Electromagnetic Induction Question 1
Question 1 - 2024 (01 Feb Shift 1)
A rectangular loop of sides $12 \mathrm{~cm}$ and $5 \mathrm{~cm}$, with its sides parallel to the $\mathrm{x}$-axis and $\mathrm{y}$-axis respectively moves with a velocity of $5 \mathrm{~cm} / \mathrm{s}$ in the positive $\mathrm{x}$ axis direction, in a space containing a variable magnetic field in the positive $\mathrm{z}$ direction. The field has a gradient of $10^{-3} \mathrm{~T} / \mathrm{cm}$ along the negative $\mathrm{x}$ 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 $\times 10^{-9} \mathrm{~W}$.
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Answer: (216)
Solution:
$\mathrm{B}_{0}$ is the magnetic field at origin
$\frac{d B}{d x}=-\frac{10^{-3}}{10^{-2}}$
$\int_{B_{0}}^{B} d B=-\int_{0}^{x} 10^{-1} d x$
$B-B_{0}=-10^{-1} x$
$B=\left(B_{0}-\frac{x}{10}\right)$
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 $\mathrm{BC}=\varepsilon_{2}=\left(\mathrm{B}_{0}+\frac{12 \times 10^{-2}}{10}\right) \ell \times \mathrm{v}$
$\varepsilon_{\text {eq }}=\varepsilon_{2}-\varepsilon_{1}=300 \times 10^{-7} \mathrm{~V}$
For time variation
$$ \begin{aligned} & \left(\varepsilon_{\mathrm{eq}}\right)^{\prime}=\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}}=60 \times 10^{-7} \mathrm{~V} \ & \left(\varepsilon_{\mathrm{eq}}\right){\text {net }}=\varepsilon{\mathrm{eq}}+\left(\varepsilon_{\mathrm{eq}}\right)^{\prime}=360 \times 10^{-7} \mathrm{~V} \ & \text { Power }=\frac{\left(\varepsilon_{\mathrm{eq}}\right)_{\text {net }}^{2}}{\mathrm{R}}=216 \times 10^{-9} \mathrm{~W} \end{aligned} $$
