An electron moving through a magnetic field [closed]
An electron, that has been accelerated from rest by a potential difference of $250 rm V$, enters a region of magnetic field of strength $0.12 rm T$. Show that speed of the electron after acceleration is $9.4 times 10^6 rm m s^{–1}.$
For this problem I first tried to find the velocity by rewriting the equation for the force an electron will experience when moving through a magnetic field B. $$F=qvB$$
$$v=frac{F}{qB}$$
I then attempted to find an equation which I could rewrite to give me $F$ and then substitute it into the equation for velocity. However, I can only find $$E=frac{F}{q}$$ for an electric field. I know the charge of an electron, but with the information given I cant find the electric field strength.
Any ideas as to which equation I could use?
homework-and-exercises electromagnetism energy magnetic-fields electrons
edited Dec 2 '18 at 15:27
Chair
3,67072034
asked Dec 2 '18 at 10:29
Pablo
1076
closed as off-topic by AccidentalFourierTransform, Chair, John Rennie, ZeroTheHero, LonelyProf Dec 3 '18 at 14:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – AccidentalFourierTransform, Chair, John Rennie, ZeroTheHero, LonelyProf
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
An electron, that has been accelerated from rest by a potential difference of $250 rm V$, enters a region of magnetic field of strength $0.12 rm T$. Show that speed of the electron after acceleration is $9.4 times 10^6 rm m s^{–1}.$
For this problem I first tried to find the velocity by rewriting the equation for the force an electron will experience when moving through a magnetic field B. $$F=qvB$$
$$v=frac{F}{qB}$$
I then attempted to find an equation which I could rewrite to give me $F$ and then substitute it into the equation for velocity. However, I can only find $$E=frac{F}{q}$$ for an electric field. I know the charge of an electron, but with the information given I cant find the electric field strength.
Any ideas as to which equation I could use?
homework-and-exercises electromagnetism energy magnetic-fields electrons
edited Dec 2 '18 at 15:27
Chair
3,67072034
asked Dec 2 '18 at 10:29
Pablo
1076
closed as off-topic by AccidentalFourierTransform, Chair, John Rennie, ZeroTheHero, LonelyProf Dec 3 '18 at 14:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – AccidentalFourierTransform, Chair, John Rennie, ZeroTheHero, LonelyProf
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
An electron, that has been accelerated from rest by a potential difference of $250 rm V$, enters a region of magnetic field of strength $0.12 rm T$. Show that speed of the electron after acceleration is $9.4 times 10^6 rm m s^{–1}.$
For this problem I first tried to find the velocity by rewriting the equation for the force an electron will experience when moving through a magnetic field B. $$F=qvB$$
$$v=frac{F}{qB}$$
I then attempted to find an equation which I could rewrite to give me $F$ and then substitute it into the equation for velocity. However, I can only find $$E=frac{F}{q}$$ for an electric field. I know the charge of an electron, but with the information given I cant find the electric field strength.
Any ideas as to which equation I could use?
homework-and-exercises electromagnetism energy magnetic-fields electrons
edited Dec 2 '18 at 15:27
Chair
3,67072034
asked Dec 2 '18 at 10:29
Pablo
1076
An electron, that has been accelerated from rest by a potential difference of $250 rm V$, enters a region of magnetic field of strength $0.12 rm T$. Show that speed of the electron after acceleration is $9.4 times 10^6 rm m s^{–1}.$
For this problem I first tried to find the velocity by rewriting the equation for the force an electron will experience when moving through a magnetic field B. $$F=qvB$$
$$v=frac{F}{qB}$$
I then attempted to find an equation which I could rewrite to give me $F$ and then substitute it into the equation for velocity. However, I can only find $$E=frac{F}{q}$$ for an electric field. I know the charge of an electron, but with the information given I cant find the electric field strength.
