What Is the Radius of the Ion's Path in the Magnetic Field?

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Homework Help Overview

The problem involves a singly charged positive ion with a specified mass that is accelerated through a potential difference before entering a magnetic field. The objective is to determine the radius of the ion's path in the magnetic field, which is a topic related to electromagnetism and motion of charged particles in magnetic fields.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss how to derive the velocity of the ion from the given potential difference and question the relationship between kinetic energy and potential energy in this context.

Discussion Status

Participants are actively exploring the relationship between kinetic energy and potential energy, with some offering insights into conservation of energy principles. There is a focus on understanding how to calculate the kinetic energy of the ion after being accelerated through the potential difference.

Contextual Notes

There is an assumption that the ion starts from rest, and the discussion includes the potential energy change associated with the electric field. Participants are also clarifying the relevance of gravitational potential energy in this scenario.

balling12
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Homework Statement


A singly charged positive ion has a mass of 2.51 X 10-26 kg. After being accelerated through a potential difference of 264 V, the ion enters a magnetic field of 0.570 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field


Homework Equations


r=mv/qB


The Attempt at a Solution


Im not sure how to get the value of v from the information in the problem.
 
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What is the kinetic energy of a singly charged ion that is accelerated (from rest we assume) through a potential difference of 264 Volts?
 
1/2mvf^2?
 
balling12 said:
1/2mvf^2?
That's definitely a true expression; but the question is, what is its value equal to?
 
the potential energy?
 
balling12 said:
the potential energy?
Strictly, no. It is equivalent to the change in potential energy of the system magnitude-wise. (recall conservation of energy)
 
1/2mvf^2+1/2mvi^2= mghi+mghf
 
balling12 said:
1/2mvf^2+1/2mvi^2= mghi+mghf
Err..no gravitational potential energy is involved here; the particle is accelerated through a (electrical) potential difference. What is the change in electrical potential energy of the particle upon passing through the electric field?
 
would it be Change In P.E.=(q)(potential difference)
(1.6X10^-19)(264)= 4.224X10^-17
 

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