The work done in moving a dipole from its most stable to most unstable position in a 0.09 T uniform magnetic field is (dipole moment of this dipole = `0.5 Am^(2))`

Options

A. 0.07J
B. 0.08J
C. 0.09J
D. 0.1J

Explanation

To solve the problem of finding the work done in moving a magnetic dipole from its most stable to its most unstable position in a uniform magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Dipole moment (m) = 0.5 Am²
- Magnetic field (B) = 0.09 T

2. **Understand the Positions**:
- The most stable position of a dipole in a magnetic field is when it is aligned with the field, which corresponds to an angle (θ) of 0 degrees.
- The most unstable position is when the dipole is anti-aligned with the field, corresponding to an angle (θ) of 180 degrees.

3. **Potential Energy Calculation**:
- The potential energy (U) of a magnetic dipole in a magnetic field is given by the formula:
\[ U = -m \cdot B \cdot \cos(\theta) \]
- For the stable position (θ = 0 degrees):
\[ U_{\text{stable}} = -m \cdot B \cdot \cos(0) = -m \cdot B \cdot 1 = -mB \]
- For the unstable position (θ = 180 degrees):
\[ U_{\text{unstable}} = -m \cdot B \cdot \cos(180) = -m \cdot B \cdot (-1) = mB \]

4. **Calculate the Work Done**:
- The work done (W) in moving the dipole from the stable position to the unstable position is the change in potential energy:
\[ W = U_{\text{unstable}} - U_{\text{stable}} = mB - (-mB) = mB + mB = 2mB \]
- Substituting the values of m and B:
\[ W = 2 \cdot (0.5 \, \text{Am}^2) \cdot (0.09 \, \text{T}) = 2 \cdot 0.5 \cdot 0.09 = 0.09 \, \text{J} \]

5. **Final Answer**:
- The work done in moving the dipole from its most stable to most unstable position is **0.09 Joules**.