \cdot B_H}} \]
where:
- \( I \) is the moment of inertia of the magnet,
- \( M \) is the magnetic moment of the magnet,
- \( B_H \) is the horizontal component of the Earth's magnetic field.

2. **Determine the Period of Oscillation in a Vertical Plane**:
When the magnet oscillates in a vertical plane coinciding with the magnetic meridian, the period \( T' \) is given by:
\[ T' = 2\pi \sqrt{\frac{I}{M \cdot B}} \]
where \( B \) is the total magnetic field strength.

3. **Relate the Horizontal and Vertical Components of the Magnetic Field**:
The horizontal component \( B_H \) can be expressed in terms of the total magnetic field \( B \) and the angle of dip \( \delta \):
\[ B_H = B \cos(\delta) \]
Given that the angle of dip \( \delta = 60^\circ \), we can substitute:
\[ B_H = B \cos(60^\circ) = B \cdot \frac{1}{2} \]

4. **Substitute \( B_H \) into the Period Formula**:
Now, substituting \( B_H \) into the equation for \( T \):
\[ T = 2\pi \sqrt{\frac{I}{M \cdot \left(B \cdot \frac{1}{2}\right)}} = 2\pi \sqrt{\frac{2I}{M \cdot B}} \]

5. **Find the Ratio of the Periods**:
To find the ratio of the periods \( \frac{T}{T'} \):
\[ \frac{T}{T'} = \sqrt{\frac{B}{B_H}} = \sqrt{\frac{B}{B \cdot \frac{1}{2}}} = \sqrt{2} \]
Therefore, we can express \( T' \) in terms of \( T \):
\[ T' = T \cdot \frac{1}{\sqrt{2}} = \frac{T}{\sqrt{2}} \]

6. **Conclusion**:
The period of oscillation of the magnet in the vertical plane is:
\[ T' = \frac{T}{\sqrt{2}} \]

### Final Answer:
The period of oscillation when the magnet oscillates in a vertical plane coinciding with the magnetic meridian is \( \frac{T}{\sqrt{2}} \).

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