Magnetic Field at the center of a square

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  • #1
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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.
 

Answers and Replies

  • #3
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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...
 
  • #4
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So, would I use Ampere's Law for static magnetic fields?

ƩB||Δl=μ0I ?
 
  • #5
rude man
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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.
 
  • #6
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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...
 
  • #7
rude man
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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.
 
  • #8
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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
 
  • #9
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Ampere's Law:
ƩBΔl=μ0I

ƩB(0.025m)=(4∏*10^-7)(1.5)

ƩB=7.54x10^-5
 
  • #10
rude man
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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
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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
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Do I add these all together?
 
  • #13
rude man
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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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