Finding the magnetic force in square loop wire.

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SUMMARY

The discussion focuses on calculating the magnetic force on a square loop wire in a magnetic field defined by B = kzx(hat). The loop, positioned in the yz plane and carrying a counterclockwise current I, experiences forces on its vertical and horizontal segments. The analysis confirms that the forces on the horizontal segments cancel each other, while the vertical segments contribute to the net force. The relationship between the magnetic field and the z-coordinate is crucial, as it determines the direction of the magnetic force based on the current's flow direction.

PREREQUISITES
  • Understanding of magnetic fields and forces, specifically Lorentz force law.
  • Familiarity with vector cross products in physics.
  • Knowledge of the Right Hand Rule for determining force directions.
  • Basic concepts of current flow in electrical circuits.
NEXT STEPS
  • Study the Lorentz force law in detail, focusing on its application to current-carrying conductors.
  • Learn about the implications of magnetic field variations in space, particularly in relation to z-coordinates.
  • Explore the Right Hand Rule and its applications in different configurations of current and magnetic fields.
  • Investigate the effects of different current directions on the resulting magnetic forces in loops.
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bfusco
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Homework Statement


Suppose that the magnetic eld in some region has the form B = kzx(hat). (where k is a
constant). Find the force on a square loop (side a), lying in the yz plane and centered
at the origin, if it carries a current I, flowing counterclockwise, when you look down
the x axis.

The Attempt at a Solution


Apparently the horizontal sides of the loop are to generate forces that cancel each other out, leaving the forces generated on the vertical parts to add.

Using the Right hand rule, i see no reason why the horizontal parts should cancel and the vertical shouldn't.
Extablishing the +z axis pointing up, and the +y axis pointing right, with the x-axis pointing out of the page.

denoting the bottom (horizontal) part of the loop as 1, the right (vertical) part of the loop as 2, the top (horizontal) part of the loop as 3, and the left (vertical) part of the loop as 4.

In 1) Idl x B = y(hat) x x(hat)=-z(hat)
In 2) Idl x B = z(hat) x x(hat)=y(hat)
In 3) Idl x B = -y(hat) x x(hat)=z(hat)
In 4) Idl x B = -z(hat) x x(hat)=-y(hat)

1 should cancel 3, and 2 should cancel 4

Or is it that B is a function of z, so when the current is going in the +z direction B is in the +x direction, and when the current is going in the -z direction B is in the -x direction? I am not sure
 
Last edited:
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bfusco said:
Or is it that B is a function of z, so when the current is going in the +z direction B is in the +x direction, and when the current is going in the -z direction B is in the -x direction? I am not sure

I'm not sure exactly what the phrase "looking down the x-axis" means in the statement of the problem, but I think you have the right idea. Since ##\vec{B} = kz\hat{x}##, B points in the positive x direction when z is positive and points in the negative x direction when z is negative.
 

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