Calculate magnetic field intensity

[DODGEVIPER13]

Homework Statement

Two infinitely long filaments are placed parallel to the x-axis as shown in Figure 1.
a)Find H at the origin
b)Find H at (-1,2,2)

Homework Equations

∫Hdl=Ienclosed
H=H1+H2

The Attempt at a Solution

Well following the equation above I get Hl= Ienc then H1=Ienc/(2piRy) and H2=Ienc/(2piRz) then Ry=(0,0,0)-(0,4,0)=(0,-4,0) then length of Ry=4 and Rz=(0,0,0)-(0,0,4)=(0,0,-4) then Rz=4 thus H1=Ienc/(8pi) and H2=Ienc/(8pi) then H=H1+H2=10/(8pi)+10/(8pi)=20/(8pi)=.795 A/m that is for part a) I only listed this as I used the same method for part b) just with (-1,2,2) instead of (0,0,0)

 

[rude man]

Homework Statement

Two infinitely long filaments are placed parallel to the x-axis as shown in Figure 1.

Figure 1?

 

[DODGEVIPER13]

whoops I was tired and forgot to add it

 

[rude man]

Use Ampere's law: ∫H*ds = i. Which you did, sort of.

What you failed to do was to
(1) pay attention to the signs of Ienc y and Ienc z. The two currents are in opposite directions.

(2) compute Ry and Rz correctly. Given two points (x1 y1 z1) and (x2 y2 z2) what is the distance between them?

 

[DODGEVIPER13]

(x1,y1,z1)-(x2,y2,z2)=(x1-x2,y1-y2,z1-z2) then Ry=(0,0,0)-(0,4,0)=(0,-4,0)

 

[DODGEVIPER13]

right or no? I think that is what I did earlier uggg

 

[rude man]

right or no? I think that is what I did earlier uggg

That's what you did, and it's wrong.

Check your analytic geometry text or the Web.

 

[rude man]

(x1,y1,z1)-(x2,y2,z2)=(x1-x2,y1-y2,z1-z2) then Ry=(0,0,0)-(0,4,0)=(0,-4,0)

That's wrong. Furthermore, Ry is a distance and cannot be described by (x,y,z).

 

[DODGEVIPER13]

well I found a thing on euclidean distance Ry=sqrt((0-0)^2+(0-4)^2+(0-0)^2)=4

 

[rude man]

well I found a thing on euclidean distance Ry=sqrt((0-0)^2+(0-4)^2+(0-0)^2)=4

Much better.

 

[DODGEVIPER13]

Ok so then Rz=sqrt((0,0,0)^2-(0,0,4)^2)=4 right?

 

[DODGEVIPER13]

Ah the directions are different one in into the page and one is out. I can't remember does x mean into the page or out of the page?

 

[rude man]

Ah the directions are different one in into the page and one is out. I can't remember does x mean into the page or out of the page?

x is into the page and . is out of the page.

 

[DODGEVIPER13]

Ok so on the current is clockwise and on the dot current is counter clockwise so in the z direction it is negative and y it is positive

 

[DODGEVIPER13]

H=H1+H2=(10/2(pi)4)-(10/2(pi)4)=0

 

[DODGEVIPER13]

That doesn't seem right?

 

[rude man]

Ok so then Rz=sqrt((0,0,0)^2-(0,0,4)^2)=4 right?

Right.

 

[DODGEVIPER13]

Ok well now that my distances are confirmed correct is my new answer correct?

 

[rude man]

Ok so on the current is clockwise and on the dot current is counter clockwise so in the z direction it is negative and y it is positive

Currents don't run in circles. Currents run in the wires which are straight.

 

[rude man]

H=H1+H2=(10/2(pi)4)-(10/2(pi)4)=0

No.
H is a vector, that's why I use bold type for it.

Take one wire at a time and determine the direction of H for it using your pix. You have the magnitudes right.

 

[DODGEVIPER13]

well it seems to be flowing into the page on the y-axis and out of the page on the z axis. So on the y-axis the H vector would go left. Then I am guessing on the z axis since they are opposite the other H field must be to the right.

