
#1
Aug2212, 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 nonstraight 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 



#2
Aug2212, 09:51 AM

P: 57

try to divide each wire into small straight segments and analyze the force on each and add them up?




#3
Aug2212, 02:14 PM

Mentor
P: 40,871





#4
Aug2212, 10:25 PM

P: 767

Magnetic force on wireFor 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 



#5
Aug2212, 10:35 PM

P: 57

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 



#6
Aug2312, 06:39 AM

Mentor
P: 40,871





#7
Aug2512, 11:01 AM

P: 767

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.... 



#8
Aug2812, 08:47 PM

P: 767

bump




#9
Aug2812, 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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