1. Magnetic Field: The magnetic field $B$ at a distance $x$ from an infinite wire carrying current $I$ is $B = \frac{\mu_0 I}{2\pi x}$.

2. Motional EMF: The conductor EF moves with velocity $v$ parallel to the wire PQ. A small element of length $dx$ at distance $x$ induces an emf $d\varepsilon = v B dx$. The total emf $\varepsilon$ across EF (from $a$ to $b$) is: $\varepsilon = \int_a^b v \frac{\mu_0 I}{2\pi x} dx = \frac{\mu_0 I v}{2\pi} [\ln x]_a^b = \frac{\mu_0 I v}{2\pi} \ln \left( \frac{b}{a} \right)$

3. Induced Current: The current $i$ in the loop is $i = \frac{\varepsilon}{R}$.

4. Magnetic Force: The magnetic force $dF$ on the element $dx$ carrying induced current $i$ is $dF = i B dx$. The total force opposing motion is: $F_{mag} = \int_a^b i B dx = i \int_a^b \frac{\mu_0 I}{2\pi x} dx = i \left[ \frac{\mu_0 I}{2\pi} \ln \left( \frac{b}{a} \right) \right]$

5. External Force: To maintain constant velocity, the external force $F$ must balance $F_{mag}$. Substituting $i = \varepsilon/R$: $F = \frac{\varepsilon}{R} \left[ \frac{\mu_0 I}{2\pi} \ln \left( \frac{b}{a} \right) \right] = \frac{1}{R} \left[ \frac{\mu_0 I v}{2\pi} \ln \left( \frac{b}{a} \right) \right] \left[ \frac{\mu_0 I}{2\pi} \ln \left( \frac{b}{a} \right) \right]$ $F = \frac{v}{R} \left[ \frac{\mu_0 I}{2\pi} \ln \left( \frac{b}{a} \right) \right]^2$

This expression can be rewritten to match Option A by multiplying numerator and denominator by $v$: $F = \frac{1}{vR} \left[ \frac{\mu_0 I v}{2\pi} \ln \left( \frac{b}{a} \right) \right]^2$