Current in a wire?

  • Thread starter fball558
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  • #1
fball558
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current in a wire??

Homework Statement



A long, straight wire carries a current of I_1 = 20 A as shown below. A rectangular coil with two sides parallel to the straight wire has sides 5 cm and 10 cm with the near side a distance 2 cm from the wire. The coil carries a current of I_2 = 4 A.

PIC IS ATTACHED!

Find the force on each segment of the rectangular coil due to the current in the long, straight wire.

i found all but right and net (if i find right i can get net)

top 2.0044e-5 N
left 8e-5 N
right
bottom 2.0044e-5 N

net

i used

((I_2 * Uo* I_1 ) / 2pi ) * ln(7/2)


i know that the force will be pointing left but dont know its magnitude?
does anyone know how this works and if so can you help me through the problem?
thanks!!
 

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Answers and Replies

  • #2
Troels
125
2


Well the forces on the top and bottom are of equal magnitude, but opposite direction, so they cancel

Along the left and right segments, the magnetic field is constant, so the force is simply the Magnitude of the magnetic field at the location of the segment multiplied by the current in the segment, and it has a direction given by the right hand rule (ie pointing to the right on the left segment, and to the left on the right segment)
 
  • #3
fball558
147
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Well the forces on the top and bottom are of equal magnitude, but opposite direction, so they cancel

Along the left and right segments, the magnetic field is constant, so the force is simply the Magnitude of the magnetic field at the location of the segment multiplied by the current in the segment, and it has a direction given by the right hand rule (ie pointing to the right on the left segment, and to the left on the right segment)



how would you find the magnetic field on the right segment?
i know the current on the right side should be the 4 A
and i know the B field should drop compared to the left because the right side is farther away than the left side.

only way i could think of to get the B field would be
B = (Uo * I)/(2pi * r) but im not sure if that is right
 
  • #4
Troels
125
2


only way i could think of to get the B field would be
B = (Uo * I)/(2pi * r) but im not sure if that is right

Seeing as

[tex]B(r)=\frac{\mu_0I}{2\pi r}[/tex]

is the field at a distance r from an infinitely long wire carrying a current I, that would be correct :)
 
  • #5
bsodmike
82
0


F (right) = 2.3e-5 N :)

Fball, can you elaborate how you came to use this for top/bottom please,
((I_2 * Uo* I_1 ) / 2pi ) * ln(7/2)

Thanks,
Mike.
 
  • #6
fball558
147
0


um... my TA helped me on this one. you have to integrate Uo*I_1 / 2pi*r
he said Uo*I_1 / 2pi*r = B(r)*I_2*dr
you integrate this with respect to r and your bounds are 2 to 7 (left and right side of the box from the long wire)
when you integrate this you get the equation that i mentioned before
 
  • #7
fball558
147
0


thanks everyone i forgot to times my answer by the length of the side of the wire (.1m)
i got the answer now.
thanks again!!
 
  • #8
bsodmike
82
0


Your values for the top and bottom will be off for the following reason. First of all, I detest working in cm as it eventually ends up throwing off my units. Integrating B along the top segment,

[itex]B_{\text{top}}=\dfrac{\mu_{0}I_1}{2\pi}\int_{0.02}^{0.07}\dfrac{1}{r}\;dr=\dfrac{\mu_{0}I_1}{2\pi}\left[ln(r)\right]_{0.02}^{0.07}=\dfrac{\mu_{0}I_1}{2\pi}ln\left(\dfrac{0.07}{0.02}\right)[/itex]

and for the Force,

[itex]F_{\text{top}}=IL\times B = I\left|L\right|\left|B\right|\sin(\pi)=ILB = I_2 \cdot (0.05) \cdot \left(\dfrac{\mu_{0}I_1}{2\pi}ln\left(\dfrac{0.07}{0.02}\right)\right)=0.05\left(\dfrac{\mu_{0}I_1 I_2}{2\pi}ln\left(\dfrac{0.07}{0.02}\right)\right)[/itex]

Have you not forgotten to include L = 0.05 m as well?
 
  • #9
fball558
147
0


i dont work in cm i just write it out that way to show where im getting my numbers from. i always try to convert my numbers to the basic S.I. units (if i remember haha)
i got all the answers but will keep those equations and study them with more examples because there will probably be something like this on the test :(
thanks again everyone
 
  • #10
bsodmike
82
0


i dont work in cm i just write it out that way to show where im getting my numbers from. i always try to convert my numbers to the basic S.I. units (if i remember haha)
i got all the answers but will keep those equations and study them with more examples because there will probably be something like this on the test :(
thanks again everyone

You need to include L = 0.05 m into your F_top and F_bottom values. Integrating B along the length of the wire does not account for it [the length of the wire that is] in F :) I'm pretty certain 'top 2.0044e-5 N' is therefore incorrect.

F_top = 1.0022e-6 N

This answer is essentially 0.05 m * 2.0044e-5 N
 
Last edited:
  • #11
Deadsion
12
0


what equation did you use to find left and right?
 
  • #12
Deadsion
12
0


never mind, im a little special
 
  • #13
bsodmike
82
0


what equation did you use to find left and right?

I used the equation, as stated previously, for B - the magnetic flux density (yes, I am an engineer :p). Physicists call it the 'Magnetic Field', and it has SI units 'Tesla' or Weber per square metre.

For the left,
[itex]B_{\text{left}}=\dfrac{\mu_0 I_1}{2\pi\left(0.02\right)}[/itex]

For the right,
[itex]B_{\text{right}}=\dfrac{\mu_0 I_1}{2\pi\left(0.07\right)}[/itex]

To get the forces, simply substitute B into,
[itex]F=I_2\cdot (0.1) \cdot B[/itex]

Both sides are 10 cm or 0.1 m tall, and remember only 4 A is flowing inside the loop.
 

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