A rectangular loop of sides 12cm and 5cm , with its sides parallel...

Let ${B}_{0}$ is the magnetic field at origin. Then,

$\dfrac{dB}{dx} = - \dfrac{{10}^{- 3}}{{10}^{- 2}}$

$⇒ {∫}_{{B}_{0}}^{B}dB = - {∫}_{0}^{x}{10}^{- 1}dx$

$⇒ B - {B}_{0} = - {10}^{- 1}x$

$⇒ B = \left({B}_{0} - \dfrac{x}{10}\right)$

Motional emf in $AB = 0$

Motional emf in $CD = 0$

Motional emf in $AD = {ε}_{1} = {B}_{0}ℓv$

Magnetic field on rod $BC$ 

$= \left({B}_{0} - \dfrac{\left(- 12 \times {10}^{- 2}\right)}{10}\right)$

Motional emf in $BC = {ε}_{2} = \left({B}_{0} + \dfrac{12 \times {10}^{- 2}}{10}\right)ℓ \times v$

Net equivalent emf ${ε}_{eq} = {ε}_{2} - {ε}_{1} = 300 \times {10}^{- 7}V$

For time variation

${\left({ε}_{eq}\right)}^{'} = A\dfrac{dB}{dt} = 60 \times {10}^{- 7}V$

${\left({ε}_{eq}\right)}_{\text{net}} = {ε}_{eq} + {\left({ε}_{eq}\right)}^{'} = 360 \times {10}^{- 7}V$

Power $r = {\left({ε}_{\text{eq}}\right)}_{\text{net}}^{2}R = 216 \times {10}^{- 9}W$