Why is there an induced EMF in a plastic ring?
If we were to pull a plastic ring across the boundary of a magnetic field, why does an emf form? If this was a metal ring I would fully understand because the electrons are free to move in a metal thus allowing it to collect together and form charged sides. However, plastic is an insulator so how can it have it's charges moving? It's even mentioned in the answers that due to the insulating nature of plastic no current will flow, but then how do the charges separate in the first place?
If it helps, I typically think of the Loretnz force acting on charges for electromagnetic induction.
Also with this induced emf, would there be a difference in magnitude of the emf between the metal and plastic ring?
electromagnetism electric-current voltage electromagnetic-induction
edited 51 secs ago
Qmechanic♦
101k121831149
asked 2 hours ago
John Hon
321210
add a comment |
If we were to pull a plastic ring across the boundary of a magnetic field, why does an emf form? If this was a metal ring I would fully understand because the electrons are free to move in a metal thus allowing it to collect together and form charged sides. However, plastic is an insulator so how can it have it's charges moving? It's even mentioned in the answers that due to the insulating nature of plastic no current will flow, but then how do the charges separate in the first place?
If it helps, I typically think of the Loretnz force acting on charges for electromagnetic induction.
Also with this induced emf, would there be a difference in magnitude of the emf between the metal and plastic ring?
electromagnetism electric-current voltage electromagnetic-induction
edited 51 secs ago
Qmechanic♦
101k121831149
asked 2 hours ago
John Hon
321210
add a comment |
If we were to pull a plastic ring across the boundary of a magnetic field, why does an emf form? If this was a metal ring I would fully understand because the electrons are free to move in a metal thus allowing it to collect together and form charged sides. However, plastic is an insulator so how can it have it's charges moving? It's even mentioned in the answers that due to the insulating nature of plastic no current will flow, but then how do the charges separate in the first place?
If it helps, I typically think of the Loretnz force acting on charges for electromagnetic induction.
Also with this induced emf, would there be a difference in magnitude of the emf between the metal and plastic ring?
electromagnetism electric-current voltage electromagnetic-induction
edited 51 secs ago
Qmechanic♦
101k121831149
asked 2 hours ago
John Hon
321210
If we were to pull a plastic ring across the boundary of a magnetic field, why does an emf form? If this was a metal ring I would fully understand because the electrons are free to move in a metal thus allowing it to collect together and form charged sides. However, plastic is an insulator so how can it have it's charges moving? It's even mentioned in the answers that due to the insulating nature of plastic no current will flow, but then how do the charges separate in the first place?
If it helps, I typically think of the Loretnz force acting on charges for electromagnetic induction.
Also with this induced emf, would there be a difference in magnitude of the emf between the metal and plastic ring?
electromagnetism electric-current voltage electromagnetic-induction
electromagnetism electric-current voltage electromagnetic-induction
edited 51 secs ago
Qmechanic♦
101k121831149
asked 2 hours ago
John Hon
321210
edited 51 secs ago
Qmechanic♦
101k121831149
asked 2 hours ago
John Hon
321210
edited 51 secs ago
Qmechanic♦
101k121831149
edited 51 secs ago
Qmechanic♦
101k121831149
edited 51 secs ago
Qmechanic♦
101k121831149
101k121831149
asked 2 hours ago
John Hon
321210
asked 2 hours ago
John Hon
321210
asked 2 hours ago
John Hon
321210
321210
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
An electromotive force doesn't require a conductor --- it doesn't even require matter. The electromagnetic field is a local property of the vacuum, governed by Maxwell's equations. The relevant one in this case is
$$
vecnabla times vec E = -fracpartial{partial t}vec B
$$
That is, at any point in space, a changing magnitude or direction for the magnetic field is inextricably associated with an electric field with nonzero curl.
Because electromagnetism obeys all the symmetries of special relativity, it doesn't matter whether the point of interest is stationary and the field there is changing, or if the point of interest is moving through a region of static but nonuniform magnetic fields.
Now, a "conductor" is some material with the property of "always" having $vec E=0$ inside. Since the changing magnetic field is associated with $vec E neq 0$, then the charges in the conductor where $partialvec B/partial t neq 0$ must move to produce $vec E=0$ by superposition. You can think of that as motion due to the Lorentz force if you like. The Lorentz force,
$$
vec F = qleft(vec vtimesvec B + vec E right),
$$
implies that a charge must be moving to experience a force from a static magnetic field. However a changing magnetic field produces a nonzero $vec E$, and can therefore exert force on stationary charges.
answered 30 mins ago
rob♦
39.1k971161
This is the correct answer. The question confuses emf and current, I think.
