If everything remains the same, how does doubling the diameter of a
solenoid affect the magnetic strength?

On May 23, 8:34 pm, soo...@webtv.net wrote:
If everything remains the same, how does doubling the diameter of a
solenoid affect the magnetic strength?

current's bigger; moment's bigger

On May 23, 8:34 pm, soo...@webtv.net wrote:
If everything remains the same, how does doubling the diameter of a
solenoid affect the magnetic strength?

current's bigger; moment's bigger

If everything remains the same, how does doubling the diameter of a
solenoid affect the magnetic strength?

Autymn D. C. wrote:
On May 23, 8:34 pm, soo...@webtv.net wrote:
If everything remains the same, how does doubling the diameter of a
solenoid affect the magnetic strength?

current's bigger; moment's bigger

I was looking for an answer that would tell me the how the magnet
intensity in the solenoid is related to the diameter. If the diameter
is doubled, is the intensity reduced by 1/2 or 1/2 times 1/2?

Autymn D. C. babbled:
On May 23, 8:34 pm, soo...@webtv.net wrote:
If everything remains the same, how does doubling the diameter of a
solenoid affect the magnetic strength?

current's bigger; moment's bigger

I was looking for an answer that would tell me the how the magnet
intensity in the solenoid is related to the diameter. If the diameter
is doubled, is the intensity reduced by 1/2 or 1/2 times 1/2?

soo...@webtv.net wrote:
Autymn D. C. babbled:
On May 23, 8:34 pm, soo...@webtv.net wrote:
If everything remains the same, how does doubling the diameter of a
solenoid affect the magnetic strength?

current's bigger; moment's bigger

One doesn't use improper contractions here. Of course we all know that
YOU are dictator of the laws of Usenet spelling. But then your answer
is totally bogus as well [What else is new?] He said "everything"
remains the same, which includes current! Remember, Bunkie?

I was looking for an answer that would tell me the how the magnet
intensity in the solenoid is related to the diameter. If the diameter
is doubled, is the intensity reduced by 1/2 or 1/2 times 1/2?

FIRST, YOU should learn how to ask a question. You really have to
define the coils you are talking about exactly to get an exact
answer. What exactly do you mean by "everything stays the same"?
What do you mean by "doubling the diameter"? This time, I'll make
these choices for you, OK?

IF you have a solenoid, and it's considered "long" (means length is
several times the outside diameter) and it has a certain number of
turns meaning it's wound with a certain gauge wire and has an inside
diameter and an outside diameter and you are putting a given current
through the wire, then you wind on more turns of the same wire until
the outside diameter of the coil is twice the original value, you'll
find that the magnetic field at the center of the solenoid for coils
of "average" size and shape is roughly two to three times what is was
originally if you apply the same current you used before Voltage and
wattage will however change. However, the values depend greatly on
coil geometry which is to say on the number of turns per unit length.
If the original winding thickness becomes small like say the
difference between a single layer coil and one with lots of turns
piled on large variations in field can result between the original and
one twice the diameter.

If on the other hand you mean a single layer solenoid which you simply
increase in diameter while keeping the current and wire gauge the
same. Then B at the center of the coil does not depend on the length
or diameter of the solenoid. B = uo ni, n = turns per unit length
(meter) , i = current. Note that since diameter does not enter in, you
can find how B changes B by comparing the TURNS per unit le
ngth for
each of the two coils regardless of geometry.
It's mostly a geometry problem finding how many turns on a given coil
of given length. Note that the formula only works for B at the center
of the coil (speaking about lengthwise here) and if the coil is
"long".

OK?
Benj

The coil is single layer. I am asking about the magnetic field
intensity inside the coil at the center. The diameter is one inch and
the length is three inches. Now double the diameter to two inches and
the length to six inches. What happens to the intensity?

I know in practice the field becomes much weaker when the diameter
and the length are doubled. I was looking for a simple formula to
predict the change.

The voltage and the capacitor bank remains the same. The coil is used
as a magnetizer. The wire gauge remains the same.

soo...@webtv.net wrote:

The coil is single layer. I am asking about the magnetic field
intensity inside the coil at the center. The diameter is one inch and
the length is three inches. Now double the diameter to two inches and
the length to six inches. What happens to the intensity?

Pretty much stays the same. A 1 to 3 ratio isn't terrifically long
but still long enough to create massive end effects.

I know in practice the field becomes much weaker when the diameter
and the length are doubled. I was looking for a simple formula to
predict the change.

Are you sure you are keeping the CURRENT the same in both cases or are
you running the coil from a power supply that keeps the VOLTAGE the
same? The larger coil is wound with a much longer wire and therefore
has more resistance and needs more voltage to drive the same current
through the wire! With the same current the field should not be "much"
weaker.

The voltage and the capacitor bank remains the same. The coil is used
as a magnetizer. The wire gauge remains the same.

Magnetizer! BINGO! Note that if, say, you wound the coil with 10 gauge
wire, the small coil would have a resistance of about .0078 Ohm while
the large one would be .031 Ohm. The large coil would take almost 4
times the voltage to drive the same current and get the same field as
the small coil! A practical solution would be to wind the large coil
with say 6 gauge wire which would give roughly the same total
resistance as the smaller coil. But all is not well in magnet land,
because the field depends not only on the current but on the number of
turns per unit length as well. With the fatter wire, the turns per
unit length drops to a roughly half the value with the original coils.
This means you are going to have to drive twice the current through
the coil to get the original field of the small coil. But at least you
don't have to drive the capacitors to four times the original voltage!
Or even better would be to use even heavier wire and use two layers or
more layers. You could probably also get away with making the larger
coil a bit shorter too unless you need a fairly uniform field over
some distance. In a magnetizer this shouldn't be important because you
are driving the field over saturation of the magnet material anyway.

