Question

A conducting rod of length L and mass m falls vertically under gravity through a region of uniform magnetic field B, directed into the plane of the page. The rod is placed on two smooth, vertical conducting rails connected at the bottom by a resistor R. Assuming no friction or air resistance, and the rod quickly reaches a constant terminal velocity, find the expression for v in terms of B, L, m, R.

• A. \( \frac{mgR}{B^2 L^2} \)
• B. \( \frac{mg}{B^2 L^2} \)
• C. \( \frac{mgR}{B^2 L} \)
• D. \( \frac{mR}{gB^2 L} \)

Hint:

In problems involving magnetic forces on moving conductors, remember that the current is induced by the motion of the rod through the magnetic field.

Answer 1

The Correct Option is A

The gravitational force on the rod is \( F_g = mg \). The magnetic force on the rod is \( F_B = BIL \), where \( I \) is the induced current. This current arises from the rod's motion through the magnetic field, and the resistance is \( R \). Ohm's law dictates the current as: \[ I = \frac{v}{R} \] At terminal velocity, the magnetic force equals the gravitational force: \[ mg = BIL \] Substituting \( I = \frac{v}{R} \) into this equation yields: \[ mg = B \frac{v}{R} L \] Solving for \( v \) gives: \[ v = \frac{mgR}{B^2 L^2} \] Therefore, the velocity is expressed as \( v = \frac{mgR}{B^2 L^2} \).