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

Step 1: Force due to magnetic field.
The force on the rod moving in a magnetic field is given by \( F = B l v I \), where \( v \) is the velocity, \( l \) is the length of the rod, and \( I \) is the current. Step 2: Induced current.
The induced EMF \( \varepsilon \) is given by \( \varepsilon = B l v \), and the current \( I \) is given by \( I = \frac{\varepsilon}{R} = \frac{B l v}{R} \). Step 3: Apply Newton's second law.
At terminal speed, the magnetic force equals the gravitational force: \[ B l v \frac{B l v}{R} = mg \] Solving for \( v \), we get the terminal speed as: \[ V_0 = \frac{mgR}{B^2 l^2} \] Final Answer: \[ \boxed{\frac{mgR}{B^2 l^2}} \]