A magnet of magnetic moment `2JT^(-1)` is aligned in the direction of magnetic field of `0.1T`. What is the net work done to bring the magnet normal to the magnrtic field?

Options

A. `0.1J`
B. `0.2 J`
C. `1 J`
D. `2 J`

Explanation

To solve the problem step by step, we will calculate the work done to bring a magnet from being aligned with a magnetic field to being perpendicular to it.

### Step 1: Understand the Initial and Final Positions
- Initially, the magnet is aligned with the magnetic field. This means the angle between the magnetic moment (M) and the magnetic field (B) is 0 degrees.
- Finally, we want to bring the magnet to a position where it is perpendicular to the magnetic field, meaning the angle between M and B will be 90 degrees.

### Step 2: Write the Formula for Potential Energy
The potential energy (U) of a magnetic moment in a magnetic field is given by:
\[ U = -\vec{M} \cdot \vec{B} = -MB \cos(\theta) \]
where:
- \( M \) is the magnetic moment,
- \( B \) is the magnetic field,
- \( \theta \) is the angle between M and B.

### Step 3: Calculate Initial Potential Energy
- For the initial position (aligned with the field, \( \theta = 0^\circ \)):
\[ U_i = -MB \cos(0^\circ) = -MB \]
- Substituting the values \( M = 2 \, \text{JT}^{-1} \) and \( B = 0.1 \, \text{T} \):
\[ U_i = - (2 \, \text{JT}^{-1}) \times (0.1 \, \text{T}) = -0.2 \, \text{J} \]

### Step 4: Calculate Final Potential Energy
- For the final position (perpendicular to the field, \( \theta = 90^\circ \)):
\[ U_f = -MB \cos(90^\circ) = -MB \cdot 0 = 0 \]
- Thus, the final potential energy \( U_f = 0 \, \text{J} \).

### Step 5: Calculate the Work Done
The work done (W) to move the magnet from the initial to the final position is equal to the change in potential energy:
\[ W = U_f - U_i \]
Substituting the values:
\[ W = 0 - (-0.2) = 0.2 \, \text{J} \]

### Conclusion
The net work done to bring the magnet normal to the magnetic field is \( 0.2 \, \text{J} \).

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