A bar magnet of magnetic moment `2.0` `A-m^2` is free to rotate about a vertical axis through its centre. The magnet is released from rest from the east west position. Find the kinetic energy of the magnet as it takes the north south position. The horizontal component of the earth's magnetic field as `B=25muT`. Earth's magnetic field is from south to north.

Explanation

To solve the problem, we need to find the kinetic energy of a bar magnet as it rotates from the east-west position to the north-south position in the presence of a magnetic field. Here's a step-by-step solution:

### Step 1: Understand the Initial and Final Positions
The bar magnet is initially positioned in the east-west direction and is released from rest. The final position is the north-south direction. The angle between the initial and final positions is 90 degrees.

### Step 2: Identify the Magnetic Moment and Magnetic Field
The magnetic moment \( M \) of the bar magnet is given as \( 2.0 \, \text{A-m}^2 \). The horizontal component of the Earth's magnetic field \( B \) is given as \( 25 \, \mu T = 25 \times 10^{-6} \, T \).

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