The magnitude of the magnetic torque on the loop

In summary, a plastic circular loop with a positive charge q distributed uniformly around its circumference is rotated around its central axis with angular speed omega. In a region with a uniform magnetic field B directed parallel to the plane of the loop, the magnitude of the magnetic torque on the loop can be calculated using the formula T = (q*omega*B*R^2)/(3*pi). However, for a more accurate calculation, the concept of magnetic dipole moment of a current loop and the torque it experiences in a magnetic field should be considered.
  • #1
marpple
30
0
A plastic circular loop of radius R and a positive charge q is distributed uniformly around the circumference of the loop. The loop is then rotated around its central axis, perpendicular to the plane of the loop, with angular speed omega.

If the loop is in a region where there is a uniform magnetic field B directed parallel to the plane of the loop, calculate the magnitude of the magnetic torque on the loop.


well, i did this
F=qvB
and know that v = w(omega).R
it gives me F=qwRB
and Torque = F. L, then i got T=qwR^2B


my answer seems not correct.., please advise.
 
Physics news on Phys.org
  • #2
marpple said:
well, i did this
F=qvB
That's the force on a moving point charge (magnitude only) where v and B are perpendicular.
and know that v = w(omega).R
OK.
it gives me F=qwRB
Don't lump all the charge together.
and Torque = F. L, then i got T=qwR^2B
Note that the force on each element of charge on the loop will give rise to a different torque contribution, since the distance to the axis (a diameter) is different.

Hint: Review the concept of magnetic dipole moment of a current loop and the torque it experiences in a magnetic field.
 
  • #3
I got close...

r varies from 0 to R
F=(q/2R*pi)Bv
v=omega*r
dtorque=(q*omega*r^2*B)/(2*pi*R)

torque= (some constant)(q*omega*B*R^2)/(3*pi)

I can't figure out the constant though. Have I only found the upper half of the loop?
 
  • #4
wait no, I lost a 2. That still doesn't give me the right answer though.
 
  • #5
supacalafrg said:
I got close...

r varies from 0 to R
F=(q/2R*pi)Bv
v=omega*r
dtorque=(q*omega*r^2*B)/(2*pi*R)
To solve it properly using dF = dq v X B requires taking the angle into consideration and integrating. I recommend the hint I gave in my last post.
 

Related to The magnitude of the magnetic torque on the loop

What is the magnetic torque on a loop?

The magnetic torque on a loop is the measure of the twisting force exerted on the loop due to the presence of a magnetic field.

What factors affect the magnitude of the magnetic torque on a loop?

The magnitude of the magnetic torque on a loop is affected by the strength of the magnetic field, the angle between the loop and the field, and the size and shape of the loop.

How is the magnitude of the magnetic torque calculated?

The magnitude of the magnetic torque on a loop can be calculated using the formula T = μ x B x A x sin(θ), where μ is the magnetic moment, B is the magnetic field strength, A is the area of the loop, and θ is the angle between the loop and the field.

What is the unit of measurement for magnetic torque?

The unit of measurement for magnetic torque is Newton-meter (N∙m) in the SI system or dyne-centimeter (dyn∙cm) in the CGS system.

How does the direction of the magnetic torque relate to the direction of the magnetic field?

The direction of the magnetic torque is perpendicular to both the direction of the magnetic field and the plane of the loop. This means that the magnetic torque can either be clockwise or counterclockwise depending on the orientation of the loop in relation to the magnetic field.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
464
  • Introductory Physics Homework Help
Replies
7
Views
393
  • Introductory Physics Homework Help
Replies
25
Views
402
  • Introductory Physics Homework Help
Replies
1
Views
708
  • Introductory Physics Homework Help
Replies
1
Views
508
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top