Magnitude and direction of current

AI Thread Summary
The discussion focuses on understanding the direction and magnitude of current in a conducting wire moving down an incline. The force opposing the wire's motion is established as directed from right to left, with the equation F = ILB confirming the relationship between current, length, and magnetic field. The gravitational force acting vertically downward does indeed have a component along the incline, contributing to the wire's motion. Clarifications on the forces involved reveal that the normal force compensates for parts of both the gravitational and magnetic forces. Overall, the participants conclude that the gravitational component along the incline is zero, resolving initial confusion.
Physicslearner500039
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Homework Statement
A straight piece of conducting wire with mass and length is placed on a friction less incline tilted at an angle Θ from the horizontal . There is a uniform, vertical magnetic field at all points. 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 and direction of the current in the wire that will cause the wire to remain at rest.
Relevant Equations
F=ILBsin(Θ) (N); F=mg (N)
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I am struggling with the angles. Since the conducting wire is moving down for it to be stand still, the force should be opposing it, hence the current should be from Right to Left.

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I am confident of the Force direction and its value is if ## I ## is the current
##F = ILB (N)## since L and B are perpendicular.
The gravitational force is ##mg (N)## which is acting vertically down. From the previous posts i assume if gravitational force is acting vertically down it does not have the other components. I am confused if the gravitational force has a component along the bar in the downward direction? Please advise.
 
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Physicslearner500039 said:
I am confused if the gravitational force has a component along the bar in the downward direction?
It has. Sketch and end view to see it: the normal force from the incline is, well, normal to the incline, and compensates part of mg and part of F. The other part of the normal force has to be compensated by the other part of F.
 
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Physicslearner500039 said:
I am confused if the gravitational force has a component along the bar in the downward direction?
What is making the wire slide down the incline?
 
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Yes now i understand
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##IBL\cos\theta = mg\sin\theta##
##I = \frac {mg\tan\theta} {BL}## Amps
The ##mg ## component along the X-axis is 0. I got confused with that. Thank You.
 
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