A long metal rod of length 'L' completes the circuit as shown. The area of the circuit is perpendicular to…

A long metal rod of length 'L' completes the circuit as shown. The area of the circuit is perpendicular to magnetic field ' $B^{\prime}$. Total resistance of the circuit is ' $\mathrm{R}$ '. The force needed to move the rod in the direction as shown with constant speed ' $V$ ' is
  1. $\frac{\mathrm{B}^{2} \mathrm{LV}}{\mathrm{R}}$
  2. $\frac{\mathrm{BLV}}{\mathrm{R}}$
  3. $\frac{\mathrm{BLV}^{2}}{\mathrm{R}}$
  4. $\frac{\mathrm{B}^{2} \mathrm{~L}^{2} \mathrm{~V}}{\mathrm{R}}$

Solution

$\begin{array}{l} \text { Power } \mathrm{P}=\frac{\mathrm{B}^{2} \mathrm{~L}^{2} \mathrm{~V}^{2}}{\mathrm{R}} \\ \therefore \mathrm{FV}=\frac{\mathrm{B}^{2} \mathrm{~L}^{2} \mathrm{~V}^{2}}{\mathrm{R}} \\ \therefore \mathrm{F} \quad=\frac{\mathrm{B}^{2} \mathrm{~L}^{2} \mathrm{~V}}{\mathrm{R}} \end{array}$

Asked in: MHT CET 2020 (19 Oct Shift 1)

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