Solving Equilibrium of Glass Cylinder w/ Coil Wrapped Around It

In summary, the question asks for the loop current I that will keep a glass cylinder with a 10-turn coil of wire wrapped lengthwise in static equilibrium on a ramp tilted at an angle θ with a uniform magnetic field of strength B pointing upward. The solution involves setting the force of gravity equal to the force of magnetism on a current and factoring in torque, resulting in the equation I = g*rho*pi*R^2/(20B).
  • #1
RichardEpic
16
0

Homework Statement


A glass cylinder of radius R, length l, and density \rho has a 10-turn coil of wire wrapped lengthwise, as seen in the figure . The cylinder is placed on a ramp tilted at angle \theta with the edge of the coil parallel to the ramp. A uniform magnetic field of strength B points upward.

For what loop current I will the cylinder rest on the ramp in static equilibrium? Assume that static friction is large enough to keep the cylinder from simply sliding down the ramp without rotating.
Express your answer in terms of the variables R, l, \rho, \theta, B, and appropriate constants.

Homework Equations



Force of gravity = mgSin(θ)

ρ = m / volume

volume = l*2∏R^2
l = length of the cylinder

B of coil = μIN/l
l = length of the solenoid

Force on a current = ILxB = ILBsin(θ)
L = length of the wire

The Attempt at a Solution



I attempted to solve this equation by solving for the mass from the density equation given above: m = ρl2∏R^2

so: Fg = gρl2∏R^2 * sin(θ)

then, I set this equal to the Force of magnetism on a current...

gρl2∏R^2 * sin(θ) = ILBsin(θ)

sin(θ)'s cancel...the l's also apparently cancel...giving me:

(gρl2∏R^2)/B = I

The homework website says that all I'm missing is an incorrect multiplier. How would I factor in torque for this problem? I believe that may be the solution, unless it has something to do with the glass cylinder and the coil wrapped around it, or perhaps there's something specific to a magnetic field acting on a solenoid...Help would be greatly appreciated! :smile:
 

Attachments

  • 33.P71.jpg
    33.P71.jpg
    7.8 KB · Views: 520
Physics news on Phys.org
  • #2


Your answer should have been I = g*rho*pi*R^2/(20B). This is due to the 2 sides affected by the magnetic force with 10 coils each making B=20IlB. This is to help anyone reading this archive and unfortunately not OP.
 

Related to Solving Equilibrium of Glass Cylinder w/ Coil Wrapped Around It

1. How do you determine the equilibrium of a glass cylinder with a coil wrapped around it?

The equilibrium of a system can be determined by analyzing the forces acting on it. In the case of a glass cylinder with a coil wrapped around it, the equilibrium is achieved when the forces exerted by the coil and the weight of the cylinder are balanced.

2. What factors affect the equilibrium of a glass cylinder with a coil wrapped around it?

The equilibrium of the system can be affected by several factors, including the weight of the cylinder, the tension in the coil, and the diameter and length of the cylinder. Additionally, the properties of the materials used for the cylinder and the coil can also play a role in determining the equilibrium.

3. How can the equilibrium of a glass cylinder with a coil wrapped around it be calculated?

To calculate the equilibrium, the forces acting on the system must be determined using equations such as Newton's Second Law and Hooke's Law. The weight of the cylinder can be calculated using its mass and the acceleration due to gravity, while the tension in the coil can be calculated using its spring constant and the displacement from its equilibrium position.

4. What are the applications of understanding the equilibrium of a glass cylinder with a coil wrapped around it?

Understanding the equilibrium of this system can have practical applications in fields such as engineering and product design. It can also be used in educational contexts to demonstrate principles of forces and equilibrium.

5. Is the equilibrium of a glass cylinder with a coil wrapped around it a stable or unstable state?

The equilibrium of this system can be either stable or unstable depending on the specific parameters and conditions. If the forces are balanced and small disturbances do not cause significant changes, the equilibrium can be considered stable. However, if small disturbances result in significant changes to the system, the equilibrium can be considered unstable.

Similar threads

  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
987
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
770
  • Introductory Physics Homework Help
Replies
23
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
959
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
495
  • Introductory Physics Homework Help
Replies
5
Views
4K
Replies
4
Views
1K
Back
Top