JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is \(l\). A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L (L > \(l\)) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is __________ m/s.
(g = acceleration due to gravity)

- A \( \frac{2mgR}{B^{2}l^{2}} \)
- B \( \frac{8mgR}{B^{2}l^{2}} \)
- C \( \frac{2mgR}{B^{2}L^{2}} \)
- D \( \frac{mgR}{B^{2}l^{2}} \)
Answer & Solution
Correct Answer
(D) \( \frac{mgR}{B^{2}l^{2}} \)
Step-by-step Solution
Detailed explanation
at equilibrium (Or for terminal velocity) \( mg=iBl\Rightarrow mg=(\frac{Bvl}{R})Bl \) \( V=\frac{mgR}{B^{2}l^{2}} \)
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