The work done in turning a magnetic dipole of magnetic moment \(M\) in a magnetic field \(B\) by an angle \(90^\circ\) from the meridian is \(n\) times the corresponding work done in turning it through an angle of \(60^\circ\) in the same field and from the same initial condition. Find the value of \(n\).
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For a magnetic dipole:
\[
U=-MB\cos\theta.
\]
Work done equals the change in potential energy.
Step 1: Work formula for turning a dipole. Turning a magnetic dipole from angle $\theta_1$ to $\theta_2$ in a field costs work $W = MB(\cos\theta_1 - \cos\theta_2)$, found from the dipole energy $U = -MB\cos\theta$. Step 2: Both rotations start from the meridian. The initial angle is $\theta_1 = 0^\circ$ in both cases, so $\cos\theta_1 = 1$. Step 3: Work to turn through $90^\circ$. \[ W_{90} = MB(\cos 0^\circ - \cos 90^\circ) = MB(1 - 0) = MB \] Step 4: Work to turn through $60^\circ$. \[ W_{60} = MB(\cos 0^\circ - \cos 60^\circ) = MB\left(1 - \frac{1}{2}\right) = \frac{MB}{2} \] Step 5: Form the ratio $n$. The problem says $W_{90} = n\,W_{60}$, so $n = \frac{W_{90}}{W_{60}}$. Step 6: Compute $n$. \[ n = \frac{MB}{\frac{MB}{2}} = 2 \] \[ \boxed{n = 2} \]