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Magnetic field and current tendency

  • Thread starter ayalala
  • Start date
  • #1
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1. Two-wire, long, straight parallel 10cm distant from each other, as shown in the chart.
Upper wire current issue I1 = 6A trend into the top level.
A. What are the strength of the current tendency I2 bottom wire, if the magnetic field equivalent at P equal Zero?
b. What is the equivalent magnetic field point? Q
C. What is the equivalent magnetic field point S?








2. not sure what to use


3. a) B[tot]=B1-B2=0

B1= μ0I1/ 2π*0.15
B2= μ0I2/ 2π*0.05

μ0I1/ 2π*0.15 = μ0 I2/ 2π*0.05
6/0.15= I2/0.05
0.3=0.15 I2
I2=2A

b) ??

c)B=√B1^2+B2^2 (?)

Thank you very much for your help.
 

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  • #2
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(b) Just do the same setup as you did for (a) except swap the distances for the two current carrying wires.

(c) Not sure what you are doing, but it is best if you draw the magnetic field lines from each wire at point S. That way you will know how to add the vectors.
 
  • #3
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Thank you.

So basically for B it should be the same answer?

As for c:

If I understand correctly I need to "break" te view of each wire at point S and then:

B1(x access)=Bcosθ
B1(y access)=Bcosθ

then find vector for B1

same for B2

and to add them both?
 
  • #4
674
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(b) Won't be the same answer, but you would solve it in a similar fashion as (a).

(c) You are right that you need to break the magnetic field up into its components for B1 and do the same for B2. Be careful with the trigonometry. I am not sure what angle you choose, but make sure you use the correct sin and cos.
 
  • #5
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Solved

Thx.
 

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