# This question is giving me a headache Biot-Savart'

## Homework Statement

The two wires shown below carry currents of 5.00 A in opposite directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic field at the following locations. (a) at a point midway between the wires

(b) at point P1, 10.0 cm to the right of the wire on the right

(c) at point P2, 20.0 cm to the left of the wire on the left

## The Attempt at a Solution

(a) Now my book says it is 40 x 10-6T. I realize that it is 40 and not 20 because there is two forces. Can someone draw me a diagram of HOW it looks like? I can only imagine one pushing the other and there is only one force from each

DOUBLE EDIT: NO I AM NOT DONE

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Also for (b) and (c) I used Ampere's Law to solve it

(b) $$\vec{B} = \frac{\mu_0 I}{2\pi d}$$

(c)$$\vec{B} = \frac{\mu_0 I}{\pi d}$$

They both gave me the incorrect answer...

ANyone?

SammyS
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Also for (b) and (c) I used Ampere's Law to solve it

(b) $$\vec{B} = \frac{\mu_0 I}{2\pi d}$$

(c)$$\vec{B} = \frac{\mu_0 I}{\pi d}$$

They both gave me the incorrect answer...
Did you get the answer to (a)?

For (b):

How did you arrive at this result?

I assume that d = 10cm.

Find B due to each wire, including direction, or at least a sign. Add them together.

For (c):

Ditto.

Also, I would expect this answer to be less than the answer to (b).

.

gneill
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There are two fields (one from each wire) to add at each point. The http://www.pa.msu.edu/courses/1997spring/phy232/lectures/ampereslaw/wire.html" [Broken] gives the directions for each field at the given point, and your expression

$$B = \frac{\mu_0 I}{2\pi d}$$

gives the magnitude of a field for a given distance d from each wire.

Perhaps you should show your detailed calculation for one of the points p1 or p2.

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Did you get the answer to (a)?
For (a) I had 20uT, but there are two forces, so I multiply it to get 40uT. But I know that it is the case, but not WHY so. I am still having the problem with the F12 = -F21 problem

For (b):

How did you arrive at this result?

I assume that d = 10cm.
Yup...I used Biot-Savart's Law's result of the B-field for wire

Find B due to each wire, including direction, or at least a sign. Add them together.

For (c):

Ditto

Also, I would expect this answer to be less than the answer to (b).

.
I'll try...

For (b) I drew

http://img864.imageshack.us/i/bfield.png/ [Broken]

Now numerically, I got it right, but I don't have the intuition pinned down.

Anyways I did

$$\vec{B} = \frac{\mu_0 I_{left}}{2\pi d}} - \frac{\mu_0 I_{right}}{4\pi d}$$

$$\vec{B} = 5.0 \cross 10^-6 T$$

The book say that the direction is out of the page. But from my drawing of the direction how do you know?

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SammyS
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For (b) I drew

http://img864.imageshack.us/i/bfield.png/ [Broken]

Now numerically, I got it right, but I don't have the intuition pinned down.

Anyways I did

$$\vec{B} = \frac{\mu_0 I_{left}}{2\pi d}} - \frac{\mu_0 I_{right}}{4\pi d}$$
Usually, out of the page is a positive direction. At point P1:
Ileft produces a B field into the page (negative).
Iright produces a B field out of the page (positive).​
So I would write: $$\vec{B} = \frac{\mu_0 I_{\text{right}}}{2\pi d}} - \frac{\mu_0 I_{\text{left}}}{4\pi d}$$

$$\vec{B} = 5.0 \cross 10^-6 T$$

The book say that the direction is out of the page. But from my drawing of the direction how do you know?

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No but from my diagram it isn't clear for the directions. Like between them they are both into the page and outside on both wires it is out of the page

SammyS
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The circle dot, $$\odot\,,$$ indicates that a vector points out of the page (up, for a book on a horizontal surface).

The circle X, $$\otimes\,,$$ indicates that a vector points into the page (down, for a book on a horizontal surface).

So, my question is: How did you put these on your figure without knowing what signs, or directions), they indicate?

No I know what those symbol means, but the picture doesn't tell me what the "net direction" is

gneill
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No I know what those symbol means, but the picture doesn't tell me what the "net direction" is
Did you look at the web page that I gave a link for in my previous post? The right hand rule allows you to determine the direction that the field surrounding a current-carrying wire circulates.

Did you look at the web page that I gave a link for in my previous post? The right hand rule allows you to determine the direction that the field surrounding a current-carrying wire circulates.
I didn't see it, I clicked it on now, but it won't load.

Also, I do know how to find the B-field direction with the right hand rule, but my question is about having two fields. How do I find the superposition based on just the diagram I have given

gneill
Mentor

Also, I do know how to find the B-field direction with the right hand rule, but my question is about having two fields. How do I find the superposition based on just the diagram I have given
It must be a problem with your browser or some network router configuration between you and the website. Ah well. Since you know how to use the right hand rule for magnetic field direction determination....

Find the direction of each field separately for a given location. You should find that they end up both in the same direction or in opposite directions. Then determine the magnitudes of both fields for the same location. Add accordingly. When the fields are in opposite directions, the biggest one "wins" for the direction of the result, and the magnitude is the difference between the field strengths.

It must be a problem with your browser or some network router configuration between you and the website. Ah well. Since you know how to use the right hand rule for magnetic field direction determination....

Find the direction of each field separately for a given location. You should find that they end up both in the same direction or in opposite directions. Then determine the magnitudes of both fields for the same location. Add accordingly. When the fields are in opposite directions, the biggest one "wins" for the direction of the result, and the magnitude is the difference between the field strengths.
I am hijacking my own thread, but....

Wat about that one? I find it almost impossible to determine the real B-field for my cross product

SammyS
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How does mentioning another thread, a thread which currently ends with you not understanding a somewhat different problem --- how does mentioning that help with this problem?

SammyS
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