Calculating the Radius of a Charged Ion's Path in a Magnetic Field

[Trista]

Here is the problem:
A singly charged positive ion has a mass of 2.50 X 10 ^-26 kg. After being accelerated through a potential difference of 250 V, the ion enters a magnetic field of 0.500 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field.

I believe the formula to use is this: r = mv/qB
and it should look like this with what is given:
r = (2.5 X 10^-26 kg) v / (1.6 X 10^-19)(.50T)

if that is correct, the only thing I'm having trouble with is the velocity. How do I find the velocity?

if that isn't correct, then I'm lost.

Thank you for your help.

 

[al_201314]

Hi :)

The formula you stated is absolutely correct. To find the velocity after being accelerated, first you'll have to find the energy the ion has gained from rest to after being accelerated. Since the ion is singly charged, it carries a charge of 1.6 X 10 ^-19.

Energy gained by the ion = qV where q is the charge of the ion and V the P.D.

This value is equivalent to the KE gained by the ion which is = 1/2(m)(v)^2. You can then find the velocity. Should work out to approx 5.66 X 10^4 ms^-1

 

[Hootenanny]

Just to add to what al_201314 said, if the velocity was to become very close to the speed of light (c), you would have to use the equation for relativistic kinetic energy. However, as in this case the velocity is much less than c, using standard kinetic energy is a good approximation.

~H

 

[Trista]

Hi :)


Energy gained by the ion = qV where q is the charge of the ion and V the P.D.

This value is equivalent to the KE gained by the ion which is = 1/2(m)(v)^2. You can then find the velocity. Should work out to approx 5.66 X 10^4 ms^-1

THANK YOU... I got it. SO MUCH HELP! thank you thank you thank you

 

[al_201314]

Just to add to what al_201314 said, if the velocity was to become very close to the speed of light (c), you would have to use the equation for relativistic kinetic energy. However, as in this case the velocity is much less than c, using standard kinetic energy is a good approximation.

Is there a way to relate the classical 1/2m(v)^2 to E=m(c)^2? Does it mean to say that if the ion is moving close to the speed of light it will possesses energy equivalent to its mass an the square of the speed of light? Why is this so since even a stationary mass will have energy according to E=m(c)^2?

Thanks for the help, these questions just got to me when I read your post and I do not have a deep understanding of relativity.

 

[Hootenanny]

Is there a way to relate the classical 1/2m(v)^2 to E=m(c)^2? Does it mean to say that if the ion is moving close to the speed of light it will possesses energy equivalent to its mass an the square of the speed of light? Why is this so since even a stationary mass will have energy according to E=m(c)^2?

Thanks for the help, these questions just got to me when I read your post and I do not have a deep understanding of relativity.

There most certainly is. The full equation for relativistic kinetic energy is given by;

$$E_{k} = \frac{m_{0}c^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m_{0}c^2$$

Where $m_{0}$ is the rest mass of the object. Regarding you question;

Why is this so since even a stationary mass will have energy according to E=m(c)^2?

See what happens to the above equation when you let v = 0.

E = mc² takes into account both rest and kinetic energy; note however, that m is not the rest mass.

~H

 

[al_201314]

Thanks H~ for the explanation.