Mixing
Where is the energy transfer from a metal ball falling from a magnet?
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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
asked Jan 6 at 13:28
yoloyolo
7181214
$endgroup$
add a comment |
$begingroup$
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
asked Jan 6 at 13:28
yoloyolo
7181214
$endgroup$
1
$begingroup$
Is the metal ball made of iron or steel?
$endgroup$
– cms
Jan 6 at 15:02
add a comment |
$begingroup$
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
asked Jan 6 at 13:28
yoloyolo
7181214
$endgroup$
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
electromagnetism gravity
asked Jan 6 at 13:28
yoloyolo
7181214
asked Jan 6 at 13:28
yoloyolo
7181214
asked Jan 6 at 13:28
yoloyolo
7181214
asked Jan 6 at 13:28
yoloyolo
7181214
asked Jan 6 at 13:28
yoloyolo
7181214
7181214
1
$begingroup$
Is the metal ball made of iron or steel?
$endgroup$
– cms
Jan 6 at 15:02
add a comment |
1
$begingroup$
Is the metal ball made of iron or steel?
$endgroup$
– cms
Jan 6 at 15:02
1
1
$begingroup$
Is the metal ball made of iron or steel?
$endgroup$
– cms
Jan 6 at 15:02
$begingroup$
Is the metal ball made of iron or steel?
$endgroup$
– cms
Jan 6 at 15:02
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
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, $Delta U_M approx 0$.
answered Jan 6 at 14:54
cmscms
3,0272415
$endgroup$
$begingroup$
Why only iron and not any metal?
$endgroup$
– yolo
Jan 6 at 18:11
1
$begingroup$
@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.
$endgroup$
– Peter Shor
Jan 6 at 18:40
$begingroup$
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?
$endgroup$
– yolo
Jan 6 at 21:56
$begingroup$
Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
$endgroup$
– yolo
Jan 6 at 22:01
add a comment |
$begingroup$
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.
answered Jan 6 at 14:54
rob♦rob
40.2k972165
$endgroup$
$begingroup$
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.
$endgroup$
– cms
Jan 6 at 15:04
1
$begingroup$
Oh, true. I'll see if the question gets clarified.
$endgroup$
– rob♦
Jan 6 at 15:05
add a comment |
$begingroup$
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.
answered Jan 6 at 14:20
Ján LalinskýJán Lalinský
14.6k1334
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add a comment |
$begingroup$
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.
answered Jan 6 at 19:43
knzhouknzhou
43.4k11118207
$endgroup$
$begingroup$
Magnetic force can do work even in classical electromagnetic theory, but only some kind of magnetic forces. The force $BIL$ acting on stationary part of current carrying wire is called magnetic force and can do work. Similarly the force on magnetized body in external field. But these magnetic forces are different thing from the Lorentz force concept in microscopic theory. Bohr - van Leeuwen theorem is rather unrelated matter about gas of free particles confined in a container, it does not imply that work can only be done on magnetized solid matter in quantum theory.
$endgroup$
– Ján Lalinský
Jan 9 at 15:48
add a comment |
$begingroup$
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.
answered Jan 6 at 14:31
Vladimir KalitvianskiVladimir Kalitvianski
10.8k11334
$endgroup$
$begingroup$
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.
$endgroup$
– rob♦
Jan 6 at 15:02
2
$begingroup$
@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).
$endgroup$
– Vladimir Kalitvianski
Jan 6 at 15:36
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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, $Delta U_M approx 0$.
answered Jan 6 at 14:54
cmscms
3,0272415
$endgroup$
$begingroup$
Why only iron and not any metal?
$endgroup$
– yolo
Jan 6 at 18:11
1
$begingroup$
@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.
$endgroup$
– Peter Shor
Jan 6 at 18:40
$begingroup$
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?
