1. Understanding the Physics

The problem involves a small circular loop placed in a time-varying magnetic field produced by a long solenoid whose axis is changing direction. This will induce a current in the loop, creating a magnetic dipole moment. The interaction of this induced magnetic moment with the solenoid's magnetic field will produce a torque on the loop.

The key principles are:

• Magnetic field of a long solenoid: The field inside is uniform, given by $\vec{B} = B \hat{n} = (\mu_0 m I) \hat{n}$, where $m$ is the turns per unit length, $I$ is the current, and $\hat{n}$ is the unit vector along the solenoid's axis.
• Magnetic Flux: The loop lies on the $xy$-plane, so its area vector is $\vec{A} = A \hat{k}$. The magnetic flux through the loop is $\Phi = \vec{B} \cdot \vec{A} = (B \hat{n}) \cdot (A \hat{k}) = BA (\hat{n} \cdot \hat{k})$.
• Faraday's Law of Induction: The induced electromotive force (EMF) is $\mathcal{E} = -\frac{d\Phi}{dt}$.
• Induced Current: The current in the loop is $i = \frac{\mathcal{E}}{R} = -\frac{1}{R} \frac{d\Phi}{dt}$.
• Magnetic Dipole Moment: The induced magnetic moment of the loop is $\vec{\mu} = i \vec{A}$.
• Torque on a Magnetic Dipole: The torque experienced by the loop is $\vec{\tau} = \vec{\mu} \times \vec{B}$.

2. Deriving a General Expression for Torque

Let's combine these principles to find a general formula for the torque.

1. Flux: $\Phi = BA (\hat{n} \cdot \hat{k})$. Since $B$ and $A$ are constant, the change in flux is due to the change in $\hat{n}$.
2. Induced EMF: $\mathcal{E} = -\frac{d}{dt} [BA (\hat{n} \cdot \hat{k})] = -BA \frac{d}{dt}(\hat{n} \cdot \hat{k})$.
3. Induced Current: $i = \frac{\mathcal{E}}{R} = -\frac{BA}{R} \frac{d}{dt}(\hat{n} \cdot \hat{k})$.
4. Magnetic Moment: $\vec{\mu} = i \vec{A} = \left( -\frac{BA}{R} \frac{d}{dt}(\hat{n} \cdot \hat{k}) \right) A \hat{k} = -\frac{BA^2}{R} \frac{d(\hat{n} \cdot \hat{k})}{dt} \hat{k}$.
5. Torque: $\vec{\tau} = \vec{\mu} \times \vec{B} = \left( -\frac{BA^2}{R} \frac{d(\hat{n} \cdot \hat{k})}{dt} \hat{k} \right) \times (B \hat{n}) = -\frac{B^2 A^2}{R} \frac{d(\hat{n} \cdot \hat{k})}{dt} (\hat{k} \times \hat{n})$.

We are given $B = \mu_0 m I$ and $\alpha = \frac{A^2 \mu_0^2 m^2 I^2 \omega}{2R} = \frac{A^2 B^2 \omega}{2R}$. This implies $\frac{A^2 B^2}{R} = \frac{2\alpha}{\omega}$. Substituting this into the torque expression: $\vec{\tau} = -\frac{2\alpha}{\omega} \frac{d(\hat{n} \cdot \hat{k})}{dt} (\hat{k} \times \hat{n})$ Let $\hat{n} = n_x \hat{\imath} + n_y \hat{\jmath} + n_z \hat{k}$. Then $\hat{n} \cdot \hat{k} = n_z$ and $\hat{k} \times \hat{n} = \hat{k} \times (n_x \hat{\imath} + n_y \hat{\jmath}) = n_x (\hat{k} \times \hat{\imath}) + n_y (\hat{k} \times \hat{\jmath}) = n_x \hat{\jmath} - n_y \hat{\imath}$. So, the general formula for torque is: $\vec{\tau} = -\frac{2\alpha}{\omega} \frac{dn_z}{dt} (n_x \hat{\jmath} - n_y \hat{\imath})$ We need to evaluate the torque at $t = \frac{\pi}{6\omega}$. At this time, $\omega t = \frac{\pi}{6}$, so $\sin(\omega t) = \frac{1}{2}$ and $\cos(\omega t) = \frac{\sqrt{3}}{2}$.

