Help finding the magnetic field of a current loop

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Homework Help Overview

The discussion revolves around finding the magnetic field generated by a current loop, specifically addressing the calculations involving cross products and angles between vectors. Participants are examining the relationship between the vectors involved and how to correctly interpret the angles in the context of the problem.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the correct angle between the vectors involved in the cross product, questioning assumptions about the geometry of the situation. There is a focus on understanding the implications of the angle being 90° and how it affects the resulting magnetic field calculation.

Discussion Status

The discussion is active, with participants providing clarifications about the angles and components involved in the magnetic field calculation. Some guidance has been offered regarding the interpretation of the cross product and its implications for the z-axis component of the field.

Contextual Notes

There is mention of a specific equation that the original poster is trying to reconcile with a textbook answer, indicating potential gaps in understanding or application of the concepts. The original poster also expresses uncertainty about the apex half angle and its relevance to their calculations.

Boltzman Oscillation
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Homework Statement
Find the magnetic field at a point due to a loop of wire.
Relevant Equations
$$ db = \frac{\vec{I}\vec{dl} \times \vec{r}}{cr^3}$$
Hi all, my work is shown on the attached image. The boxed equation is what I get to but I do not understand how to go from there to what the book has. I am guessing that the problem arises when trying to solve the cross product. I understand that I will need to find the value of the sine of the angle between the arguments of the cross product but I do not see how I could do that, all I know is that the direction will be in the z axis. Any help is appreciated as always.
16369316997682310877457715037074.jpg
 
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Theta is not the angle between the vectors ##\vec{d l}, \vec r##. What is the angle between them?
 
haruspex said:
Theta is not the angle between the vectors ##\vec l, \vec r##. What is the angle between them?
Yes, theta is not the angle between the vectors dl and r, I forgot to label the angle on the drawing. I think I can assume that angle to be close to 90° if the point is very far away but that would still not get me the answer that was provided.
 
Boltzmann Oscillation said:
Yes, theta is not the angle between the vectors dl and r, I forgot to label the angle on the drawing. I think I can assume that angle to be close to 90° if the point is very far away but that would still not get me the answer that was provided.
It is not just approximately 90°, it is exactly 90°, regardless of the distance to the point. E.g. consider the point at the centre of the circle.

I'm not following how you took the z axis component of the field. Looks like you forgot to multiply by the sine of the apex half angle.
 
haruspex said:
It is not just approximately 90°, it is exactly 90°, regardless of the distance to the point. E.g. consider the point at the centre of the circle.

I'm not following how you took the z axis component of the field. Looks like you forgot to multiply by the sine of the apex half angle.
Im sorry but what is the apex half angle? From what I've gathered so far my current equation is
$$\frac{I*a*d \theta *r}{r^3}\vec{d}$$

I would somehow have to show that r = a which it isn't so I know somewhere a long the way I did something wrong.

I thought that by solving the cross product then I would get the z axis component. Hence:

$$\vec{I}\vec{dl} \times \vec{r} = I*dl*r*sin(\theta)\vec{z} $$
 
haruspex said:
It is not just approximately 90°, it is exactly 90°, regardless of the distance to the point. E.g. consider the point at the centre of the circle.

I'm not following how you took the z axis component of the field. Looks like you forgot to multiply by the sine of the apex half angle.
Ah, okay now I understand. The cross product does not show that the dB is only on the z axis so I must make that change myself. Once I took this into effect then my equation equaled that equation which I was supposed to have, many thanks!
 

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