Question 81 of 153

A rectangular loop of sides $12cm$ and $5cm$, with its sides parallel to the $x$-axis and $y$-axis respectively moves with a velocity of $5cm{s}^{-1}$ in the positive $x$ axis direction, in a space containing a variable magnetic field in the positive $z$ direction. The field has a gradient of ${10}^{-3}T{cm}^{-1}$ along the negative $x$ direction and it is decreasing with time at the rate of ${10}^{-3}T{s}^{-1}.$ If the resistance of the loop is $6m\Omega$, the power dissipated by the loop as heat is ______ $\times {10}^{-9}W.$

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Answer: 216

Let ${B}_{0}$ is the magnetic field at origin. Then,
$\dfrac{dB}{dx}=-\dfrac{{10}^{-3}}{{10}^{-2}}$
$\Rightarrow {\int }_{{B}_{0}}^{B}dB=-{\int }_{0}^{x}{10}^{-1}dx$
$\Rightarrow B-{B}_{0}=-{10}^{-1}x$
$\Rightarrow B=\left({B}_{0}-\dfrac{x}{10}\right)$
Motional emf in $AB=0$
Motional emf in $CD=0$
Motional emf in $AD={\epsilon }_{1}={B}_{0}\ell v$
Magnetic field on rod $BC$
$=\left({B}_{0}-\dfrac{\left(-12\times {10}^{-2}\right)}{10}\right)$
Motional emf in $BC={\epsilon }_{2}=\left({B}_{0}+\dfrac{12\times {10}^{-2}}{10}\right)\ell \times v$
Net equivalent emf ${\epsilon }_{eq}={\epsilon }_{2}-{\epsilon }_{1}=300\times {10}^{-7}V$
For time variation
${\left({\epsilon }_{eq}\right)}^{'}=A\dfrac{dB}{dt}=60\times {10}^{-7}V$
${\left({\epsilon }_{eq}\right)}_{\text{net}}={\epsilon }_{eq}+{\left({\epsilon }_{eq}\right)}^{'}=360\times {10}^{-7}V$
Power $r=\dfrac{{\left({\epsilon }_{\text{eq}}\right)}_{\text{net}}^{2}}{R}=216\times {10}^{-9}W$