$endgroup$
– yolo
Jan 6 at 21:56
$begingroup$
Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
$endgroup$
– yolo
Jan 6 at 22:01
add a comment |
$begingroup$
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, $Delta U_M approx 0$.
answered Jan 6 at 14:54
cmscms
3,0272415
$endgroup$
$begingroup$
Why only iron and not any metal?
$endgroup$
– yolo
Jan 6 at 18:11
1
$begingroup$
@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.
$endgroup$
– Peter Shor
Jan 6 at 18:40
$begingroup$
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?
$endgroup$
– yolo
Jan 6 at 21:56
$begingroup$
Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
$endgroup$
– yolo
Jan 6 at 22:01
add a comment |
$begingroup$
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, $Delta U_M approx 0$.
answered Jan 6 at 14:54
cmscms
3,0272415
$endgroup$
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, $Delta U_M approx 0$.
answered Jan 6 at 14:54
cmscms
3,0272415
edited Jan 7 at 16:04
answered Jan 6 at 14:54
cmscms
3,0272415
answered Jan 6 at 14:54
cmscms
3,0272415
answered Jan 6 at 14:54
cmscms
3,0272415
3,0272415
$begingroup$
Why only iron and not any metal?
$endgroup$
– yolo
Jan 6 at 18:11
1
$begingroup$
@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.
$endgroup$
– Peter Shor
Jan 6 at 18:40
$begingroup$
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?
$endgroup$
– yolo
Jan 6 at 21:56
$begingroup$
Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
$endgroup$
– yolo
Jan 6 at 22:01
add a comment |
$begingroup$
Why only iron and not any metal?
$endgroup$
– yolo
Jan 6 at 18:11
1
$begingroup$
@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.
$endgroup$
– Peter Shor
Jan 6 at 18:40
$begingroup$
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?
$endgroup$
– yolo
Jan 6 at 21:56
$begingroup$
Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
$endgroup$
– yolo
Jan 6 at 22:01
$begingroup$
Why only iron and not any metal?
$endgroup$
– yolo
Jan 6 at 18:11
$begingroup$
Why only iron and not any metal?
$endgroup$
– yolo
Jan 6 at 18:11
1
1
$begingroup$
@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.
$endgroup$
– Peter Shor
Jan 6 at 18:40
$begingroup$
@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.
$endgroup$
– Peter Shor
Jan 6 at 18:40
$begingroup$
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?
$endgroup$
– yolo
Jan 6 at 21:56
$begingroup$
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?
$endgroup$
– yolo
Jan 6 at 21:56
$begingroup$
Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
$endgroup$
– yolo
Jan 6 at 22:01
$begingroup$
Note that those above equations are relative to their initial state before the experiment. I know earth does not have 0 energy
$endgroup$
– yolo
Jan 6 at 22:01
add a comment |
$begingroup$
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.
answered Jan 6 at 14:54
rob♦rob
40.2k972165
$endgroup$
$begingroup$
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.
$endgroup$
– cms
Jan 6 at 15:04
1
$begingroup$
Oh, true. I'll see if the question gets clarified.
$endgroup$
– rob♦
Jan 6 at 15:05
add a comment |
$begingroup$
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.
answered Jan 6 at 14:54
rob♦rob
40.2k972165
$endgroup$
$begingroup$
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.
$endgroup$
– cms
Jan 6 at 15:04
1
$begingroup$
Oh, true. I'll see if the question gets clarified.
$endgroup$
– rob♦
Jan 6 at 15:05
add a comment |
$begingroup$
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.
answered Jan 6 at 14:54
rob♦rob
40.2k972165
$endgroup$
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.
answered Jan 6 at 14:54
rob♦rob
40.2k972165
answered Jan 6 at 14:54
rob♦rob
40.2k972165
answered Jan 6 at 14:54
rob♦rob
40.2k972165
answered Jan 6 at 14:54
rob♦rob
40.2k972165
40.2k972165
$begingroup$
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.
$endgroup$
– cms
Jan 6 at 15:04
1
$begingroup$
Oh, true. I'll see if the question gets clarified.
