1. The problem statement, all variables and given/known data four parallel infinite wires are arranged as shown in the figure: (see attachment) a) calculate the value of the magnetic field produced along the central axis of the configuration (magnitude and direction) b) a fifth wires is placed along a central axis, parallel to the four wires and equidistant from all four. if a current I_2 is sent through this wire, what is the force per unit length on the fifth wire? c) if one wished to used the four outer wires to magnetically levitate the fifth central wire, what would be the relation between the linear mass density of the central wire (mass/length) and the currents I and I_2? also, in which direction would I_2 have to flow for the levitation to be possible? assume gravity acts downward, towards the bottom of the page, in the end-on view shown (see attachment). 2. Relevant equations magnetic field, ampere's law, B = mu_0(I)/2pi(r) where mu_0 is constant = 4pi*10^-7, I is current, r is radius magnetic field, biot-savart law, B = mu_0/4pi[integral(IdL/r^2)], where dL is change in length force F = IL X B where x indicates cross product, L is length, I is current, B is net magnetic field 3. The attempt at a solution part a, use amperes law to find the magnetic field from each wire at the center of the square formed by the four wires. where r is the center, so r = sqrt((a/2)^2 + (a/2^2). how will i factor in the directions of the current flow, wouldn't they cancel each other out, such that the net magnetic field at the center would be B = 0? or do i just take the magnitude(absolute value)? part b, use force equation and use net magnetic field from part a for B, use I = I_2, and L as constant to find force/unit length on fifth wire. part c, i am lost here, ideas/tips appreciated. were my attempt for part a and b along the right lines?
Not having seen the diagram, take this with a grain of salt, but yeah, I think part a is a trick question, and the magnetic field there is just zero. Part B, I think you're on the right track. Part C, think about it conceptually. If two wires in the same direction with the same current produce forces that cancel each other out directly inbetween, what does that mean the current of this fifth wire has to be? (hint: let's say the 4 wires are all +I)
part b, so if i use net magnetic field from a, which seems to just cancel out to zero, than the force F = IL X B = I_2(L) X B_net = I_2(L) X 0 = I_2(L)0sin(theta) = 0sin(theta) = 0? correct? part c, if the net magnetic force of the four wires originally there cancel out to zero, than the magnetic field must only come from the fifth wire, correct? thus the only current must be that of the fifth wire, which is given to be I_2, right?
Field isnt 0 :) This can be guessed easily - take only 2 wires. If wires cancel each other out when current is flowing in the same direction, then they would add up when current is going in different directions. Therefore - these 4 wires cancel each other out either in left-right direction, or up-down direction. Now to a proper solution. Use ampere law. Calculating B in the middle point of the each wire is simple. Problem is that you do not have direction of B. Use right hand rule for that (if the thumb is pointing in the direction of the current, then fingers show the direction of the field). So, B from the lower right wire is pointing in the direction of upper left wire... You see that up-down components of the wires cancel out and only 4 times the field of one wire in left direction remains. That component can be calculated easily using Pythagorean theorem. Calculating F - note cross product. Current has to flow inside (same as upper two) if you want the force to point up.
i thought opposite directions cancel out, thus as you said zizy 'left-right' cancel out, how is that possible they have the same direction? i understand that 'top-bottom' wires cancel out as they have opposite directions. so for part a, field would be zero as each wire is the same distance from the center, and each wire is canceled out by their opposite-direction pair. as for part b, are you saying the direction of the fifth wire is in the I_in direction? thus wouldn't there be a net I_in current since only two pairs cancel out, leaving the central fifth wire?
You dont have field cancelled by antiparallel wires but by parallel ones. Lower left wire would cancel lower right, if you were calculating filed in the middle of the distance between those wires. But in this case point is further up, so you will be adding two vectors, one in the direction of the upper left corner and other in the direction of the upper right corner. Same with the upper wires. B almost cancels out if you are looking antiparallel wires from a point that is very far away from both. Its doubled in the middle of the distance.