– Jerry Schirmer
21 mins ago
add a comment |
Something is very funny here. "Why does an emf form?"
As far as I understand physics it doesn't. And shouldn't
"thus allowing it to collect together and form charged sides"
This seems to describe something like the amber rod rubbed with wool. Only there you can have charges accumulating. Not in a metal.
answered 1 hour ago
Dr S T Lakshmikumar
212
add a comment |
The electron shells of atoms get distorted ie move a small amount relative to the nuclei, so that electric dipoles are induced.
answered 1 hour ago
Farcher
47.3k33696
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
An electromotive force doesn't require a conductor --- it doesn't even require matter. The electromagnetic field is a local property of the vacuum, governed by Maxwell's equations. The relevant one in this case is
$$
vecnabla times vec E = -fracpartial{partial t}vec B
$$
That is, at any point in space, a changing magnitude or direction for the magnetic field is inextricably associated with an electric field with nonzero curl.
Because electromagnetism obeys all the symmetries of special relativity, it doesn't matter whether the point of interest is stationary and the field there is changing, or if the point of interest is moving through a region of static but nonuniform magnetic fields.
Now, a "conductor" is some material with the property of "always" having $vec E=0$ inside. Since the changing magnetic field is associated with $vec E neq 0$, then the charges in the conductor where $partialvec B/partial t neq 0$ must move to produce $vec E=0$ by superposition. You can think of that as motion due to the Lorentz force if you like. The Lorentz force,
$$
vec F = qleft(vec vtimesvec B + vec E right),
$$
implies that a charge must be moving to experience a force from a static magnetic field. However a changing magnetic field produces a nonzero $vec E$, and can therefore exert force on stationary charges.
answered 30 mins ago
rob♦
39.1k971161
This is the correct answer. The question confuses emf and current, I think.
– Jerry Schirmer
21 mins ago
add a comment |
An electromotive force doesn't require a conductor --- it doesn't even require matter. The electromagnetic field is a local property of the vacuum, governed by Maxwell's equations. The relevant one in this case is
$$
vecnabla times vec E = -fracpartial{partial t}vec B
$$
That is, at any point in space, a changing magnitude or direction for the magnetic field is inextricably associated with an electric field with nonzero curl.
Because electromagnetism obeys all the symmetries of special relativity, it doesn't matter whether the point of interest is stationary and the field there is changing, or if the point of interest is moving through a region of static but nonuniform magnetic fields.
Now, a "conductor" is some material with the property of "always" having $vec E=0$ inside. Since the changing magnetic field is associated with $vec E neq 0$, then the charges in the conductor where $partialvec B/partial t neq 0$ must move to produce $vec E=0$ by superposition. You can think of that as motion due to the Lorentz force if you like. The Lorentz force,
$$
vec F = qleft(vec vtimesvec B + vec E right),
$$
implies that a charge must be moving to experience a force from a static magnetic field. However a changing magnetic field produces a nonzero $vec E$, and can therefore exert force on stationary charges.
answered 30 mins ago
rob♦
39.1k971161
This is the correct answer. The question confuses emf and current, I think.
– Jerry Schirmer
21 mins ago
add a comment |
An electromotive force doesn't require a conductor --- it doesn't even require matter. The electromagnetic field is a local property of the vacuum, governed by Maxwell's equations. The relevant one in this case is
$$
vecnabla times vec E = -fracpartial{partial t}vec B
$$
That is, at any point in space, a changing magnitude or direction for the magnetic field is inextricably associated with an electric field with nonzero curl.
Because electromagnetism obeys all the symmetries of special relativity, it doesn't matter whether the point of interest is stationary and the field there is changing, or if the point of interest is moving through a region of static but nonuniform magnetic fields.
