Force between an infinitely long wire and a square loop

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SUMMARY

The discussion focuses on calculating the net force, F, on a square loop of wire carrying current I1, positioned a distance d from an infinitely long wire carrying current I2. The derived formula for the force is F = I1 * a * μ0 * I2 * ((1/2π(d - 0.5a)) - (1/2π(d + 0.5a))). Additionally, the magnetic moment, m, of the loop is defined as m = I * A, where A equals the area of the loop. The final expression for the force in terms of the magnetic moment is F = (m/a^2) * a * μ0 * I2 * ((1/2π(d - 0.5a)) - (1/2π(d + 0.5a))).

PREREQUISITES
  • Understanding of electromagnetic theory, specifically Ampère's Law.
  • Familiarity with the concept of magnetic moment in current loops.
  • Knowledge of the Biot-Savart Law for calculating magnetic fields.
  • Basic algebra and calculus for manipulating equations.
NEXT STEPS
  • Study the derivation of the Biot-Savart Law for magnetic fields around current-carrying conductors.
  • Learn about the applications of magnetic moments in various physical systems.
  • Explore the effects of varying distances on the force between current-carrying wires.
  • Investigate the implications of magnetic forces in electromagnetic devices.
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Students and professionals in physics, electrical engineering, and anyone interested in the principles of electromagnetism and magnetic forces between current-carrying conductors.

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Homework Statement



A square loop of wire with side length a carries a current I1 . The center of the loop is located a distance d from an infinite wire carrying a current I2. The infinite wire and loop are in the same plane; two sides of the square loop are parallel to the wire and two are perpendicular as shown.

1. What is the magnitude, F , of the net force on the loop?
I got that one, which is
F= I1*a*uo*I2*((1/2pi(d-0.5a)-(1/2pi(d+0.5a))

2. The magnetic moment m of a current loop is defined as the vector whose magnitude equals the area of the loop times the magnitude of the current flowing in it (m=IA), and whose direction is perpendicular to the plane in which the current flows. Find the magnitude, F , of the force on the loop from Part A in terms of the magnitude of its magnetic moment.
Express in terms of m,I2 ,a ,d , and uo .


Homework Equations





The Attempt at a Solution


F= I1*a*uo*I2*((1/2pi(d-0.5a)-(1/2pi(d+0.5a))
since m=I/A, then I= m/A
F = (m/A)*a*uo*I2*((1/2pi(d-0.5a)-(1/2pi(d+0.5a))

Did I do it correct?

Thank you..
 

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Get rid of the A. It equals a^2.
 
thank you Doc Al.. :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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