3. Case-by-Case Analysis

(I) $\hat{n} = \frac{1}{\sqrt{2}}(\sin \omega t \hat{\jmath}+\cos \omega t \hat{k})$

• $n_x=0$, $n_y = \frac{1}{\sqrt{2}} \sin \omega t$, $n_z = \frac{1}{\sqrt{2}} \cos \omega t$.
• $\frac{dn_z}{dt} = \frac{d}{dt} \left(\frac{1}{\sqrt{2}} \cos \omega t\right) = -\frac{\omega}{\sqrt{2}} \sin \omega t$.
• At $t = \frac{\pi}{6\omega}$: $n_y = \frac{1}{\sqrt{2}} (\frac{1}{2}) = \frac{1}{2\sqrt{2}}$, and $\frac{dn_z}{dt} = -\frac{\omega}{\sqrt{2}} (\frac{1}{2}) = -\frac{\omega}{2\sqrt{2}}$.
• $\vec{\tau} = -\frac{2\alpha}{\omega} \left(-\frac{\omega}{2\sqrt{2}}\right) \left(0 \cdot \hat{\jmath} - \frac{1}{2\sqrt{2}} \hat{\imath}\right) = \frac{\alpha}{\sqrt{2}} \left(-\frac{1}{2\sqrt{2}} \hat{\imath}\right) = -\frac{\alpha}{4} \hat{\imath}$.
• This matches (Q). So, I $\rightarrow$ Q.

(II) $\hat{n} = \frac{1}{\sqrt{2}}(\sin \omega t \hat{\imath}+\cos \omega t \hat{\jmath})$

• $n_x = \frac{1}{\sqrt{2}} \sin \omega t$, $n_y = \frac{1}{\sqrt{2}} \cos \omega t$, $n_z = 0$.
• $\frac{dn_z}{dt} = 0$.
• Since $\frac{dn_z}{dt} = 0$, the torque $\vec{\tau}$ is zero.
• This matches (P). So, II $\rightarrow$ P.

(III) $\hat{n} = \frac{1}{\sqrt{2}}(\sin \omega t \hat{\imath}+\cos \omega t \hat{k})$

• $n_x = \frac{1}{\sqrt{2}} \sin \omega t$, $n_y = 0$, $n_z = \frac{1}{\sqrt{2}} \cos \omega t$.
• $\frac{dn_z}{dt} = -\frac{\omega}{\sqrt{2}} \sin \omega t$.
• At $t = \frac{\pi}{6\omega}$: $n_x = \frac{1}{\sqrt{2}} (\frac{1}{2}) = \frac{1}{2\sqrt{2}}$, and $\frac{dn_z}{dt} = -\frac{\omega}{\sqrt{2}} (\frac{1}{2}) = -\frac{\omega}{2\sqrt{2}}$.
• $\vec{\tau} = -\frac{2\alpha}{\omega} \left(-\frac{\omega}{2\sqrt{2}}\right) \left(\frac{1}{2\sqrt{2}} \hat{\jmath} - 0 \cdot \hat{\imath}\right) = \frac{\alpha}{\sqrt{2}} \left(\frac{1}{2\sqrt{2}} \hat{\jmath}\right) = \frac{\alpha}{4} \hat{\jmath}$.
• The list has $\frac{\alpha}{4} \hat{\jmath}$ as (S). Note: The provided image shows option (S) as $\frac{\alpha}{4} \hat{\jmath}$. Assuming this is a typo in the question text and it should be $\hat{j}$ not $\hat{i}$. Given the options, this is a reasonable assumption. Let's proceed.
• Correction: The image for the problem correctly shows (S) is $\frac{\alpha}{4} \hat{\jmath}$. The text description was a typo. Thus, it matches (S). So, III $\rightarrow$ S.

(IV) $\hat{n} = \frac{1}{\sqrt{2}}(\cos \omega t \hat{\jmath}+\sin \omega t \hat{k})$

• $n_x=0$, $n_y = \frac{1}{\sqrt{2}} \cos \omega t$, $n_z = \frac{1}{\sqrt{2}} \sin \omega t$.
• $\frac{dn_z}{dt} = \frac{d}{dt} \left(\frac{1}{\sqrt{2}} \sin \omega t\right) = \frac{\omega}{\sqrt{2}} \cos \omega t$.
• At $t = \frac{\pi}{6\omega}$: $n_y = \frac{1}{\sqrt{2}} (\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2\sqrt{2}}$, and $\frac{dn_z}{dt} = \frac{\omega}{\sqrt{2}} (\frac{\sqrt{3}}{2}) = \frac{\omega\sqrt{3}}{2\sqrt{2}}$.
• $\vec{\tau} = -\frac{2\alpha}{\omega} \left(\frac{\omega\sqrt{3}}{2\sqrt{2}}\right) \left(0 \cdot \hat{\jmath} - \frac{\sqrt{3}}{2\sqrt{2}} \hat{\imath}\right) = -\frac{\alpha\sqrt{3}}{\sqrt{2}} \left(-\frac{\sqrt{3}}{2\sqrt{2}} \hat{\imath}\right) = \frac{3\alpha}{4} \hat{\imath}$.
• This matches (R). So, IV $\rightarrow$ R.

4. Conclusion

Summarizing the matches:

• I $\rightarrow$ Q
• II $\rightarrow$ P
• III $\rightarrow$ S
• IV $\rightarrow$ R

This combination corresponds to option C.