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:
As the rod moves on the rails in a uniform magnetic field, an emf is induced due to change of magnetic flux. This induces a current in the closed circuit, which produces a magnetic force opposing the motion (Lenz’s law). At terminal velocity, the magnetic retarding force balances the weight of the rod. Key relations:
Motional emf: \( \varepsilon = B\ell v \)
Current: \( I=\dfrac{\varepsilon}{R} \)
Magnetic force on rod: \( F = B I \ell \)
Step 1: Calculate the induced emf. If the rod moves with speed \(v\), \[ \varepsilon = B\ell v \]
Step 2: Find the induced current. \[ I=\frac{\varepsilon}{R}=\frac{B\ell v}{R} \]
Step 3: Determine the magnetic force on the rod. The force on a current-carrying conductor in a magnetic field is: \[ F = B I \ell \] Substitute \(I\): \[ F = B\ell \left(\frac{B\ell v}{R}\right) = \frac{B^2\ell^2 v}{R} \] This force acts upward, opposing the downward motion of the rod.
Step 4: Apply the condition for terminal velocity. At terminal velocity \(v_t\), net force is zero: \[ mg = \frac{B^2\ell^2 v_t}{R} \] Solve for \(v_t\): \[ v_t=\frac{mgR}{B^2\ell^2} \] \[ \boxed{v_t=\dfrac{mgR}{B^2\ell^2}} \]