Why is there an induced EMF in a plastic ring?
$begingroup$
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 Dec 30 '18 at 15:35
Qmechanic♦
108k122001253
asked Dec 30 '18 at 12:37
John HonJohn Hon
501412
$endgroup$
add a comment |
$begingroup$
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 Dec 30 '18 at 15:35
Qmechanic♦
108k122001253
asked Dec 30 '18 at 12:37
John HonJohn Hon
501412
$endgroup$
add a comment |
$begingroup$
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 Dec 30 '18 at 15:35
Qmechanic♦
108k122001253
asked Dec 30 '18 at 12:37
John HonJohn Hon
501412
$endgroup$
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 Dec 30 '18 at 15:35
Qmechanic♦
108k122001253
asked Dec 30 '18 at 12:37
John HonJohn Hon
501412
edited Dec 30 '18 at 15:35
Qmechanic♦
108k122001253
asked Dec 30 '18 at 12:37
John HonJohn Hon
501412
edited Dec 30 '18 at 15:35
Qmechanic♦
108k122001253
edited Dec 30 '18 at 15:35
Qmechanic♦
108k122001253
edited Dec 30 '18 at 15:35
Qmechanic♦
108k122001253
108k122001253
asked Dec 30 '18 at 12:37
John HonJohn Hon
501412
asked Dec 30 '18 at 12:37
John HonJohn Hon
501412
asked Dec 30 '18 at 12:37
John HonJohn Hon
501412
501412
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 Dec 30 '18 at 15:05
rob♦rob
41.6k1076173
$endgroup$
1
$begingroup$
This is the correct answer. The question confuses emf and current, I think.
$endgroup$
– Jerry Schirmer
Dec 30 '18 at 15:14
$begingroup$
As a layman, I would read that as "just as there would be an emf induced in no plastic ring at all"?
$endgroup$
– rackandboneman
Dec 31 '18 at 2:00
$begingroup$
@rackandboneman Yes. The emf is a property of the changing magnetic field, whether the space of interest is occupied by a plastic ring, a metal ring, a calico cat, or nothing at all.
$endgroup$
– rob♦
Dec 31 '18 at 4:15
$begingroup$
@rob I'm still a bit confused. I can accept the fact that a changing magnetic field intrinsically creates a changing electric field with it. What I don't get is how can this changing E field create an emf or voltage if there is no separation of charge? EDIT: Upon thinking more closely I can actually understand why there is an EMF (by summing the loop of electric field) but now I can't seem to understand why there should be a circular electric field produced. At the risk of being blasted for asking metaphysics, could you explain this?
$endgroup$
– John Hon
Dec 31 '18 at 6:48
$begingroup$
@JohnHon The summing-of-fields-over-loop that you like is related to the differential equation for $vecnablatimesvec E$ by Stokes' Theorem.
$endgroup$
– rob♦
Dec 31 '18 at 13:16
|
show 2 more comments
$begingroup$
The electron shells of atoms get distorted ie move a small amount relative to the nuclei, so that electric dipoles are induced.
answered Dec 30 '18 at 14:31
FarcherFarcher
52.5k340110
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f451166%2fwhy-is-there-an-induced-emf-in-a-plastic-ring%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Dec 30 '18 at 15:05
rob♦rob
41.6k1076173
$endgroup$
1
$begingroup$
This is the correct answer. The question confuses emf and current, I think.
$endgroup$
– Jerry Schirmer
Dec 30 '18 at 15:14
$begingroup$
As a layman, I would read that as "just as there would be an emf induced in no plastic ring at all"?
$endgroup$
– rackandboneman
Dec 31 '18 at 2:00
$begingroup$
@rackandboneman Yes. The emf is a property of the changing magnetic field, whether the space of interest is occupied by a plastic ring, a metal ring, a calico cat, or nothing at all.
$endgroup$
– rob♦
Dec 31 '18 at 4:15
$begingroup$
@rob I'm still a bit confused. I can accept the fact that a changing magnetic field intrinsically creates a changing electric field with it. What I don't get is how can this changing E field create an emf or voltage if there is no separation of charge? EDIT: Upon thinking more closely I can actually understand why there is an EMF (by summing the loop of electric field) but now I can't seem to understand why there should be a circular electric field produced. At the risk of being blasted for asking metaphysics, could you explain this?
