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

[balling12]

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.

 

[kuruman]

What is the kinetic energy of a singly charged ion that is accelerated (from rest we assume) through a potential difference of 264 Volts?

 

[balling12]

1/2mvf^2?

 

[Fightfish]

1/2mvf^2?

That's definitely a true expression; but the question is, what is its value equal to?

 

[balling12]

the potential energy?

 

[Fightfish]

the potential energy?

Strictly, no. It is equivalent to the change in potential energy of the system magnitude-wise. (recall conservation of energy)

 

[balling12]

1/2mvf^2+1/2mvi^2= mghi+mghf

 

[Fightfish]

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?

 

[balling12]

would it be Change In P.E.=(q)(potential difference)
(1.6X10^-19)(264)= 4.224X10^-17