The units of the magnetic intensity field H are amps/meter, but what
does this mean? For the E field, if you place a wire 1m long in a
field of X volts/m, you will generate X volts between the ends of the
wire. What's the analogy for H? A closed loop 1m long in an H field
of Y amps/m will have Y amps induced? That doesn't seem right.

Thanks for any insight.

"WillWorkForNeurons" <ger...@exality.com> wrote in message

news:1188414709.874814.293750@x35g2000prf.googlegroups.com...

The units of the magnetic intensity field H are amps/meter, but what
does this mean? For the E field, if you place a wire 1m long in a
field of X volts/m, you will generate X volts between the ends of the
wire. What's the analogy for H? A closed loop 1m long in an H field
of Y amps/m will have Y amps induced? That doesn't seem right.

Thanks for any insight.

close, integrate over the enclosed surface, differentiate by time, multiply
by mu-0... oh go read it for yourself athttp://en.wikipedia.org/wiki/Maxwell%27s_equations look for faraday's law
and figure out how B and H are related by mu and mu-0.

For the E field, if you place a wire 1m long in a
field of X volts/m, you will generate X volts between the ends of
the
wire.

"WillWorkForNeurons" <gerrit@exality.com> wrote in message
news:1188414709.874814.293750@x35g2000prf.googlegroups.com...


For the E field, if you place a wire 1m long in a
field of X volts/m, you will generate X volts between the ends of
the
wire.


I don't agree with this? Perhaps: if you have a voltage drop of X
volts along a 1m length of a conductor, the E field along that
length averages 1V/m.

For the A/m, if this is true it may be what you're after. The H in
a circle centered around a straight infinitely long current carrying
conductor is H=I/2*pi*r. So the A/m is the 'Amps' in the conductor
divided by the 'meters' of the path.

"WillWorkForNeurons" <ger...@exality.com> wrote in message

news:1188414709.874814.293750@x35g2000prf.googlegroups.com...

For the E field, if you place a wire 1m long in a
field of X volts/m, you will generate X volts between the ends of
the
wire.

I don't agree with this? Perhaps: if you have a voltage drop of X
volts along a 1m length of a conductor, the E field along that length
averages 1V/m.

For the A/m, if this is true it may be what you're after. The H in a
circle centered around a straight infinitely long current carrying
conductor is H=I/2*pi*r. So the A/m is the 'Amps' in the conductor
divided by the 'meters' of the path.

On Aug 29, 6:26 pm, "operator jay" <n...@none.none> wrote:
"WillWorkForNeurons" <ger...@exality.com> wrote in message

news:1188414709.874814.293750@x35g2000prf.googlegroups.com...

For the E field, if you place a wire 1m long in a
field of X volts/m, you will generate X volts between the ends of
the
wire.

I don't agree with this? Perhaps: if you have a voltage drop of X
volts along a 1m length of a conductor, the E field along that length
averages 1V/m.

For the A/m, if this is true it may be what you're after. The H in a
circle centered around a straight infinitely long current carrying
conductor is H=I/2*pi*r. So the A/m is the 'Amps' in the conductor
divided by the 'meters' of the path.

Yes, you're right about the E field. A wire would short out the
field, so it must be that a high impedance measurement separated by 1m
would show X volts in a field of X volts/m.

H = 2*pi*R is the formula for H at a *point* R meters from a wire, so
H is therefore the same at all points of a circle of radius R around
the wire, so therefore a circle of radius R is subject to a constant H
field around its circumference and produces a current I = 2*pi*R*H.
This does seem to say that a loop of length X meters in a field of H
amps/m will have H * X amps induced in it.

But this only applies when the H field is tangential to the loop at
all points. This is Ampere's Law: the line integral of H (i.e.
integral of tangential H) around a loop equals the current passing
through (penetrating the plane of) the loop. So H * X amps induced in
a loop only applies for a circular H field tangential to the loop.

Here's a clue from Krauss & Fleisch: the H field inside an N-turn
solenoid is H = N * I / L, where L is *the length of the solenoid*
along its axis. Maybe this is the way to think of it then: placing an
N-turn solenoid of length L in a field H will produce a current I = H
* L / N. (The solenoid axis would have to be parallel to the H field
vector.) The length part of H then refers to the *axial length of a
test solenoid*, and you could say that placing an N-turn solenoid 1m
long in a field of H would produce a current H / N.

This says nothing about the diameter of the solenoid however, so it
seems that you could have a zero-diameter solenoid of 1 turn (i.e. a
straight wire), and it would produce a current of H * L amps when
aligned parallel with the field. Now THIS is a nice analogy with the
electric field!

