Question 298: 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)

Choose the correct answer: