Question

A conducting rod of mass \( m \) and length \( l \) is moving on an infinite pair of conducting rails as shown. Conducting rails are connected to a resistance \( R \) at one end. Motion is in the vertical plane and horizontal magnetic field in the region is \( B \). Find the terminal speed of the rod.

• A. \( V_0 = \frac{3mgR}{2B^2 l^2} \)
• B. \( V_0 = \frac{mgR}{2B^2 l^2} \)
• C. \( V_0 = \frac{mgR}{B^2 l^2} \)
• D. \( V_0 = \frac{2mgR}{B^2 l^2} \)

Hint:

In problems involving magnetic fields and induced currents, always apply Newton's second law and solve for velocity at equilibrium (terminal speed).

Answer 1

The Correct Option is C

To find the terminal speed of the rod moving on conducting rails in a magnetic field, we analyze the forces acting on the rod and the induced electromotive force (emf).

1. When the rod moves with velocity \(v\) in a magnetic field \(B\), an emf is induced across its length due to electromagnetic induction. This emf \(\epsilon\) is given by: 
   \(\epsilon = B \cdot l \cdot v\)
2. The induced emf causes a current \(I\) to flow through the resistance \(R\): 
   \(I = \frac{\epsilon}{R} = \frac{B \cdot l \cdot v}{R}\)
3. This current interacts with the magnetic field to produce a magnetic force (Lorentz force) on the rod, which opposes its motion: 
   \(F_{\text{magnetic}} = I \cdot l \cdot B = \frac{B \cdot l \cdot v}{R} \cdot l \cdot B = \frac{B^2 \cdot l^2 \cdot v}{R}\)
4. The gravitational force acting on the rod is: 
   \(F_{\text{gravity}} = m \cdot g\)
5. At terminal speed, these forces balance each other: 
   \(F_{\text{gravity}} = F_{\text{magnetic}}\) 
   \(m \cdot g = \frac{B^2 \cdot l^2 \cdot v_0}{R}\)
6. Solving for the terminal speed \(v_0\):  
   \(v_0 = \frac{m \cdot g \cdot R}{B^2 \cdot l^2}\)

Thus, the terminal speed of the rod is given by the correct option:

\(< V_0 = \frac{mgR}{B^2 l^2} \)