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

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In summary, the conversation discussed the relationship between the radius of a circular path of two particles with the same charge and different masses, accelerated by the same electric potential and entering a magnetic field. The relevant equation is given by ##R=\dfrac{mE}{|q|B}##, and it was concluded that the ratio of the radii, ##R_1/R_2##, is equal to the ratio of the masses, ##m_1/m_2##. The conversation also briefly touched on the possibility of the radius being proportional to the square root of the mass, but it was determined that it can only be one or the other. The correct answer to the multiple choice question is c).
  • #1
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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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  • #2
mcastillo356 said:
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##?
 
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  • #3
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?
 
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  • #4
mcastillo356 said:
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!
 
  • #5
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}##
 
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  • #6
mcastillo356 said:
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?
 
  • #7
No, ##v## is not the same. I must deal with ##r=\dfrac{1}{B}\times{\left(\sqrt{\dfrac{2Vm}{q}}\right)}##
 
  • #8
mcastillo356 said:
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?
 
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  • #9
hmm.. That's wright. Is it a contradition to state ##r\propto\sqrt m## and ##r\propto m##?
 
  • #10
mcastillo356 said:
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.
 
  • #11
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
 
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  • #12
:smile:
 

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

1. What is the difference between accelerating charges with a voltage and with a magnetic field?

Accelerating charges with a voltage, also known as a potential difference, involves applying an electric field to the charges. This results in a change in the charges' kinetic energy. On the other hand, accelerating charges with a magnetic field involves applying a magnetic force to the charges, which causes them to move in a circular path. This results in a change in the charges' direction of motion, but not their kinetic energy.

2. How does the acceleration of charges with a voltage compare to the acceleration of charges with a magnetic field?

The acceleration of charges with a voltage is directly proportional to the voltage applied. This means that the higher the voltage, the greater the acceleration. However, the acceleration of charges with a magnetic field is dependent on the strength of the magnetic field, the charge of the particles, and their velocity. It is also perpendicular to the direction of motion of the charges.

3. Can charges be accelerated by both a voltage and a magnetic field at the same time?

Yes, charges can be accelerated by both a voltage and a magnetic field at the same time. This is known as a crossed field configuration, where the electric and magnetic fields are perpendicular to each other. In this case, the charges will experience a combined force that will result in both a change in their direction and kinetic energy.

4. How does the mass of the charges affect their acceleration when accelerated by a magnetic field?

The mass of the charges does not affect their acceleration when accelerated by a magnetic field. This is because the acceleration is dependent on the charge and velocity of the particles, not their mass. However, the mass can affect the radius of the circular path the charges will follow, with heavier particles having a larger radius than lighter particles.

5. Is the acceleration of charges with a magnetic field affected by the orientation of the magnetic field?

Yes, the orientation of the magnetic field can affect the acceleration of charges. The magnetic field must be perpendicular to the direction of motion of the charges in order for them to experience a force and be accelerated. If the magnetic field is not perpendicular, the charges will not experience a force and will not be accelerated.

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