Question

A rod of mass \(m\) and length \(L\) is released on a rail placed in a uniform magnetic field \(B\) as shown. The circuit has resistance \(R\). What will be the terminal velocity of the rod? v

• A. \( \dfrac{mgR}{B^2L^2} \)
• B. \( \dfrac{mgR}{B^2\ell^2} \)
• C. \( \dfrac{mgR}{B\ell^2} \)
• D. \( \dfrac{mg}{B^2\ell^2R} \)

Hint:

For electromagnetic damping problems:
Induced current always opposes motion (Lenz’s law)
Terminal velocity occurs when magnetic force balances gravity
Speed depends on resistance: higher \(R\) \(\Rightarrow\) larger terminal speed

Answer 1

The Correct Option is B

Concept:  
When the rod moves along the rails in a uniform magnetic field, it induces an electromotive force (emf) due to the change in magnetic flux. This emf results in a current flowing through the closed circuit, which creates a magnetic force opposing the rod's motion (as per Lenz’s law). At the terminal velocity, the magnetic retarding force balances the weight of the rod. Key relations: 
Motional emf: \( \varepsilon = B \ell v \) 
Current: \( I = \frac{\varepsilon}{R} \) 
Magnetic force on the rod: \( F = B I \ell \) 
Step 1: Calculate the induced emf. If the rod moves with a velocity \( v \), the induced emf is: \[ \varepsilon = B \ell v \] 
Step 2: Find the induced current. The induced current is given by: \[ I = \frac{\varepsilon}{R} = \frac{B \ell v}{R} \] 
Step 3: Determine the magnetic force on the rod. The force on a current-carrying conductor in a magnetic field is: \[ F = B I \ell \] Substituting the expression for \( I \): \[ F = B \ell \left( \frac{B \ell v}{R} \right) = \frac{B^2 \ell^2 v}{R} \] This magnetic force acts upwards, opposing the downward motion of the rod. 
Step 4: Apply the condition for terminal velocity. At terminal velocity \( v_t \), the net force becomes zero: \[ mg = \frac{B^2 \ell^2 v_t}{R} \] Solving for \( v_t \): \[ v_t = \frac{mg R}{B^2 \ell^2} \] 
Final Answer: \[ \boxed{v_t = \frac{mg R}{B^2 \ell^2}} \]