 

[rude man]

well it seems to be flowing into the page on the y-axis and out of the page on the z axis. So on the y-axis the H vector would go left. Then I am guessing on the z axis since they are opposite the other H field must be to the right.

Why left on the z axis? Do you know the right-hand rule?
Same comment for the z wire.
Draw circles around each wire intersecting with the origin, then use the rt-hand rule.

 

[DODGEVIPER13]

Well for the one into the page the B field will rotate clockwise and out of the page the B field will go counterclockwise. I said left and right because you have said something about the wires being straight and not circular so I guess I did not understand what you meant?

 

[rude man]

Well for the one into the page the B field will rotate clockwise and out of the page the B field will go counterclockwise.

That is correct. So, taking the one into the page first, what direction does the H field point at the origin?

 

[DODGEVIPER13]

well since H is the magnetic field strength then it should turn in the same direction as the B field right so wouldn't it point in the same direction as the B field I am confused?

 

[rude man]

well since H is the magnetic field strength then it should turn in the same direction as the B field right so wouldn't it point in the same direction as the B field I am confused?

Yes, same direction as the B field. Which direction is that? I need it in terms of unit vectors or at least a verbal description.

 

[DODGEVIPER13]

well the one going into the page is curling up so it is in the direction of the z-axis?

 

[rude man]

well the one going into the page is curling up so it is in the direction of the z-axis?

YES! The +z direction, let's call it +k, k the unit vector along the z axis.

Now, the other one?

 

[DODGEVIPER13]

the -Z direction so -k

 

[rude man]

the -Z direction so -k

No. Look more carefully at the H circle where it intersects with the origin.

 

[DODGEVIPER13]

I guess I used RHR wrong the second time would it be +Z again or is it on the Y-axis? Is there an easier way for me to see this?

 

[DODGEVIPER13]

would it be in the direction of the Y-axis?

 

[rude man]

would it be in the direction of the Y-axis?

Possibly. Which direction, +y or -y?

 

[DODGEVIPER13]

+y heh I am just guessing based on RHR

 

[DODGEVIPER13]

Well I think I found and appropriate equation setup for the into the page part H=I/2(pi)rho=(5/(pi)(4))k

 

[DODGEVIPER13]

H2=10/(2(pi)(4))j

 

[rude man]

H2=10/(2(pi)(4))j

Good. We have part (a) done.

 

[DODGEVIPER13]

sweet then H=(10/2(pi)4)j+(10/2(pi)4)k right?

 

[rude man]

sweet then H=(10/2(pi)4)j+(10/2(pi)4)k right?

Right.

 

[DODGEVIPER13]

so that would be an acceptable answer then accept that I must add units A/m or should I get the magnitude and angle and put in phasor form

 

[DODGEVIPER13]

whoops forgot about part b heh well assuming that it is worked the same then ry=sqrt((-1,2,2)-(0,4,0))=3 and rz=sqrt((-1,2,2)-(0,0,4))=3 and H=H1+H2=(10/(2(pi)3))k+(10/(2(pi)3))j

 

[rude man]

so that would be an acceptable answer then accept that I must add units A/m or should I get the magnitude and angle and put in phasor form

I would leave it in terms of i and j.

For part (b) it's different.

The first question you should ask yourself: does it matter what the x compoent of the observation point is?

 

[DODGEVIPER13]

for the first part no because it is 0 howeverI sense that it is different for part b

 

[DODGEVIPER13]

well I drew this thing on paper if I did it right the one that is into the page is on the x axis

 

[DODGEVIPER13]

I believe the other is on the y but I am not sure?

 

[rude man]

Both wires run parallel to the x axis.

Think again about the observation point. Does it matter where on the x-axis it lies?

 

[DODGEVIPER13]

it would lie on the negative side of the x-axis right?

 

[DODGEVIPER13]

I guess it doesn't matter as the two wires both have a 0 x component

 

[DODGEVIPER13]

The point is the only thing that has an x component of -1

 

[rude man]

I guess it doesn't matter as the two wires both have a 0 x component

The two wires run from x = -∞ to x = + ∞.
So that's not the reason.


Think again: as you move along the x-axis a fixed distance away from either wire, does the H field change?