Now, a "conductor" is some material with the property of "always" having $vec E=0$ inside. Since the changing magnetic field is associated with $vec E neq 0$, then the charges in the conductor where $partialvec B/partial t neq 0$ must move to produce $vec E=0$ by superposition. You can think of that as motion due to the Lorentz force if you like. The Lorentz force,
$$
vec F = qleft(vec vtimesvec B + vec E right),
$$
implies that a charge must be moving to experience a force from a static magnetic field. However a changing magnetic field produces a nonzero $vec E$, and can therefore exert force on stationary charges.
answered 30 mins ago
rob♦
39.1k971161
An electromotive force doesn't require a conductor --- it doesn't even require matter. The electromagnetic field is a local property of the vacuum, governed by Maxwell's equations. The relevant one in this case is
$$
vecnabla times vec E = -fracpartial{partial t}vec B
$$
That is, at any point in space, a changing magnitude or direction for the magnetic field is inextricably associated with an electric field with nonzero curl.
Because electromagnetism obeys all the symmetries of special relativity, it doesn't matter whether the point of interest is stationary and the field there is changing, or if the point of interest is moving through a region of static but nonuniform magnetic fields.
Now, a "conductor" is some material with the property of "always" having $vec E=0$ inside. Since the changing magnetic field is associated with $vec E neq 0$, then the charges in the conductor where $partialvec B/partial t neq 0$ must move to produce $vec E=0$ by superposition. You can think of that as motion due to the Lorentz force if you like. The Lorentz force,
$$
vec F = qleft(vec vtimesvec B + vec E right),
$$
implies that a charge must be moving to experience a force from a static magnetic field. However a changing magnetic field produces a nonzero $vec E$, and can therefore exert force on stationary charges.
answered 30 mins ago
rob♦
39.1k971161
answered 30 mins ago
rob♦
39.1k971161
answered 30 mins ago
rob♦
39.1k971161
answered 30 mins ago
rob♦
39.1k971161
39.1k971161
This is the correct answer. The question confuses emf and current, I think.
– Jerry Schirmer
21 mins ago
add a comment |
This is the correct answer. The question confuses emf and current, I think.
– Jerry Schirmer
21 mins ago
This is the correct answer. The question confuses emf and current, I think.
– Jerry Schirmer
21 mins ago
This is the correct answer. The question confuses emf and current, I think.
– Jerry Schirmer
21 mins ago
add a comment |
Something is very funny here. "Why does an emf form?"
As far as I understand physics it doesn't. And shouldn't
"thus allowing it to collect together and form charged sides"
This seems to describe something like the amber rod rubbed with wool. Only there you can have charges accumulating. Not in a metal.
answered 1 hour ago
Dr S T Lakshmikumar
212
add a comment |
Something is very funny here. "Why does an emf form?"
As far as I understand physics it doesn't. And shouldn't
"thus allowing it to collect together and form charged sides"
This seems to describe something like the amber rod rubbed with wool. Only there you can have charges accumulating. Not in a metal.
answered 1 hour ago
Dr S T Lakshmikumar
212
add a comment |
Something is very funny here. "Why does an emf form?"
As far as I understand physics it doesn't. And shouldn't
"thus allowing it to collect together and form charged sides"
This seems to describe something like the amber rod rubbed with wool. Only there you can have charges accumulating. Not in a metal.
answered 1 hour ago
Dr S T Lakshmikumar
212
Something is very funny here. "Why does an emf form?"
As far as I understand physics it doesn't. And shouldn't
"thus allowing it to collect together and form charged sides"
This seems to describe something like the amber rod rubbed with wool. Only there you can have charges accumulating. Not in a metal.
answered 1 hour ago
Dr S T Lakshmikumar
212
answered 1 hour ago
Dr S T Lakshmikumar
212
answered 1 hour ago
Dr S T Lakshmikumar
212
answered 1 hour ago
Dr S T Lakshmikumar
212
212
add a comment |
add a comment |
The electron shells of atoms get distorted ie move a small amount relative to the nuclei, so that electric dipoles are induced.
answered 1 hour ago
Farcher
47.3k33696
add a comment |
The electron shells of atoms get distorted ie move a small amount relative to the nuclei, so that electric dipoles are induced.
answered 1 hour ago
Farcher
47.3k33696
add a comment |
The electron shells of atoms get distorted ie move a small amount relative to the nuclei, so that electric dipoles are induced.
answered 1 hour ago
Farcher
47.3k33696
The electron shells of atoms get distorted ie move a small amount relative to the nuclei, so that electric dipoles are induced.
answered 1 hour ago
Farcher
47.3k33696
answered 1 hour ago
Farcher
47.3k33696
answered 1 hour ago
Farcher
47.3k33696
answered 1 hour ago
Farcher
47.3k33696
47.3k33696
add a comment |
add a comment |
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