Question

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)

• A. \(\frac{B^2l^2R^2}{mg}\)
• B. \(\frac{mgR}{B^2l^2}\)
• C. \(\frac{mgl}{B^2R^2}\)
• D. \(\frac{mgR^2}{Bl}\)

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.

Answer 1

The Correct Option is B

Step 1: Energy approach
For constant speed, the power lost by gravity must equal the electrical power turned to heat in the loop.

Step 2: Power balance
Gravity does work at rate $P=mgv$. The emf is $Blv$, so the heating rate is $\frac{(Blv)^2}{R}$.

Step 3: Equate
$mgv=\frac{B^2l^2v^2}{R}$. Cancel one $v$: $mg=\frac{B^2l^2v}{R}$.

Step 4: Solve
$v=\frac{mgR}{B^2l^2}$. This is option (B). The other options do not give the dimensions of speed.

Final Answer:
Terminal speed is $\frac{mgR}{B^2l^2}$. \[ \boxed{\dfrac{mgR}{B^2l^2}} \]