JEE MAIN - Physics (2024 - 1st February Morning Shift - No. 21)
A rectangular loop of sides $12 \mathrm{~cm}$ and $5 \mathrm{~cm}$, with its sides parallel to the $x$-axis and $y$-axis respectively, moves with a velocity of $5 \mathrm{~cm} / \mathrm{s}$ 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 $10^{-3} \mathrm{~T} / \mathrm{cm}$ along the negative $x$ direction and it is decreasing with time at the rate of $10^{-3} \mathrm{~T} / \mathrm{s}$. If the resistance of the loop is $6 \mathrm{~m} \Omega$, the power dissipated by the loop as heat is __________ $\times 10^{-9} \mathrm{~W}$.
Answer
216
Explanation
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$\mathrm{B}_0$ is the magnetic field at origin
$$ \begin{aligned} & \frac{d B}{d x}=-\frac{10^{-3}}{10^{-2}} \\\\ & \int_{B_0}^B d B=-\int_0^x 10^{-1} d x \\\\ & B-B_0=-10^{-1} x \\\\ & B=\left(B_0-\frac{x}{10}\right) \end{aligned} $$
Motional emf in $\mathrm{AB}=0$
Motional emf in $\mathrm{CD}=0$
Motional emf in $\mathrm{AD}=\varepsilon_1=\mathrm{B}_0 \ell \mathrm{v}$
Magnetic field on rod BC B
$$ =\left(\mathrm{B}_0-\frac{\left(-12 \times 10^{-2}\right)}{10}\right) $$
Motional emf in $\mathrm{BC}=\varepsilon_2=\left(\mathrm{B}_0+\frac{12 \times 10^{-2}}{10}\right) \ell \times \mathrm{v}$
$$ \varepsilon_{\mathrm{eq}}=\varepsilon_2-\varepsilon_1=300 \times 10^{-7} \mathrm{~V} $$
For time variation
$$ \begin{aligned} & \left(\varepsilon_{\text {eq }}\right)^{\prime}=\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}}=60 \times 10^{-7} \mathrm{~V} \\\\ & \left(\varepsilon_{\mathrm{eq}}\right)_{\text {net }}=\varepsilon_{\mathrm{eq}}+\left(\varepsilon_{\mathrm{eq}}\right)^{\prime}=360 \times 10^{-7} \mathrm{~V} \\\\ & \text { Power }=\frac{\left(\varepsilon_{\text {eq }}\right)_{\text {net }}^2}{\mathrm{R}}=216 \times 10^{-9} \mathrm{~W} \end{aligned} $$
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