Two charges accelerated first by ##\Delta{V}##, then by ##\vec{B}##

[mcastillo356]

Homework Statement

Two particles of the same charge and masses $m_1$ and $m_2$ are accelerated by the same difference of electric potential $\Delta{V}$. Then they enter at a place where there is a magnetic field $\vec{B}$ perpendicular to their movement direction. What relation is between the radius of the circular path drawn by both particles?
a)$R_1/R_2=[m_1/m_2]^{1/2}$
b)$R_1/R_2=m_1\Delta{V}/m_2 B$
c)$R_1/R_2=m_1/m_2$
d$R_1/R_2=m_2/m_1$

Relevant Equations

$R=\dfrac{m E}{|q|B}$

$$R_1=\dfrac{m_1 E}{|q|B}$$
$$R_2=\dfrac{m_2 E}{|q|B}$$
$$\therefore{\dfrac{R_1}{R_2}=\dfrac{m_1}{m_2}}$$
In my opinion, the answer to a this multiple choice question is c)

 

[PeroK]

Homework Statement:: Two particles of the same charge and masses $m_1$ and $m_2$ are accelerated by the same difference of electric potential $\Delta{V}$. Then they enter at a place where there is a magnetic field $\vec{B}$ perpendicular to their movement direction. What relation is between the radius of the circular path drawn by both particles?
a)$R_1/R_2=[m_1/m_2]^{1/2}$
b)$R_1/R_2=m_1\Delta{V}/m_2 B$
c)$R_1/R_2=m_1/m_2$
d$R_1/R_2=m_2/m_1$
Relevant Equations:: $R=\dfrac{m E}{|q|B}$

$$R_1=\dfrac{m_1 E}{|q|B}$$
$$R_2=\dfrac{m_2 E}{|q|B}$$
$$\therefore{\dfrac{R_1}{R_2}=\dfrac{m_1}{m_2}}$$
In my opinion, the answer to a this multiple choice question is c)

Are you sure about the dependence on $E$?

 

[mcastillo356]

No, not at all. What about $r=\dfrac{mv}{qB}=\dfrac{\sqrt{2qVm}}{qB}=\dfrac{1}{B}\times{\left(\sqrt{\dfrac{2Vm}{q}}\right)}$? Could be a way?

 

[PeroK]

No, not at all. What about $r=\dfrac{mv}{qB}=\dfrac{\sqrt{2qVm}}{qB}=\dfrac{1}{B}\times{\left(\sqrt{\dfrac{2Vm}{q}}\right)}$? Could be a way?

That looks better!

 

[mcastillo356]

Well, I infer that the answer is the same, because $v$, $q$ and $B$ are the same for both particles: $\dfrac{R_1}{R_2}=\dfrac{m_1}{m_2}$

 

[PeroK]

Well, I infer that the answer is the same, because $v$, $q$ and $B$ are the same for both particles: $\dfrac{R_1}{R_2}=\dfrac{m_1}{m_2}$

What happened to the square root?

 

[mcastillo356]

No, $v$ is not the same. I must deal with $r=\dfrac{1}{B}\times{\left(\sqrt{\dfrac{2Vm}{q}}\right)}$

 

[PeroK]

No, $v$ is not the same. I must deal with $r=\dfrac{1}{B}\times{\left(\sqrt{\dfrac{2Vm}{q}}\right)}$

I don't understand. This looks right. $B, V$ and $q$ are constant. That gives you $r \propto \sqrt m$, no?

 

[mcastillo356]

hmm.. That's wright. Is it a contradition to state $r\propto\sqrt m$ and $r\propto m$?

 

[PeroK]

hmm.. That's wright. Is it a contradition to state $r\propto\sqrt m$ and $r\propto m$?

It can only be one or the other.

 

[mcastillo356]

So I must say rather this one: $\dfrac{R_1}{R_2}=\dfrac{\sqrt{m_1}}{\sqrt{m_2}}$. This is $R_1/R_2=[m_1/m_2]^{1/2}$: the a) choice

 

[mcastillo356]