A bar magnet with a magnetic moment `5.0Am^(2)` is placed in parallel position relative to a magnetic field of `0.4T`.The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is

Options

A. `4J`
B. `1J`
C. `2J`
D. zero

Answer

To solve the problem, we need to calculate the work done in turning a bar magnet from a parallel position to an antiparallel position in a magnetic field. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Work Done in a Magnetic Field
The work done (W) in turning a magnetic moment (m) in a magnetic field (B) can be expressed as:
\[ W = \int \tau \, d\theta \]
where \( \tau \) is the torque acting on the magnetic moment.

### Step 2: Calculate the Torque
The torque (\( \tau \)) on a magnetic moment in a magnetic field is given by:
\[ \tau = mB \sin \theta \]
where:
- \( m \) is the magnetic moment,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the magnetic moment and the magnetic field.

### Step 3: Set Up the Integral
To find the work done in turning the magnet from parallel (0 degrees) to antiparallel (180 degrees), we set up the integral:
\[ W = \int_0^{\pi} mB \sin \theta \, d\theta \]

### Step 4: Factor Out Constants
Since \( m \) and \( B \) are constants, we can factor them out of the integral:
\[ W = mB \int_0^{\pi} \sin \theta \, d\theta \]

### Step 5: Evaluate the Integral
The integral of \( \sin \theta \) from 0 to \( \pi \) is:
\[ \int_0^{\pi} \sin \theta \, d\theta = [-\cos \theta]_0^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 2 \]

### Step 6: Substitute Back into the Work Equation
Now substituting back into the work equation:
\[ W = mB \cdot 2 \]
\[ W = 2mB \]

### Step 7: Substitute the Values
Given:
- \( m = 5.0 \, \text{Am}^2 \)
- \( B = 0.4 \, \text{T} \)

Substituting these values:
\[ W = 2 \times 5.0 \, \text{Am}^2 \times 0.4 \, \text{T} \]
\[ W = 2 \times 5.0 \times 0.4 = 4.0 \, \text{J} \]

### Final Answer
The work done in turning the magnet from parallel to antiparallel is:
\[ \boxed{4 \, \text{J}} \]