The figure shows a rod of length L = 10.0 cm that is forced to move at constant

Question

The figure shows a rod of length L = 10.0 cm that is forced to move at constant speed v =3.98 m/s along horizontal rails. The rod, rails, and connecting strip at the right form a conducting loop. The rod has resistance 0.403 ohm; the rest of the loop has negligible resistance. A current I = 80.4 A through the long straight wire at distance a = 9.27 mm from the loop sets up a (nonuniform) magnetic field throughout loop. Find the (a) magnitude of the emf in volts and (b) current in amperes induced in the loop. (c) At what rate is thermal energy generated in the rod? (d) What is the magnitude of the force that must be applied to the rod to make it move at constant speed? (e) At what rate does this force do work on the rod?

Explanation / Answer

Solution:

The rate at which the amount of fertilizer (N) in the tank changes with time is given by:

dN/dt = rate fertilizer comes in - rate fertilizer goes out

The rate that fertilizer comes in is given by the rate of inflow (f_in) multiplied by the fertilizer concentration in the inflow (C_in), which in this case is constant with time:

rate fertilizer comes in = f_in*C_in = (1 gal/min)*(2 lb/gal) = 2 lb/min

The rate fertilizer goes out is given by the rate of outflow, multiplied by the concentration in the tank at that moment:

rate fertilizer goes out = f_out * C_tank(t)

The concentration of fertilizer in the tank at any moment is given by the amount of fertilizer in the tank at that moment, divided by the volume of solution in the tank at that moment. Note that because the flow rates in and out of the tank are not equal, then the volume of solution in the tank is a function of time.

C_tank(t) = N(t)/V(t)

If Vo is the initial volume of solution in the tank, then the volume of solution as a function of time is given by:

V(t) = Vo + (f_in - f_out)*t

in this case, we have:

V(t) = 100 gal - (2 gal/min)*t, where 0 < t < 50 min.

so

rate fertilizer goes out = (3 gal/min)*N(t)/((100 gal - 2 gal/min)*t), 0 < t < 50 min.

Putting all this together, we get that:

dN/dt = 2 lb/min - (3 gal/min)*N(t)/(100 gal - (2 gal/min)*t), 0 < t < 50 min.

This is a first-order linear equation. Putting it into standard form, we have:

dN/dt + (3 gal/min)*N(t)/(100 gal - t*2 gal/min) = 2 lb/min

This can be solved using an integrating factor to get:

N(t) = a*(50 min - t)^(3/2) + (2 lb/min)*( 50 min - t)

where a is the constant of integration.

In this case, we know that at t = 0, the tank contained pure water, so N(0) = 0, and:

0 = a*(50 min)^3/2 + 100lb

a = -sqrt(2)/5 (lb/min^3/2)

N(t) = (2 lb/min)*( 50 min - t) - lb*(sqrt(2)/5)lb*(t/min - 50)^(3/2)

The maximum amount of fertilizer in the tank occurs when dN(t)/dt = 0.

dN(t)/dt = (-2 + (3/10)*sqrt(2)*sqrt(50 - t))lb/min

Setting dN/dt = 0 and solving for t_max, we find that

t_max = (250/9) min ~= 27.78 min

This is when the tank contains the maximum amount of fertilizer. At this time, the amount of fertilizer is given by:

N_max = N((250/9) min) = 14.815 lbs

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