Question 285: Electromagnetic Induction

Question

If current '$I$' is flowing in the closed circuit with collective resistance '$R$', the rate of production of heat energy in the loop as we pull it along with a constant speed '$\mathrm{V}$' is ( $\mathrm{L}=$ length of conductor, $\mathrm{B}=$ magnetic field)

Accepted Answer

From motional emf,
$\mathrm{e}_{\max }=\mathrm{BLV}$
$\begin{array}{ll}
	herefore \quad & 	ext { Heat produced }=\frac{\mathrm{V}^2}{\mathrm{r}}=\frac{\mathrm{B}^2 \mathrm{~L}^2 \mathrm{~V}^2}{\mathrm{R}} \
& \mathrm{i}=\frac{\mathrm{BLV}}{\mathrm{R}} \
& |\mathrm{F}|=\mathrm{BiL} 	ext { and } \mathrm{P}=\mathrm{F} \cdot \mathrm{V} \
	herefore \quad & \mathrm{P}=\mathrm{B}\left(\frac{\mathrm{BLV}}{\mathrm{R}}ight) \mathrm{LV} \
& =\frac{\mathrm{B}^2 \mathrm{~L}^2 \mathrm{~V}^2}{\mathrm{R}}
\end{array}$

Answer

$\frac{\mathrm{BLV}}{\mathrm{R}}$

Answer

$\frac{\mathrm{B}^2 \mathrm{~L}^2 \mathrm{~V}^2}{\mathrm{R}^2}$

Answer

$\frac{B L V}{R^2}$

Answer

$\frac{\mathrm{B}^2 \mathrm{~L}^2 \mathrm{~V}^2}{\mathrm{R}}$

Electromagnetic Induction

If current 'I' is flowing in the closed circuit with collective resistance 'R', the rate of production of heat energy in the loop as we pull it along with a constant speed '\mathrm{V}' is ( \mathrm{L}= length of conductor, \mathrm{B}= magnetic field)

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