A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60^(@) is W. Now the torrue required to keep the magnet in this new position is

Options

A. `(W)/(sqrt3)`
B. `sqrt3W`
C. `(sqrt3W)/(2)`
D. `(2W)/(sqrt3)`

Answer

To solve the problem step by step, we will analyze the situation of a bar magnet hung in a uniform magnetic field and determine the torque required to maintain its position after being rotated by 60 degrees.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- The bar magnet is initially in equilibrium in a uniform horizontal magnetic field. At this position, the potential energy (U_initial) of the magnet can be expressed as:
\[ U_i = -m \cdot B \cdot h \]
where \(m\) is the magnetic moment of the magnet, \(B\) is the magnetic field strength, and \(h\) is the height from the reference point.

2. **Energy After Rotation**:
- When the magnet is rotated by 60 degrees, the new potential energy (U_final) becomes:
\[ U_f = -m \cdot B \cdot h \cdot \cos(60^\circ) \]
- Since \(\cos(60^\circ) = \frac{1}{2}\), we can rewrite the equation as:
\[ U_f = -m \cdot B \cdot h \cdot \frac{1}{2} = -\frac{m \cdot B \cdot h}{2} \]

3. **Calculating the Work Done (W)**:
- The work done (W) to rotate the magnet from its initial position to the new position is given by the change in potential energy:
\[ W = U_f - U_i \]
- Substituting the expressions for \(U_f\) and \(U_i\):
\[ W = \left(-\frac{m \cdot B \cdot h}{2}\right) - \left(-m \cdot B \cdot h\right) \]
- Simplifying this gives:
\[ W = -\frac{m \cdot B \cdot h}{2} + m \cdot B \cdot h = \frac{m \cdot B \cdot h}{2} \]

4. **Finding the Torque (τ)**:
- The torque (τ) required to maintain the magnet in the new position can be calculated using the formula:
\[ \tau = m \cdot B \cdot h \cdot \sin(\theta) \]
- Here, \(\theta = 60^\circ\), and \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\). Thus, we have:
\[ \tau = m \cdot B \cdot h \cdot \frac{\sqrt{3}}{2} \]

5. **Relating Torque to Work Done**:
- We already found that \(m \cdot B \cdot h = 2W\). Substituting this into the torque equation gives:
\[ \tau = 2W \cdot \frac{\sqrt{3}}{2} = W \cdot \sqrt{3} \]

6. **Final Answer**:
- Therefore, the torque required to keep the magnet in the new position is:
\[ \tau = \sqrt{3}W \]