How Does Biot-Savart Law Calculate Magnetic Field at Point P?

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SUMMARY

The discussion focuses on calculating the net magnetic field at point P due to two long straight wires carrying a current of 3.19 A using Biot-Savart's Law. The formula applied is { B }_{ long,\quad straight\quad wire } = \frac { { \mu }_{ 0 }I }{ 2\pi r }, where r is determined as r = \sqrt { { (\frac { { d }_{ 1 } }{ 2 } ) }^{ 2 }\quad +\quad { { d }_{ 2 } }^{ 2 } }. The total magnetic field is calculated as { B }_{ total } = \frac { { \mu }_{ 0 }I }{ \pi r }, resulting in a magnitude of 1.7695e-07 T. The discussion emphasizes the importance of considering the direction of the magnetic fields produced by each wire, which are not aligned in the same direction at point P.

PREREQUISITES
  • Understanding of Biot-Savart's Law
  • Knowledge of magnetic field calculations
  • Familiarity with vector components in physics
  • Basic concepts of current and magnetic fields
NEXT STEPS
  • Study the application of Biot-Savart's Law in different configurations of current-carrying wires
  • Learn about the superposition principle for magnetic fields
  • Explore vector decomposition in physics for analyzing magnetic field directions
  • Investigate the effects of varying current magnitudes on magnetic field strength
USEFUL FOR

Physics students, educators, and anyone interested in electromagnetism and magnetic field calculations will benefit from this discussion.

Sho Kano
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Homework Statement


The figure shows two very long straight wires (in cross section) that each carry a current of 3.19 A directly out of the page. Distance d1 = 6.00 m and distance d2 = 4.00 m. What is the magnitude of the net magnetic field at point P, which lies on a perpendicular bisector to the wires?
HW9Q7.png

Homework Equations


Biot-Savart's Law

The Attempt at a Solution


{ B }_{ long,\quad straight\quad wire }\quad =\quad \frac { { \mu }_{ 0 }I }{ 2\pi r } \\ r\quad =\quad \sqrt { { (\frac { { d }_{ 1 } }{ 2 } ) }^{ 2 }\quad +\quad { { d }_{ 2 } }^{ 2 } } \\ { B }_{ total }\quad =\quad 2B\quad =\quad \frac { { \mu }_{ 0 }I }{ \pi r } \\ { B }_{ total }\quad =\quad 1.7695e-07

I'm aware that I'm missing a sin(theta) in my multiplication, but why would I need to only specify a vertical component? The magnitude isn't only the vertical component.
 
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What is the direction of B for each current?
 
TSny said:
What is the direction of B for each current?
Both act upwards on the point
 
The two fields at P are not in the same direction.
 
TSny said:
The two fields at P are not in the same direction.
Is this what you're visualizing:
Upper field points in the positive x and y direction
Lower field points in the negative x but positive y direction
?
 
Yes.
 
TSny said:
Yes.
Alright that makes sense now to put the y component in. I was visualizing both fields pointing up at the same time, but that's not possible at all with the orientation of the field.
 
OK.
 

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