If $v_0 = 1.5 K L$ , the loop will stop before it enters completely inside the region of magnetic field
When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.
If $v_0 = \frac{K L}{10}$ , the loop comes to rest at $t = \left( \frac{1}{K} \right) \ln \left( \frac{5}{2} \right)$
If $v_0 = 3 K L$ , the complete loop enters inside the region of magnetic field at time $t = \left( \frac{1}{K} \right) \ln \left( \frac{3}{2} \right)$

Solution:

K=\frac{B_0^2 L^2}{R M}

\varepsilon=v B L

I=\frac{v B L}{R}

F_{(\text{drag})} = - \frac{v B L}{R} \cdot L B

m \frac{d v}{d t} = - \frac{(B^2 L^2) v}{R}

m \frac{d v}{d x} \cdot v = - \frac{B^2 L^2}{R} \cdot v

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

m \left. v \right|_{v_0}^{v} = \left. \frac{-B^2 L^2}{R} \cdot x \right|_{0}^{x}

\Rightarrow m (v - v_0) = \frac{-B^2 L^2 x}{R}

\Rightarrow v = v_0 - \frac{B^2 L^2}{m R} \cdot x

\text{If } v_0 = 1.5 K L \text{ then for } x = L

v = \frac{3}{2} K L - K L = \frac{1}{2} K L

So, the loop will have some velocity and it will not stop.

If the loop completely gets inside then further no current flows and no force acts.

v_0 = \frac{K L}{10}

then for

v=0

\frac{d v}{d t} = - \frac{B^2 L^2}{m R} v

\int_{v_0}^{v} \frac{d v}{v} = \int\limits_{0}^{t-K} \cdot dt \Rightarrow \ln \frac{v}{v_0} = -K t

\frac{v}{v_0} = e^{-K t}

it will never stop

v_0 = 3 K L

then

v

x = L

v = 3 K L - K L = 2 K L

\Rightarrow \frac{2}{3} = e^{-K t} \Rightarrow \frac{1}{K} \ln \left( \frac{3}{2} \right) = t

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