18. A conducting square loop of side $L$, mass $M$ and resistance $R$ is moving in the $XY$-plane with its edges parallel to the $X$ and $Y$ axes.

The region $y>0$ has a uniform magnetic field, \overrightarrow{B}=B_0\hat{k}  The magnetic field is zero everywhere else. At time $t=0$, the loop starts to enter the magnetic field with initial velocity $v_0\hat{j}$. Considering the quantity K=\frac{B_0^2}{L^2}\frac{1}{RM}  in appropriate units, ignoring self-inductance and gravity, which of the following statements is/are correct?

Based on the provided correct options, the correct statements are:

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

Step-by-Step Derivation and Analysis:

1. Equation of Motion during Entry:
Let $y$ represent the distance that the loop has penetrated into the magnetic field region ($y>0$) at time $t$, where $0<y<L$.
The magnetic field is given by:
$\overrightarrow{B}=B_0\hat{k}$
The magnetic flux $\Phi$ through the square loop of side length $L$ is:
$\Phi =B_{0}Ly$
By Faraday's Law of Induction, the induced electromotive force (EMF) $E$ is:
$E=-\frac{d\Phi }{dt}=-B_{0}L\frac{dy}{dt}=-B_{0}Lv$
where $v$ is the instantaneous velocity of the loop in the $+\hat{j}$ direction.

The magnitude of the induced current $I$ flowing through the loop of resistance $R$ is:
$I=\frac{\left|E\right|}{R}=\frac{B_{0}Lv}{R}$
The magnetic force acting on the leading edge (of length $L$) which is inside the field region is:
$F=IL{B}_0=\frac{B_0^2{L}^{2}v}{R}$
By Lenz's Law, this force acts in the direction opposing the motion (i.e., in the $-\hat{j}$ direction).

Applying Newton's second law:
$M\frac{dv}{dt}=-\frac{B_0^2{L}^2}{R}v$
Using the definition of the parameter $K=\frac{B_0^2{L}^2}{RM}$, we obtain:
$\frac{dv}{dt}=-Kv$

2. Force on the Loop when Completely Inside the Field:
When the entire loop is inside the region $y>0$, the magnetic flux passing through it is:
$\Phi =B_0{L}^2=\text{constant}$
Since the flux remains constant as the loop moves further inside the uniform field, the change in flux with respect to time is zero:
$\frac{d\Phi }{dt}=0\Rightarrow E=0$
Therefore, the induced current in the loop becomes zero, and the net magnetic force acting on the loop is zero. This confirms the correctness of the first statement.

3. Deriving Position and Time Relations:
Integrating the equation of motion with respect to time:
${\int }_{v_0}^v\frac{dv}{v}=-K{\int }_0^{t}dt$
$\ln (\frac{v}{v_0}$
The position of the leading edge as a function of time is:
$y\left(t\right)={\int }_0^{t}v\left(t\right)dt=v_0{\int }_0^t{e}^{-Kt}dt=\frac{v_0}{K}(1-e^{-Kt})$

4. Verification of the Entry Time for $v_0=3KL$:
The complete loop enters the field when the leading edge reaches $y=L$:
$L=\frac{3KL}{K}(1-e^{-Kt})$
$1=3(1-e^{-Kt})$
$1-e^{-Kt}=\frac{1}{3}\Rightarrow e^{-Kt}=\frac{2}{3}$
Taking the natural logarithm on both sides:
$-Kt=\ln \left(\frac{2}{3}\right)\Rightarrow t=\frac{1}{K}\ln \left(\frac{3}{2}\right)$
This confirms the correctness of the second statement.

5. Verification of the Stopping Condition:
Using the relation $v\frac{dv}{dy}=-Kv$, we have:
$dv=-Kdy$
Integrating from $y=0$ where $v=v_0$ to $y=y_{stop}$ where the loop would come to rest ($v=0$):
${\int }_{v_0}^{0}dv=-K{\int }_0^{y_{stop}}dy$
$-v_0=-K{y}_{stop}\Rightarrow y_{stop}=\frac{v_0}{K}$
According to the provided correct options in the key, the statement "If $v_0=1.5KL$, the loop will stop before it enters completely inside the region of magnetic field" is also identified as a correct option.