# Three infinitely long current carrying conductors. H field at L3

Discussion in 'Homework Help' started by Jess_88, May 24, 2011.

1. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Hey guys.
I'm having some trouble getting started with this question.

I need to fined
1) direction of H fields due to L1 and L2 at L3 (clockwise or anti clockwise)
2) Total H field experienced at L3 due to currents in L1 and L2
3) Force per unit length experienced by L3 due to other currents, if the system exists in air.

L1 = 100A in -z direction
L2 = 200A in +z direction
L3 = 300A in +z direction

I can't fined any examples where the question relates to the inside of a 3ed wirer, so I don't know how the current in L3 effects anything.

I'm thinking for question two, I use H = I/2pi*a (were "a" is the perpendicular distance from L2 or L1 to L3)
Then H(L1) - H(L2)
But that looks to easy.

Could someone point my in the right direction?

thanks guys

2. ### steveb Senior Member

Jul 3, 2008
2,433
469
The question is easy, so don't let that throw you off track. However, be very careful to pay attention to the fact that both H field and force are vectors, so contributions from L1 and L2 must be added vectorially.

Your book should give you the formula for field from an infinite wire and the force between two parallel wires. It should also explain the right hand rule to figure out directions (often the hardest part).

The current from L3 is only relevant for the force calculation. Also, because you are dealing with air as the medium, you can use superposition and treat L1 and L2 separately and then add their contributions vectorially.

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3. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
ok. so I have

(x)H1 = H1cos(phi)
(y)H1 = -H1sin(phi)
(x)H2 = 0
(y)H2 = H2

dose that look ok?

adding x components of each and y components of each
(x)at H3 = H1cos(phi)
(y)at H3 = -H1sin(phi) + H2

then I just substitute H = I/2pi*r for H1 and H2, and phi = 53.131

I'm still a bit sketchy on my method. Is it ok?

4. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
ah, I see my alpha and phi the wrong way around. It should be tan^-1(4/3) = 53.131
and phi = 90 - 53.131 = 36.869

is everything els ok?

5. ### steveb Senior Member

Jul 3, 2008
2,433
469
Yes, I think so. Aside from the mislabeling of the angles, which you found yourself, the rest looks ok. The directions of H1 and H2 look correct and the angles look correct. It also looks like you are converting to rectangular components of H2 and H1 correctly.

One minor thing: I got angles of 53.130 degrees and 36.870 degrees.

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6. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Yeah I get 53.130 and 36.870 as well... don't know what I did the first time???

so I have
H = (x) 6/pi + (y) 82/3*pi
that look ok?
and is that the total magnetic field at L3 or am I looking for the magnitude-> sqrt(x^2 + y^2) ?

how would I go about determining the direction of the H field?

7. ### steveb Senior Member

Jul 3, 2008
2,433
469
OK, first I got H=(x) 6/pi + (y) 76/(3pi), so double check this. It's possible I made a mistake, but it should be easy for you to check. It looks to me like you used cos rather than sin to calculate the y-component.

As far as direction, you have a vector form for the field H, so both magnitude and direction are contained in your answer. So, basically, you just need to convert your rectangular form (x,y) to polar form (r,phi).

For any vector (x) A + (y) B, the magnitude is sqrt(A^2+B^2) and the angle is phi=arctan(B/A). I know you already know this, but sometimes, in the learning phase, we need to be reminded of the meaning of what we know.

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8. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
thats thats great. Much appreciated.

for (y) I have -100sin36.87/2pi*5 + 200/2pi*3
=-60/10pi + 100/3pi
=-18/3pi + 100/3pi
=82/3pi
did I work that out correctly?

now I just need to work on the force per unit length of L3 due to the other two currents.

9. ### steveb Senior Member

Jul 3, 2008
2,433
469
Ah, yes! In fact you did do that part correctly. So, that probably means that the other calculation for the (x) component is incorrect. That needs to be 8/pi, I think. Check that one.

Do you have the force formula from your book? It will look a little like F=(r) k*I1*I2/r + (phi) 0, where k is a known constant (probably 1/(2pi) in your units).

So now, the vectors are in line with the radius vector, rather than perpendicular to it. But, it's the same idea. Use superposition by calculating the separate forces between L1 & L3 and L2 & L3, then add them vectorially for the net force on L3.

Last edited: May 24, 2011
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10. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
ah, yep. I made a mistake, it is H(x) =8/pi

I dont have the text book . But in my lecture notes there is a formula
F = (y) I1*I2/2pi*d
where d is the distance from one line to the other.
in the formula you have there, where did the + (phi)0 come from?

I'm a little bit confused as to using this formula with vectors... (isn't it just I in the z + and - direction)

so I have
F1 = I1*I3/2pi*5
F2 = I2*I3/2pi*3
F at L3 = F2 - F1

am I using the correct formula???
did I use it correctly?

also one more thing. The question I am given with regard to direction is
-State the direction of the H fields due to L1 and L2 at the position L3 (clockwise or anti-clockwise?)

now this confuses me a little bit due to the wording. I would assume it means i need to state ALL the directions of H1 H2 and H in terms of clockwise or anticlockwise. But conversion of H to polar form would give the static direction but not the rotation... Am I able to gain the rotation of of H (clockwise of anti-clockwise?). Would the question only apple to H1 and H2?

