The bar of mass m in the figure below is pulled horizontally across parallel, fr

Question

The bar of mass m in the figure below is pulled horizontally across parallel, frictionless rails by a massless string that passes over a light, frictionless pulley and is attached to a suspended mass M. The uniform upward magnetic field has a magnitude B, and the distance between the rails is scripted l. The only significant electrical resistance is the load resistor R shown connecting the rails at one end. Assuming the suspended object is released with the bar at rest at t = 0, derive an expression that gives the bar's horizontal speed as a function of time. (Use any variable or symbol stated above along with the following as necessary: g.)

Explanation / Answer

emf =induced voltage =VBL

i= current = emf/R

F = force on bar due to magnetic fieled =iLB = emf*L*B/R

a = acceleration of M = g -emf*L*B/(MR)

a = g - VBL*L*B/(MR)

dv/dt = g - (LB)^2*V/(MR)

dv/[g - (LB)^2*V/(MR)] =dt

integrate ln[g - (LB)^2*V/(MR)] = -t

g - (LB)^2*V/(MR) = Ae^-t , where A =constant

V = (g -Ae^-t)MR/(LB)^2

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