"WillWorkForNeurons" <ger...@exality.com> wrote in message

news:1188679174.139482.176290@r29g2000hsg.googlegroups.com...



On Aug 29, 6:26 pm, "operator jay" <n...@none.none> wrote:
"WillWorkForNeurons" <ger...@exality.com> wrote in message

news:1188414709.874814.293750@x35g2000prf.googlegroups.com...

For the E field, if you place a wire 1m long in a
field of X volts/m, you will generate X volts between the ends of
the
wire.

I don't agree with this? Perhaps: if you have a voltage drop of X
volts along a 1m length of a conductor, the E field along that length
averages 1V/m.

For the A/m, if this is true it may be what you're after. The H in a
circle centered around a straight infinitely long current carrying
conductor is H=I/2*pi*r. So the A/m is the 'Amps' in the conductor
divided by the 'meters' of the path.

Yes, you're right about the E field. A wire would short out the
field, so it must be that a high impedance measurement separated by 1m
would show X volts in a field of X volts/m.

H = 2*pi*R is the formula for H at a *point* R meters from a wire, so
H is therefore the same at all points of a circle of radius R around
the wire, so therefore a circle of radius R is subject to a constant H
field around its circumference and produces a current I = 2*pi*R*H.
This does seem to say that a loop of length X meters in a field of H
amps/m will have H * X amps induced in it.

But this only applies when the H field is tangential to the loop at
all points. This is Ampere's Law: the line integral of H (i.e.
integral of tangential H) around a loop equals the current passing
through (penetrating the plane of) the loop. So H * X amps induced in
a loop only applies for a circular H field tangential to the loop.

Here's a clue from Krauss & Fleisch: the H field inside an N-turn
solenoid is H = N * I / L, where L is *the length of the solenoid*
along its axis. Maybe this is the way to think of it then: placing an
N-turn solenoid of length L in a field H will produce a current I = H
* L / N. (The solenoid axis would have to be parallel to the H field
vector.) The length part of H then refers to the *axial length of a
test solenoid*, and you could say that placing an N-turn solenoid 1m
long in a field of H would produce a current H / N.

This says nothing about the diameter of the solenoid however, so it
seems that you could have a zero-diameter solenoid of 1 turn (i.e. a
straight wire), and it would produce a current of H * L amps when
aligned parallel with the field. Now THIS is a nice analogy with the
electric field!

with only one problem... its the time derivative of the B or H field that
induces current in the wire... you can have a constant field and a
non-moving wire forever and not get any current... but move the wire or
change the field and you start the current.

This says nothing about the diameter of the solenoid however, so it
seems that you could have a zero-diameter solenoid of 1 turn (i.e. a
straight wire), and it would produce a current of H * L amps when

The units of the magnetic intensity field H are amps/meter, but what
does this mean? For the E field, if you place a wire 1m long in a
field of X volts/m, you will generate X volts between the ends of the
wire. What's the analogy for H? A closed loop 1m long in an H field
of Y amps/m will have Y amps induced? That doesn't seem right.

Thanks for any insight.

WillWorkForNeurons wrote:
The units of the magnetic intensity field H are amps/meter, but what
does this mean? For the E field, if you place a wire 1m long in a
field of X volts/m, you will generate X volts between the ends of the
wire. What's the analogy for H? A closed loop 1m long in an H field
of Y amps/m will have Y amps induced? That doesn't seem right.

Thanks for any insight.

With the E field strength and Faraday's law (rot E = -dB/dt) the induced
voltage can be calculated. Faraday's Law is one of Maxwells equations.

The analogon for the magnetic field strength is using Ampere's Law (rot
H = J + dD/dt) to calculate the ampere turns. That means that a closed
loop in a H field describes the current in that loop. But there is no
dependency to the frequency like in Faraday's law.

Another analogon is the relationship between electric and magnetic
circuits, that can be derived from the formulas above.
So, for electric circuits there are Ohms law
(resistance=voltage/current). For magnetic there are also such a
relationship (reluctance=ampere turns/flux). Using that magnetic
circuits can be calculated.

electric circuit magnetic circuit
voltage <-> ampere turns (theta)
current <-> flux (phi)
resistance <-> reluctannce

Regards
Frank

On Sep 3, 12:42 am, Frank Deicke <frankdei...@usenet.cnntp.org> wrote:

WillWorkForNeurons wrote:

The units of the magnetic intensity field H are amps/meter, but what
does this mean? For the E field, if you place a wire 1m long in a
field of X volts/m, you will generate X volts between the ends of the
wire. What's the analogy for H? A closed loop 1m long in an H field
of Y amps/m will have Y amps induced? That doesn't seem right.

