Magnetic Field at the center of a square

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Homework Help Overview

The discussion revolves around calculating the magnetic field at the center of a square configuration of wires, involving concepts from electromagnetism, specifically the Biot-Savart law and Ampere's law.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of Biot-Savart law and Ampere's law for calculating the magnetic field. There are questions about the correct application of these laws, particularly regarding the integration path and the radius used in calculations.

Discussion Status

Some participants have provided calculations using Ampere's law, while others question the correctness of the approach and the assumptions made regarding the radius and integration path. There is an ongoing examination of the individual contributions of each wire to the total magnetic field.

Contextual Notes

There is confusion regarding the radius used in calculations, with some participants noting discrepancies in previous posts. The discussion also highlights the need for clarity on the integration path when applying Ampere's law.

aChordate
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Homework Statement


photo[1].jpg



Homework Equations



not sure

The Attempt at a Solution



I am not sure how to start this problem.
 
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I would start with Biot-Savart.
 
You've probably seen how to obtain the field of a single line, whether through the Biot-Savart law or, more simply, through Ampere's law. Then just remember that the fields of individual components of a configuration add-up...
 
So, would I use Ampere's Law for static magnetic fields?

ƩB||Δl=μ0I ?
 
aChordate said:
So, would I use Ampere's Law for static magnetic fields?

ƩB||Δl=μ0I ?

Yes, since the wires are "long", you can use Ampere's law individually for each wire and then add the results vectorially.
 
Ok, so I used ƩB||Δl=μ0I

and got:

ƩB||=3.8x10^-5
ƩB||=4.3x10^-5
ƩB||=5.0x10^-5
ƩB||=3.0x10^-5

For x,
(3.8x10^-5)+(-5.0x10^-5)=-1.2x10^-5

For y,
(-4.3x10^-5)+(-3.0x10^-5)=-7.3x10^-5

I have a feeling I didn't do this correctly...
 
aChordate said:
Ok, so I used ƩB||Δl=μ0I

and got:

ƩB||=3.8x10^-5
ƩB||=4.3x10^-5
ƩB||=5.0x10^-5
ƩB||=3.0x10^-5

For x,
(3.8x10^-5)+(-5.0x10^-5)=-1.2x10^-5

For y,
(-4.3x10^-5)+(-3.0x10^-5)=-7.3x10^-5

I have a feeling I didn't do this correctly...

Let's take one wire at a time, and show all your work.
For the wire carrying the 1.2A, write ampere's law and determine B at the center of the square.
 
Ok, so I used ƩB||Δl=μ0I

and got:

ƩB||=[(4∏*10^-7T*m/A)(1.5A)]/0.05m
ƩB||=3.8x10^-5

ƩB||=[(4∏*10^-7T*m/A)(1.7A)]/0.05m
ƩB||=4.3x10^-5

ƩB||=[(4∏*10^-7T*m/A)(2.0A)]/0.05m
ƩB||=5.0x10^-5

ƩB||=[(4∏*10^-7T*m/A)(1.2A)]/0.05m
ƩB||=3.0x10^-5
 
Ampere's Law:
ƩBΔl=μ0I

ƩB(0.025m)=(4∏*10^-7)(1.5)

ƩB=7.54x10^-5
 
  • #10
aChordate said:
Ampere's Law:
ƩBΔl=μ0I

ƩB(0.025m)=(4∏*10^-7)(1.5)

ƩB=7.54x10^-5

Ampere's law refers to the integration of a path around the current-carrying wire, i.e in this case you want the circumference, not the radius.
 
  • #11
ƩB(0.050m)=(4∏*10^-7)(1.5A)
ƩB=3.77x10^-5

ƩB(0.050m)=(4∏*10^-7)(1.7A)
ƩB=4.27x10^-5

ƩB(0.050m)=(4∏*10^-7)(2.0A)
ƩB=5.03x10^-5

ƩB(0.050m)=(4∏*10^-7)(1.2A)
ƩB=3.02x10^-5
 
  • #12
Do I add these all together?
 
  • #13
aChordate said:
Do I add these all together?

They were wrong before & are still wrong.
In your post #9 you seemed to realize that the radius is 0.025m, not 0.05m. Now you're back to 0.05m?

And you still haven't understood that the path of integration is not along the radius but the circumference.
 

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