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

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SUMMARY

The discussion focuses on calculating the force exerted on a circular loop of wire carrying current I when placed in a magnetic field B at an angle θ. The force F on the loop is derived using the equation F = I * L * B * sin(θ), where L is the loop's circumference (L = 2πr). The analysis emphasizes the need to integrate the forces across the loop due to the varying angle of the magnetic field, concluding that the net force acts downwards, expanding the loop while maintaining horizontal equilibrium.

PREREQUISITES
  • Understanding of magnetic fields and forces on current-carrying conductors
  • Familiarity with vector calculus and integration techniques
  • Knowledge of the right-hand rule for cross products
  • Basic principles of electromagnetism, particularly Lorentz force
NEXT STEPS
  • Study the derivation of the Lorentz force law in electromagnetism
  • Learn about the integration of forces in non-uniform magnetic fields
  • Explore applications of magnetic fields in engineering, particularly in motors
  • Investigate the effects of varying magnetic field angles on different geometries of current-carrying loops
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Students and professionals in physics, electrical engineering, and anyone involved in the study of electromagnetism and its applications in current-carrying conductors.

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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