Calculating magnetic force on circular coil carrying current

Click For Summary
SUMMARY

The discussion focuses on calculating the magnetic force acting on a circular coil carrying current in a magnetic field with axial symmetry. The relevant equation used is $$\vec{F_B} = i \vec{dl} \times \vec{B}$$, where the magnetic force is derived as $$F = 2\pi irB\cos\theta$$. However, the expected result from the textbook is $$2\pi irB\sin\theta$$, leading to confusion regarding the definition of the angle θ. The angle θ is clarified as the angle between the magnetic field B and the z-axis component of the field.

PREREQUISITES
  • Understanding of magnetic force calculations using the Lorentz force law.
  • Familiarity with circular coils and their properties in magnetic fields.
  • Knowledge of vector calculus, particularly cross products.
  • Basic understanding of axial symmetry in magnetic fields.
NEXT STEPS
  • Review the derivation of the Lorentz force law in electromagnetic theory.
  • Study the properties of circular coils in varying magnetic fields.
  • Learn about vector calculus applications in physics, focusing on cross products.
  • Examine axial symmetry in magnetic fields and its implications in coil design.
USEFUL FOR

Students studying electromagnetism, physics educators explaining magnetic forces, and engineers designing electromagnetic devices involving circular coils.

cseil
Messages
28
Reaction score
0

Homework Statement


A circular coil with center on the z axis and orthogonal to the xy axis carries a current. The coil is in a magnetic field B with axial symmetry compared to the x axis. B forms an angle θ with the z axis. Calculate the magnetic force acting on the coil.

exerc.png


2. Homework Equations

$$\vec{F_B} = i \vec{dl} x \vec{B}$$

The Attempt at a Solution


[/B]
I tried considering $$ dl = r d\phi $$
So F = irB_zd\phi
where Bz is Bcos\theta

By is null because it has opposite verse on opposite points of the coil. So I can consider just Bz

$$ F = irBcos\theta d\phi = irBcos\theta \int_0^{2\pi} d\phi = 2\pi irBcos\theta$$

The result, according to the book, is 2\pi irBsin\theta
Why?
 
Physics news on Phys.org
How does your book define ##\theta##? Is it the angle the magnetic field makes with the horizontal(i.e. ##B_y##)?
 
Vagn said:
How does your book define ##\theta##? Is it the angle the magnetic field makes with the horizontal(i.e. ##B_y##)?

It is the angle between B and Bz!
 
cseil said:
The coil is in a magnetic field B with axial symmetry compared to the x axis.
What does this mean?
 
rude man said:
What does this mean?

I can't understand what does it mean, actually.
 
cseil said:
I can't understand what does it mean, actually.
So ask! The problem is not you, it's the instructor who can't write understandably!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
1
Views
1K
Replies
3
Views
525
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
850
Replies
8
Views
1K
Replies
4
Views
1K
Replies
5
Views
1K
Replies
1
Views
2K
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
6K