A long rectangular conducting loop of width '\(l\)' mass '\(m\)' and resistance '\(R\)' is placed partly in a perpendicular magnetic field '\(B\)'. With what velocity should it be pushed downwards so that it may continue to fall without any acceleration? (\(g\) = acceleration due to gravity)
Show Hint
At constant speed the upward magnetic force on the side in the field equals the weight. Use emf = Blv and force = B I l.
Step 1: Understand the figure
A rectangular loop of width \(l\), mass \(m\) and resistance \(R\) falls with velocity \(v\) downward. The magnetic field \(B\) points into the page and covers part of the loop, so a horizontal side of length \(l\) is inside the field.
Step 2: Find the induced emf
As the loop falls, the flux through it changes. The motional emf is \(\varepsilon=Blv\).
Step 3: Find the current and the force
Current: \(I=\frac{\varepsilon}{R}=\frac{Blv}{R}\). This current runs through the side of length \(l\) in the field, so the magnetic force is \[ F=IlB=\frac{B^2l^2v}{R} \] By Lenz's law this force acts upward and opposes the fall.
Step 4: Apply the no acceleration condition
The loop falls at constant velocity when the net force is zero: \(F=mg\). \[ \frac{B^2l^2v}{R}=mg\ \Rightarrow\ v=\frac{mgR}{B^2l^2} \]
Step 5: Check the options
Options (A), (C) and (D) have \(R\), \(B\) or \(l\) to the wrong power. A quick unit check helps: only \(\frac{mgR}{B^2l^2}\) comes from \(F=\frac{B^2l^2v}{R}\) and so has units of speed.
Final Answer:
The required speed is \(\frac{mgR}{B^2l^2}\), option (B).
\[ \boxed{v=\dfrac{mgR}{B^2l^2}} \]