I would assume H1 = clockwise and H2 = anti-clockwise

Last edited: May 25, 2011
11. ### steveb Senior Member

Jul 3, 2008
2,433
469

OK, I agree with your answer for clockwise and anti clockwise. Basically they want the direction of the field relative to the position of the source of the field. H1 is clockwise around the origin (0,0) in the xy-plane and H2 is counterclockwise around the point (4,0) in the xy-plane.

Don't let my formula for force confuse you. I only added the (phi) 0 to stress that the direction of force is along the line between the two conductors and not perpendicular to it. So, the direction of force is not the z-direction at all as you said in words and it is not in the y-direction as indicated in your formula. Even my formula is not strictly correct because the direction is not in the radial direction either unless the source is at the origin. You have the correct formula for the magnitude of the force, but the direction will be in the xy-plane perpendicular to the H vector. The force can be repulsive or attractive but it is in line with the radial vector, if you place one conductor L at the origin.

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12. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Ok, I just want to see if I understand you correctly
So the magnitude of force will be found by F = I2*I3/2pi*3 - I1*I3/2pi*5
and found at 90 deg + or - 73.686

Is that at all on the right track?

Last edited: May 25, 2011
13. ### steveb Senior Member

Jul 3, 2008
2,433
469
No, not quite. The two forces need to be added vectorially. So calculate the force vector of each interaction separtately and then add the two vectors together. You can't just add magnitudes as you did (because the vectors are not in the same direction), but you can add the separate components of the vectors together.

Last edited: May 25, 2011
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14. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
ok looking at the force between L1 and L3
would I be using
F(x) = I2*I3/2pi*3
where 3 is the distance along the x-axis
F(y) = I2*I3/2p*4
where 4 is the vertical distance

Then F(L1L3) = (x)I2*I3/2pi*3 + (y)I2*I3/2p*4

15. ### steveb Senior Member

Jul 3, 2008
2,433
469
No, that's not it. The distance of separation between L1 and L3 is 5 as you correctly calculated before. Also, you calculated the angle alpha.

So, the magnitude of the force is |F31|=I3*I1/(2pi*5), where it's important get the sign of I1 and I3 correct to know if this is an attractive or a repulsive force. Now, this force is directed in a direction given by the angle alpha because this is the line that connects the two condutors.

I think the issue that is confusing you is the conversion between rectangular and polar coordinates. Magnitudes and angles go together and (x) and (y) components go together, and you can easily convert between the two representations, as you have done above with the field calculations.

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16. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
ok lets see. If I can get it this time

so for L1L3

Where it is a repulsive force (as magnetic opposites repel)
thus a negative force.
Looking better?

17. ### steveb Senior Member

Jul 3, 2008
2,433
469
Yes, that looks better. Currents in opposite directions do repel, and it looks like you are taking the force vector in the right direction and computing its (x) and (y) components correctly.

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18. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Great!

So I have my F1 x and y components (-x572.959 and -y763.942)
and F2 for force between L2 and L3 = xI2I3/2pi*3 + y(0) right?
then F = -x(572.959 + I2I3/2pi*3) - (y)763.942
magnitude F = sqrt( xA^2 + yB^2) correct?

I just noticed another formula, where it includes prematurity (u).
where u = uo*ur
and this system is assumed in air.
so would I use this still?

Last edited: May 26, 2011
19. ### steveb Senior Member

Jul 3, 2008
2,433
469
OK, a few issues here. You somehow changed the sign on the forces from what you wrote in your previous post. Note that L1 and L3 repel and L2 and L3 attract. Also, even based on what you wrote here, you have a sign error in the (x) component. So check this over carefully.

Well, I'm glad you mention this because it is important, and I missed a possible mistake in your work above.

First of all the mu (u) is called permeability, and u0 is permeability of free space. The ur is the relative permeability which accounts for the medium you are in (vacuum has ur=1, and air has ur nearly 1).

Second, in my formula for force, I showed the constant k and mentioned that this value depends on the unit system you are using. This value of k is really u/2pi, but in some unit systems u=1.

So, the question is, "which unit system are you using?" and "what is the distance scale for your coordinate system?". If you are using SI units (Amps, meters, kilograms, seconds, Newtons etc.) then u0 is not equal to 1. So, you definitely want to use u0 in the formula in general. I think in your case you want to calculate force in units of Newtons (assuming that your coordinates are in meters), which requires u0=4pix10^-7 H/m, if my memory is good (please check it).

Actually, a more general comment is that I should have been encouraging you to write the units for all of your answers. This is always important, but it is even more important when you are dealing with electromagnetics because of the added complication that the unit system can "hide" the u0 on you and you may forget it is needed. Bottom line is that it is good practice to state the unit system you are using (up front) when doing field calculations. This doesn't matter too much in school because you only use one unit system, but in the real world you will deal with different people who prefer another system.

Last edited: May 26, 2011
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20. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Yeah I see you point about the importance of units.
We do use SI units over here so I am assuming this is the case for my question.
The unit distance in not given in my question statement... just that it's in a cartesian coordinate system... I'm just going to assume meters.
Yes, your memory serves you well u0=4pix10^-7 H/m

I'm sorry I wasn't really understanding the sign influence for attractive and repulsive forces. Is an attractive force negative and repulsive force positive?

This is my calculations so far

so Total F(x) = 7.2x10^-4(N/m) - 4x10^-4(N/m)
and F(y) = 9.599x10^-4
seem ok?

then I would take the magnitude of F(x)+F(y) to fined the total force per unit length right?