# Find the induction of the magnetic field at the point O

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1. Mar 23, 2016

1. The problem statement, all variables and given/known data
In a wire passes the current I=21A. A part of the wire is bent to form three consecutive sides of a square with sides 40 cm. Find the induction of the magnetic field in the point O which is the center of the square.
2. Relevant equations
B=I*μ0*/2*pi*d

3. The attempt at a solution
I want to know if my sketch is correct. I think I know the rule of the Right Hand but I don't know to draw it correctly in a plan. Here's what I did.
http://postimg.org/image/ynx280cdb/
Can you correct it?

2. Mar 24, 2016

### BvU

$\vec B$ is perpendicular to $\vec i$. So if $\vec i$ is in the plane of the paper, $\vec B$ is into or out of the paper

3. Mar 24, 2016

### Hesch

I don't know if your sketch is correct ( 3 or 4 sides of the square ) because how do you get 21A through 3 sides of the square? Some piece of conductor is missing, which will add B-field to the point O.

The equation used ( B=I*μ0*/2*pi*d ) is not correct because it comes from Ampere's law:

circulationH⋅ds = N * I

which states that the mean value of H = N * I / (2πr) along a circulation path. You may use this law under absolutely symmetrical conditions, such as an infinite long straight wire. You have made a circulation path around 40cm wire, but say the length of wire were 1cm: Would the H-field be the same?

You must use the Biot-Savart law instead, and integrate along the 40cm wire.

4. Mar 24, 2016

I don't know how to show that on a plan. I just tried to do it somehow

5. Mar 24, 2016

What does circulationH⋅ds = N * I mean?

6. Mar 24, 2016

### BvU

A circle with a dot in the middle represents an arrow coming towards you, out of the paper.
A circle with a cross in the middle represents an arrow going away from you, in to the paper.
This comes from the middle ages, when arrows had sharp points on the harmful end and guide vanes on the other.
Not necesarily: if the current comes from the left and leaves to the right on a line through O, that current does not contribute to the magnetic field.  Sorry, wrong O !! So these contributions do have to be drawn and taken into account !

7. Mar 24, 2016

### Staff: Mentor

Often these types of problems are more pedagogical than practical. I think it's safe to assume that the figure is comprised of three finite current carrying wire segments and disregard where the current comes from or goes to.

I suggest doing a web search on the magnetic field due to a finite current carrying wire segment. Of particular interest might be the field along the perpendicular bisector of such a wire segment.

8. Mar 24, 2016

### BvU

Alteratively, would this figure be a hint ?

9. Mar 24, 2016

### Hesch

Search for "Ampere's law" and you will find a lot of videos.

In these videos they speak of

c B⋅dL = N * I * μ0 , ( N = number of wires through the loop )

but since this formulation is only valid in vacuum, I prefer the more general formulation:

B = μ * H →

c H⋅dL = N * I

B is the magnetic induction [Tesla], H is the magnetic field strength [ A/m ].

Please note that they speak of Ampere's law to be used under symmetrical conditions.

10. Apr 12, 2016

Does it result 6.3*10^-5 T using this method (because we haven't learnt L yet)

11. Apr 12, 2016

12. Apr 12, 2016

### BvU

If you can find the field at O for the two parts on the right, all you have to do is add them up to get the field from the one part on the left.

And be sure to follow Gneill's advice: your relevant equation is for an infinitely long wire, so Ok for the lower part on the right. But not for the upper part.

Last edited: Apr 12, 2016
13. Apr 12, 2016

How
http://postimg.org/image/stsjq1a5z/

14. Apr 12, 2016

### BvU

Doesn't make sense.

• You removed the left and right wires that carry in the current and carry it off again.

• The direction of the B field from a current is perpendicular to the direction of the current. All current lines lie in the paper, so all B lines are perpendicular to the paper. See here .

15. Apr 12, 2016