Where is the energy transfer from a metal ball falling from a magnet?












7














A ball falls from a magnet but the magnet still exerts an upwards force against gravity yet the ball falls anyway. However, the ball slows down and thus the sound when it hits the floor is less signalling that some energy has been lost during its descent. What I'm wondering is where that energy has gone? Has the magnet gained magnet energy? Or has earth gained energy. Or has the ball not lost energy but its remaining energy just wasn't turned into sound upon contact with the ground? Or is it something else?










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  • 1




    Is the metal ball made of iron or steel?
    – cms
    yesterday
















7














A ball falls from a magnet but the magnet still exerts an upwards force against gravity yet the ball falls anyway. However, the ball slows down and thus the sound when it hits the floor is less signalling that some energy has been lost during its descent. What I'm wondering is where that energy has gone? Has the magnet gained magnet energy? Or has earth gained energy. Or has the ball not lost energy but its remaining energy just wasn't turned into sound upon contact with the ground? Or is it something else?










share|cite|improve this question


















  • 1




    Is the metal ball made of iron or steel?
    – cms
    yesterday














7












7








7


2





A ball falls from a magnet but the magnet still exerts an upwards force against gravity yet the ball falls anyway. However, the ball slows down and thus the sound when it hits the floor is less signalling that some energy has been lost during its descent. What I'm wondering is where that energy has gone? Has the magnet gained magnet energy? Or has earth gained energy. Or has the ball not lost energy but its remaining energy just wasn't turned into sound upon contact with the ground? Or is it something else?










share|cite|improve this question













A ball falls from a magnet but the magnet still exerts an upwards force against gravity yet the ball falls anyway. However, the ball slows down and thus the sound when it hits the floor is less signalling that some energy has been lost during its descent. What I'm wondering is where that energy has gone? Has the magnet gained magnet energy? Or has earth gained energy. Or has the ball not lost energy but its remaining energy just wasn't turned into sound upon contact with the ground? Or is it something else?







electromagnetism gravity






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asked yesterday









yoloyolo

6981212




6981212








  • 1




    Is the metal ball made of iron or steel?
    – cms
    yesterday














  • 1




    Is the metal ball made of iron or steel?
    – cms
    yesterday








1




1




Is the metal ball made of iron or steel?
– cms
yesterday




Is the metal ball made of iron or steel?
– cms
yesterday










5 Answers
5






active

oldest

votes


















4














My favorite version of this demonstration is dropping a magnet through a conducting tube. In that case the gravitational potential energy in the system is converted, by the electromagnetic field, into thermal energy in the tube.
The mechanism is Joule heating by the induced currents as the magnetic field at different parts of the tube changes.
The terminal velocity decreases if the conductivity of the tube is improved; a superconducting tube would prevent the magnet from falling altogether, such is another well-known demonstration.



In your case, with a fixed magnet and a falling metal ball, the mechanism is the same: eddy currents in the metal cause resistive heating, for which the energy comes from the electromagnetic field. But the geometry there makes computations trickier.






share|cite|improve this answer





















  • What if the “metal ball” is a ferrous material like iron or steel? Simple induced magnetization would be a much better explanation then for the force exerted on the ball.
    – cms
    yesterday






  • 1




    Oh, true. I'll see if the question gets clarified.
    – rob
    yesterday



















4














If the ball is made of iron



You put “magnetic” potential energy into the system when you brought the metal ball up to the magnet. The sign of that potential energy is negative indicating an attractive force.



When you drop the ball, the ball loses gravitational potential energy, but adds magnetic potential energy and kinetic energy. Therefore less energy was available for kinetic energy and the corresponding impact speed was less.



edit:



We can write this as an energy relationship:
$$
begin{align}
Delta E &= Delta K + Delta U \
0 &= (K_f-K_i) + Delta U_G + Delta U_{M} \
K_f &= -Delta U_G - Delta U_M
end{align}
$$

Where $K_i = Delta E = 0$. In the situation where the ball falls the gravitational potential energy decreases ($Delta U_G < 0$) but the magnetic potential energy increases ($Delta U_M > 0$), since the metal ball has moved further from the magnet. The type of forces two magnets experience is a conservative force (since it's path independent) and so it makes sense to talk about a magnetic potential energy.