Any ideas as to which equation I could use?
homework-and-exercises electromagnetism energy magnetic-fields electrons
homework-and-exercises electromagnetism energy magnetic-fields electrons
edited Dec 2 '18 at 15:27
Chair
3,67072034
asked Dec 2 '18 at 10:29
Pablo
1076
edited Dec 2 '18 at 15:27
Chair
3,67072034
asked Dec 2 '18 at 10:29
Pablo
1076
edited Dec 2 '18 at 15:27
Chair
3,67072034
edited Dec 2 '18 at 15:27
Chair
3,67072034
edited Dec 2 '18 at 15:27
Chair
3,67072034
3,67072034
asked Dec 2 '18 at 10:29
Pablo
1076
asked Dec 2 '18 at 10:29
Pablo
1076
asked Dec 2 '18 at 10:29
Pablo
1076
1076
closed as off-topic by AccidentalFourierTransform, Chair, John Rennie, ZeroTheHero, LonelyProf Dec 3 '18 at 14:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – AccidentalFourierTransform, Chair, John Rennie, ZeroTheHero, LonelyProf
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by AccidentalFourierTransform, Chair, John Rennie, ZeroTheHero, LonelyProf Dec 3 '18 at 14:28
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – AccidentalFourierTransform, Chair, John Rennie, ZeroTheHero, LonelyProf
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
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votes
You need to observe that magnetic field never does any work on a moving charge so it can't accelerate or decelerate the electron. All the kinetic energy that the electron must have gained must come from the initial accelerating potential. So as the the electron is accelerated through a potential say V we can denote its kinetic energy as $ rm eV$. Now also we know that $KE=frac{1}{2}mv^2$ by equating the two you can obtain the velocity of the accelerated electron.
edited Dec 2 '18 at 15:24
Chair
3,67072034
answered Dec 2 '18 at 10:40
Aditya Garg
357110
Thanks for the answer. Where do you get $eV$ from? From which expression of kinetic energy is that from?
– Pablo
Dec 2 '18 at 11:22
Work done by any electric field on a charge particle is given as (charge)(potential difference between the two points) in this case charge is e and the P.D is V
– Aditya Garg
Dec 2 '18 at 11:45
@PabloeVis the customary unit of energy in particle physics. It's used because it's got such a nice direct relation to the experimental setup: A particle with the charge of 1e (electrons, protons, deuterons, etc.) gains/looses exactly 1eV when it's accelerated by an electric field of 1V. I.e. You want 42eV electrons? Just put 42 volts on your electron gun. You want 100keV X-Rays, apply 100kV to your X-Ray emitter (ok, not precisely because of conversion losses, but you get the idea...). In your question, the potential is 250V, you have electrons, so 250eV. Simple, isn't it?
– cmaster
Dec 2 '18 at 12:47
Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
– Rob Jeffries
Dec 2 '18 at 18:44
The acceleration I want to talk about here is the change in speed and not velocity ......... Please understand and sorry @Rob Jeffries
– Aditya Garg
Dec 2 '18 at 19:38
add a comment |
Since magnetic fields do no work, the electron's kinetic energy is 250 eV, giving the desired speed.
answered Dec 2 '18 at 10:36
d_b
1,435715
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You need to observe that magnetic field never does any work on a moving charge so it can't accelerate or decelerate the electron. All the kinetic energy that the electron must have gained must come from the initial accelerating potential. So as the the electron is accelerated through a potential say V we can denote its kinetic energy as $ rm eV$. Now also we know that $KE=frac{1}{2}mv^2$ by equating the two you can obtain the velocity of the accelerated electron.
edited Dec 2 '18 at 15:24
Chair
3,67072034
answered Dec 2 '18 at 10:40
Aditya Garg
357110
Thanks for the answer. Where do you get $eV$ from? From which expression of kinetic energy is that from?
– Pablo
Dec 2 '18 at 11:22
Work done by any electric field on a charge particle is given as (charge)(potential difference between the two points) in this case charge is e and the P.D is V
– Aditya Garg
Dec 2 '18 at 11:45
@PabloeVis the customary unit of energy in particle physics. It's used because it's got such a nice direct relation to the experimental setup: A particle with the charge of 1e (electrons, protons, deuterons, etc.) gains/looses exactly 1eV when it's accelerated by an electric field of 1V. I.e. You want 42eV electrons? Just put 42 volts on your electron gun. You want 100keV X-Rays, apply 100kV to your X-Ray emitter (ok, not precisely because of conversion losses, but you get the idea...). In your question, the potential is 250V, you have electrons, so 250eV. Simple, isn't it?
