How Does Magnetic Field Angle Affect Force on a Current-Carrying Loop?

AI Thread Summary
The discussion focuses on determining the force on a circular loop of wire carrying current I in a magnetic field B that diverges at an angle θ. The force is expressed using the equation F = I*L*B*sinθ, where L is the loop's circumference. Participants suggest integrating the force over the loop's volume, considering the geometry of the magnetic field's angle. The analysis indicates that while the forces act to expand the loop and pull it downward, the horizontal forces cancel out, resulting in no net horizontal movement. The overall conclusion emphasizes the importance of the magnetic field angle in calculating the force on the current-carrying loop.
Angie K.
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Homework Statement


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HW13_2.jpg


A circular loop of wire, of radius r, carries current I. It is placed in a magnetic field B whose straight lines seem to diverge from a point a distance d below the ring on its axis. (That is, the field makes an angle θ with the loop at all points, where tan(θ) = r/d.) Determine the force F on the loop. Express your answer in terms of the given quantities.

Homework Equations



F = I*L*B*sinθ

The Attempt at a Solution


Not sure how to approach this.
 
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You should probably integrate it over the whole volume since dFb = I * dsXB (cross product), then the Fb =I* ∫dsXB. In this case ds will be dv - a small volume which I assume the volume of cone will be. (not sure, but maybe lighten your solution). If you consider only ds (surface) then from a triangle ds will be r/sinθ * cosθ.
∫r cosθ/sinθ dr
 
F = L * ( I × B ). So the forces will try to expand the loop, and will pull the loop downwards. So assuming that the given value of B is at the location of the wire, Fdown = I*L*B*sinθ.

L = 2πr.

Σ(radial forces) = 0
 
Last edited:
Hesch said:
Σ(radial forces) = 0

What do you mean by that?
 
Angie K. said:
What do you mean by that?

If you look at a piece of wire with the length dL, there will be an opposite piece of wire, dL, in the loop, pulled in the opposite horizontal direction. The sum of the horizontal forces = 0. The loop as a whole will not be pulled horizontally.
 
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