Physics - EMI/AC Question with Solution | TestHub

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

Motional emf in EF is ${\mathrm{e}}_1=\left(\frac{{\text{B}}_0\text{y}}{\text{a}}\right)\left(\text{a}\right)\mathrm{v}={\text{B}}_0\text{y}\mathrm{v}$

Similarly, motional emf in GH will be
 
${\text{e}}_2=\left\{\frac{{\text{B}}_0\left(\text{y}+\text{a}\right)}{\text{a}}\right\}\left(\text{a}\right)\left(\mathrm{v}\right)={\text{B}}_0\left(\text{y}+\text{a}\right)\mathrm{v}$
 
Polarities of e₁and e₂ are shown in adjoining figures.

So the net emf is

$$\begin{gathered} \mathrm{e}={\mathrm{e}}_2-{\mathrm{e}}_1 \\ \mathrm{e} = {\mathrm{B}}_0\mathrm{av} \end{gathered}$$
$\text{i}=\frac{\text{e}}{\text{R}}=\frac{{\text{B}}_0\text{a}\mathrm{v}}{\text{R}}$ 

${\text{F}}_{\text{EF}}=\left(\frac{{\text{B}}_0\text{a}\mathrm{v}}{\text{R}}\right)\left(\text{a}\right)\left(\frac{{\text{B}}_0\text{y}}{\text{a}}\right)$ (downwards)

${\text{F}}_{\text{GH}}=\left(\frac{{\text{B}}_0\text{a}\mathrm{v}}{\text{R}}\right)\left(\text{a}\right)\left(\frac{{\text{B}}_0\left(\text{y}+\text{a}\right)}{\text{a}}\right)$ 
 
${\text{F}}_{\text{GH}}=\left(\frac{{\text{B}}_0^2\text{a}\mathrm{v}}{\text{R}}\right)\left(\text{y}+\text{a}\right)$  (upwards)

Net Lorentz force on the loop

${\mathrm{F}}_{\mathrm{net}}={\text{F}}_{\text{GH}}-{\text{F}}_{\text{EF}}=\frac{{\text{B}}_0^2{\text{a}}^2\mathrm{v}}{\text{R}}$

$F=\frac{B_0^2{a}^{2}v}{R}$