– cmaster
Dec 2 '18 at 12:47
Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
– Rob Jeffries
Dec 2 '18 at 18:44
The acceleration I want to talk about here is the change in speed and not velocity ......... Please understand and sorry @Rob Jeffries
– Aditya Garg
Dec 2 '18 at 19:38
add a comment |
You need to observe that magnetic field never does any work on a moving charge so it can't accelerate or decelerate the electron. All the kinetic energy that the electron must have gained must come from the initial accelerating potential. So as the the electron is accelerated through a potential say V we can denote its kinetic energy as $ rm eV$. Now also we know that $KE=frac{1}{2}mv^2$ by equating the two you can obtain the velocity of the accelerated electron.
edited Dec 2 '18 at 15:24
Chair
3,67072034
answered Dec 2 '18 at 10:40
Aditya Garg
357110
Thanks for the answer. Where do you get $eV$ from? From which expression of kinetic energy is that from?
– Pablo
Dec 2 '18 at 11:22
Work done by any electric field on a charge particle is given as (charge)(potential difference between the two points) in this case charge is e and the P.D is V
– Aditya Garg
Dec 2 '18 at 11:45
@PabloeVis the customary unit of energy in particle physics. It's used because it's got such a nice direct relation to the experimental setup: A particle with the charge of 1e (electrons, protons, deuterons, etc.) gains/looses exactly 1eV when it's accelerated by an electric field of 1V. I.e. You want 42eV electrons? Just put 42 volts on your electron gun. You want 100keV X-Rays, apply 100kV to your X-Ray emitter (ok, not precisely because of conversion losses, but you get the idea...). In your question, the potential is 250V, you have electrons, so 250eV. Simple, isn't it?
– cmaster
Dec 2 '18 at 12:47
Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
– Rob Jeffries
Dec 2 '18 at 18:44
The acceleration I want to talk about here is the change in speed and not velocity ......... Please understand and sorry @Rob Jeffries
– Aditya Garg
Dec 2 '18 at 19:38
add a comment |
You need to observe that magnetic field never does any work on a moving charge so it can't accelerate or decelerate the electron. All the kinetic energy that the electron must have gained must come from the initial accelerating potential. So as the the electron is accelerated through a potential say V we can denote its kinetic energy as $ rm eV$. Now also we know that $KE=frac{1}{2}mv^2$ by equating the two you can obtain the velocity of the accelerated electron.
edited Dec 2 '18 at 15:24
Chair
3,67072034
answered Dec 2 '18 at 10:40
Aditya Garg
357110
You need to observe that magnetic field never does any work on a moving charge so it can't accelerate or decelerate the electron. All the kinetic energy that the electron must have gained must come from the initial accelerating potential. So as the the electron is accelerated through a potential say V we can denote its kinetic energy as $ rm eV$. Now also we know that $KE=frac{1}{2}mv^2$ by equating the two you can obtain the velocity of the accelerated electron.
edited Dec 2 '18 at 15:24
Chair
3,67072034
answered Dec 2 '18 at 10:40
Aditya Garg
357110
edited Dec 2 '18 at 15:24
Chair
3,67072034
edited Dec 2 '18 at 15:24
Chair
3,67072034
edited Dec 2 '18 at 15:24
Chair
3,67072034
3,67072034
answered Dec 2 '18 at 10:40
Aditya Garg
357110
answered Dec 2 '18 at 10:40
Aditya Garg
357110
answered Dec 2 '18 at 10:40
Aditya Garg
357110
357110
Thanks for the answer. Where do you get $eV$ from? From which expression of kinetic energy is that from?
– Pablo
Dec 2 '18 at 11:22
Work done by any electric field on a charge particle is given as (charge)(potential difference between the two points) in this case charge is e and the P.D is V
– Aditya Garg
Dec 2 '18 at 11:45
@PabloeVis the customary unit of energy in particle physics. It's used because it's got such a nice direct relation to the experimental setup: A particle with the charge of 1e (electrons, protons, deuterons, etc.) gains/looses exactly 1eV when it's accelerated by an electric field of 1V. I.e. You want 42eV electrons? Just put 42 volts on your electron gun. You want 100keV X-Rays, apply 100kV to your X-Ray emitter (ok, not precisely because of conversion losses, but you get the idea...). In your question, the potential is 250V, you have electrons, so 250eV. Simple, isn't it?
– cmaster
Dec 2 '18 at 12:47
Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
– Rob Jeffries
Dec 2 '18 at 18:44
The acceleration I want to talk about here is the change in speed and not velocity ......... Please understand and sorry @Rob Jeffries
– Aditya Garg
Dec 2 '18 at 19:38
add a comment |
Thanks for the answer. Where do you get $eV$ from? From which expression of kinetic energy is that from?
