Question

A bar magnet of magnetic moment \(1.8\ \mathrm{Am^2}\) is free to rotate about a vertical axis passing through its centre at a place where the vertical component of earth's magnetic field is \[ 0.3\times10^{-4}\text{ T} \] and the dip angle is \(45^\circ\). If the magnet at rest in east-west direction is released, then the kinetic energy (in \(\mu\)J) of the magnet when it reaches north-south direction is

• A. \(72\)
• B. \(54\)
• C. \(27\)
• D. \(36\)

Hint:

For a freely suspended magnet, \[ \boxed{ U=-MB\cos\theta. } \] If the magnet rotates about a vertical axis, only the horizontal component of Earth's magnetic field contributes: \[ \boxed{ B_H=\frac{B_V}{\tan\delta}. } \]

Answer 1

The Correct Option is B

Step 1: Find the horizontal component of Earth's magnetic field. Given, \[ B_V = 0.3\times10^{-4}\text{ T}, \] and \[ \delta=45^\circ. \] Since \[ \tan\delta = \frac{B_V}{B_H}, \] we get \[ B_H = \frac{B_V}{\tan45^\circ} = 0.3\times10^{-4}\text{ T}. \]

Step 2: Use the change in magnetic potential energy. The magnet rotates in the horizontal plane, so only the horizontal component \(B_H\) is effective. Magnetic potential energy is \[ U = -MB_H\cos\theta. \] Initially, the magnet is along the east-west direction, \[ \theta=90^\circ, \] so \[ U_i=0. \] Finally, the magnet aligns along the north-south direction, \[ \theta=0^\circ, \] thus \[ U_f = -MB_H. \] Hence, the gain in kinetic energy is \[ K = U_i-U_f = MB_H. \]

Step 3: Calculate the kinetic energy. Substituting, \[ K = 1.8 \times 0.3\times10^{-4} = 5.4\times10^{-5}\text{ J}. \] Converting into \[ \mu\text{J}, \] \[ K = 5.4\times10^{-5}\times10^6 = 54\ \mu\text{J}. \] Hence, \[ \boxed{54\ \mu\text{J}}. \] Thus, \[ \boxed{(B)} \] is the correct answer.