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Determining the force on a loop cause by an infinite line

  • #1
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Homework Statement


An infinitely long line of current $I_1=6[A]$ is following along the positive z-axis in the direction of +$\hat{a_z}$. Another current is following a triangular loop counter clockwise from the points A(0,2,2), B(0,6,2) and C(0,6,6).

Homework Equations


To start I applied Ampere's Law since an amperian path could be applied to current $I_1$ which results in
$$\vec{H}=\frac{I_1}{2\pi\rho} \hat{a_{\phi}}$$

Applying the force on two conductors formula $$F=\int{I_2 \vec{dl_2}} \times{\vec{B_1}}$$

The Attempt at a Solution


Where I am stuck is at segment AB, I have determined that the following for this segment $$dl=dy\,\hat{a_y}$$

The problem that I am having is determining the direction of the magnetic field vector, is it it directed in the positive $ \hat{a_z} $ because of how the problem is described or is it another direction all together? The teacher provided a solution but did not outline why they did certain steps.
 

Answers and Replies

  • #2
TSny
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Welcome to PF!
The problem that I am having is determining the direction of the magnetic field vector, is it it directed in the positive ##\hat{a_z} ## because of how the problem is described or is it another direction all together?
You know ##\vec{H}=\frac{I_1}{2\pi\rho} \hat{a_{\phi}}##. Thus, ##\hat{a_{\phi}}## gives you the direction.

Are you familiar with the magnetic field pattern of a long straight current? (Right hand rule)
 
  • #3
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Welcome to PF!

You know ##\vec{H}=\frac{I_1}{2\pi\rho} \hat{a_{\phi}}##. Thus, ##\hat{a_{\phi}}## gives you the direction.

Are you familiar with the magnetic field pattern of a long straight current? (Right hand rule)
Do you mean that since the current is moving in the positive [itex] \hat{a_x} [/itex] that I apply the right hand rule? If so then the result is that the force is directed in the negative [itex] \hat{a_x} [/itex] direction. Therefore, [tex] \vec{B}=\frac{-{\mu}_0\,I_1}{2\pi\,y} \hat{a_x}[/tex]

At this point I integrate the and solve for segments AB and BC which results in $$\vec{F_{1,AB}}= 6{\mu}_0\,ln(3) \,\hat{a_z}$$ $$\vec{F_{1,BC}}= -4\mu \,\hat{a_y}$$

Up to this point I understand and did correctly according to the given solution, however, the final integral confounds me...After the cross product is applied and simplified my result is this $$6{\mu}_0[\int\frac{-dz}{y}\,\hat{a_y} + \int\frac{dy}{y}\,\hat{a_z} ]$$

I believe the second integration should result in [itex] 6{\mu}_0\,ln(1/3)\,\hat{a_z} [/itex] but I dont know about the about the first integral since there is the is "y" term in the denominator. Do I make that y=6 because that is where the integration started at or y=2?
 
  • #4
TSny
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Do you mean that since the current is moving in the positive [itex] \hat{a_x} [/itex] that I apply the right hand rule? If so then the result is that the force is directed in the negative [itex] \hat{a_x} [/itex] direction.
Typo? The current is in the direction of [itex] \hat{a_z} [/itex]. The field is in the negative x direction.

Therefore, [tex] \vec{B}=\frac{-{\mu}_0\,I_1}{2\pi\,y} \hat{a_x}[/tex]
OK

At this point I integrate the and solve for segments AB and BC which results in $$\vec{F_{1,AB}}= 6{\mu}_0\,ln(3) \,\hat{a_z}$$ $$\vec{F_{1,BC}}= -4\mu \,\hat{a_y}$$
I can't check the above, since ##I_2## was not specified.

Up to this point I understand and did correctly according to the given solution, however, the final integral confounds me...After the cross product is applied and simplified my result is this $$6{\mu}_0[\int\frac{-dz}{y}\,\hat{a_y} + \int\frac{dy}{y}\,\hat{a_z} ]$$

I believe the second integration should result in [itex] 6{\mu}_0\,ln(1/3)\,\hat{a_z} [/itex] but I dont know about the about the first integral since there is the is "y" term in the denominator. Do I make that y=6 because that is where the integration started at or y=2?
For the first integral you'll need to relate ##y## and ##z## along the path CA.
 
  • #5
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Typo? The current is in the direction of [itex] \hat{a_z} [/itex]. The field is in the negative x direction.


OK


I can't check the above, since ##I_2## was not specified.


For the first integral you'll need to relate ##y## and ##z## along the path CA.

Ok I will see what I can do, sorry I forgot but [itex] I_2 = 2 [A] and I_1=6\pi [A][/itex]. I am still trying to get a handle on latex
 
  • #6
TSny
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[itex] I_2 = 2 [A] and I_1=6\pi [A][/itex].
OK. Your results so far look good.
 

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