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A horizontal conducting wire of mass m and length l can slide without friction down two vertical conducting rails of total circuit resistance R. A uniform magnetic field B is perpendicular to the plane of the rails. Find the terminal speed of the wire as it falls under gravity.
- vₜ = m*g*R/(B²*l²)
- vₜ = m*g*R/(B*l)
- vₜ = B²*l²/(m*g*R)
- vₜ = m*g*R²/(B²*l²)
Correct answer: vₜ = m*g*R/(B²*l²)
Solution
Falling at speed v gives emf = B*l*v, current I = B*l*v/R, and an opposing magnetic force F = B*I*l = B²*l²*v/R. Terminal speed is reached when this equals gravity: m*g = B²*l²*vₜ/R, so vₜ = m*g*R/(B²*l²).
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