Magnetic force on wire


by songoku
Tags: force, magnetic, wire
songoku
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#1
Aug22-12, 07:46 AM
P: 767
1. The problem statement, all variables and given/known data


Which wire will experience the biggest force?
a. 1
b. 2
c. 3
d. all wires experience the same force
e. can't be determined without additional information


2. Relevant equations
F = BIL sin θ


3. The attempt at a solution

Can the above formula be applied to a non-straight wire? I only know that the formula is used for straight conductor.

My guess: the answer is wire 2 because for wire 1 and 3, there will be vertical and horizontal components of magnetic force and some of the horizontal components will eliminate each other. The sin or cos term of the force will reduce its magnitude hence the biggest force will be experienced by wire 2.

Am I correct? Thanks
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bigerst
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#2
Aug22-12, 09:51 AM
P: 57
try to divide each wire into small straight segments and analyze the force on each and add them up?
Doc Al
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#3
Aug22-12, 02:14 PM
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Quote Quote by songoku View Post
Can the above formula be applied to a non-straight wire? I only know that the formula is used for straight conductor.
You can still use it for each small segment of wire. Replace L with dL. As bigerst suggested, analyze the force on each segment and add them up.

My guess: the answer is wire 2 because for wire 1 and 3, there will be vertical and horizontal components of magnetic force and some of the horizontal components will eliminate each other. The sin or cos term of the force will reduce its magnitude hence the biggest force will be experienced by wire 2.
If you look at all the segments of the wire, it is true that they will have various vertical and horizontal components. But exactly which components add and which cancel? Think it over again.

songoku
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#4
Aug22-12, 10:25 PM
P: 767

Magnetic force on wire


Quote Quote by Doc Al View Post
You can still use it for each small segment of wire. Replace L with dL. As bigerst suggested, analyze the force on each segment and add them up.
So I cannot just change the "L" with length of the circular wire, which is its circumference?

If you look at all the segments of the wire, it is true that they will have various vertical and horizontal components. But exactly which components add and which cancel? Think it over again.
For wire 1, all the horizontal components will cancel each other.

For wire 3, maybe not all the horizontal components cancel. But how to can we compare the magnitude of force on wire 3 and 2?

Thanks
bigerst
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#5
Aug22-12, 10:35 PM
P: 57
Quote Quote by songoku View Post
So I cannot just change the "L" with length of the circular wire, which is its circumference?
indeed, F=BL (sin(θ)) works for straight segments only

however if u divide it up into several segments, for each segment

F= B sin(θ) ΔL since B is constant, then u can sum this up




Quote Quote by songoku View Post
For wire 1, all the horizontal components will cancel each other.

For wire 3, maybe not all the horizontal components cancel. But how to can we compare the magnitude of force on wire 3 and 2?

Thanks
what makes you think that? do a detailed mathematical analysis of the wires (split it into tiny, essentially straight segments and add the force contributions to each segment) and see what happens. it might be useful to separate each ΔL into vertical and horizontal components, rearrange which ones you add first, and see what happens.
Doc Al
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#6
Aug23-12, 06:39 AM
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Quote Quote by songoku View Post
So I cannot just change the "L" with length of the circular wire, which is its circumference?
No, since the force on each segment acts in a different direction.
songoku
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#7
Aug25-12, 11:01 AM
P: 767
Quote Quote by bigerst View Post
indeed, F=BL (sin(θ)) works for straight segments only

however if u divide it up into several segments, for each segment

F= B sin(θ) ΔL since B is constant, then u can sum this up






what makes you think that? do a detailed mathematical analysis of the wires (split it into tiny, essentially straight segments and add the force contributions to each segment) and see what happens. it might be useful to separate each ΔL into vertical and horizontal components, rearrange which ones you add first, and see what happens.
Sorry I don't know how to do a detailed mathematical analysis. But I am pretty sure that for wire 1, all the horizontal components will cancel each other because the direction of force at each segment will be like "radially outward".

For wire 3, I divide it to 4 segments:
segment 1 --> from (a) until the letter "i"
segment 2 --> the "lower curve"
segment 3 --> the straight part until number "3"
segment 4 --> from "3" to "b"

Assume segment 1 and 4 are symmetry, the horizontal components will cancel out each other, as the case of wire 1

Assuming segment 2 has a smooth and symmetry curve, the horizontal components will also cancel out each other, as the case of wire 1

Assuming segment 3 is straight line, there will be force directed upwards.

But I think my analysis is very weak....
songoku
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#8
Aug28-12, 08:47 PM
P: 767
bump
bigerst
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#9
Aug28-12, 09:39 PM
P: 57
since i have yet to figure out how to draw on physics forums i will use texts and hope it doesnt get too confusing. (you may find it helpful to draw on a piece of paper)
let us establish a 3d cartesian plane, with x axis points to the right., y axis points up, and the z axis points out of the page. let us suppose the magnetic field B, points along the z axis, out of the page, and the wires solely lie in the xy plane. suppose a wire goes from (0,0,0) to (1,1,0) and carries current I. since this is a straight wire F=BlIsin(θ) applies. find the force on the wire along with its direction, name this F1 (bold denoting it a vector)
now consider a wire that goes from O(0,0,0) to A(1,0,0) then to B(1,1,0) which also carries current I. use F=IBLsin θ on the two segments OA and OB separately (since each segment is straight the formula applies) to find the net force F2, yup another vector. you should be able to show F1=F2 in both magnitude and direction.
now comes the fruit of the argument, you should be able to reason that any small straight wire could be separate into x and y components with the net force on the segment unchanged.
also notice that once you divide the wires into components the order at which the compoenents appear doesnt matter since the field is uniform.
apply the above argument to the 3 wires in the problem by dividing each wire into infinitisimal (so essentially straight) segments, and separate each into x,y compoenents, by rearranging you should be able to see clearly which components cancel and which dont. (you dont have to do very sophisticated math, just an argument)


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