Question

The block of mass (M) is connected by thread which is wound on a pulley, free to rotate about fixed horizontal axis as shown. A uniform magnetic field B exists in a horizontal plane. The disc is connected with the resistance R as shown. Calculate the terminal velocity of the block if it was released from rest. Treat pulley as uniform metallic disc of radius L.

• A. $\frac{4mgR}{B^2L^2}$
• B. $\frac{3mgR}{4B^2L^2}$
• C. $\frac{2mgR}{B^2L^2}$
• D. $\frac{3mgR}{2B^2L^2}$

Answer 1

Answer: (A)

(A) $\frac{4mgR}{B^2L^2}$

Answer 2

Solution Explanation

1. When the disc rotates with angular speed ω, a potential difference is induced between its center and rim given by
     ε = ∫₀ᴸ B ω r dr = (½)B ω L².
   Since the block unwinds the thread at linear speed v, we have ω = v/L, so
     ε = (½)B v L.

2. The resulting current in the circuit is
     I = ε/R = (½ B v L)/R.

3. The electrical power dissipated in the resistor is
     P = I²R = [(½ B v L)²/R²]·R = (B² v² L²)/(4R).

4. At terminal velocity, the gravitational power input equals the electrical power dissipated:
     mgv = (B² v² L²)/(4R).
   Cancelling v (v > 0) gives
     mg = (B² v L²)/(4R),
   so
     v = (4mgR)/(B² L²).