Problem to find magnetic force?

Click For Summary

Discussion Overview

The discussion revolves around a problem related to the calculation of magnetic force, specifically focusing on the interpretation of the integral of a length of wire represented as a vector. Participants explore the implications of using vector notation in the integral and how it relates to the physical length of the wire.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes that the integral involves the vector dl, suggesting that it represents a sum of vectors rather than just a magnitude.
  • Another participant emphasizes that the integral can be viewed as a vector sum, where the start of the first vector is A1 and the end of the last vector is B1, leading to the conclusion that the integral represents the vector A1B1.
  • A later reply reiterates the importance of recognizing the integral as a sum of vectors aligned head to tail, indicating that the resultant is the line A1B1.
  • One participant clarifies that if the integral were simply dl, it would equal the length of the wire, but since it is ∫ dl, it equals the vector A1B1, highlighting the distinction between infinitesimal vectors and line segments.

Areas of Agreement / Disagreement

Participants generally agree on the interpretation of the integral involving vector dl, but there are nuances in how they articulate the implications of this interpretation, leading to some contestation regarding the physical representation of the wire's length.

Contextual Notes

There are assumptions regarding the straightness of the wire and the interpretation of vector components that remain unresolved. The discussion does not clarify how these factors influence the overall understanding of the problem.

Hardik Batra
Messages
130
Reaction score
5
I have attached a problem.

∫dl = A1B1 = 1

Here dl is a length of wire but in problem it takes distance between the point A1 and B1

Why does this happen ?
 

Attachments

  • Untitled.png
    Untitled.png
    36.2 KB · Views: 496
Physics news on Phys.org
its important to notice that the integral contains vector dl and not just the magnitude of dl.

You can view that integral as a sum of vectors where the end of each vector is the start of the next one. We know from vector addition that the sum is the vector that has as start the start of the first vector and end the end of the last vector (for example for 3 vectors AB, BC and CD its easy to se that the vector sum AB+BC+CD=AD). The start of the "first dl" vector is A1 and the end of the "last dl" is B1. Hence that integral is the vector A1B1.
 
Delta² said:
its important to notice that the integral contains vector dl and not just the magnitude of dl.

You can view that integral as a sum of vectors where the end of each vector is the start of the next one. We know from vector addition that the sum is the vector that has as start the start of the first vector and end the end of the last vector (for example for 3 vectors AB, BC and CD its easy to se that the vector sum AB+BC+CD=AD). The start of the "first dl" vector is A1 and the end of the "last dl" is B1. Hence that integral is the vector A1B1.

From figure you can see than...
The length of wire is greater than 1m.
because A1B1 = 1m. and wire is not straight.
 
Hardik Batra said:
From figure you can see than...
The length of wire is greater than 1m.
because A1B1 = 1m. and wire is not straight.
All true, but if you view things as Delta² suggests, you can view that integral as a sum of vectors aligned head to tail. The sum of those vectors, the resultant, is just the line A1B1.

You can also realize that the components of the vectors perpendicular to the line A1B1 cancel out.
 
If it was [tex]\int dl[/tex] then it would be equal to length of the wire. But it is

[tex]\int \vec{dl}[/tex] so it is equal to the vector [tex]\vec{A_1B_1}[/tex]. The reason is as i said in my first post , you can view Doc Al's reason as well. I think the critical thing you "filtering" in your mind is that we have an integral of infinitesimal vectors [tex]\vec{dl}[/tex] and not an integral of infinitesimal line segments [tex]dl[/tex]
 
Last edited:

Similar threads

  • · Replies 11 ·
Replies
11
Views
886
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K