The ball would need to be iron (or one of its alloys like steel) because in terms of everday materials, iron is the only one that can be temporarily (induced) magnetized. For most other materials, $Delta U_M approx 0$.






share|cite|improve this answer























  • Why only iron and not any metal?
    – yolo
    yesterday






  • 1




    @yolo: because not all metals are attracted to magnets. The ball would need to be made of iron, cobalt, nickel, or maybe a few other alloys.
    – Peter Shor
    yesterday












  • So the ball gains some sort of 'MPE' (x) as well as GPE (y) when put into the system so currently- in general we have this; M=-x, B=x+y , E=0 where e is earth, b is ball and m is the magnet (and whatever lifted the ball has -y to keep it balanced. What were you saying is the resultant energy difference?
    – yolo
    yesterday












  • Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
    – yolo
    yesterday



















3














The magnetic force acting on the ball as a whole (ponderomotive force due to magnet) slows down the fall, thus does negative work on the ball. This means energy is taken away from the mechanical energy of the system (ball+Earth). Since the force is due to electromagnetic field, the only obvious destination where it goes is the electromagnetic energy of the system (ball+magnet). The electromagnetic field is everywhere, and due to descent of the ball, it changes everywhere. In some places, the EM energy density increases, in some other places, it decreases. If there are no losses of energy to heat, net EM energy has to increase by the same amount that mechanical energy decreased.






share|cite|improve this answer





























    3














    The magnet does negative work on the ball. The energy becomes internal energy in the magnet/ball.



    You might have heard that magnetic forces do no work. However, this is only true for magnetic fields acting on classical particles. The permanent magnetic dipoles in the atoms of the magnet and ball, which are responsible for the force between them, can only be explained in quantum mechanics, as shown by the Bohr--van Leeuwen theorem, and in the quantum case work can be performed. The end result is that the quantum energy levels of the ball and magnet will be higher than they were before. This internal energy may then be manifested as heat.






    share|cite|improve this answer





























      2














      Magnetic force makes work, for example, in electromotors, in electrogenertors, etc. There are many devices that use magnetic force to do some useful work. The energy conservation is applicable to the sum of all different forms of energy of the system.






      share|cite|improve this answer





















      • I like the clear intent of this answer. However it will cause confusion later when the student discovers that the differential work $dW = vec Fcdot dx = dtvec Fcdotvec v$ is exactly zero for the magnetic force $qvec vtimesvec B$ due to a magnetic field. A little more explanation would be valuable.
        – rob
        yesterday






      • 2




        @rob: But the original question is not about interaction of an external magnetic field with a single moving charge; rather, it is about interaction of magnets (which involve many other forces existing inside them).
        – Vladimir Kalitvianski
        yesterday











      Your Answer





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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      My favorite version of this demonstration is dropping a magnet through a conducting tube. In that case the gravitational potential energy in the system is converted, by the electromagnetic field, into thermal energy in the tube.
      The mechanism is Joule heating by the induced currents as the magnetic field at different parts of the tube changes.
      The terminal velocity decreases if the conductivity of the tube is improved; a superconducting tube would prevent the magnet from falling altogether, such is another well-known demonstration.



      In your case, with a fixed magnet and a falling metal ball, the mechanism is the same: eddy currents in the metal cause resistive heating, for which the energy comes from the electromagnetic field. But the geometry there makes computations trickier.






      share|cite|improve this answer





















      • What if the “metal ball” is a ferrous material like iron or steel? Simple induced magnetization would be a much better explanation then for the force exerted on the ball.
        – cms
        yesterday






      • 1




        Oh, true. I'll see if the question gets clarified.
        – rob
        yesterday
















      4














      My favorite version of this demonstration is dropping a magnet through a conducting tube. In that case the gravitational potential energy in the system is converted, by the electromagnetic field, into thermal energy in the tube.
      The mechanism is Joule heating by the induced currents as the magnetic field at different parts of the tube changes.
      The terminal velocity decreases if the conductivity of the tube is improved; a superconducting tube would prevent the magnet from falling altogether, such is another well-known demonstration.



