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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)
image

  1. A \( \frac{2mgR}{B^{2}l^{2}} \)
  2. B \( \frac{8mgR}{B^{2}l^{2}} \)
  3. C \( \frac{2mgR}{B^{2}L^{2}} \)
  4. D \( \frac{mgR}{B^{2}l^{2}} \)
Verified Solution

Answer & Solution

Correct Answer

(D) \( \frac{mgR}{B^{2}l^{2}} \)

Step-by-step Solution

Detailed explanation

at equilibrium (Or for terminal velocity) \( mg=iBl\Rightarrow mg=(\frac{Bvl}{R})Bl \) \( V=\frac{mgR}{B^{2}l^{2}} \)