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Can a magnetic field be induced without an electric field?

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2

1

$begingroup$

Can a magnetic field be induced without an electric field?
Because, as far as I know, a time varying electric field induces a magnetic field an vice versa.
But in the case of conductors carrying currennt, it doesn't seem that electric field varies with time, then how is a magnetic field induced?

electromagnetism

share|cite|improve this question

asked 8 hours ago

Kothapalli SanthoshKothapalli Santhosh

133

New contributor

Kothapalli Santhosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you hold a permanent magnet over a peace of metal, will that induce a magnetic field?
  $endgroup$
  – HolgerFiedler
  8 hours ago
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Can a magnetic field be induced without an electric field?
Because, as far as I know, a time varying electric field induces a magnetic field an vice versa.
But in the case of conductors carrying currennt, it doesn't seem that electric field varies with time, then how is a magnetic field induced?

electromagnetism

share|cite|improve this question

asked 8 hours ago

Kothapalli SanthoshKothapalli Santhosh

133

New contributor

Kothapalli Santhosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you hold a permanent magnet over a peace of metal, will that induce a magnetic field?
  $endgroup$
  – HolgerFiedler
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Can a magnetic field be induced without an electric field?
Because, as far as I know, a time varying electric field induces a magnetic field an vice versa.
But in the case of conductors carrying currennt, it doesn't seem that electric field varies with time, then how is a magnetic field induced?

electromagnetism

share|cite|improve this question

asked 8 hours ago

Kothapalli SanthoshKothapalli Santhosh

133

New contributor

Kothapalli Santhosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 8 hours ago

Kothapalli SanthoshKothapalli Santhosh

133

New contributor

Kothapalli Santhosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 8 hours ago

Kothapalli SanthoshKothapalli Santhosh

133

New contributor

Kothapalli Santhosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

Kothapalli SanthoshKothapalli Santhosh

133

New contributor

Kothapalli Santhosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

Kothapalli SanthoshKothapalli Santhosh

133

2 Answers
2

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4

$begingroup$

One of Maxwell’s four equations for electromagnetism in a vacuum shows how magnetic fields are produced:

$$nablatimesmathbf{B}=frac{1}{c}left(4pimathbf{J}+frac{partialmathbf{E}}{partial t}right).$$

(I’ve written it in Gaussian units.)

From this equation you can see that there are two different sources for magnetic fields: the first is a current density, and the second is a changing electric field.

So to have a magnetic field you do not need to have a time-varying electric field. You can just have moving charge. But when a magnetic field is produced by moving charge, physicists don’t call it “induced”.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

G. SmithG. Smith

8,01211425

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  But aren't moving charges producing an electric field? And what about spin? Can a neutron for instance create a B field while no electric field?
  $endgroup$
  – thermomagnetic condensed boson
  3 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

From Griffiths, Electrodynamics, Jefimenko’s equations are given as
$${bf E}({bf r},t) = frac{1}{4 pi epsilon_0} int [ frac{rho ({bf r}',t_r)}{{mathfrak r}^2} {bf hat{mathfrak r}} + frac{dot{rho} ({bf r}',t_r)}{c {mathfrak r}} {bf hat{mathfrak r}} - frac{{bf {dot J}} ({bf r}',t_r)}{c^2 {mathfrak r}}] d tau',$$
$${bf B}({bf r},t) = frac{mu_0}{4 pi} int [frac{{bf {J}} ({bf r}',t_r)}{{mathfrak r}^2} + frac{{bf {dot J}} ({bf r}',t_r)}{c {mathfrak r}} ] times {bf hat{mathfrak r}} d tau'.$$

These equations show that to create a magnetic field you require either a steady current or/as well a changing current. If the current density is steady (so that ${bf {dot J}} equiv 0$) then you can see that you can arrange for no electric field by having the charge density $rho$ vanish everywhere. Another way to create a magnetic field is to have a time varying current density, which necessarily creates an electric field.

share|cite|improve this answer

answered 1 hour ago

jimjim

2,507721

$endgroup$

add a comment | 

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4

$begingroup$

One of Maxwell’s four equations for electromagnetism in a vacuum shows how magnetic fields are produced:

$$nablatimesmathbf{B}=frac{1}{c}left(4pimathbf{J}+frac{partialmathbf{E}}{partial t}right).$$

(I’ve written it in Gaussian units.)

From this equation you can see that there are two different sources for magnetic fields: the first is a current density, and the second is a changing electric field.

So to have a magnetic field you do not need to have a time-varying electric field. You can just have moving charge. But when a magnetic field is produced by moving charge, physicists don’t call it “induced”.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

G. SmithG. Smith

8,01211425

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  But aren't moving charges producing an electric field? And what about spin? Can a neutron for instance create a B field while no electric field?
  $endgroup$
  – thermomagnetic condensed boson
  3 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

One of Maxwell’s four equations for electromagnetism in a vacuum shows how magnetic fields are produced:

$$nablatimesmathbf{B}=frac{1}{c}left(4pimathbf{J}+frac{partialmathbf{E}}{partial t}right).$$

(I’ve written it in Gaussian units.)

