Question

A long rectangular conducting loop of width $l$, mass $m$ and resistance $R$ is placed partly in a perpendicular magnetic field $B$. It is pushed downwards with velocity $V$ so that it may continue to fall freely. The velocity $V$ is

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

Hint:

Terminal velocity occurs when net force on the object becomes zero.

Answer 1

The Correct Option is C

Step 1: Concept
For the loop to fall "freely" at constant velocity (terminal velocity), the magnetic force must balance the gravitational force.

Step 2: Analysis
- Upward Magnetic Force $F_m = BIl$.
- Induced EMF $e = BlV$.
- Induced Current $I = e/R = BlV/R$.

Step 3: Calculation
$F_m = B(BlV/R)l = \frac{B^2 l^2 V}{R}$.
Setting $F_m = mg \implies \frac{B^2 l^2 V}{R} = mg$.
$V = \frac{mgR}{B^2 l^2}$.

Step 4: Conclusion
Hence, the constant velocity $V$ is $\frac{mgR}{B^{2}l^{2}}$. Final Answer: (C)