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

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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)

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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 and Delta2
Gold Member
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?

Delta2
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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!

Gold Member
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
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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?

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

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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?

Delta2
Gold Member
hmm.. That's wright. Is it a contradition to state ##r\propto\sqrt m## and ##r\propto m##?