Magnetic Field & 2 wires with same current

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
sami23
Messages
69
Reaction score
1

Homework Statement


Two long, straight, parallel wires, 10.0 cm apart carry equal 3.78A currents in the same direction. Find the magnitude and direction of the magnetic field at point 15.4 cm above one wire.


Homework Equations


B= ([tex]\mu[/tex]0I)/(2[tex]\pi[/tex]r)
[tex]\mu[/tex]0 = 4[tex]\pi[/tex] * 10-7
Use the right hand rule to find the direction of magnetic field

The Attempt at a Solution


(4[tex]\pi[/tex]*10-7)*3.78 / (2[tex]\pi[/tex]0.154) * cos([tex]\theta[/tex])*2
I'm not sure on how to go about solving this problem
 

Attachments

  • c10.jpg
    c10.jpg
    1.2 KB · Views: 516
Physics news on Phys.org
That's the correct formula for finding the flux density B at a point a distance r from a long straight wire carrying a current I
A diagram helps a lot here.
The two wires are L and R and we are looking along them, the current flowing into the page/screen.
The point we are interested in is at P.
The distances are shown.
Can you calculate the distance LP which you need to find the field due to L?
Mag-BField2.png

The magnitude and direction of the field at P is found by adding the two vectors for the forces due to L and R. These are shown as BL and BR.
Do you know how to add vectors?


Remember, the magnetic field around a wire is a circle with the wire at the centre.
If the current is into the page the field is clockwise around the wire.
 
LP = [tex]\sqrt{0.154^2+0.1^2}[/tex] = 0.1836 m

B = (4[tex]\pi[/tex]*10-7)(3.78) / (2[tex]\pi[/tex]*0.1836) = 4.12 T
 
You have calculated BL
The answer should be 4.12μT not 4.12T
You also need to calculate BR
(It's the same formula but with r=0.154m)

When you have both BL and BR you need to add the vectors. Do you know how to do that?
 
Yes, Thanks so much.