Question

A square conducting loop of mass \(m\), side \(l\) and resistance \(R\) is dropped into a region with a uniform horizontal magnetic field \(B\) whose direction is perpendicular to the plane of the falling loop. The loop will reach a terminal velocity \(v\) given by:

• A. \(V = \dfrac{mgR}{(Bl)^2}\)
• B. \(V = \dfrac{2mgR}{(Bl)^2}\)
• C. \(V = \dfrac{mgR}{2(Bl)^2}\)
• D. None of these

Hint:

At terminal velocity the retarding force \(B^2 l^2 v/R\) equals the weight \(mg\). Solve for \(v\).

Answer 1

The Correct Option is A

Step 1: As the loop falls with speed \(v\), the flux through it changes as a horizontal side of length \(l\) cuts field lines. The induced EMF is \[\varepsilon = Blv.\]
Step 2: The induced current in the loop is \[i = \frac{\varepsilon}{R} = \frac{Blv}{R}.\]
Step 3: This current, in the field \(B\), feels a force on the side of length \(l\). By Lenz's law this magnetic force opposes the fall (acts upward): \[F_{mag} = Bil = B\left(\frac{Blv}{R}\right)l = \frac{B^2 l^2 v}{R}.\]
Step 4: At terminal velocity the net force is zero, so the upward magnetic force balances the weight: \[mg = \frac{B^2 l^2 v}{R}.\]
Step 5: Solve for \(v\): \[v = \frac{mgR}{B^2 l^2} = \frac{mgR}{(Bl)^2}.\] \[\boxed{v = \frac{mgR}{(Bl)^2}}\]