Parallel wires carrying current out of the page, find B

Click For Summary
SUMMARY

The discussion focuses on calculating the magnetic field B along the y-axis due to two parallel wires carrying current I out of the plane. The magnetic field is derived using the equation B = μI/(2πr), where r is the distance from the wire to the point on the y-axis, defined as r = √(b² + y²). The total magnetic field is expressed as B_total = 2Bcos(θ), with cos(θ) determined as b/√(b² + y²). The confusion arises from the answer key stating cos(θ) = y/(b² + y²), which is clarified as relating to the angle between the magnetic field and the horizontal axis.

PREREQUISITES
  • Understanding of Ampère's Law and magnetic fields
  • Familiarity with trigonometric functions in physics
  • Knowledge of vector components in magnetic field calculations
  • Basic calculus for integrating magnetic field equations
NEXT STEPS
  • Review Ampère's Law and its applications in magnetic field calculations
  • Study vector decomposition in physics to clarify component analysis
  • Explore the Biot-Savart Law for magnetic field calculations from current-carrying wires
  • Practice problems involving magnetic fields from multiple sources of current
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding magnetic fields generated by current-carrying conductors.

Dextrine
Messages
102
Reaction score
7

Homework Statement


Two parallel wires are a distance 2 b apart from the origin and carry current I out of the plane of the paper.
Find the magnitude and direction of the magnetic field B(y) along the y axis.

Homework Equations


∫B⋅dl=μI[/B]

The Attempt at a Solution


If we take ∫dl = 2πr
B=μI/(2πr)
r=√(b^2+y^2)

Clearly the y components cancel so we are left with only the horizontal components, that is

B_total=2Bcos(θ)

cos(θ)=b/√(b^2+y^2)

However, in the answer key provided, it says that cos(θ)=y/(b^2+y^2)

I don't understand why this would give this horizontal component. It seems like this would give the vertical component.
 
Physics news on Phys.org
Nevermind, I see how. The angle made by the B field from each wire and the horizontal is not the same as the angle between r and b
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
4K
Replies
3
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 14 ·
Replies
14
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
2
Views
3K