From this equation you can see that there are two different sources for magnetic fields: the first is a current density, and the second is a changing electric field.

So to have a magnetic field you do not need to have a time-varying electric field. You can just have moving charge. But when a magnetic field is produced by moving charge, physicists don’t call it “induced”.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

G. SmithG. Smith

8,01211425

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  But aren't moving charges producing an electric field? And what about spin? Can a neutron for instance create a B field while no electric field?
  $endgroup$
  – thermomagnetic condensed boson
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

One of Maxwell’s four equations for electromagnetism in a vacuum shows how magnetic fields are produced:

$$nablatimesmathbf{B}=frac{1}{c}left(4pimathbf{J}+frac{partialmathbf{E}}{partial t}right).$$

(I’ve written it in Gaussian units.)

From this equation you can see that there are two different sources for magnetic fields: the first is a current density, and the second is a changing electric field.

So to have a magnetic field you do not need to have a time-varying electric field. You can just have moving charge. But when a magnetic field is produced by moving charge, physicists don’t call it “induced”.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

G. SmithG. Smith

8,01211425

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

G. SmithG. Smith

8,01211425

answered 8 hours ago

G. SmithG. Smith

8,01211425

answered 8 hours ago

G. SmithG. Smith

8,01211425

0

$begingroup$

From Griffiths, Electrodynamics, Jefimenko’s equations are given as
$${bf E}({bf r},t) = frac{1}{4 pi epsilon_0} int [ frac{rho ({bf r}',t_r)}{{mathfrak r}^2} {bf hat{mathfrak r}} + frac{dot{rho} ({bf r}',t_r)}{c {mathfrak r}} {bf hat{mathfrak r}} - frac{{bf {dot J}} ({bf r}',t_r)}{c^2 {mathfrak r}}] d tau',$$
$${bf B}({bf r},t) = frac{mu_0}{4 pi} int [frac{{bf {J}} ({bf r}',t_r)}{{mathfrak r}^2} + frac{{bf {dot J}} ({bf r}',t_r)}{c {mathfrak r}} ] times {bf hat{mathfrak r}} d tau'.$$

These equations show that to create a magnetic field you require either a steady current or/as well a changing current. If the current density is steady (so that ${bf {dot J}} equiv 0$) then you can see that you can arrange for no electric field by having the charge density $rho$ vanish everywhere. Another way to create a magnetic field is to have a time varying current density, which necessarily creates an electric field.

share|cite|improve this answer

answered 1 hour ago

jimjim

2,507721

$endgroup$

add a comment | 

0

$begingroup$

From Griffiths, Electrodynamics, Jefimenko’s equations are given as
$${bf E}({bf r},t) = frac{1}{4 pi epsilon_0} int [ frac{rho ({bf r}',t_r)}{{mathfrak r}^2} {bf hat{mathfrak r}} + frac{dot{rho} ({bf r}',t_r)}{c {mathfrak r}} {bf hat{mathfrak r}} - frac{{bf {dot J}} ({bf r}',t_r)}{c^2 {mathfrak r}}] d tau',$$
$${bf B}({bf r},t) = frac{mu_0}{4 pi} int [frac{{bf {J}} ({bf r}',t_r)}{{mathfrak r}^2} + frac{{bf {dot J}} ({bf r}',t_r)}{c {mathfrak r}} ] times {bf hat{mathfrak r}} d tau'.$$

These equations show that to create a magnetic field you require either a steady current or/as well a changing current. If the current density is steady (so that ${bf {dot J}} equiv 0$) then you can see that you can arrange for no electric field by having the charge density $rho$ vanish everywhere. Another way to create a magnetic field is to have a time varying current density, which necessarily creates an electric field.

share|cite|improve this answer

answered 1 hour ago

jimjim

2,507721

$endgroup$

add a comment | 

$begingroup$

From Griffiths, Electrodynamics, Jefimenko’s equations are given as
$${bf E}({bf r},t) = frac{1}{4 pi epsilon_0} int [ frac{rho ({bf r}',t_r)}{{mathfrak r}^2} {bf hat{mathfrak r}} + frac{dot{rho} ({bf r}',t_r)}{c {mathfrak r}} {bf hat{mathfrak r}} - frac{{bf {dot J}} ({bf r}',t_r)}{c^2 {mathfrak r}}] d tau',$$
$${bf B}({bf r},t) = frac{mu_0}{4 pi} int [frac{{bf {J}} ({bf r}',t_r)}{{mathfrak r}^2} + frac{{bf {dot J}} ({bf r}',t_r)}{c {mathfrak r}} ] times {bf hat{mathfrak r}} d tau'.$$

These equations show that to create a magnetic field you require either a steady current or/as well a changing current. If the current density is steady (so that ${bf {dot J}} equiv 0$) then you can see that you can arrange for no electric field by having the charge density $rho$ vanish everywhere. Another way to create a magnetic field is to have a time varying current density, which necessarily creates an electric field.

share|cite|improve this answer

answered 1 hour ago

jimjim

2,507721

$endgroup$

share|cite|improve this answer

answered 1 hour ago

jimjim

2,507721

answered 1 hour ago

jimjim

2,507721

answered 1 hour ago

jimjim

2,507721