$endgroup$
– John Hon
Dec 31 '18 at 6:48
$begingroup$
@JohnHon The summing-of-fields-over-loop that you like is related to the differential equation for $vecnablatimesvec E$ by Stokes' Theorem.
$endgroup$
– rob♦
Dec 31 '18 at 13:16
|
show 2 more comments
$begingroup$
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 Dec 30 '18 at 15:05
rob♦rob
41.6k1076173
$endgroup$
1
$begingroup$
This is the correct answer. The question confuses emf and current, I think.
$endgroup$
– Jerry Schirmer
Dec 30 '18 at 15:14
$begingroup$
As a layman, I would read that as "just as there would be an emf induced in no plastic ring at all"?
$endgroup$
– rackandboneman
Dec 31 '18 at 2:00
$begingroup$
@rackandboneman Yes. The emf is a property of the changing magnetic field, whether the space of interest is occupied by a plastic ring, a metal ring, a calico cat, or nothing at all.
$endgroup$
– rob♦
Dec 31 '18 at 4:15
$begingroup$
@rob I'm still a bit confused. I can accept the fact that a changing magnetic field intrinsically creates a changing electric field with it. What I don't get is how can this changing E field create an emf or voltage if there is no separation of charge? EDIT: Upon thinking more closely I can actually understand why there is an EMF (by summing the loop of electric field) but now I can't seem to understand why there should be a circular electric field produced. At the risk of being blasted for asking metaphysics, could you explain this?
$endgroup$
– John Hon
Dec 31 '18 at 6:48
$begingroup$
@JohnHon The summing-of-fields-over-loop that you like is related to the differential equation for $vecnablatimesvec E$ by Stokes' Theorem.
$endgroup$
– rob♦
Dec 31 '18 at 13:16
|
show 2 more comments
$begingroup$
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 Dec 30 '18 at 15:05
rob♦rob
41.6k1076173
$endgroup$
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 Dec 30 '18 at 15:05
rob♦rob
41.6k1076173
answered Dec 30 '18 at 15:05
rob♦rob
41.6k1076173
answered Dec 30 '18 at 15:05
rob♦rob
41.6k1076173
answered Dec 30 '18 at 15:05
rob♦rob
41.6k1076173
41.6k1076173
1
$begingroup$
This is the correct answer. The question confuses emf and current, I think.
$endgroup$
– Jerry Schirmer
Dec 30 '18 at 15:14
$begingroup$
As a layman, I would read that as "just as there would be an emf induced in no plastic ring at all"?
$endgroup$
– rackandboneman
Dec 31 '18 at 2:00
$begingroup$
@rackandboneman Yes. The emf is a property of the changing magnetic field, whether the space of interest is occupied by a plastic ring, a metal ring, a calico cat, or nothing at all.
$endgroup$
– rob♦
Dec 31 '18 at 4:15
$begingroup$
@rob I'm still a bit confused. I can accept the fact that a changing magnetic field intrinsically creates a changing electric field with it. What I don't get is how can this changing E field create an emf or voltage if there is no separation of charge? EDIT: Upon thinking more closely I can actually understand why there is an EMF (by summing the loop of electric field) but now I can't seem to understand why there should be a circular electric field produced. At the risk of being blasted for asking metaphysics, could you explain this?
$endgroup$
– John Hon
Dec 31 '18 at 6:48
$begingroup$
@JohnHon The summing-of-fields-over-loop that you like is related to the differential equation for $vecnablatimesvec E$ by Stokes' Theorem.
$endgroup$
– rob♦
Dec 31 '18 at 13:16
|
show 2 more comments
1
$begingroup$
This is the correct answer. The question confuses emf and current, I think.
$endgroup$
– Jerry Schirmer
Dec 30 '18 at 15:14
$begingroup$
As a layman, I would read that as "just as there would be an emf induced in no plastic ring at all"?
$endgroup$
– rackandboneman
Dec 31 '18 at 2:00
$begingroup$
@rackandboneman Yes. The emf is a property of the changing magnetic field, whether the space of interest is occupied by a plastic ring, a metal ring, a calico cat, or nothing at all.
$endgroup$
– rob♦
Dec 31 '18 at 4:15
$begingroup$
@rob I'm still a bit confused. I can accept the fact that a changing magnetic field intrinsically creates a changing electric field with it. What I don't get is how can this changing E field create an emf or voltage if there is no separation of charge? EDIT: Upon thinking more closely I can actually understand why there is an EMF (by summing the loop of electric field) but now I can't seem to understand why there should be a circular electric field produced. At the risk of being blasted for asking metaphysics, could you explain this?