      In your case, with a fixed magnet and a falling metal ball, the mechanism is the same: eddy currents in the metal cause resistive heating, for which the energy comes from the electromagnetic field. But the geometry there makes computations trickier.






      share|cite|improve this answer





















      • What if the “metal ball” is a ferrous material like iron or steel? Simple induced magnetization would be a much better explanation then for the force exerted on the ball.
        – cms
        yesterday






      • 1




        Oh, true. I'll see if the question gets clarified.
        – rob
        yesterday














      4












      4








      4






      My favorite version of this demonstration is dropping a magnet through a conducting tube. In that case the gravitational potential energy in the system is converted, by the electromagnetic field, into thermal energy in the tube.
      The mechanism is Joule heating by the induced currents as the magnetic field at different parts of the tube changes.
      The terminal velocity decreases if the conductivity of the tube is improved; a superconducting tube would prevent the magnet from falling altogether, such is another well-known demonstration.



      In your case, with a fixed magnet and a falling metal ball, the mechanism is the same: eddy currents in the metal cause resistive heating, for which the energy comes from the electromagnetic field. But the geometry there makes computations trickier.






      share|cite|improve this answer












      My favorite version of this demonstration is dropping a magnet through a conducting tube. In that case the gravitational potential energy in the system is converted, by the electromagnetic field, into thermal energy in the tube.
      The mechanism is Joule heating by the induced currents as the magnetic field at different parts of the tube changes.
      The terminal velocity decreases if the conductivity of the tube is improved; a superconducting tube would prevent the magnet from falling altogether, such is another well-known demonstration.



      In your case, with a fixed magnet and a falling metal ball, the mechanism is the same: eddy currents in the metal cause resistive heating, for which the energy comes from the electromagnetic field. But the geometry there makes computations trickier.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered yesterday









      robrob

      39.7k971164




      39.7k971164












      • What if the “metal ball” is a ferrous material like iron or steel? Simple induced magnetization would be a much better explanation then for the force exerted on the ball.
        – cms
        yesterday






      • 1




        Oh, true. I'll see if the question gets clarified.
        – rob
        yesterday


















      • What if the “metal ball” is a ferrous material like iron or steel? Simple induced magnetization would be a much better explanation then for the force exerted on the ball.
        – cms
        yesterday






      • 1




        Oh, true. I'll see if the question gets clarified.
        – rob
        yesterday
















      What if the “metal ball” is a ferrous material like iron or steel? Simple induced magnetization would be a much better explanation then for the force exerted on the ball.
      – cms
      yesterday




      What if the “metal ball” is a ferrous material like iron or steel? Simple induced magnetization would be a much better explanation then for the force exerted on the ball.
      – cms
      yesterday




      1




      1




      Oh, true. I'll see if the question gets clarified.
      – rob
      yesterday




      Oh, true. I'll see if the question gets clarified.
      – rob
      yesterday











      4














      If the ball is made of iron



      You put “magnetic” potential energy into the system when you brought the metal ball up to the magnet. The sign of that potential energy is negative indicating an attractive force.



      When you drop the ball, the ball loses gravitational potential energy, but adds magnetic potential energy and kinetic energy. Therefore less energy was available for kinetic energy and the corresponding impact speed was less.



      edit:



      We can write this as an energy relationship:
      $$
      begin{align}
      Delta E &= Delta K + Delta U \
      0 &= (K_f-K_i) + Delta U_G + Delta U_{M} \
      K_f &= -Delta U_G - Delta U_M
      end{align}
      $$

      Where $K_i = Delta E = 0$. In the situation where the ball falls the gravitational potential energy decreases ($Delta U_G < 0$) but the magnetic potential energy increases ($Delta U_M > 0$), since the metal ball has moved further from the magnet. The type of forces two magnets experience is a conservative force (since it's path independent) and so it makes sense to talk about a magnetic potential energy.