$endgroup$
– rob♦
Jan 6 at 15:05
add a comment |
$begingroup$
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.
$endgroup$
– cms
Jan 6 at 15:04
1
$begingroup$
Oh, true. I'll see if the question gets clarified.
$endgroup$
– rob♦
Jan 6 at 15:05
$begingroup$
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.
$endgroup$
– cms
Jan 6 at 15:04
$begingroup$
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.
$endgroup$
– cms
Jan 6 at 15:04
1
1
$begingroup$
Oh, true. I'll see if the question gets clarified.
$endgroup$
– rob♦
Jan 6 at 15:05
$begingroup$
Oh, true. I'll see if the question gets clarified.
$endgroup$
– rob♦
Jan 6 at 15:05
add a comment |
$begingroup$
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.
answered Jan 6 at 14:20
Ján LalinskýJán Lalinský
14.6k1334
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add a comment |
$begingroup$
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.
answered Jan 6 at 14:20
Ján LalinskýJán Lalinský
14.6k1334
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 6 at 14:20
Ján LalinskýJán Lalinský
14.6k1334
$endgroup$
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.
answered Jan 6 at 14:20
Ján LalinskýJán Lalinský
14.6k1334
answered Jan 6 at 14:20
Ján LalinskýJán Lalinský
14.6k1334
answered Jan 6 at 14:20
Ján LalinskýJán Lalinský
14.6k1334
answered Jan 6 at 14:20
Ján LalinskýJán Lalinský
14.6k1334
14.6k1334
add a comment |
add a comment |
$begingroup$
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.
answered Jan 6 at 19:43
knzhouknzhou
43.4k11118207
$endgroup$
$begingroup$
Magnetic force can do work even in classical electromagnetic theory, but only some kind of magnetic forces. The force $BIL$ acting on stationary part of current carrying wire is called magnetic force and can do work. Similarly the force on magnetized body in external field. But these magnetic forces are different thing from the Lorentz force concept in microscopic theory. Bohr - van Leeuwen theorem is rather unrelated matter about gas of free particles confined in a container, it does not imply that work can only be done on magnetized solid matter in quantum theory.
$endgroup$
– Ján Lalinský
Jan 9 at 15:48
add a comment |
$begingroup$
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.
answered Jan 6 at 19:43
knzhouknzhou
43.4k11118207
$endgroup$
$begingroup$
Magnetic force can do work even in classical electromagnetic theory, but only some kind of magnetic forces. The force $BIL$ acting on stationary part of current carrying wire is called magnetic force and can do work. Similarly the force on magnetized body in external field. But these magnetic forces are different thing from the Lorentz force concept in microscopic theory. Bohr - van Leeuwen theorem is rather unrelated matter about gas of free particles confined in a container, it does not imply that work can only be done on magnetized solid matter in quantum theory.
$endgroup$
– Ján Lalinský
Jan 9 at 15:48
add a comment |
$begingroup$
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.
answered Jan 6 at 19:43
knzhouknzhou
43.4k11118207
$endgroup$
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.
answered Jan 6 at 19:43
knzhouknzhou
43.4k11118207
answered Jan 6 at 19:43
knzhouknzhou
43.4k11118207
answered Jan 6 at 19:43
knzhouknzhou
43.4k11118207
answered Jan 6 at 19:43
knzhouknzhou
43.4k11118207
43.4k11118207
$begingroup$
Magnetic force can do work even in classical electromagnetic theory, but only some kind of magnetic forces. The force $BIL$ acting on stationary part of current carrying wire is called magnetic force and can do work. Similarly the force on magnetized body in external field. But these magnetic forces are different thing from the Lorentz force concept in microscopic theory. Bohr - van Leeuwen theorem is rather unrelated matter about gas of free particles confined in a container, it does not imply that work can only be done on magnetized solid matter in quantum theory.
