Solving Cathode Ray Tube Homework with Mass, Speed & Magnetic Field

In summary, a positively charged particle with a mass of 7.2 * 10^-8kg and a speed of 85 m/s enters a 0.31 T uniform magnetic field and exits one-quarter south with the same speed after 2.2 * 10^-3 seconds. The equations used for this problem are Fc=Fm, mv^2/r=qvB, and Fm=qvB. To find the charge of the particle and the magnetic force, it is necessary to calculate the period for a complete circular orbit using the given time and then use this to find the radius. From there, the charge and magnetic force can be solved for using the equations.
  • #1
soul5
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0

Homework Statement


A positively charged particle of mass 7.2 * 10^-8kg is traveling east with speed of 85 m/s. The particle enters a 0.31 T uniform magnetic field, and 2.2 * 10^-3 secounds later leaves the field one-quarter south with speed of 85 m/s. What is the charge of the particle and magnetic force?


Homework Equations



Fc=Fm

mv^2/r=qvB

Fm=qvB

The Attempt at a Solution



I tried but I don't have radius and charge what do I do?
 
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  • #2
I think because they give you a time here you can calculate the period for what would be a complete circular orbit, and then extrapolate the radius. From there, I think you should be able to solve it.
 
  • #3


I understand that the equations used in this problem involve mass, speed, and magnetic field, but not radius. It is possible that the radius of the particle's path was not given in the problem, or it may not be relevant to the solution. To solve for the charge of the particle, we can rearrange the equation Fm=qvB to solve for q. We can also use the equation Fc=mv^2/r to solve for the radius, if needed. However, based on the given information, it seems that the charge of the particle is not necessary to solve for the magnetic force. We can use the equation Fm=qvB to solve for the magnetic force, since we have values for the mass, speed, and magnetic field. It is also important to note that the magnetic force does not depend on the charge of the particle, only on its velocity and the strength of the magnetic field. Therefore, the charge of the particle may not be relevant to this specific problem.
 

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