      The ball would need to be iron (or one of its alloys like steel) because in terms of everday materials, iron is the only one that can be temporarily (induced) magnetized. For most other materials, $Delta U_M approx 0$.






      share|cite|improve this answer























      • Why only iron and not any metal?
        – yolo
        yesterday






      • 1




        @yolo: because not all metals are attracted to magnets. The ball would need to be made of iron, cobalt, nickel, or maybe a few other alloys.
        – Peter Shor
        yesterday












      • So the ball gains some sort of 'MPE' (x) as well as GPE (y) when put into the system so currently- in general we have this; M=-x, B=x+y , E=0 where e is earth, b is ball and m is the magnet (and whatever lifted the ball has -y to keep it balanced. What were you saying is the resultant energy difference?
        – yolo
        yesterday












      • Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
        – yolo
        yesterday
















      4














      If the ball is made of iron



      You put “magnetic” potential energy into the system when you brought the metal ball up to the magnet. The sign of that potential energy is negative indicating an attractive force.



      When you drop the ball, the ball loses gravitational potential energy, but adds magnetic potential energy and kinetic energy. Therefore less energy was available for kinetic energy and the corresponding impact speed was less.



      edit:



      We can write this as an energy relationship:
      $$
      begin{align}
      Delta E &= Delta K + Delta U \
      0 &= (K_f-K_i) + Delta U_G + Delta U_{M} \
      K_f &= -Delta U_G - Delta U_M
      end{align}
      $$

      Where $K_i = Delta E = 0$. In the situation where the ball falls the gravitational potential energy decreases ($Delta U_G < 0$) but the magnetic potential energy increases ($Delta U_M > 0$), since the metal ball has moved further from the magnet. The type of forces two magnets experience is a conservative force (since it's path independent) and so it makes sense to talk about a magnetic potential energy.



      The ball would need to be iron (or one of its alloys like steel) because in terms of everday materials, iron is the only one that can be temporarily (induced) magnetized. For most other materials, $Delta U_M approx 0$.






      share|cite|improve this answer























      • Why only iron and not any metal?
        – yolo
        yesterday






      • 1




        @yolo: because not all metals are attracted to magnets. The ball would need to be made of iron, cobalt, nickel, or maybe a few other alloys.
        – Peter Shor
        yesterday












      • So the ball gains some sort of 'MPE' (x) as well as GPE (y) when put into the system so currently- in general we have this; M=-x, B=x+y , E=0 where e is earth, b is ball and m is the magnet (and whatever lifted the ball has -y to keep it balanced. What were you saying is the resultant energy difference?
        – yolo
        yesterday












      • Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
        – yolo
        yesterday














      4












      4








      4






      If the ball is made of iron



      You put “magnetic” potential energy into the system when you brought the metal ball up to the magnet. The sign of that potential energy is negative indicating an attractive force.



      When you drop the ball, the ball loses gravitational potential energy, but adds magnetic potential energy and kinetic energy. Therefore less energy was available for kinetic energy and the corresponding impact speed was less.



      edit:



      We can write this as an energy relationship:
      $$
      begin{align}
      Delta E &= Delta K + Delta U \
      0 &= (K_f-K_i) + Delta U_G + Delta U_{M} \
      K_f &= -Delta U_G - Delta U_M
      end{align}
      $$

      Where $K_i = Delta E = 0$. In the situation where the ball falls the gravitational potential energy decreases ($Delta U_G < 0$) but the magnetic potential energy increases ($Delta U_M > 0$), since the metal ball has moved further from the magnet. The type of forces two magnets experience is a conservative force (since it's path independent) and so it makes sense to talk about a magnetic potential energy.



      The ball would need to be iron (or one of its alloys like steel) because in terms of everday materials, iron is the only one that can be temporarily (induced) magnetized. For most other materials, $Delta U_M approx 0$.






      share|cite|improve this answer














      If the ball is made of iron



      You put “magnetic” potential energy into the system when you brought the metal ball up to the magnet. The sign of that potential energy is negative indicating an attractive force.



