Find the induction of the magnetic field at the point O

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SUMMARY

The discussion focuses on calculating the magnetic field induction at point O, the center of a square formed by three segments of a wire carrying a current of 21A. The participants clarify that the equation B=I*μ0/(2*pi*d) is inappropriate for this scenario, as it stems from Ampere's law, which applies only under symmetrical conditions. Instead, they recommend using the Biot-Savart law to accurately integrate the magnetic field contributions from the finite wire segments. The importance of correctly visualizing the current's direction and the resulting magnetic field orientation is emphasized.

PREREQUISITES
  • Understanding of Biot-Savart law for magnetic field calculations
  • Familiarity with Ampere's law and its limitations
  • Knowledge of magnetic field direction conventions (right-hand rule)
  • Basic principles of electromagnetism and current-carrying conductors
NEXT STEPS
  • Study the application of the Biot-Savart law for finite wire segments
  • Learn about the right-hand rule for determining magnetic field direction
  • Research the differences between Ampere's law and Biot-Savart law
  • Explore video tutorials on magnetic fields from current-carrying wires on platforms like YouTube
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone involved in electrical engineering or related fields seeking to deepen their understanding of magnetic fields generated by currents in wires.

zade70
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Homework Statement


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.

Homework Equations


B=I*μ0*/2*pi*d

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?
 
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##\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
 
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.
 
BvU said:
##\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
I don't know how to show that on a plan. I just tried to do it somehow
 
Hesch said:
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.
What does ∫circulationH⋅ds = N * I mean?
 
zade70 said:
I don't know how to show that on a plan. I just tried to do it somehow
upload_2016-3-24_14-43-42.png
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, into the paper.
This comes from the middle ages, when arrows had sharp points on the harmful end and guide vanes on the other.
Hesch said:
Some piece of conductor is missing, which will add B-field to the point O.
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. [edit] Sorry, wrong O o:) ! So these contributions do have to be drawn and taken into account !
 
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.
 
Alteratively, would this figure be a hint ?

upload_2016-3-24_15-5-23.png
 
Go to Youtube.com

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
Hesch said:
Go to Youtube.com

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.
Does it result 6.3*10^-5 T using this method (because we haven't learned L yet)
 
  • #11
BvU said:
Alteratively, would this figure be a hint ?

View attachment 97878
Could you explain something more?
 
  • #12
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:
  • #13
How
BvU said:
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.
How about now?
http://postimg.org/image/stsjq1a5z/
 
  • #14
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
BvU said:
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 .
Then how would the sketch be with the wires and inductivity?
 
  • #16
For you to propose something; I don't want to rob you of the opportunity to practice what you have learned by doing the exercise for you :smile: -- it's also against PF culture (and even against the rules ! -- but like Gibbs: We figured they were more actual *guidelines* ).
 

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