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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
Options:
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
Solution:
2420 Upvotes
Verified Answer
The correct answer is:
$\frac{\mathrm{B}^{2} \mathrm{~L}^{2} \mathrm{~V}}{\mathrm{R}}$
(A)
$\begin{array}{l}
\text { Power } \mathrm{P}=\frac{\mathrm{B}^{2} \mathrm{~L}^{2} \mathrm{~V}^{2}}{\mathrm{R}} \\
\therefore \mathrm{FV}=\frac{\mathrm{B}^{2} \mathrm{~L}^{2} \mathrm{~V}^{2}}{\mathrm{R}} \\
\therefore \mathrm{F} \quad=\frac{\mathrm{B}^{2} \mathrm{~L}^{2} \mathrm{~V}}{\mathrm{R}}
\end{array}$
$\begin{array}{l}
\text { Power } \mathrm{P}=\frac{\mathrm{B}^{2} \mathrm{~L}^{2} \mathrm{~V}^{2}}{\mathrm{R}} \\
\therefore \mathrm{FV}=\frac{\mathrm{B}^{2} \mathrm{~L}^{2} \mathrm{~V}^{2}}{\mathrm{R}} \\
\therefore \mathrm{F} \quad=\frac{\mathrm{B}^{2} \mathrm{~L}^{2} \mathrm{~V}}{\mathrm{R}}
\end{array}$
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