Question

You have a summer job at a company that developed systems to safely lower large loads down ramps. Your team is investigating a magnetic system by modeling it inthe laboratory. The safety system is a conducting bar that slides on two parallel conducting rails thatrun down the ramp (similar to the one in the previous problem). The bar is perpendicular to the railsand is in contact with them. At the bottom of the ramp, the two rails are connected together. The box slides down the rails through a uniform vertical magnetic field. The magnetic field is supposed to causethe bar to slide down the ramp at a constant velocity even when friction between the bar and the rails negligible.Before setting up the laboratory model, your task is to calculate the constant velocity of the bar (sliding down the ramp on rails in a vertical magnetic field) as a function of the mass of the bar, the strength ofthe magnetic field, the angle of the ramp from the horizontal, the length of the bar (which is the same asthe distance between the rails), and the resistance of the bar. Assume that all of the other conductors inthe system have a much smaller resistance than the bar.a) If the force due to the changing flux exactly cancells out the net force due to the combination of gravity and normal force, then the bare will cease to accelerate and instead move at a constant velocity. Please solve for this velocity algebraically.b)Write out the units for each of your variables and show (by cancellation and substitution) that the units for your veloctiy will be m/s on both left and right side of your equation.

Answer 1

Note that the emf induced is

emf = B d v cos (A)

---> v = emf / [B d cos (A)]

where

B = magnetic field

d = distance of two rails

v = constant speed

A = angle of rails with respect to the horizontal

Also, note that

I = emf/R

where R = resistance of the bar

Thus,

I = B d v cos (A) / R

Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

Thus, the component of this parallel to the incline is

F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R

As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)] [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s [DONE! WE SHOWED THE UNITS ARE CORRECT! ]