Force of one wire due to the other

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The discussion focuses on calculating the magnetic field and force between two parallel wires carrying current. The magnetic field B produced by each wire at a distance of 1.0 cm is calculated to be 2.0 x 10^-5 T. The relationship between the force on a wire and the magnetic field is established using the equation F = BIL, where L represents the length of the wire. Participants clarify that the force is being calculated per unit length, resulting in F/L = 2 x 10^-5 N/m. The conversation emphasizes the importance of understanding both the magnetic field generation and the subsequent force calculation.
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Homework Statement



3.jpg


D=1.0cm=0.01m
I=1.0A


Homework Equations


B=(μ0I)/(2∏r)

The Attempt at a Solution



B1=(4∏x10^-7 * 1.0A)/(2∏ * 0.01m)= 2.0x10^-5

B2=(4∏x10^-7 * 1.0A)/(2∏ * 0.01m)= 2.0x10^-5

Not sure what to do with these two numbers.
 
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How does the force on a wire relate to the current through the wire and the B field?
 
I would use the Right Hand Rule to find the relation. Since the current is going upward (thumb), the magnetic field B would be going into the page (fingers) and the force would be pointing to the left (direction the palm faces).
 
The right hand rule will give the direction of the force but not the magnitude. You have found B, the magnetic field produced by a wire carying 1 amp at a distance of 10 cm. Now How do you find the magnitude of the force on the wire that is in this field?
 
Would I use F=|q0|vBsinθ ?
 
Wait, nevermind. That would be for a moving particle.
 
F = BIL, remember this?
 
F=ILBsinθ would be the equation, I think.
 
Oh, saw your post after I posted. Oops. What would be L? I thought that L was the length of the wire, but that was not a given variable.
 
  • #10
If you have studied Vectors i would suggest you use the more general equation force that is i*L(cross)B
 
  • #11
F=ILBsinθ is technically correct and is used if the wire is not perpendicular to the B field which it is in this case. s far as what is L, look it up in your text. It will do you some good :-) Remember the force requested is per unit length.
 
  • #12
Oh ok, so...

F/L=(1.0A)(2x10^-5)

and sin90=1

So F/L=2*10^-5
 
  • #13
would the units be Gauss/meter?
 
  • #14
What is the unit of force?
 
  • #15
oops, N/m!
 
  • #16
Remember that this is a two step process. Step 1 is to determine the B
field in the vicinity of the wire, and step 2 is to determine the force on the wire due to the B field generated by the other wire..
 

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