The motion of a charged particle in a magnetic field

[peace]

Homework Statement

A particle has entered a magnetic field. Show the radius of the particle's motion as follows:
R= pc/qB

p= Particle momentum
q= Particle electric charge

Relevant Equations

Magnetic force=qvB
Centripetal force=mv^2/R

qvB=mv^2/R
R=mv/qB= p/qB !
As you can see, the difference between this relation and the relation in question is in 'c'.
Maybe my way is wrong. Maybe I should get help from relativity because the speed of light is involved here.
Please help. Thankful

 

[kuruman]

The Lorentz force in c.g.s. units is ##\vec F=\dfrac{q\vec v \times \vec B}{c}.~## Does this help?

 

[peace]

using this relationship, the result is correct. But I'm still not sure.
Does this mean that the radius of motion is expressed in c.g.s. ?

 

[kuruman]

using this relationship, the result is correct. But I'm still not sure.
Does this mean that the radius of motion is expressed in c.g.s. ?

Yes, all quantities in the expression, including the speed of light, must be in c.g.s. The speed of light may be independent of the frame of reference but it is not independent of the systems of units.

 

[peace]

So the Centripetal force also changes and must be in the c.g.s. Because I'm going to put both forces equal and get the radius through it. Yes?
How is this force in c.g.s ?

 

[Steve4Physics]

Hi. I’d like to add a few comments.

SI units are used 90-something % of the time these days. But CGS units are still occasionally used.

The equation [itex] R = \frac {pc}{qB} [/itex] is correct in CGS units. The CGS units to be used are:
length (radius): cm
momentum: g·cm/s
speed: cm/s
charge: statcoulomb
magnetic field: gauss
(If required, the CGS unit of force is the dyne, which is equivlaent to 1g·cm/s²)

The question as set is poor because it should tell you what system of units is being used.

 

[peace]

Hi. I’d like to add a few comments.

SI units are used 90-something % of the time these days. But CGS units are still occasionally used.

The equation [itex] R = \frac {pc}{qB} [/itex] is correct in CGS units. The CGS units to be used are:
length (radius): cm
momentum: g·cm/s
speed: cm/s
charge: statcoulomb
magnetic field: gauss
(If required, the CGS unit of force is the dyne, which is equivlaent to 1g·cm/s²)

The question as set is poor because it should tell you what system of units is being used.

hi. Thanks for your explanation.
Yes, so the required radius is in CGS and not SI. Of course, unfortunately, the question did not say anything about this.
So I just have to write the Lorentz force in CGS and then I get the result. yes?

 

[etotheipi]

So I just have to write the Lorentz force in CGS and then I get the result. yes?

Yes that's good. If you want a challenge, you might notice that you haven't actually shown the particle performs circular motion, you just assumed it. If you want, you could try to solve$$\frac{1}{c}q\vec{v} \times \vec{B} = m\ddot{\vec{r}}$$which, if we take ##\vec{B} = B\hat{z}##, is actually the coupled set of differential equations$$\dot{y} = \frac{mc}{Bq} \ddot{x}, \quad\dot{x} = \frac{mc}{Bq} \ddot{y}$$

 

[Steve4Physics]

hi. Thanks for your explanation.
Yes, so the required radius is in CGS and not SI. Of course, unfortunately, the question did not say anything about this.
So I just have to write the Lorentz force in CGS and then I get the result. yes?Ye

Yes. I would start a written answer by noting that [itex]R =\frac{pc}{qB} [/itex] applies when CGS units are used, Then just state the magnetic force is F = Bqv/c in CGS units.

For information, the Lorentz force is actually the sum of the electric and magnetic forces. Since there is no electric field in this question , only the magnetic force is present. So I wouldn't refer to the force as the Lorentz force.

 

[etotheipi]

So I wouldn't refer to the force as the Lorentz force.

This is a fun question. Electromagnetism is Lorentz invariant, but in the limit ##c\rightarrow \infty## the Galilean transformation of the EM field is*$$\begin{align*}\vec{E} &\rightarrow \vec{E}' = \vec{E} + \vec{u} \times \vec{B} \\ \vec{B} &\rightarrow \vec{B}' = \vec{B} \end{align*}$$To see how it plays out in this non-relativistic regime ##|\vec{v}| \ll c##, take the Lorentz force law in the frame where the thing does circular motion, where ##\vec{E} = 0##,$$\vec{F} = q\vec{v} \times \vec{B}$$Now let's Galilean boost to a new frame so that ##\vec{v} = \vec{u} + \vec{v}'##,$$\begin{align*}\vec{F}' = q\vec{E}' + q\vec{v}' \times \vec{B}' &= \left[ q\vec{E} + q\vec{u} \times \vec{B} \right] + \left[ q(\vec{v} - \vec{u})\times \vec{B} \right] \\ &= q\vec{E} + q\vec{v} \times \vec{B} \\ &= q\vec{v} \times \vec{B} \\ &= \vec{F} \end{align*}$$We notice that the Lorentz force is invariant under this Galilean boost, ##\vec{F} = \vec{F}'##, however now it has an electric contribution ##q\vec{u} \times \vec{B}##, and a magnetic contribution of ##q\vec{v}' \times \vec{B}##!

A special case is where you take ##\vec{u} \parallel \vec{B}##, in which case ##\vec{u} \times \vec{B} = \vec{0}## and both fields are completely unchanged under a Galilean boost, i.e. ##\vec{E} = \vec{E}'## and ##\vec{B} = \vec{B}'##. That's a point which is often brushed under the rug when dealing with e.g. helical motion in a magnetic field, where a common method of solution is to transform into a frame where the thing is doing circular motion. It's not often explained that it's only valid because it's a special case where the fields are unchanged!

*to see where this comes from, the Lorentz transformation of the EM field is$$\vec{E}' = \gamma \left( \vec{E} + \vec{u} \times \vec{B} \right ) - \left ({\gamma-1} \right ) ( \vec{E} \cdot \hat{u} ) \hat{u}$$ $$\vec{B}' = \gamma \left( \vec{B} - \frac {\vec{u} \times \vec{E}}{c^2} \right ) - \left ({\gamma-1} \right ) ( \vec{B} \cdot \hat{u} ) \hat{u}$$now take the limit ##c\rightarrow \infty##, or ##\gamma \rightarrow 1## to obtain the EM transformation in the Galilean regime!