Question 159: Electromagnetic Induction

Question

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 __________ $	imes 10^{-9} \mathrm{~W}$.

Accepted Answer

$\mathrm{B}_0$ is the magnetic field at origin
$$
\begin{aligned}
& \frac{d B}{d x}=-\frac{10^{-3}}{10^{-2}} \\
& \int_{B_0}^B d B=-\int_0^x 10^{-1} d x \\
& B-B_0=-10^{-1} x \\
& B=\left(B_0-\frac{x}{10}ight)
\end{aligned}
$$
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
$$
=\left(\mathrm{B}_0-\frac{\left(-12 	imes 10^{-2}ight)}{10}ight)
$$
Motional emf in $\mathrm{BC}=\varepsilon_2=\left(\mathrm{B}_0+\frac{12 	imes 10^{-2}}{10}ight) \ell 	imes \mathrm{v}$
$$
\varepsilon_{\mathrm{eq}}=\varepsilon_2-\varepsilon_1=300 	imes 10^{-7} \mathrm{~V}
$$
For time variation
$$
\begin{aligned}
& \left(\varepsilon_{	ext {eq }}ight)^{\prime}=\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}}=60 	imes 10^{-7} \mathrm{~V} \\
& \left(\varepsilon_{\mathrm{eq}}ight)_{	ext {net }}=\varepsilon_{\mathrm{eq}}+\left(\varepsilon_{\mathrm{eq}}ight)^{\prime}=360 	imes 10^{-7} \mathrm{~V} \\
& 	ext { Power }=\frac{\left(\varepsilon_{	ext {eq }}ight)_{	ext {net }}^2}{\mathrm{R}}=216 	imes 10^{-9} \mathrm{~W}
\end{aligned}
$$

Electromagnetic Induction

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