What is the Proof for Kinetic Energy in a Circular Magnetic Field?

[joej]

this should probably go in the HS section since it is review of some material covered there but... I'm guessing the same peole brose the two forums so I might as well post it here since it came up in college.

now the questions seems quite straight forward:

for a particle of mass m and charge q, moving in a circular path in a magnetic field B, show that its kinetic energe is proportional to the square of the radius of curvature of its path.

so we have:

r = (mv^2)/qvB
r = mv^2/F

F = ma
F = m((v^2)/r)

r = (mv^2)/(m((v^2)/r))

now... it all evens out and I'm stuck with nothing... could someon tell me where I'm going wrong with this

 

[Justin Lazear]

When you get a result like 1 = 1 or r = r, it means you're substituting an equation into itself, which should naturally take you nowhere.

--J

 

[dextercioby]

1.Do you know the expression of KE in terms of velocity??

2.Can u compute/find a relation between the radius of trajectory and the velocity the particle has...?

Daniel.

 

[joej]

KE = 1/2 mv^2

yeah I know that

still don't see how I can go from r = (mv^2)/F to r^2 = 1/2 mv^2

 

[dextercioby]

KE = 1/2 mv^2

yeah I know that

still don't see how I can go from r = (mv^2)/F to r^2 = 1/2 mv^2

Can u prove that:
$$r_{traj.}=\frac{mv}{qB}$$

Daniel.

 

[learningphysics]

this should probably go in the HS section since it is review of some material covered there but... I'm guessing the same peole brose the two forums so I might as well post it here since it came up in college.

now the questions seems quite straight forward:

for a particle of mass m and charge q, moving in a circular path in a magnetic field B, show that its kinetic energe is proportional to the square of the radius of curvature of its path.

so we have:

r = (mv^2)/qvB

This is all you need. As well as KE=1/2 mv^2

Use the two equations you've got:
r=(mv^2)/qvB and
KE=1/2mv^2

to get a relationship between KE and r.

 

[joej]

This is all you need. As well as KE=1/2 mv^2

Use the two equations you've got:
r=(mv^2)/qvB and
KE=1/2mv^2

to get a relationship between KE and r.

Forgive my thickness but I can't figure out how to go from those two to proving that (r^2) is proportional to (1/2 mv^2)

Can u prove that:
$$r_{traj.}=\frac{mv}{qB}$$

Daniel.

well yeah that's just the v at the bottom is canceled out by v^2 and the square is thus also gone, but where do I go from there

 

[dextercioby]

If u can prove that,it means u can prove it is valid even when it's squared:
$$r_{traj.}^{2}=\frac{m^{2}v^{2}}{q^{2}B^{2}}=\frac{2m}{q^{2}B^{2}}\frac{mv^{2}}{2}=C\cdot (KE)$$

Voilà.

Daniel.

 

[learningphysics]

Forgive my thickness but I can't figure out how to go from those two to proving that (r^2) is proportional to (1/2 mv^2)

Taking your first equation
r=(mv^2)/qvB

if I solve for v I get
v=rqB/m

Then I plug it into the KE equation
KE=(1/2)mv^2

$$KE=\frac{1}{2}m(\frac{rqB}{m})^2$$

$$KE=\frac{q^2B^2}{2m}r^2$$

since $$\frac{q^2B^2}{2m}$$ is a constant, this proves KE is proportional to r^2.

You needed an expression for KE in terms of r, and constants (v is not a constant here as it changes with r), so you needed to get rid of that v... so solve for v in the first equation and plug it into the second... so all that is left is constants and r's.

 

[joej]

ah okay I finally get it thnx to both of you, great help, really appreciate it ;)