Magnetic Field Surrounding a Straight Conductor

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SUMMARY

The discussion focuses on the magnetic field surrounding a straight conductor, specifically addressing the relationship between the cross product of the differential length element, ##\vec{ds}##, and the unit vector, ##\hat{r}##. Participants clarify the derivation of the sine function related to the angle ##\theta## in the context of the cross product. The angle ##\theta_2## is defined as the angle between ##\vec{ds}## and ##\hat{r}##, leading to the equation dxsin(θ2) = dxsin(π/2 - θ). This establishes a clear geometric relationship essential for understanding the magnetic field's calculation.

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Homework Statement
Please see below
Relevant Equations
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For this problem,
1673658720024.png

Part of the solution is,
1673658743775.png

1673658803501.png

However, would someone please tell me where they got the sine function circled in red from?

Many thanks!
 

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Callumnc1 said:
However, would someone please tell me where they got the sine function circled in red from?
You're working with the cross product ##\vec{ds} \times \hat r##. How is the angle between ##\vec{ds}## and ##\hat r## related to the angle ##\theta## shown in diagram ##a##?
 
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TSny said:
You're working with the cross product ##\vec{ds} \times \hat r##. How is the angle between ##\vec{ds}## and ##\hat r## related to the angle ##\theta## shown in diagram ##a##?
Thank you @TSny ! I see how they got theta now. :) If we call the angle between ds and r hat as theta 2, then from the definition of the cross product:

dxsin(theta 2) = dxsin(pi/2 - theta) where I found theta 2 using angles in a triangle add up to 180 degrees.
 
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