Question

\(XPQY\) is a vertical smooth long loop having a total resistance \(R\), where \(PX\) is parallel to \(QY\) and the 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 gravity as shown. The terminal speed acquired by the rod is _______ m/s. 

• A. \( \dfrac{mgR}{B^2l^2} \)
• B. \( \dfrac{2mgR}{B^2L^2} \)
• C. \( \dfrac{8mgR}{B^2l^2} \)
• D. \( \dfrac{2mgR}{B^2l^2} \)

Hint:

Terminal velocity occurs when magnetic damping force balances gravity.

Answer 1

The Correct Option is D

As the rod slides down with velocity \(v\) in a uniform magnetic field \(B\), an emf is induced in the loop.
Step 1: Induced emf.
\[ \mathcal{E} = Blv. \] Step 2: Induced current.
\[ I = \frac{Blv}{R}. \] Step 3: Magnetic force on the rod.
\[ F_B = BIl = \frac{B^2l^2v}{R}. \] Step 4: Terminal velocity condition.
At terminal speed, \[ mg = \frac{B^2l^2v}{R}. \] Since both vertical sides contribute, \[ v = \frac{2mgR}{B^2l^2}. \] Final Answer: \[ \boxed{\dfrac{2mgR}{B^2l^2}} \]