In tan A position two short bar magnets of magnetic moments in ratio of `sqrt3 : 1` are placed at same distance respectively . If the deflection produced for the first magnet be `60^(@)` then what will be deflection for second magnet ?

Options

A. A. `45^(@)`
B. B. `60^(@)`
C. C. `30^(@)`
D. D. `90^(@)`

Answer

To solve the problem, we will use the relationship between the magnetic moments of the bar magnets and the angles of deflection they produce. The key points are:

1. The deflection angle produced by a bar magnet is related to its magnetic moment and the distance from the point of observation.
2. The ratio of the magnetic moments is given, and we can use this to find the deflection angle for the second magnet.

### Step-by-Step Solution:

1. **Identify Given Information:**
- The magnetic moments of the two bar magnets are in the ratio \( M_1 : M_2 = \sqrt{3} : 1 \).
- The deflection angle for the first magnet (\( \theta_1 \)) is \( 60^\circ \).

2. **Use the Formula for Magnetic Field:**
The magnetic field \( B_H \) produced by a bar magnet at a distance \( d \) along its axis is given by:
\[ B_H \propto \frac{M}{d^3} \]
where \( M \) is the magnetic moment.

3. **Set Up the Equations for Each Magnet:**
For the first magnet:
\[ B_H \tan \theta_1 = k \cdot \frac{M_1}{d^3} \]
For the second magnet:
\[ B_H \tan \theta_2 = k \cdot \frac{M_2}{d^3} \]
Here, \( k \) is a constant that includes factors like \( \mu_0 \) and \( 4\pi \).

4. **Divide the Two Equations:**
By dividing the equation for the first magnet by the equation for the second magnet:
\[ \frac{\tan \theta_1}{\tan \theta_2} = \frac{M_1}{M_2} \]

5. **Substitute Known Values:**
We know \( \tan \theta_1 = \tan 60^\circ = \sqrt{3} \) and the ratio of magnetic moments \( \frac{M_1}{M_2} = \sqrt{3} \):
\[ \frac{\sqrt{3}}{\tan \theta_2} = \sqrt{3} \]

6. **Solve for \( \tan \theta_2 \):**
From the equation:
\[ \tan \theta_2 = 1 \]

7. **Find the Angle \( \theta_2 \):**
The angle whose tangent is 1 is:
\[ \theta_2 = 45^\circ \]

### Final Answer:
The deflection for the second magnet will be \( 45^\circ \).