      When you drop the ball, the ball loses gravitational potential energy, but adds magnetic potential energy and kinetic energy. Therefore less energy was available for kinetic energy and the corresponding impact speed was less.



      edit:



      We can write this as an energy relationship:
      $$
      begin{align}
      Delta E &= Delta K + Delta U \
      0 &= (K_f-K_i) + Delta U_G + Delta U_{M} \
      K_f &= -Delta U_G - Delta U_M
      end{align}
      $$

      Where $K_i = Delta E = 0$. In the situation where the ball falls the gravitational potential energy decreases ($Delta U_G < 0$) but the magnetic potential energy increases ($Delta U_M > 0$), since the metal ball has moved further from the magnet. The type of forces two magnets experience is a conservative force (since it's path independent) and so it makes sense to talk about a magnetic potential energy.



      The ball would need to be iron (or one of its alloys like steel) because in terms of everday materials, iron is the only one that can be temporarily (induced) magnetized. For most other materials, $Delta U_M approx 0$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 8 hours ago

























      answered yesterday









      cmscms

      3,0022415




      3,0022415












      • Why only iron and not any metal?
        – yolo
        yesterday






      • 1




        @yolo: because not all metals are attracted to magnets. The ball would need to be made of iron, cobalt, nickel, or maybe a few other alloys.
        – Peter Shor
        yesterday












      • So the ball gains some sort of 'MPE' (x) as well as GPE (y) when put into the system so currently- in general we have this; M=-x, B=x+y , E=0 where e is earth, b is ball and m is the magnet (and whatever lifted the ball has -y to keep it balanced. What were you saying is the resultant energy difference?
        – yolo
        yesterday












      • Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
        – yolo
        yesterday


















      • Why only iron and not any metal?
        – yolo
        yesterday






      • 1




        @yolo: because not all metals are attracted to magnets. The ball would need to be made of iron, cobalt, nickel, or maybe a few other alloys.
        – Peter Shor
        yesterday












      • So the ball gains some sort of 'MPE' (x) as well as GPE (y) when put into the system so currently- in general we have this; M=-x, B=x+y , E=0 where e is earth, b is ball and m is the magnet (and whatever lifted the ball has -y to keep it balanced. What were you saying is the resultant energy difference?
        – yolo
        yesterday












      • Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
        – yolo
        yesterday
















      Why only iron and not any metal?
      – yolo
      yesterday




      Why only iron and not any metal?
      – yolo
      yesterday




      1




      1




      @yolo: because not all metals are attracted to magnets. The ball would need to be made of iron, cobalt, nickel, or maybe a few other alloys.
      – Peter Shor
      yesterday






      @yolo: because not all metals are attracted to magnets. The ball would need to be made of iron, cobalt, nickel, or maybe a few other alloys.
      – Peter Shor
      yesterday














      So the ball gains some sort of 'MPE' (x) as well as GPE (y) when put into the system so currently- in general we have this; M=-x, B=x+y , E=0 where e is earth, b is ball and m is the magnet (and whatever lifted the ball has -y to keep it balanced. What were you saying is the resultant energy difference?
      – yolo
      yesterday






      So the ball gains some sort of 'MPE' (x) as well as GPE (y) when put into the system so currently- in general we have this; M=-x, B=x+y , E=0 where e is earth, b is ball and m is the magnet (and whatever lifted the ball has -y to keep it balanced. What were you saying is the resultant energy difference?
      – yolo
      yesterday














      Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
      – yolo
      yesterday




      Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
      – yolo
      yesterday











      3














      The magnetic force acting on the ball as a whole (ponderomotive force due to magnet) slows down the fall, thus does negative work on the ball. This means energy is taken away from the mechanical energy of the system (ball+Earth). Since the force is due to electromagnetic field, the only obvious destination where it goes is the electromagnetic energy of the system (ball+magnet). The electromagnetic field is everywhere, and due to descent of the ball, it changes everywhere. In some places, the EM energy density increases, in some other places, it decreases. If there are no losses of energy to heat, net EM energy has to increase by the same amount that mechanical energy decreased.






