Force due to three magnetic fields

In summary: Actually, Ampere's Law is not going to help with this one. Only infinitely long, straight wires have the appropriate cylindrical symmetry. But you can solve this by using Biot–Savart law, and integrating over the length of each straight-line segment. There is some symmetry involved, so you might not have to integrate over every segment. But you'll probably need to break up the problem into at least two integrals (3 integrals if you ignore the symmetry).
  • #1
Frostfire
50
0

Homework Statement




A wire comes in from in finity (left side) and forms a shape by going up a length A, over a length A and back down a length A.(forming three sides of a square that is attached to this infinite wire), then
takes a turn and leaves the page opposite to the side it came in on, headed to infinity.

There is a particle along the horizontal axis of this, a length 1/2 A from the left or right side of the shape, what is the magnetic field at that point,




Homework Equations





The Attempt at a Solution



the answer is (sqrt 5)/5 * (Mu * I)/pi A But I have no idea how to get there
 
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  • #2
Try Ampere's Law. [tex]\oint \vec{B} \cdot d\vec{l}=\mu_{0}I_{enclosed}[/tex]
 
  • #3
americanforest said:
Try Ampere's Law. [tex]\oint \vec{B} \cdot d\vec{l}=\mu_{0}I_{enclosed}[/tex]

Would this qualify as high sym.?

The way I was going to approach it was calculate the force on the particle due to the two verticle lengths of wires, since it would just be the force due to 1*2, then add the effect of the upper horizontal wire to that. But how do you calculate the first part? Can I use Ampere's law ?
 
  • #4
Frostfire said:
Would this qualify as high sym.?

The way I was going to approach it was calculate the force on the particle due to the two verticle lengths of wires, since it would just be the force due to 1*2, then add the effect of the upper horizontal wire to that. But how do you calculate the first part? Can I use Ampere's law ?

Superpose the fields from each of the sides of interest. The wire has cylindrical symmetry so you can use Ampere's law.
 
  • #5
americanforest said:
Superpose the fields from each of the sides of interest. The wire has cylindrical symmetry so you can use Ampere's law.

Actually, Ampere's Law is not going to help with this one. Only infinitely long, straight wires have the appropriate cylindrical symmetry.

But you can solve this by using Biot–Savart law, and integrating over the length of each straight-line segment. There is some symmetry involved, so you might not have to integrate over every segment. But you'll probably need to break up the problem into at least two integrals (3 integrals if you ignore the symmetry).
 
Last edited:

Related to Force due to three magnetic fields

What is the formula for calculating the force due to three magnetic fields?

The formula for calculating the force due to three magnetic fields is F = qvB1sinθ1 + qvB2sinθ2 + qvB3sinθ3, where q is the charge of the particle, v is the velocity, B1, B2, and B3 are the magnetic field strengths, and θ1, θ2, and θ3 are the angles between the particle's velocity and the direction of each magnetic field.

How does the direction of the magnetic fields affect the force on a charged particle?

The direction of the magnetic fields affects the force on a charged particle by changing the angle θ between the particle's velocity and the direction of each field. The force is strongest when the angle is 90 degrees, and weakest when the angle is 0 degrees. If the angles of all three fields are the same, the particle will experience no net force.

What happens when the magnetic fields are parallel to each other?

When the magnetic fields are parallel to each other, the force on a charged particle is zero. This is because the angle θ between the particle's velocity and the direction of each field is 0 degrees, resulting in a sinθ value of 0 in the formula for force.

What is the relationship between the strength of the magnetic fields and the force on a charged particle?

The strength of the magnetic fields directly affects the force on a charged particle. The stronger the fields, the greater the force on the particle. This is because a stronger magnetic field will exert a larger force on the moving charged particles passing through it.

How is the direction of the force determined in a three magnetic field scenario?

The direction of the force on a charged particle in a three magnetic field scenario is determined by the right-hand rule. If the particle's velocity is in the direction of your thumb and the magnetic fields are in the direction of your fingers, the force will be perpendicular to both and can be determined by the direction your palm is facing.

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