$endgroup$
– Ján Lalinský
Jan 9 at 15:48
add a comment |
$begingroup$
Magnetic force can do work even in classical electromagnetic theory, but only some kind of magnetic forces. The force $BIL$ acting on stationary part of current carrying wire is called magnetic force and can do work. Similarly the force on magnetized body in external field. But these magnetic forces are different thing from the Lorentz force concept in microscopic theory. Bohr - van Leeuwen theorem is rather unrelated matter about gas of free particles confined in a container, it does not imply that work can only be done on magnetized solid matter in quantum theory.
$endgroup$
– Ján Lalinský
Jan 9 at 15:48
$begingroup$
Magnetic force can do work even in classical electromagnetic theory, but only some kind of magnetic forces. The force $BIL$ acting on stationary part of current carrying wire is called magnetic force and can do work. Similarly the force on magnetized body in external field. But these magnetic forces are different thing from the Lorentz force concept in microscopic theory. Bohr - van Leeuwen theorem is rather unrelated matter about gas of free particles confined in a container, it does not imply that work can only be done on magnetized solid matter in quantum theory.
$endgroup$
– Ján Lalinský
Jan 9 at 15:48
$begingroup$
Magnetic force can do work even in classical electromagnetic theory, but only some kind of magnetic forces. The force $BIL$ acting on stationary part of current carrying wire is called magnetic force and can do work. Similarly the force on magnetized body in external field. But these magnetic forces are different thing from the Lorentz force concept in microscopic theory. Bohr - van Leeuwen theorem is rather unrelated matter about gas of free particles confined in a container, it does not imply that work can only be done on magnetized solid matter in quantum theory.
$endgroup$
– Ján Lalinský
Jan 9 at 15:48
add a comment |
$begingroup$
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.
answered Jan 6 at 14:31
Vladimir KalitvianskiVladimir Kalitvianski
10.8k11334
$endgroup$
$begingroup$
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.
$endgroup$
– rob♦
Jan 6 at 15:02
2
$begingroup$
@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).
$endgroup$
– Vladimir Kalitvianski
Jan 6 at 15:36
add a comment |
$begingroup$
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.
answered Jan 6 at 14:31
Vladimir KalitvianskiVladimir Kalitvianski
10.8k11334
$endgroup$
$begingroup$
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.
$endgroup$
– rob♦
Jan 6 at 15:02
2
$begingroup$
@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).
$endgroup$
– Vladimir Kalitvianski
Jan 6 at 15:36
add a comment |
$begingroup$
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.
answered Jan 6 at 14:31
Vladimir KalitvianskiVladimir Kalitvianski
10.8k11334
$endgroup$
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.
answered Jan 6 at 14:31
Vladimir KalitvianskiVladimir Kalitvianski
10.8k11334
answered Jan 6 at 14:31
Vladimir KalitvianskiVladimir Kalitvianski
10.8k11334
answered Jan 6 at 14:31
Vladimir KalitvianskiVladimir Kalitvianski
10.8k11334
answered Jan 6 at 14:31
Vladimir KalitvianskiVladimir Kalitvianski
10.8k11334
10.8k11334
$begingroup$
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.
$endgroup$
– rob♦
Jan 6 at 15:02
2
$begingroup$
@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).
$endgroup$
– Vladimir Kalitvianski
Jan 6 at 15:36
add a comment |
$begingroup$
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.
$endgroup$
– rob♦
Jan 6 at 15:02
2
$begingroup$
@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).
$endgroup$
– Vladimir Kalitvianski
Jan 6 at 15:36
$begingroup$
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.
$endgroup$
– rob♦
Jan 6 at 15:02
$begingroup$
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.
$endgroup$
– rob♦
Jan 6 at 15:02
2
2
$begingroup$
@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).
$endgroup$
– Vladimir Kalitvianski
Jan 6 at 15:36
$begingroup$
@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).
$endgroup$
– Vladimir Kalitvianski
Jan 6 at 15:36
add a comment |
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Is the metal ball made of iron or steel?
$endgroup$
– cms
Jan 6 at 15:02