Magnetic Field at the center of a square

AI Thread Summary
The discussion focuses on calculating the magnetic field at the center of a square formed by current-carrying wires using Ampere's Law and the Biot-Savart Law. Participants emphasize the importance of correctly applying Ampere's Law, particularly noting that the integration path should be around the circumference rather than along the radius. Several calculations are presented, but there are concerns about the accuracy of the radius used in the calculations, with a correct radius of 0.025m being highlighted. The need for a clear step-by-step approach is stressed, as well as the importance of vectorially adding the magnetic fields from individual wires. Overall, the conversation aims to clarify the proper methodology for solving the problem.
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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