Help finding the magnetic field of a current loop

AI Thread Summary
The discussion focuses on calculating the magnetic field of a current loop, with particular emphasis on understanding the cross product involved in the calculations. The original poster struggles to relate their derived equation to the book's solution, specifically regarding the angle between vectors. Clarifications reveal that the angle in question is exactly 90°, not approximately, which simplifies the calculations. The importance of considering the apex half angle in the cross product is highlighted, leading to a realization that adjustments are needed to accurately represent the z-axis component. Ultimately, the poster successfully aligns their equation with the expected result after addressing these points.
Boltzman Oscillation
Messages
233
Reaction score
26
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
 
Physics news on Phys.org
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.
 
Boltzman 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!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top