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! ]
