Charged mass connected to spring, swung in circle in mag. field

AI Thread Summary
The discussion revolves around calculating the maximum radius of a mass attached to a spring, moving in a circle within a magnetic field. The mass is 2 kg, has a charge of 3.0 C, and is moving at 5 m/s in a 1.5-T magnetic field. The participant attempts to apply centripetal force, Hooke's Law, and magnetic force equations but is unsure about the elongation of the spring and its relationship to the radius. Clarification is sought on whether the elongation equals the radius and how to correctly set up the equations to solve for the radius. The participant is looking for guidance on resolving their calculations to find a numerical solution.
clubguppy
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Homework Statement


A spring with an unstretched length of 20 cm and a force constant of 100 N/m is attached to a 2-kg mass with a charge of 3.0 C. The mass is swung in a circle in a zero gravity environment, so that the spring is perfectly horizontal and is parallel to the radius of the circle, as shown at right. A vertical 1.5-T magnetic field permeates the entire region.


If the mass is moving at 5 m/s, what is the maximum radius of its motion?

Homework Equations


Centripetal Force: F= \frac{mv^{2}}{r}
Hooke's Law: F=kx where k is the spring constant and x is elongation (how long the string is stretched)
Magnetic Force: F = qvBsin \theta

The Attempt at a Solution


I tried qvBsin(90 degrees) = qvB =F= \frac{mv^{2}}{r} + kr
I assumed the force on the spring itself is the tenion (the constant * how long it stretched)

I'm not sure if I did it correctly...please help.
 
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What is x in F = kx ?
 
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clubguppy said:

Homework Statement


A spring with an unstretched length of 20 cm and a force constant of 100 N/m is attached to a 2-kg mass with a charge of 3.0 C. The mass is swung in a circle in a zero gravity environment, so that the spring is perfectly horizontal and is parallel to the radius of the circle, as shown at right. A vertical 1.5-T magnetic field permeates the entire region.


If the mass is moving at 5 m/s, what is the maximum radius of its motion?

Homework Equations


Centripetal Force: F= \frac{mv^{2}}{r}
Hooke's Law: F=kx where k is the spring constant and x is elongation (how long the string is stretched)
Magnetic Force: F = qvBsin \theta

The Attempt at a Solution


I tried qvBsin(90 degrees) = qvB =F= \frac{mv^{2}}{r} + kr
I assumed the force on the spring itself is the tenion (the constant * how long it stretched)

I'm not sure if I did it correctly...please help.

I have updated the information in the question.
 
Last edited by a moderator:
is the elongation is equal to radius of the circular path?
 
response

mukundpa said:
is the elongation is equal to radius of the circular path?

Yes, that is what I assumed. But, I can't get the radius in real number. What did I do wrong? Do you have suggestions for what I should do?

Should I change something about qvB =F= \frac{mv^{2}}{r} + kr or do something different?
 
Last edited:
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