Question

a conducint light string is wound around the rim of a metal ring of radius r and mass m. the free end of the string is fixed to the ceiling. a verticl infinite conducitng smooth plane is always tangent to the ring. the ring is released in a uniform magnetic fieldon strength B perpendicular to plane of ring. the string and the plane are cinnected by wire o essistance r. ring is always in magneric field. terminal velocity is?

Answer 1

Terminal speed v = $\displaystyle \frac{4mgR}{B^2r^2}$.

Answer 2

We analyze the situation in analogy with the similar question. For the ring of mass m and radius r (which is analogous to the disc of mass M and radius L) connected in series with a resistor of resistance R, as it falls with terminal speed v:

1. Induced emf

A point on the rim at distance r from the center has a linear speed v giving an angular speed

ω = v/r.

The induced emf (by integrating over the radius) is

ε = ∫₀ʳ B ω r′ dr′ = B ω (r²/2) = (½)B (v/r) r² = (½)B v r.

2. Current in the Circuit

The emf drives a current

I = ε/R = (½ B v r)/R.

3. Electrical Power Dissipation

The power dissipated is

P = I² R = [(½ B v r)²/R²] R = (B² v² r²)/(4R).

4. Balance with Gravitational Power

At terminal speed, gravitational power input equals the electrical dissipation, so

mg v = (B² v² r²)/(4R).

Canceling v (v > 0) yields

mg = (B² v r²)/(4R) ⟹ v = (4mgR)/(B² r²).

---

Summary of the Response

• Explanation (Minimal):

The induced emf is ε = (½)Bv r. With I = ε/R, the dissipated power is (B²v²r²)/(4R). Equate mg v = (B²v²r²)/(4R) and solve to get

v = (4mgR)/(B²r²).