A 15.1-g conducting rod of length 1.37 m is free to slide downward between two v

Question

A 15.1-g conducting rod of length 1.37 m is free to slide downward between two vertical rails without friction. The rails are connected to an 7.02-Ohm resistor, and the entire apparatus A. placed in a 0.436-T uniform magnetic field. Ignore the resistance of the rod and rails. (a) What is the terminal velocity of the rod? m/s (b) At this terminal velocity, compare the magnitude of the change In gravitational potential energy per second with the power dissipated in the resistor. Delta/Deltat= W, PR= W

Explanation / Answer

(a) The weight is m*g = 0.0151kg*9.8m/s^2 = 0.1479N
Now this force = I*L*B where I = E/R and E = v*B*L

so F = v*(B*L)^2/R
so v = F*R/(B*L)^2 = 0.1479*7.02/(0.436*1.37)^2 = 2.91m/s

terminal velocity =2.91 m/s

(b) dU/dt = m*g*dy/dt = 0.0151*9.8*(-2.91) = -0.4306J/s

PR = dW/dt = -dU/dt = 0.4306J

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