Question

A metal bar with a length L has a mass of m. Initially it is placed on an incline θ above the horizontal. There is a magnetic field T that is directed upward around the bar and the metal contacts along the incline allow a current through the bar. If the coefficient of static friction is μ,

a). What are the minimum magnitude and the direction of the current that must pass through the bar to keep it from moving once released?

b). Is there a maximum current that can be used in that same direction that will also hold the bar in place?

Answer 1

Here is my thought process for a):
\[F=ILB \sin(\theta)\]\[F_f= \mu N = \mu mg \cos(\theta)\]
Thus I can set them equal to each other and solve for I\[ILB \sin(\theta) = \mu mg \cos(\theta) \rightarrow I=\frac{ \mu mg \cos(\theta) }{ LB \sin(\theta) }\]

Answer 2

i think that for \(F=ILB \sin(\theta)\), \(\theta = 90^o\). the current will flow across the wedge, the field acts vertically upwards.

the force will follow the right hand rule