Thanks for any insight.

With the E field strength and Faraday's law (rot E = -dB/dt) the induced
voltage can be calculated. Faraday's Law is one of Maxwells equations.

The analogon for the magnetic field strength is using Ampere's Law (rot
H = J + dD/dt) to calculate the ampere turns. That means that a closed
loop in a H field describes the current in that loop. But there is no
dependency to the frequency like in Faraday's law.

Another analogon is the relationship between electric and magnetic
circuits, that can be derived from the formulas above.
So, for electric circuits there are Ohms law
(resistance=voltage/current). For magnetic there are also such a
relationship (reluctance=ampere turns/flux). Using that magnetic
circuits can be calculated.

electric circuit magnetic circuit
voltage <-> ampere turns (theta)
current <-> flux (phi)
resistance <-> reluctannce

Regards
Frank


Thank you Frank. I know these equations, but I am trying to determine
the practical effect of the H field. What sort of test probe would be
used, and what would the effect be? Placing a loop in an arbitrary H
field will not create a current in the loop unless there is current
which passes through the plane of the loop, so that is not it. By the
mag circuit rules you relate, a given H field would produce a certain
flux when passing through a given reluctance, but that's not a direct
measurement of amps/m and doesn't relate to the electric field.
Generating a current of H * L / N when a solenoid is placed in the
field is the best I can come up with, but I'm still not happy with
that. For one thing, a coil in a mag field is nothing more than the
secondary of a transformer. Yet a transformer induces *voltage* based
on total *flux*, so how do these two relate? Maybe if I work through
the units of B = H * mu that will be clear.

PS your English is great (much better than my Dutch), but it's
'analogy', not 'analogon'. There's no English word 'analogon'.

WillWorkForNeurons wrote:
On Sep 3, 12:42 am, Frank Deicke <frankdei...@usenet.cnntp.org> wrote:

WillWorkForNeurons wrote:

The units of the magnetic intensity field H are amps/meter, but what
does this mean? For the E field, if you place a wire 1m long in a
field of X volts/m, you will generate X volts between the ends of the
wire. What's the analogy for H? A closed loop 1m long in an H field
of Y amps/m will have Y amps induced? That doesn't seem right.

Thanks for any insight.

With the E field strength and Faraday's law (rot E = -dB/dt) the induced
voltage can be calculated. Faraday's Law is one of Maxwells equations.

The analogon for the magnetic field strength is using Ampere's Law (rot
H = J + dD/dt) to calculate the ampere turns. That means that a closed
loop in a H field describes the current in that loop. But there is no
dependency to the frequency like in Faraday's law.

Another analogon is the relationship between electric and magnetic
circuits, that can be derived from the formulas above.
So, for electric circuits there are Ohms law
(resistance=voltage/current). For magnetic there are also such a
relationship (reluctance=ampere turns/flux). Using that magnetic
circuits can be calculated.

electric circuit magnetic circuit
voltage <-> ampere turns (theta)
current <-> flux (phi)
resistance <-> reluctannce

Regards
Frank

Thank you Frank. I know these equations, but I am trying to determine
the practical effect of the H field. What sort of test probe would be
used, and what would the effect be? Placing a loop in an arbitrary H
field will not create a current in the loop unless there is current
which passes through the plane of the loop, so that is not it. By the
mag circuit rules you relate, a given H field would produce a certain
flux when passing through a given reluctance, but that's not a direct
measurement of amps/m and doesn't relate to the electric field.
Generating a current of H * L / N when a solenoid is placed in the
field is the best I can come up with, but I'm still not happy with
that. For one thing, a coil in a mag field is nothing more than the
secondary of a transformer. Yet a transformer induces *voltage* based
on total *flux*, so how do these two relate? Maybe if I work through
the units of B = H * mu that will be clear.

Use Faradays Law, I think that describes the relation between *flux* and
*voltage* on a coil.

rot E = -dB/dt

if integrals are used:

V = int(E,ds) = - dflux/dt = - d(int(B,dA))/dt ...



PS your English is great (much better than my Dutch), but it's
'analogy', not 'analogon'. There's no English word 'analogon'.

I looked at different dictonaries and there I find the english word
analogon. So, it seems that it exists.

