In a uniform magnetic field of strength 0.15 T, a short bar magnet of magnetic `m=0.32JT^(-1)` is placed. The potential energy at equilibrium position position is

Options

A. `4.8xx10^(-2)J`
B. `9.6xx10^(-2)J`
C. `2.8xx10^(-2)J`
D. `1.2xx10^(-2)J`

Answer

To find the potential energy at the equilibrium position of a short bar magnet placed in a uniform magnetic field, we can follow these steps:

### Step 1: Understand the System
A short bar magnet has a magnetic moment \( m \) and is placed in a uniform magnetic field \( B \). The potential energy \( U \) of a magnetic dipole in a magnetic field is given by the formula:
\[ U = -\vec{m} \cdot \vec{B} \]
In terms of magnitudes, this can be expressed as:
\[ U = -mB \cos \theta \]
where \( \theta \) is the angle between the magnetic moment vector \( \vec{m} \) and the magnetic field vector \( \vec{B} \).

### Step 2: Determine the Equilibrium Position
The equilibrium position occurs when the torque acting on the magnet is zero. This happens when the magnetic moment \( \vec{m} \) is aligned with the magnetic field \( \vec{B} \), meaning \( \theta = 0^\circ \). At this angle, \( \cos \theta = \cos 0^\circ = 1 \).

### Step 3: Substitute Values
Given:
- \( m = 0.32 \, \text{JT}^{-1} \)
- \( B = 0.15 \, \text{T} \)

Now substituting these values into the potential energy formula:
\[ U = -mB \cos 0 \]
\[ U = -0.32 \times 0.15 \times 1 \]

### Step 4: Calculate Potential Energy
Now, perform the multiplication:
\[ U = -0.32 \times 0.15 = -0.048 \, \text{J} \]

### Conclusion
The potential energy at the equilibrium position is:
\[ U = -0.048 \, \text{J} \]