A 12.0 -g conducting rod of length 1.30 m is free to slide downward between two

Question

A 12.0-g conducting rod of length 1.30 m is free to slide downward between two vertical rails without friction. The rails are connected to an 8.00 resistor, and the entire apparatus is placed in a 0.460 T uniform magnetic field. Ignore the resistance of the rod and rails.

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(a) What is the terminal velocity of the rod? (in m/s)

(b) At this terminal velocity, calculate the rate of change in gravitational potential energy (include sign) and the power dissipated in the resistor.

W (rate of change in gravitational potential energy)
A 12.0-g conducting rod of length 1.30 m is free to slide downward between two vertical rails without friction. The rails are connected to an 8.00 resistor, and the entire apparatus is placed in a 0.460 T uniform magnetic field. Ignore the resistance of the rod and rails. What is the terminal velocity of the rod? (in m/s) At this terminal velocity, calculate the rate of change in gravitational potential energy (include sign) and the power dissipated in the resistor.

Explanation / Answer

a) The weight is m*g = 0.012kg*9.8m/s^2 = 0.1176N
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 = 0.1176*8.00/(0.460*1.3)^2 = 2.63m/s

b) dU/dt = m*g*dy/dt = 0.012*9.8*(-2.63) = -0.309J/s

c) P = dW/dt = -dU/dt = 0.309J

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