A square loop EFGH of side a , mass m and total resistance R is...

Motional emf in EH and FG = 0 as $\overset{\to}{\text{v}} || \overset{\to}{\text{I}}$

Motional emf in EF is ${e}_{1} = \left(\dfrac{{\text{B}}_{0}\text{y}}{\text{a}}\right)\left(\text{a}\right)v = {\text{B}}_{0}\text{y}v$

Similarly, motional emf in GH will be

${\text{e}}_{2} = \left\{\dfrac{{\text{B}}_{0}\left(\text{y} + \text{a}\right)}{\text{a}}\right\}\left(\text{a}\right)\left(v \right) = {\text{B}}_{0}\left(\text{y} + \text{a}\right)v$

Polarities of e₁and e₂ are shown in adjoining figures.

So the net emf is

$e = {e}_{2} - {e}_{1}e = {B}_{0}av$
$\text{i} = \dfrac{\text{e}}{\text{R}} = \dfrac{{\text{B}}_{0} \text{a} v }{\text{R}}$

${\text{F}}_{\text{EF}} = \left(\dfrac{{\text{B}}_{0} \text{a} v }{\text{R}}\right) \left(\text{a}\right) \left(\dfrac{{\text{B}}_{0} \text{y}}{\text{a}}\right)$ (downwards)

${\text{F}}_{\text{GH}} = \left(\dfrac{{\text{B}}_{0} \text{a} v }{\text{R}}\right) \left(\text{a}\right) \left(\dfrac{{\text{B}}_{0} \left(\text{y} + \text{a}\right)}{\text{a}}\right)$

${\text{F}}_{\text{GH}} = \left(\dfrac{{\text{B}}_{0}^{2}\text{a}v }{\text{R}}\right)\left(\text{y} + \text{a}\right)$ (upwards)

Net Lorentz force on the loop

${F}_{net} = {\text{F}}_{\text{GH}} - {\text{F}}_{\text{EF}} = \dfrac{{\text{B}}_{0}^{2}{\text{a}}^{2}v }{\text{R}}$

$F = \dfrac{{B}_{0}^{2}{a}^{2}v}{R}$