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Step-by-Step Solution
Step 1: Identify the Known Quantities
ā¢ External magnetic field, $B = 0.06\text{ T}$
ā¢ Angle between bar magnet axis and the field, $\theta = 30^\circ$
ā¢ Torque experienced by the bar magnet, $\tau = 0.018\text{ NĀ·m}$
Step 2: Write the Torque Formula for a Bar Magnet in a Uniform Field
The torque $\tau$ on a bar magnet of magnetic dipole moment $M$ placed in a uniform magnetic field $B$ at an angle $\theta$ is given by:
$$\tau = M B \sin \theta.$$
Step 3: Solve for the Magnetic Dipole Moment, $M$
Substitute the given values of $\tau$, $B$, and $\theta = 30^\circ$ into the equation:
$$0.018 = M \times 0.06 \times \sin(30^\circ).$$
Since $\sin(30^\circ) = 0.5$, we get:
$$0.018 = M \times 0.06 \times 0.5.$$
Hence,
$$M = \frac{0.018}{0.06 \times 0.5} = 0.6\text{ AĀ·m}^2.$$
Step 4: Determine the Potential Energy Difference Between Stable and Unstable Positions
The work required to rotate the magnet from its stable to its unstable equilibrium position corresponds to the change in potential energy:
$$W = U_f - U_i = M B (\cos \theta_i - \cos \theta_f).$$
ā¢ In the stable equilibrium position, the magnet is aligned with the field, so $\theta_i = 0^\circ$, and $\cos(0^\circ) = 1.$
ā¢ In the unstable equilibrium position, the magnet is aligned opposite to the field, so $\theta_f = 180^\circ$, and $\cos(180^\circ) = -1.$
Step 5: Compute the Work Done
Substitute $M = 0.6\text{ AĀ·m}^2$, $B = 0.06\text{ T}$, and the cosines for the two angles:
$$W = 0.6 \times 0.06 \left( \cos 0^\circ - \cos 180^\circ \right).$$
Since $\cos 0^\circ = 1$ and $\cos 180^\circ = -1$, we have:
$$W = 0.6 \times 0.06 \, (1 - (-1)) = 0.6 \times 0.06 \times 2 = 7.2 \times 10^{-2}\text{ J}.$$
Final Answer
The minimum work required to rotate the magnet from its stable to its unstable equilibrium position is:
$$7.2 \times 10^{-2}\,\text{J}.$$
