How Do You Calculate Terminal Velocity of a Metal Ring in a Magnetic Field?

  • Thread starter Thread starter accountdracula
  • Start date Start date
  • Tags Tags
    Area Emf
AI Thread Summary
To calculate the terminal velocity of a metal ring dropped through a magnetic field, the magnetic force generated by the ring's induced current must equal its weight. The relevant equations include the electromotive force (emf) and the relationship between current, resistance, and magnetic force. The area through which the ring moves is defined as A = πr²vdt, leading to the expression for emf as emf = dBπr²v. The confusion arises regarding whether to use the diameter or circumference of the ring for the length L in the magnetic force equation. Clarification is needed on the uniformity of the magnetic field and the meaning of "self-parallel" in this context.
accountdracula
Messages
21
Reaction score
0

Homework Statement


I'm answering a question which describes a situation in which a metal ring is dropped through a magnetic field such that, when it falls, its area is perpendicular to the magnetic field.
I need to find its terminal velocity given:
Mass : 2.66 x 10-4 kg
Magnetic flux density : 2.00 T
Radius : 2.00 cm
Resistance : 2.48 m(ohms)

Homework Equations


Emf = dBA/dt
V = IR
F = BILsin(theta)

The Attempt at a Solution



At terminal velocity, the magnetic force as a result of the ring's current must equal its weight:

mg = BIL

I'm confused about how to introduce v into the equation E.m.f = dBA/dt.
My thoughts were as follows:

If A is the area through which the ring moves in time dt then
A = pi r2vdt

e.m.f = (dBpi r2vdt)/dt
e.m.f = dBpi r2v

Dividing both sides by R :

I = (dBpi r2)/ R
I would then set this equal to mg / BL to find v.
However, in previous questions 'L' has always been a straight wire. Would you use the diameter of this metal ring or its circumference? My feeling is the circumference but I'm not 100% sure.
Also, is the way I've approached this question right?
 
Physics news on Phys.org
accountdracula said:
when it falls, its area is perpendicular to the magnetic field.
What does this mean? Do you mean that the disc that has the ring as boundary has a normal that is perpendicular to the mag field lines?

Is the magnetic field uniformly linear and self-parallel within the region of interest?
 
andrewkirk said:
What does this mean? Do you mean that the disc that has the ring as boundary has a normal that is perpendicular to the mag field lines?

Is the magnetic field uniformly linear and self-parallel within the region of interest?

I mean the plane of the area of the metal ring is perpendicular to the plane of the magnetic field lines. The magnetic field is uniform. I don't have a clue what self parallel within the region of interest means.
 
Bump
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top