My practical choice would be to wind the larger coil with fatter wire,
and shorten it by half and put two layers on it. Should be close to
working. There is also the problem of the inductance of the coil
since a magnetizer is a pulsed device, but we don't want to open that
can of worms, do we?

Ok?

Benj

soo...@webtv.net wrote:

The coil is single layer. I am asking about the magnetic field
intensity inside the coil at the center. The diameter is one inch and
the length is three inches. Now double the diameter to two inches and
the length to six inches. What happens to the intensity?

Pretty much stays the same. A 1 to 3 ratio isn't terrifically long
but still long enough to create massive end effects.

I know in practice the field becomes much weaker when the diameter
and the length are doubled. I was looking for a simple formula to
predict the change.

Are you sure you are keeping the CURRENT the same in both cases or are
you running the coil from a power supply that keeps the VOLTAGE the
same? The larger coil is wound with a much longer wire and therefore
has more resistance and needs more voltage to drive the same current
through the wire! With the same current the field should not be "much"
weaker.

The voltage and the capacitor bank remains the same. The coil is used
as a magnetizer. The wire gauge remains the same.

Magnetizer! BINGO! Note that if, say, you wound the coil with 10 gauge
wire, the small coil would have a resistance of about .0078 Ohm while
the large one would be .031 Ohm. The large coil would take almost 4
times the voltage to drive the same current and get the same field as
the small coil! A practical solution would be to wind the large coil
with say 6 gauge wire which would give roughly the same total
resistance as the smaller coil. But all is not well in magnet land,
because the field depends not only on the current but on the number of
turns per unit length as well. With the fatter wire, the turns per
unit length drops to a roughly half the value with the original coils.
This means you are going to have to drive twice the current through
the coil to get the original field of the small coil. But at least you
don't have to drive the capacitors to four times the original voltage!
Or even better would be to use even heavier wire and use two layers or
more layers. You could probably also get away with making the larger
coil a bit shorter too unless you need a fairly uniform field over
some distance. In a magnetizer this shouldn't be important because you
are driving the field over saturation of the magnet material anyway.

My practical choice would be to wind the larger coil with fatter wire,
and shorten it by half and put two layers on it. Should be close to
working. There is also the problem of the inductance of the coil
since a magnetizer is a pulsed device, but we don't want to open that
can of worms, do we?

Ok?

Benj

Benj wrote:
soo...@webtv.net wrote:

The coil is single layer. I am asking about the magnetic field
intensity inside the coil at the center. The diameter is one inch and
the length is three inches. Now double the diameter to two inches and
the length to six inches. What happens to the intensity?

Pretty much stays the same. A 1 to 3 ratio isn't terrifically long
but still long enough to create massive end effects.

I know in practice the field becomes much weaker when the diameter
and the length are doubled. I was looking for a simple formula to
predict the change.

Are you sure you are keeping the CURRENT the same in both cases or are
you running the coil from a power supply that keeps the VOLTAGE the
same? The larger coil is wound with a much longer wire and therefore
has more resistance and needs more voltage to drive the same current
through the wire! With the same current the field should not be "much"
weaker.

The voltage and the capacitor bank remains the same. The coil is used
as a magnetizer. The wire gauge remains the same.

Magnetizer! BINGO! Note that if, say, you wound the coil with 10 gauge
wire, the small coil would have a resistance of about .0078 Ohm while
the large one would be .031 Ohm. The large coil would take almost 4
times the voltage to drive the same current and get the same field as
the small coil! A practical solution would be to wind the large coil
with say 6 gauge wire which would give roughly the same total
resistance as the smaller coil. But all is not well in magnet land,
because the field depends not only on the current but on the number of
turns per unit length as well. With the fatter wire, the turns per
unit length drops to a roughly half the value with the original coils.
This means you are going to have to drive twice the current through
the coil to get the original field of the small coil. But at least you
don't have to drive the capacitors to four times the original voltage!
Or even better would be to use even heavier wire and use two layers or
more layers. You could probably also get away with making the larger
coil a bit shorter too unless you need a fairly uniform field over
some distance. In a magnetizer this shouldn't be important because you
are driving the field over saturation of the magnet material anyway.

My practical choice would be to wind the larger coil with fatter wire,
and shorten it by half and put two layers on it. Should be close to
working. There is also the problem of the inductance of the coil
since a magnetizer is a pulsed device, but we don't want to open that
can of worms, do we?

Ok?

Benj

Sorry that I
was not clearer, the voltage remains the same not the current and
resistance.

It looks like there is no clear and simple formula because of all the
variables, just some ballpark estimates. That is OK. It is easy enough
to build different coils. Your explanations have been very helpful.

In coils of this size is there any inductive kickback? In the circuit
I have there is no diode across the coil.

It looks like there is no clear and simple formula because of all the
variables, just some ballpark estimates. That is OK. It is easy enough
to build different coils. Your explanations have been very helpful.

In coils of this size is there any inductive kickback? In the circuit
I have there is no diode across the coil.

soo...@webtv.net wrote:
Autymn D. C. babbled:
On May 23, 8:34 pm, soo...@webtv.net wrote:
If everything remains the same, how does doubling the diameter of a
solenoid affect the magnetic strength?

current's bigger; moment's bigger

One doesn't use improper contractions here. Of course we all know that
YOU are dictator of the laws of Usenet spelling. But then your answer

is totally bogus as well [What else is new?] He said "everything"
remains the same, which includes current! Remember, Bunkie?