If current ' $I$ ' is flowing in the closed circuit with collective resistance ' $R$ ', the rate of…

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)
  1. $\frac{B L V}{R}$
  2. $\frac{\mathrm{B}^2 \mathrm{~L}^2 \mathrm{~V}^2}{\mathrm{R}^2}$
  3. $\frac{B L V}{R^2}$
  4. $\frac{\mathrm{B}^2 \mathrm{~L}^2 V^2}{\mathrm{R}}$

Solution

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

Asked in: MHT CET 2023 (09 May Shift 2)

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