Solving Sliding Wire Problem: Determine Magnitude & Direction of Current

[EvanQ]

Homework Statement

A straight piece of conducting wire with mass M and length L is placed on a frictionless incline tilted at an angle theta from the horizontal. There is a uniform, vertical magnetic field vecB at all points (produced by an arrangement of magnets not shown in the figure). To keep the wire from sliding down the incline, a voltage source is attached to the ends of the wire. When just the right amount of current flows through the wire, the wire remains at rest.

Determine the magnitude of the current in the wire that will cause the wire to remain at rest.
Express your answer in terms of the variables M, theta, L, B, and appropriate constants.

Determine the direction of the current in the wire that will cause the wire to remain at rest.

In addition viewing the wire from its left-hand end, show in a free-body diagram all the forces that act on the wire.

Homework Equations

dF = Idl ×B
or possible the bio-stavart law.. I'm not sure

The Attempt at a Solution

the current must be directed from the right to the left i think?

other than that, unsure how to start, just chasing a hint or beginning to work from.

thanks heaps.

 

[Astronuc]

See if this helps - http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html

The force pulling the wire down the incline is just the component of the weight parallel with the plane of the incline.

 

[EvanQ]

ok so force pulling the wire down:

F=masinθ
= -9.8Msinθ

so for the rest:

F=ILBsinθ
9.8Msinθ=ILBsinθ
9.8M=ILB
I=9.8M/LB ??

 

[EvanQ]

and the current would go to the left, determined by the right hand rule?

 

[Astronuc]

and the current would go to the left, determined by the right hand rule?

Yes, but be careful with the angles. The angle between current and magnetic field is 90°, and so the resulting force points horizontally. One has to ensure that the force on the wire parallel to incline matches the weight component down the incline.

In this geometry, the angle between magnetic field and current is not the angle of the incline.

 

[EvanQ]

9.8Msin180=ILBsin90??

 

[EvanQ]

really confused sorry :(

 

[chaoseverlasting]

Think in 3D. You know the force $$F_b=\vec{B}\times \vec{i}L$$. Here the force is perpendicular to both the current and the magnetic field. Using the left hand rule, find out the direction of the current, and the equate the appropriate components.