Finding the magnetic force in square loop wire.

AI Thread Summary
The discussion revolves around calculating the magnetic force on a square loop carrying a current in a magnetic field defined as B = kzx. The participants analyze how the forces on the horizontal and vertical sides of the loop interact, suggesting that the horizontal forces cancel each other while the vertical forces add up. There is some confusion about the implications of the magnetic field's dependence on the z-coordinate and how it affects the direction of the force on the loop. Clarification is provided that B points in the positive x direction for positive z and negative x for negative z. The conversation emphasizes the importance of understanding the geometry and orientation of the loop in relation to the magnetic field.
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
 
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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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