      share|cite|improve this answer


























        3














        The magnetic force acting on the ball as a whole (ponderomotive force due to magnet) slows down the fall, thus does negative work on the ball. This means energy is taken away from the mechanical energy of the system (ball+Earth). Since the force is due to electromagnetic field, the only obvious destination where it goes is the electromagnetic energy of the system (ball+magnet). The electromagnetic field is everywhere, and due to descent of the ball, it changes everywhere. In some places, the EM energy density increases, in some other places, it decreases. If there are no losses of energy to heat, net EM energy has to increase by the same amount that mechanical energy decreased.






        share|cite|improve this answer
























          3












          3








          3






          The magnetic force acting on the ball as a whole (ponderomotive force due to magnet) slows down the fall, thus does negative work on the ball. This means energy is taken away from the mechanical energy of the system (ball+Earth). Since the force is due to electromagnetic field, the only obvious destination where it goes is the electromagnetic energy of the system (ball+magnet). The electromagnetic field is everywhere, and due to descent of the ball, it changes everywhere. In some places, the EM energy density increases, in some other places, it decreases. If there are no losses of energy to heat, net EM energy has to increase by the same amount that mechanical energy decreased.






          share|cite|improve this answer












          The magnetic force acting on the ball as a whole (ponderomotive force due to magnet) slows down the fall, thus does negative work on the ball. This means energy is taken away from the mechanical energy of the system (ball+Earth). Since the force is due to electromagnetic field, the only obvious destination where it goes is the electromagnetic energy of the system (ball+magnet). The electromagnetic field is everywhere, and due to descent of the ball, it changes everywhere. In some places, the EM energy density increases, in some other places, it decreases. If there are no losses of energy to heat, net EM energy has to increase by the same amount that mechanical energy decreased.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Ján LalinskýJán Lalinský

          14.3k1334




          14.3k1334























              3














              The magnet does negative work on the ball. The energy becomes internal energy in the magnet/ball.



              You might have heard that magnetic forces do no work. However, this is only true for magnetic fields acting on classical particles. The permanent magnetic dipoles in the atoms of the magnet and ball, which are responsible for the force between them, can only be explained in quantum mechanics, as shown by the Bohr--van Leeuwen theorem, and in the quantum case work can be performed. The end result is that the quantum energy levels of the ball and magnet will be higher than they were before. This internal energy may then be manifested as heat.






              share|cite|improve this answer


























                3














                The magnet does negative work on the ball. The energy becomes internal energy in the magnet/ball.



                You might have heard that magnetic forces do no work. However, this is only true for magnetic fields acting on classical particles. The permanent magnetic dipoles in the atoms of the magnet and ball, which are responsible for the force between them, can only be explained in quantum mechanics, as shown by the Bohr--van Leeuwen theorem, and in the quantum case work can be performed. The end result is that the quantum energy levels of the ball and magnet will be higher than they were before. This internal energy may then be manifested as heat.






                share|cite|improve this answer
























                  3












                  3








                  3






                  The magnet does negative work on the ball. The energy becomes internal energy in the magnet/ball.



                  You might have heard that magnetic forces do no work. However, this is only true for magnetic fields acting on classical particles. The permanent magnetic dipoles in the atoms of the magnet and ball, which are responsible for the force between them, can only be explained in quantum mechanics, as shown by the Bohr--van Leeuwen theorem, and in the quantum case work can be performed. The end result is that the quantum energy levels of the ball and magnet will be higher than they were before. This internal energy may then be manifested as heat.






                  share|cite|improve this answer












                  The magnet does negative work on the ball. The energy becomes internal energy in the magnet/ball.



                  You might have heard that magnetic forces do no work. However, this is only true for magnetic fields acting on classical particles. The permanent magnetic dipoles in the atoms of the magnet and ball, which are responsible for the force between them, can only be explained in quantum mechanics, as shown by the Bohr--van Leeuwen theorem, and in the quantum case work can be performed. The end result is that the quantum energy levels of the ball and magnet will be higher than they were before. This internal energy may then be manifested as heat.







