You have a summer job working at a company developing systems to safely lower la
ID: 1261208 • Letter: Y
Question
You have a summer job working at a company developing systems to safely lower large loads down ramps. Your team is investigating a magnetic system by modeling it in the laboratory.
The safety system is a conducting bar that slides on two parallel conducting rails that run down the ramp. The bar is perpendicular to the rails and is in contact with them. At the bottom of the ramp, the two rails are connected together.
The bar slides down the rails through a vertical uniform magnetic field. The magnetic field is supposed to cause the bar to slide down the ramp at a constant velocity even when friction between the bar and the rails is negligible.
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 of the magnetic field, the angle of the ramp from the horizontal, the length of the bar which is the same as the distance between the tracks, and the resistance of the bar. Assume that all of the other conductors in the system have a much smaller resistance than the bar.
1a) 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.
1b)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.
Explanation / Answer
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! ]
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