– Pablo
Dec 2 '18 at 11:22
Work done by any electric field on a charge particle is given as (charge)(potential difference between the two points) in this case charge is e and the P.D is V
– Aditya Garg
Dec 2 '18 at 11:45
@PabloeVis the customary unit of energy in particle physics. It's used because it's got such a nice direct relation to the experimental setup: A particle with the charge of 1e (electrons, protons, deuterons, etc.) gains/looses exactly 1eV when it's accelerated by an electric field of 1V. I.e. You want 42eV electrons? Just put 42 volts on your electron gun. You want 100keV X-Rays, apply 100kV to your X-Ray emitter (ok, not precisely because of conversion losses, but you get the idea...). In your question, the potential is 250V, you have electrons, so 250eV. Simple, isn't it?
– cmaster
Dec 2 '18 at 12:47
Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
– Rob Jeffries
Dec 2 '18 at 18:44
The acceleration I want to talk about here is the change in speed and not velocity ......... Please understand and sorry @Rob Jeffries
– Aditya Garg
Dec 2 '18 at 19:38
Thanks for the answer. Where do you get $eV$ from? From which expression of kinetic energy is that from?
– Pablo
Dec 2 '18 at 11:22
Thanks for the answer. Where do you get $eV$ from? From which expression of kinetic energy is that from?
– Pablo
Dec 2 '18 at 11:22
Work done by any electric field on a charge particle is given as (charge)(potential difference between the two points) in this case charge is e and the P.D is V
– Aditya Garg
Dec 2 '18 at 11:45
Work done by any electric field on a charge particle is given as (charge)(potential difference between the two points) in this case charge is e and the P.D is V
– Aditya Garg
Dec 2 '18 at 11:45
@Pablo
eV is the customary unit of energy in particle physics. It's used because it's got such a nice direct relation to the experimental setup: A particle with the charge of 1e (electrons, protons, deuterons, etc.) gains/looses exactly 1eV when it's accelerated by an electric field of 1V. I.e. You want 42eV electrons? Just put 42 volts on your electron gun. You want 100keV X-Rays, apply 100kV to your X-Ray emitter (ok, not precisely because of conversion losses, but you get the idea...). In your question, the potential is 250V, you have electrons, so 250eV. Simple, isn't it?– cmaster
Dec 2 '18 at 12:47
@Pablo
eV is the customary unit of energy in particle physics. It's used because it's got such a nice direct relation to the experimental setup: A particle with the charge of 1e (electrons, protons, deuterons, etc.) gains/looses exactly 1eV when it's accelerated by an electric field of 1V. I.e. You want 42eV electrons? Just put 42 volts on your electron gun. You want 100keV X-Rays, apply 100kV to your X-Ray emitter (ok, not precisely because of conversion losses, but you get the idea...). In your question, the potential is 250V, you have electrons, so 250eV. Simple, isn't it?– cmaster
Dec 2 '18 at 12:47
Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
– Rob Jeffries
Dec 2 '18 at 18:44
Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
– Rob Jeffries
Dec 2 '18 at 18:44
The acceleration I want to talk about here is the change in speed and not velocity ......... Please understand and sorry @Rob Jeffries
– Aditya Garg
Dec 2 '18 at 19:38
The acceleration I want to talk about here is the change in speed and not velocity ......... Please understand and sorry @Rob Jeffries
– Aditya Garg
Dec 2 '18 at 19:38
add a comment |
Since magnetic fields do no work, the electron's kinetic energy is 250 eV, giving the desired speed.
answered Dec 2 '18 at 10:36
d_b
1,435715
add a comment |
Since magnetic fields do no work, the electron's kinetic energy is 250 eV, giving the desired speed.
answered Dec 2 '18 at 10:36
d_b
1,435715
add a comment |
Since magnetic fields do no work, the electron's kinetic energy is 250 eV, giving the desired speed.
answered Dec 2 '18 at 10:36
d_b
1,435715
Since magnetic fields do no work, the electron's kinetic energy is 250 eV, giving the desired speed.
answered Dec 2 '18 at 10:36
d_b
1,435715
answered Dec 2 '18 at 10:36
d_b
1,435715
answered Dec 2 '18 at 10:36
d_b
1,435715
answered Dec 2 '18 at 10:36
d_b
1,435715
1,435715
add a comment |
add a comment |
