Magnetic field strength between two wires homework

[shanie]

Homework Statement

Two straight parallel wires are placed vertically on a wall. The distance between the wires is 12 cm and the current through each one is 7.5A. In the left wire the current is directed upwards and in the right one the current points downwards. Determine the magnetic field strength (or flux density), and its direction B in the following:
a) A point between the wires.
b) A point on the wall 12 cm to the left of the left wire.

Homework Equations

a) B = k*I/d giving 12.5 μT from each wire.
b) B = k*I/d giving 12.5 μT from the left wire and 6.25 μT from the right.

The Attempt at a Solution

a) Now my first guess was to treat the magnetic field strength like waves; since the magnetic fields are in opposite directions they may at certain points strengthen each other and at some points even cancel out. So I thought that 25 μT (12.5+12.5) would be the maximum field strength and 0 μT would be the minimum.. However I'm also quite sure that the field between the wires is homogeneous giving a more specific value for B, perhaps the field strength is just 0 μT because they cancel out everywhere?
b) This question works on the same principle as a), so do I subtract or add the field strengths? I'm guessing subtract because they are in opposite directions?

 

[Gear300]

You're more or less getting the right idea. Take into account the direction of the magnetic fields due to the current in the wires at a certain point and add them as vectors (if they are in opposite directions, you subtract).

 

[shanie]

But if the location of the point isn't specified, do I just assume that the two magnetic fields affect the point equally? That is, assuming that the point is in the exact middle of the two wires. That would essentially give me a magnetic field of 0 T for question a)? And in b), the correct answer would then be 12.5-6.25=6.25 μT?

 

[alphysicist]

But if the location of the point isn't specified, do I just assume that the two magnetic fields affect the point equally? That is, assuming that the point is in the exact middle of the two wires. That would essentially give me a magnetic field of 0 T for question a)? And in b), the correct answer would then be 12.5-6.25=6.25 μT?

For part a, I think you are correct in considering the point exactly between the wires. Now, what is the direction of the left wire's field at that point? and what is the direction of the right wire's field at that point? The directions will tell you how to find the total field.

Also, in your original post you got 12.5 \mu \mbox{T} as the field from each wire, but I don't believe that is correct. What numbers did you use?

 

[shanie]

The numbers I used where the current I=7.5A and the radius from each wire r=0.12m and then the constant k=2*10^-7. Giving: B=kI/r => B = (7.5*2*10^-7)/0.12=1.25*10^-5 T..? Is this incorrect?

The direction for the left wire (current going upwards) is anticlockwise, and for the right it is clockwise according to the rule of thumb..? Doesn't this mean that they cancel out so the answer for a) is 0T?

I read a previous post in this forum about a similar question and from what I understood I'm actually supposed to add the field strengths. Could someone please explain why?

 

[mal4mac]

So I thought that 25 μT (12.5+12.5) would be the maximum field strength and 0 μT would be the minimum.

Assuming for the moment that 12.5 is correct, where exactly would the field strength take these values? At which points do the fields strengthen each other? Do you have an unclear visualisation of the whole of the magnetic field between and around the two wires? To get happy with these types of problems I think you need to do this visualisation (step by step).

Tricky visualisations these because they are inherently three dimensional, unlike cannon shots which you can easily visualise in two dimensions. That is you need to *clearly* see the wires going up, and the B fields circling round in the other two dimensions. Even Stephen Hawking, in "Universe in a Nutshell" says he has trouble visualising in three dimensions, adding that he can just about manage two and a half!

(Hawking, as with the God concept and too many other concepts, doesn't explain what he means by 21/2 D. Fortunately I've studied cognitive psychology so I think I know what he means. Two and half dimensions is what you get on a computer screen -- a window is two dimensional, when one overlaps another you get two a half dimensions, i.e. an impression of three dimensions without seeing the extension.)

So if you feel your mind warping and your 3D image of this situation being less than clear, don't panic! Hawking (and me) have physics degrees and we both struggle.

(Wise acres who say it's perfectly clear are either: (i) kidding themselves (ii) lucky to have a great talent and should now try it in the 11 dimensions of string theory :-)

With this problem you can't easily devolve it into two and half dimensions, so to visualise it I suggest taking a two dimensional slice. For instance, visualise the two wires. Imagine two points between the wires, the left represents the magnetic field (B) for the left wire, the right point the B field for the right wire. That is, the magnetic force is represented by a line going into the paper. You can only see its cross-section (a point).

