Calculate magnetic field intensity

[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?

 

[DODGEVIPER13]

If the distance from either wire is fixed then no

 

[rude man]

If the distance from either wire is fixed then no

OK, so how can we make life easier by altering the second observation point (-1,2,2) without changing H at any given (x,y,z)?

 

[DODGEVIPER13]

Ok so what you are saying is that it does matter

 

[DODGEVIPER13]

im really sorry man for dragging this out I am just trying to understand. That being said if you change positions of course you will change the strength, right?

 

[rude man]

im really sorry man for dragging this out I am just trying to understand. That being said if you change positions of course you will change the strength, right?

As x changes, does H change?

 

[DODGEVIPER13]

I think so when I said no you said I was wrong?

 

[rude man]

I think so when I said no you said I was wrong?

Why would H change as you move along it, keeping the same perpendicular distance from it? The wire is infinitely long!

 

[DODGEVIPER13]

ok so i was correct before and it doesn't change

 

[rude man]

ok so i was correct before and it doesn't change

Yes. I didn't feel youunderstood why. Just gussing right does not help you understand.

Anyway, OK, so how do we write (-1,2,2) if we want to put the observation point in the yz plane? This step is not really necessary but helps to visualize things.

 

[DODGEVIPER13]

well your right I am not understanding. Do you have any tips on that? As to your question you should draw the x-axis line so that it is parallel to the two wires. One side of that line will lie in the yz plane which is where it should be marked

 

[DODGEVIPER13]

Whoa I flubbed that up that wasnt what you asked at all heh. Well would it (0,2,2) so that the x part is eliminated?

 

[DODGEVIPER13]

Ok you will be happy I think i found a solution online that is right, I appreciate all the help on the problem. I am a little iffy on whether the solution is correct but I am just growing tired of this problem, and will come back to it later on.

 

[DODGEVIPER13]

Is it against the rules to ask if you can assist me with other issues I don't want an infraction?Sorry hope I am not being pushy or anything just wondering.