$endgroup$
– John Hon
Dec 31 '18 at 6:48
$begingroup$
@JohnHon The summing-of-fields-over-loop that you like is related to the differential equation for $vecnablatimesvec E$ by Stokes' Theorem.
$endgroup$
– rob♦
Dec 31 '18 at 13:16
1
1
$begingroup$
This is the correct answer. The question confuses emf and current, I think.
$endgroup$
– Jerry Schirmer
Dec 30 '18 at 15:14
$begingroup$
This is the correct answer. The question confuses emf and current, I think.
$endgroup$
– Jerry Schirmer
Dec 30 '18 at 15:14
$begingroup$
As a layman, I would read that as "just as there would be an emf induced in no plastic ring at all"?
$endgroup$
– rackandboneman
Dec 31 '18 at 2:00
$begingroup$
As a layman, I would read that as "just as there would be an emf induced in no plastic ring at all"?
$endgroup$
– rackandboneman
Dec 31 '18 at 2:00
$begingroup$
@rackandboneman Yes. The emf is a property of the changing magnetic field, whether the space of interest is occupied by a plastic ring, a metal ring, a calico cat, or nothing at all.
$endgroup$
– rob♦
Dec 31 '18 at 4:15
$begingroup$
@rackandboneman Yes. The emf is a property of the changing magnetic field, whether the space of interest is occupied by a plastic ring, a metal ring, a calico cat, or nothing at all.
$endgroup$
– rob♦
Dec 31 '18 at 4:15
$begingroup$
@rob I'm still a bit confused. I can accept the fact that a changing magnetic field intrinsically creates a changing electric field with it. What I don't get is how can this changing E field create an emf or voltage if there is no separation of charge? EDIT: Upon thinking more closely I can actually understand why there is an EMF (by summing the loop of electric field) but now I can't seem to understand why there should be a circular electric field produced. At the risk of being blasted for asking metaphysics, could you explain this?
$endgroup$
– John Hon
Dec 31 '18 at 6:48
$begingroup$
@rob I'm still a bit confused. I can accept the fact that a changing magnetic field intrinsically creates a changing electric field with it. What I don't get is how can this changing E field create an emf or voltage if there is no separation of charge? EDIT: Upon thinking more closely I can actually understand why there is an EMF (by summing the loop of electric field) but now I can't seem to understand why there should be a circular electric field produced. At the risk of being blasted for asking metaphysics, could you explain this?
$endgroup$
– John Hon
Dec 31 '18 at 6:48
$begingroup$
@JohnHon The summing-of-fields-over-loop that you like is related to the differential equation for $vecnablatimesvec E$ by Stokes' Theorem.
$endgroup$
– rob♦
Dec 31 '18 at 13:16
$begingroup$
@JohnHon The summing-of-fields-over-loop that you like is related to the differential equation for $vecnablatimesvec E$ by Stokes' Theorem.
$endgroup$
– rob♦
Dec 31 '18 at 13:16
|
show 2 more comments
$begingroup$
The electron shells of atoms get distorted ie move a small amount relative to the nuclei, so that electric dipoles are induced.
answered Dec 30 '18 at 14:31
FarcherFarcher
52.5k340110
$endgroup$
add a comment |
$begingroup$
The electron shells of atoms get distorted ie move a small amount relative to the nuclei, so that electric dipoles are induced.
answered Dec 30 '18 at 14:31
FarcherFarcher
52.5k340110
$endgroup$
add a comment |
$begingroup$
The electron shells of atoms get distorted ie move a small amount relative to the nuclei, so that electric dipoles are induced.
answered Dec 30 '18 at 14:31
FarcherFarcher
52.5k340110
$endgroup$
The electron shells of atoms get distorted ie move a small amount relative to the nuclei, so that electric dipoles are induced.
answered Dec 30 '18 at 14:31
FarcherFarcher
52.5k340110
answered Dec 30 '18 at 14:31
FarcherFarcher
52.5k340110
answered Dec 30 '18 at 14:31
FarcherFarcher
52.5k340110
answered Dec 30 '18 at 14:31
FarcherFarcher
52.5k340110
52.5k340110
add a comment |
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f451166%2fwhy-is-there-an-induced-emf-in-a-plastic-ring%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
