Step-by-step Derivation

1. Analyze the physical principle: The problem states that the closed wire loop is made of a special metal S that conducts electricity without any resistance (a superconductor). A key property of a superconductor is that it does not allow any change in the magnetic flux passing through it. This is a consequence of Lenz's law and zero resistance, which allows a persistent current to flow to counteract any change in external flux.

2. Determine the initial and final flux conditions:

   • Initially, the magnetic dipole m is at infinity. The magnetic field from the dipole at the loop is zero, so the initial magnetic flux through the loop is zero ($Φ_{initial} = 0$).
   • Since the loop is superconducting, the total magnetic flux through it must remain constant. Therefore, the net flux through the loop must always be zero.
   • When the dipole is brought to a distance r from the center of the loop, it produces a magnetic flux, let's call it $Φ_{dipole}$, through the loop.
   • To keep the net flux zero, the loop must induce a current I that generates its own magnetic flux, $Φ_{induced}$, such that it exactly cancels the flux from the dipole.

3. Formulate the flux balance equation: The condition for the net flux to remain zero is: $Φ_{net} = Φ_{dipole} + Φ_{induced} = 0$ This implies that the magnitude of the induced flux must be equal to the magnitude of the flux from the dipole: $|Φ_{induced}| = |Φ_{dipole}|$

4. Calculate the flux due to the magnetic dipole ($Φ_{dipole}$):

   • The problem gives the magnitude of the magnetic field of a dipole m at a point on its axis at a distance r as: $B = \frac{\mu_0}{2\pi} \frac{m}{r^3}$
   • The loop has a radius a, and it is given that r >> a. This condition allows us to approximate the magnetic field as being uniform over the entire area of the loop, with the value it has at the center.
   • The area of the loop is $A = \pi a^2$.
   • The magnetic flux through the loop due to the dipole is the product of the magnetic field and the area: $Φ_{dipole} = B \cdot A = \left( \frac{\mu_0}{2\pi} \frac{m}{r^3} \right) (\pi a^2)$ $Φ_{dipole} = \frac{\mu_0 a^2 m}{2 r^3}$

5. Relate the induced flux to the induced current:

   • The magnetic flux generated by a current I flowing in a loop through the loop itself is given by its self-inductance L. $Φ_{induced} = L \cdot I$
   • Here, L is the self-inductance of the circular loop, which is a constant that depends only on the geometry of the loop (its radius a and the thickness of the wire).

6. Solve for the induced current I:

   • Using the flux balance equation from Step 3: $|Φ_{induced}| = |Φ_{dipole}|$ $L \cdot I = \frac{\mu_0 a^2 m}{2 r^3}$
   • Solving for the current I: $I = \frac{\mu_0 a^2 m}{2 L r^3}$

7. Determine the proportionality:

   • In the expression for the current I$, the terms $\mu_0$, $a (radius of the loop), and L (self-inductance of the loop) are all constants for the given setup.
   • Therefore, the induced current I is proportional to the variable parts of the expression, which are m and r. $I \propto \frac{m}{r^3}$

8. Compare with the given options:

   • A: m/r³ - This matches our derived proportionality.
   • B: m²/r² - Incorrect.
   • C: m/r² - Incorrect.
   • D: m²/r - Incorrect.

The information about the force between two dipoles is extra information not required to solve for the induced current's proportionality.

Conclusion

The current induced in the loop is proportional to $m/r^3$. Therefore, option A is the correct answer.