sci.physics.electromag

"WillWorkForNeurons" wrote in message
news: @ ...
> On Sep 6, 2:22 pm, "Dave" wrote:
>> "WillWorkForNeurons" wrote in message
>>
>> news: @ ...
>>
>>
>>
>> > On Sep 4, 4:26 pm, "Dave" wrote:
>> >> "WillWorkForNeurons" wrote in message
>>
>> >>news: @ ...
>>
>> >> > On Sep 3, 12:42 am, Frank Deicke
>> >> > 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.
>>
>> > What I'm thinking of is a loop entirely within the field. In that
>> > case, by Ampere's Law the current in the loop equals the line integral
>> > of H around the loop, which is zero unless current passes through the
>> > plane of the loop ( . the loop links the current). You're proposing
>> > a more practical solution, where only a part of the loop is in the
>> > field, and the loop is broken to bring it outside the field to an
>> > ammeter. In this case, yes, you've got a mag field detector which
>> > responds to changing flux as you say. But what is the current, for a
>> > given H and loop configuration? We're back to the original question:
>> > with this test loop in an H field, what does the 'per meter' part
>> > mean? It's easy to calculate the *voltage* induced in this test loop
>> > as a function of the *B* field through it and its area, but how is
>> > this a direct probe for H? Again I come to the solenoid (also
>> > connected to an ammeter outside the field) which produces a current of
>> > H * L / N. If you allow the solenoid to degenerate to 1 turn with
>> > solenoid diameter zero, this becomes just a wire in the field. I
>> > think that means a wire 1 m long in an H field of X amps/m, parallel
>> > to the lines of H, will generate X amps (if you connect it to an
>> > ammeter outside the field). I think that's the probe analogy to
>> > measuring the E field with high-impedance probes 1m apart.
>>
>> nope, you still haven't got the idea. it is not the being in/out of the
>> field that makes the current flow, nor is it current through the loop.
>> you
>> don't need current through the loop at all. what you need to detect the
>> field with a wire loop is a changing field. it has to have a time
>> varying
>> component, and you can only detect the time varying component of it...
>> unless you move the loop through the field. if you want to measure a
>> static
>> field without moving the loop you need something like a hall effect
>> device
>> that measures the field by moving electrons through it. you can also do
>> it
>> by shooting charged particles into the field and measuring the deflection
>> and other things like that, but you need the relative movement to do it.
>
> OK, sorry, I was misremembering Ampere's Law. It's *mmf* in a closed
> loop which is zero unless the loop links current. You are right --
> all it takes is changing flux through a loop to induce voltage.
> That's the way a transformer works, after all. But most of of my
> previous post remains -- just cross out the first seven lines.
>

/wiki/Maxwell%27s_equations


B=u_0*H
u_0=4*pi*10^-7 henries/m
H units = amp/m
B units = weber/m^2
E units = volt/m

para (3) from above seems most useful for demonstration... faraday's law in
integral form:

integral of E*dl around loop = -d/dt integral of B*dA over the enclosed
surface
but just for grins, replace B with u_0*H
integral of E*dl around loop = -d/dt integral of u_0*H*dA over the enclosed
surface

just looking at units...

E*dl = volt/m * m = volts

d/dt * B * dA = 1/sec * henries/m*amp/m * m^2
reduced to
= henries*amp/sec

/wiki/Henries
the henry unit H = m^2*kg/s^2*A^2 = Wb/A = V*s/A = m^2*kg/C^2

using V*s/A as most easily plugged in you reduce the above to:

henries*amp/sec = Volts*sec/Amp * amp/sec = Volts
QED... well, not really, just showing that units match up across faraday's
law in integral form.

the physical interpretation is that the rate of change of the field through
a loop of area A induces a voltage around the loop... various conversions
are needed depending on units chosen and materials in the loop (above
assumes free space so using u_0, but same analysis works using permeability
of materials placed inside the loop).

so instead of just looking at the E field along the path between 2 points to
get the voltage, with the magnetic field you have to first define a closed
loop, then integrate the magnetic field passing through the loop, then
measure how fast that total field is changing, multiply by the appropriate
constants and you get voltage. some interesting aspects are that it doesn't
matter what the absolute field strength is, only how quickly it is changing.
and it doesn't matter if the change is caused by the loop moving or the
field changing while the loop is stationary.