An electron moving through a magnetic field [closed]

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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. $$F=qvB$$
$$v=fracFqB$$
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 $$E=fracFq$$ 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?










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closed as off-topic by AccidentalFourierTransform, Chair, John Rennie, ZeroTheHero, LonelyProf Dec 3 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.
















    up vote
    0
    down vote

    favorite













    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. $$F=qvB$$
    $$v=fracFqB$$
    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 $$E=fracFq$$ 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?










    share|cite|improve this question















    closed as off-topic by AccidentalFourierTransform, Chair, John Rennie, ZeroTheHero, LonelyProf Dec 3 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.














      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      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. $$F=qvB$$
      $$v=fracFqB$$
      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 $$E=fracFq$$ 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?










      share|cite|improve this question
















      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. $$F=qvB$$
      $$v=fracFqB$$
      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 $$E=fracFq$$ 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






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      edited Dec 2 at 15:27









      Chair

      3,62572034




      3,62572034










      asked Dec 2 at 10:29









      Pablo

      1076




      1076




      closed as off-topic by AccidentalFourierTransform, Chair, John Rennie, ZeroTheHero, LonelyProf Dec 3 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 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.




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          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 $KE=frac12mv^2$ by equating the two you can obtain the velocity of the accelerated electron.






          share|cite|improve this answer






















          • Thanks for the answer. Where do you get $eV$ from? From which expression of kinetic energy is that from?
            – Pablo
            Dec 2 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 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 at 12:47










          • Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
            – Rob Jeffries
            Dec 2 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 at 19:38


















          up vote
          4
          down vote













          Since magnetic fields do no work, the electron's kinetic energy is 250 eV, giving the desired speed.






          share|cite|improve this answer



























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            4
            down vote



            accepted










            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 $KE=frac12mv^2$ by equating the two you can obtain the velocity of the accelerated electron.






            share|cite|improve this answer






















            • Thanks for the answer. Where do you get $eV$ from? From which expression of kinetic energy is that from?
              – Pablo
              Dec 2 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 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 at 12:47










            • Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
              – Rob Jeffries
              Dec 2 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 at 19:38















            up vote
            4
            down vote



            accepted










            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 $KE=frac12mv^2$ by equating the two you can obtain the velocity of the accelerated electron.






            share|cite|improve this answer






















            • Thanks for the answer. Where do you get $eV$ from? From which expression of kinetic energy is that from?
              – Pablo
              Dec 2 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 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 at 12:47










            • Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
              – Rob Jeffries
              Dec 2 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 at 19:38













            up vote
            4
            down vote



            accepted







            up vote
            4
            down vote



            accepted






            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 $KE=frac12mv^2$ by equating the two you can obtain the velocity of the accelerated electron.






            share|cite|improve this answer














            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 $KE=frac12mv^2$ by equating the two you can obtain the velocity of the accelerated electron.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 2 at 15:24









            Chair

            3,62572034




            3,62572034










            answered Dec 2 at 10:40









            Aditya Garg

            333110




            333110











            • Thanks for the answer. Where do you get $eV$ from? From which expression of kinetic energy is that from?
              – Pablo
              Dec 2 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 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 at 12:47










            • Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
              – Rob Jeffries
              Dec 2 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 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 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 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 at 12:47










            • Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
              – Rob Jeffries
              Dec 2 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 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 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 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 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 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 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 at 12:47












            Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
            – Rob Jeffries
            Dec 2 at 18:44




            Magnetic fields certainly do "accelerate or decelerate" charged particles. Perhaps you should rephrase that.
            – Rob Jeffries
            Dec 2 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 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 at 19:38











            up vote
            4
            down vote













            Since magnetic fields do no work, the electron's kinetic energy is 250 eV, giving the desired speed.






            share|cite|improve this answer
























              up vote
              4
              down vote













              Since magnetic fields do no work, the electron's kinetic energy is 250 eV, giving the desired speed.






              share|cite|improve this answer






















                up vote
                4
                down vote










                up vote
                4
                down vote









                Since magnetic fields do no work, the electron's kinetic energy is 250 eV, giving the desired speed.






                share|cite|improve this answer












                Since magnetic fields do no work, the electron's kinetic energy is 250 eV, giving the desired speed.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 at 10:36









                d_b

                1,250715




                1,250715












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