On Sep 3, 12:42 am, Frank Deicke <frankdei...@usenet.cnntp.org> wrote:
WillWorkForNeurons wrote:
The units of the magnetic intensity field H are amps/meter, but what
does this mean? For the E field, if you place a wire 1m long in a
field of X volts/m, you will generate X volts between the ends of the
wire. What's the analogy for H? A closed loop 1m long in an H field
of Y amps/m will have Y amps induced? That doesn't seem right.

Thanks for any insight.

With the E field strength and Faraday's law (rot E = -dB/dt) the induced
voltage can be calculated. Faraday's Law is one of Maxwells equations.

The analogon for the magnetic field strength is using Ampere's Law (rot
H = J + dD/dt) to calculate the ampere turns. That means that a closed
loop in a H field describes the current in that loop. But there is no
dependency to the frequency like in Faraday's law.

Another analogon is the relationship between electric and magnetic
circuits, that can be derived from the formulas above.
So, for electric circuits there are Ohms law
(resistance=voltage/current). For magnetic there are also such a
relationship (reluctance=ampere turns/flux). Using that magnetic
circuits can be calculated.

electric circuit magnetic circuit
voltage <-> ampere turns (theta)
current <-> flux (phi)
resistance <-> reluctannce

Regards
Frank

Thank you Frank. I know these equations, but I am trying to determine
the practical effect of the H field. What sort of test probe would be
used, and what would the effect be? Placing a loop in an arbitrary H
field will not create a current in the loop unless there is current
which passes through the plane of the loop, so that is not it. By the
mag circuit rules you relate, a given H field would produce a certain
flux when passing through a given reluctance, but that's not a direct
measurement of amps/m and doesn't relate to the electric field.
Generating a current of H * L / N when a solenoid is placed in the
field is the best I can come up with, but I'm still not happy with
that. For one thing, a coil in a mag field is nothing more than the
secondary of a transformer. Yet a transformer induces *voltage* based
on total *flux*, so how do these two relate? Maybe if I work through
the units of B = H * mu that will be clear.

PS your English is great (much better than my Dutch), but it's
'analogy', not 'analogon'. There's no English word 'analogon'.

"WillWorkForNeurons" <ger...@exality.com> wrote in message

news:1188853007.780323.289230@g4g2000hsf.googlegroups.com...



On Sep 3, 12:42 am, Frank Deicke <frankdei...@usenet.cnntp.org> wrote:
WillWorkForNeurons wrote:
The units of the magnetic intensity field H are amps/meter, but what
does this mean? For the E field, if you place a wire 1m long in a
field of X volts/m, you will generate X volts between the ends of the
wire. What's the analogy for H? A closed loop 1m long in an H field
of Y amps/m will have Y amps induced? That doesn't seem right.

Thanks for any insight.

With the E field strength and Faraday's law (rot E = -dB/dt) the induced
voltage can be calculated. Faraday's Law is one of Maxwells equations.

The analogon for the magnetic field strength is using Ampere's Law (rot
H = J + dD/dt) to calculate the ampere turns. That means that a closed
loop in a H field describes the current in that loop. But there is no
dependency to the frequency like in Faraday's law.

Another analogon is the relationship between electric and magnetic
circuits, that can be derived from the formulas above.
So, for electric circuits there are Ohms law
(resistance=voltage/current). For magnetic there are also such a
relationship (reluctance=ampere turns/flux). Using that magnetic
circuits can be calculated.

electric circuit magnetic circuit
voltage <-> ampere turns (theta)
current <-> flux (phi)
resistance <-> reluctannce

Regards
Frank

Thank you Frank. I know these equations, but I am trying to determine
the practical effect of the H field. What sort of test probe would be
used, and what would the effect be? Placing a loop in an arbitrary H
field will not create a current in the loop unless there is current
which passes through the plane of the loop, so that is not it. By the
mag circuit rules you relate, a given H field would produce a certain
flux when passing through a given reluctance, but that's not a direct
measurement of amps/m and doesn't relate to the electric field.
Generating a current of H * L / N when a solenoid is placed in the
field is the best I can come up with, but I'm still not happy with
that. For one thing, a coil in a mag field is nothing more than the
secondary of a transformer. Yet a transformer induces *voltage* based
on total *flux*, so how do these two relate? Maybe if I work through
the units of B = H * mu that will be clear.

PS your English is great (much better than my Dutch), but it's
'analogy', not 'analogon'. There's no English word 'analogon'.

a passive probe is just a loop of wire with an ammeter. it will only
register if the field is changing, or if you move it through the field. you
do not need a current to pass through the plane of the loop, only for the
flux through the loop to change.

on the other hand, check out hall effect detectors and sensors. Also SQUID
detectors.