This force is given by

$F=-BIa=-B\left(\frac{E}{R}\right) \cdot a$

$=-B\left(\frac{B a v}{R}\right) a=\frac{-B^{2} a^{2} v}{R}$

(Negative sign shows retarding force)

So, deacceleration $A$ of loop is

$A =\frac{F}{m}=\frac{-B^{2} a^{2} v}{m R}$

$\Rightarrow \quad \frac{d v}{d t} =\frac{-B^{2} a^{2}}{m R} v \Rightarrow \frac{d v}{d x} \cdot \frac{d x}{d t}=\frac{-B^{2} a^{2} v}{m R}$

$\Rightarrow v \frac{d v}{d x}=\frac{-B^{2} a^{2}}{m R} \cdot v$

$\Rightarrow d v=\frac{-B^{2} a^{2}}{m R} \cdot d x$

Integrating between limits, we have

$\int \limits_{v_{0}}^{v} d v=\int \limits_{0}^{x}-\frac{B^{2} a^{2}}{m R} \cdot d x$

$v-v_{0}=\frac{-B^{2} a^{2}}{m R} \cdot x$

So, velocity of loop at distance $x$ is

$\Rightarrow \quad v=v_{0}-\frac{B^{2} a^{2}}{m R} \cdot x$