MHT CET · Physics · Magnetic Effects of Current
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
- A \(\frac{\mathrm{B}^{2} \mathrm{LV}}{\mathrm{R}}\)
- B \(\frac{\mathrm{BLV}}{\mathrm{R}}\)
- C \(\frac{\mathrm{BLV}^{2}}{\mathrm{R}}\)
- D \(\frac{\mathrm{B}^{2} \mathrm{~L}^{2} \mathrm{~V}}{\mathrm{R}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{B}^{2} \mathrm{~L}^{2} \mathrm{~V}}{\mathrm{R}}\)
Step-by-step Solution
Detailed explanation
(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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