That's A Weird Frame

When the rod is at a distance $x$ metres from the origin moving with velocity v, the area of the loop is,

$|A| = \displaystyle \int_{0}^{x}z^{3}dz$

The magnetic flux through the loop $\phi = B\cdot A = |B||A|\cos0 = B\cdot \displaystyle \int_{0}^{x}z^{3}dz$
According to Faraday's laws an EMF is induced in the coil, $E = -\dfrac{d\phi}{dt} = -Bx^{3}\dfrac{dx}{dt} = -Bx^{3}v$ $V$

The instantaneous current through the coil, $I = \dfrac{E}{R} = \dfrac{-Bx^{3}v}{\sigma x^{3}} = \dfrac{-Bv}{\sigma_{0}(1+x^{4})} A$
A current carrying conductor moving in a magnetic field experiences a force,
$F = ilB$
$F = \dfrac{-Bv}{\sigma_{0}(1+x^{4})} \cdot x^{3} \cdot B = \dfrac{-B^{2}vx^{3}}{\sigma_{0}(1+x^{4})} N$
$ma =\dfrac{-B^{2}vx^{3}}{\sigma_{0}(1+x^{4})}$
$v\dfrac{dv}{dx} =\dfrac{-B^{2}vx^{3}}{m\sigma_{0}(1+x^{4})}$

$-dv =\dfrac{B^{2}x^{3}}{m\sigma_{0}(1+x^{4})}dx$

$\displaystyle \int_{u}^{0} -dv = \int_{0}^{L} \dfrac{B^{2}4x^{3}}{4m\sigma_{0}(1+x^{4})}dx$
$u = \dfrac{B^{2}}{4m\sigma_{0}}\cdot \left[ \ln(1+x^{4}) \right ]_{0}^{L} =\dfrac{B^{2}}{4m\sigma_{0}}\cdot \ln(1+L^{4})$

$\therefore L^{4} = e^{\dfrac{4m\sigma_{0}u}{B^{2}}} -1$
$\therefore L =\left(e^{\dfrac{4m\sigma_{0}u}{B^{2}}} -1\right)^{\dfrac{1}{4}}$

$\left \lfloor L \right \rfloor = \left \lfloor \left(e^{\dfrac{4m\sigma_{0}u}{B^{2}}} -1\right)^{\dfrac{1}{4}} \right\rfloor$

Substituting the values

$\therefore \left \lfloor L \right \rfloor = \left \lfloor \left(e^{\dfrac{4\cdot1\cdot1\cdot2}{1^{2}}} - 1\right)^{\dfrac{1}{4}} \right\rfloor$
$\therefore \left \lfloor L \right \rfloor = \left \lfloor \left(e^{8} - 1\right)^{\dfrac{1}{4}} \right\rfloor = \left \lfloor \left(e^{8}\right)^{\dfrac{1}{4}} \right\rfloor = \left \lfloor e^{2} \right \rfloor$
$\therefore \left \lfloor L \right \rfloor = \left \lfloor 7.389 \right \rfloor = 7$