The direction of the current and that of the resulting magnetic field are related to each other as the forward travel of a corkscrew and the direction in which it is rotated. So for the left wire imagine the point changing into a cross to indicate the field line is going into the page, and keep the right point as a point (or a small circle) to indicate the field coming out of the paper. So - left wire, cross, point, right wire, even Hawking could visualise that :-)

To get a feel for the strength of the field at various points gets a bit hairy to visualise in the "mind's eye", so get your paper and pen out and draw crosses and circles with radius proportional to the field strength at various interesting points ( I'll leave alphy to help you with calculating the field, if necessary!)

 

[shanie]

Thank you :-)
I drew a picture like you said and I also drew the arrows of the magnetic field according to the right hand rule. Now I realized that the field arrows coming from each wire points in the same direction once they reach the area in the exact middle of the wire (the area where the point in a) is located). So essentially this means that I add the forces of 12.5 μT (assuming this is the right field strength) up to make 25 μT in that point.. Did I get it right now or am I still wandering off..?

 

[alphysicist]

The numbers I used where the current I=7.5A and the radius from each wire r=0.12m and then the constant k=2*10^-7. Giving: B=kI/r => B = (7.5*2*10^-7)/0.12=1.25*10^-5 T..? Is this incorrect?

In the formula, r is the distance from the wire to the point that you are calculating the field at, so it will not be 0.12m (0.12m is the distance that separates the two wires).

 

[shanie]

Oh of course, it's 0.06m which gives B=25μT.
For
a) this would make 50μT (25+25) in the point (with B pointing in which direction? since each field goes its own direction after having left the middle..)
b) it would make 12.5μT from the left (12cm away from the wire as specified), and 6.25 μT (12+12=24cm away) from the right wire, and subtracting to give a total of B=6.25μT, with B pointing inwards perpendicular to the surface of the paper..

 

[alphysicist]

Oh of course, it's 0.06m which gives B=25μT.
For
a) this would make 50μT (25+25) in the point (with B pointing in which direction? since each field goes its own direction after having left the middle..)
b) it would make 12.5μT from the left (12cm away from the wire as specified), and 6.25 μT (12+12=24cm away) from the right wire, and subtracting to give a total of B=6.25μT, with B pointing inwards perpendicular to the surface of the paper..

That's right, but there are two direction perpendicular to the paper: into the paper and out of the paper (or the wall as the problem states). Which are the directions of the individual fields? In other words, for points to the left of the current going vertically upward, what direction is its field? and for points to the left of the current going vertically downward, what is the direction of its field?

 

[shanie]

For points to the left of the current going vertically upward, the field is anticlockwise according to the right hand rule. Vertically downward gives clockwise direction.. So for b) when the point is closer to the wire with current going anticlockwise, B will be going out of the paper (point).

Left-outwards, point
Right-inward, cross

B in the point P will then be O (pointing vertically out of the paper). A point P1 where both directions going upwards.. so B will be pointing in towards the paper (cross)?

 

[alphysicist]

For points to the left of the current going vertically upward, the field is anticlockwise according to the right hand rule. Vertically downward gives clockwise direction.. So for b) when the point is closer to the wire with current going anticlockwise, B will be going out of the paper (point).

That's right; the field from the left wire is out of the paper, and the field from the right wire is into the paper. Since the outward field happens to be larger than the inward field, the total net field will be out of the paper.

 

[merryjman]

As someone posted earlier, it's hard to visualize this stuff. Some people find the right hand curl rule really useful for wires. You curl your hand around the wire, and point your thumb in the direction of the current (like http://physicsed.buffalostate.edu/SeatExpts/resource/rhr/CNB.JPG ), and whatever direction your fingers are pointing will show you how the field curls around the wire. So like for an example, on the right side of a wire carrying current upwards, your fingers will curl into the page so the B field points in.

This way, it can be easier to draw your exes and dots for fields pointing in and out, and you can easily see how parallel and antiparallel wires behave near each other. You'll see sometimes that wires carrying current out to an electronic device and back are twisted together, and a few minutes with this type of problem tells you why.

 

[shanie]

Thanks so much for your help!