- #1

daselocution

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## Homework Statement

**First part of the problem:**

Newton’s second law is given by F=dp/dt. If the force

is always perpendicular to the velocity, show that F=gamma*m*a, where a is the acceleration.

**Second part of the problem:**Use the result of the previous problem to show that

the radius of a particle’s circular path having charge q traveling with speed v in a magnetic field perpendicular to the particle’s path is r = p/qB. What happens to the radius as the speed increases as in a cyclotron?

## Homework Equations

p=gamma*m*v

F

_{magnetic field}= qv x B = (in this case b/c of θ=90º) qvB

## The Attempt at a Solution

**The first part:**I am thinking that since the force is perpendicular to the path of motion, that the

**speed**of the particle will not change, only it's direction--is this logical? If this were indeed the case, then I would solve as follows (and get the answer as directed):

F=dp/dt=d(m*gamma*v)/dt = m*gamma*d(v)/dt

=m*gamma*a

where dv/dt=a and where speed is unchanging so gamma should be constant

**Second part of the problem:**

F=m*gamma*acceleration=qvB

At this point I'm entirely unsure of how to proceed. I remember that in classical physics a=v

^{2}/r, but I don't know if that applies here.

If it does indeed apply here, then the answer seems to be straightforward:

F=m*gamma*acceleration=m*gamma*v

^{2}/r=qvB

r=m*v*gamma/(qb)=p/(qB)

Thus, as the speed increases in a cyclotron, the radius should increase as well