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                  share|cite|improve this answer










                  answered yesterday









                  knzhouknzhou

                  42.5k11117203




                  42.5k11117203























                      2














                      Magnetic force makes work, for example, in electromotors, in electrogenertors, etc. There are many devices that use magnetic force to do some useful work. The energy conservation is applicable to the sum of all different forms of energy of the system.






                      share|cite|improve this answer





















                      • I like the clear intent of this answer. However it will cause confusion later when the student discovers that the differential work $dW = vec Fcdot dx = dtvec Fcdotvec v$ is exactly zero for the magnetic force $qvec vtimesvec B$ due to a magnetic field. A little more explanation would be valuable.
                        – rob
                        yesterday






                      • 2




                        @rob: But the original question is not about interaction of an external magnetic field with a single moving charge; rather, it is about interaction of magnets (which involve many other forces existing inside them).
                        – Vladimir Kalitvianski
                        yesterday
















                      2














                      Magnetic force makes work, for example, in electromotors, in electrogenertors, etc. There are many devices that use magnetic force to do some useful work. The energy conservation is applicable to the sum of all different forms of energy of the system.






                      share|cite|improve this answer





















                      • I like the clear intent of this answer. However it will cause confusion later when the student discovers that the differential work $dW = vec Fcdot dx = dtvec Fcdotvec v$ is exactly zero for the magnetic force $qvec vtimesvec B$ due to a magnetic field. A little more explanation would be valuable.
                        – rob
                        yesterday






                      • 2




                        @rob: But the original question is not about interaction of an external magnetic field with a single moving charge; rather, it is about interaction of magnets (which involve many other forces existing inside them).
                        – Vladimir Kalitvianski
                        yesterday














                      2












                      2








                      2






                      Magnetic force makes work, for example, in electromotors, in electrogenertors, etc. There are many devices that use magnetic force to do some useful work. The energy conservation is applicable to the sum of all different forms of energy of the system.






                      share|cite|improve this answer












                      Magnetic force makes work, for example, in electromotors, in electrogenertors, etc. There are many devices that use magnetic force to do some useful work. The energy conservation is applicable to the sum of all different forms of energy of the system.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered yesterday









                      Vladimir KalitvianskiVladimir Kalitvianski

                      10.5k11234




                      10.5k11234












                      • I like the clear intent of this answer. However it will cause confusion later when the student discovers that the differential work $dW = vec Fcdot dx = dtvec Fcdotvec v$ is exactly zero for the magnetic force $qvec vtimesvec B$ due to a magnetic field. A little more explanation would be valuable.
                        – rob
                        yesterday






                      • 2




                        @rob: But the original question is not about interaction of an external magnetic field with a single moving charge; rather, it is about interaction of magnets (which involve many other forces existing inside them).
                        – Vladimir Kalitvianski
                        yesterday


















                      • I like the clear intent of this answer. However it will cause confusion later when the student discovers that the differential work $dW = vec Fcdot dx = dtvec Fcdotvec v$ is exactly zero for the magnetic force $qvec vtimesvec B$ due to a magnetic field. A little more explanation would be valuable.
                        – rob
                        yesterday






                      • 2




                        @rob: But the original question is not about interaction of an external magnetic field with a single moving charge; rather, it is about interaction of magnets (which involve many other forces existing inside them).
                        – Vladimir Kalitvianski
                        yesterday
















                      I like the clear intent of this answer. However it will cause confusion later when the student discovers that the differential work $dW = vec Fcdot dx = dtvec Fcdotvec v$ is exactly zero for the magnetic force $qvec vtimesvec B$ due to a magnetic field. A little more explanation would be valuable.
                      – rob
                      yesterday




                      I like the clear intent of this answer. However it will cause confusion later when the student discovers that the differential work $dW = vec Fcdot dx = dtvec Fcdotvec v$ is exactly zero for the magnetic force $qvec vtimesvec B$ due to a magnetic field. A little more explanation would be valuable.
                      – rob
                      yesterday




                      2




                      2




                      @rob: But the original question is not about interaction of an external magnetic field with a single moving charge; rather, it is about interaction of magnets (which involve many other forces existing inside them).
                      – Vladimir Kalitvianski
                      yesterday




                      @rob: But the original question is not about interaction of an external magnetic field with a single moving charge; rather, it is about interaction of magnets (which involve many other forces existing inside them).
                      – Vladimir Kalitvianski
                      yesterday


















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