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: Understanding the Question:
The loop falls through a magnetic field. As it moves, the magnetic flux through it changes, inducing an emf and a current. This current experiences a magnetic force that opposes the motion. "Falling freely" with a constant velocity \( V \) implies terminal velocity where net force is zero.

Step 2: Key Formula or Approach:
1. Induced emf: \( \epsilon = B l V \)
2. Induced current: \( I = \frac{\epsilon}{R} = \frac{BlV}{R} \)
3. Upward magnetic force on the horizontal segment: \( F_m = I l B \)
4. At terminal velocity: \( mg = F_m \)

Step 3: Detailed Explanation:
Substitute the expression for \( I \) into the magnetic force formula:
\[ F_m = \left( \frac{BlV}{R} \right) l B = \frac{B^2 l^2 V}{R} \] Equate this to the gravitational force (\( mg \)) because the velocity is constant:
\[ mg = \frac{B^2 l^2 V}{R} \] Rearranging for velocity \( V \):
\[ V = \frac{mgR}{B^2 l^2} \]

Step 4: Final Answer:
The velocity \( V \) is \( \frac{mgR}{B^2 l^2} \).