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Re: electromagnetism



sorry for the delay in sending the rest of that electromagnetism question as
i have been busy studying and trying to do this damn question.
for part b
since the frequency is not told, i converted the 240 volts in the primary
into 60volts in the secondary(400 turns versus 100 turns)((240/400)*100)
this 60v is terminated by 50 ohms which gives a secondary current of 1.2
Amps(I=v/r)
this is where i got stuck.
any suggestions, if i am right in what i have done.

p.s i don't know if this has been posted already but was the B for magnetic
flux density not due to Biot-Savarts,s law. i was studying today and
noticed something about Biot-Savart's law. this has probably been posted as
i vaguely remember someone posting about Biot.
goodnight
paul


----- Original Message -----
From: "Brian Whatcott" <inet@INTELLISYS.NET>
To: <PHYS-L@lists.nau.edu>
Sent: Wednesday, April 03, 2002 5:07 AM
Subject: Re: electromagnetism


At 03:14 PM 4/2/02, Paul Giusti wrote:
sorry but who is harnwell and bleaney and bleaney?

These are the authors of some texts used in American schools.


i have tried to do some
of it but as you can see, this is why i got stuck.:

firstly, i tried to calculate the total reluctance using S=flux
length/mu*A
= 5*10^-3/ 1200 * mu0 * 16*10^-3
=207.2 At/Wb

You determined the flux length was 5 mm and the cross section area was 16
mm^2
and mu0 was 1.257E-6 N/A^2 Your handling of scientific notation was
unusual.
You wrote 5*10^-3 for 50 mm instead of 5*10^-2.

But later, when expressing the numbers in decimal fractions, the numbers
looked better.
There you expressed the same length as 0.05

i then calculated the self inductance using L=N^2/S
=400^2/207.2
= 772.2 H
this is where i got confused and started e-mailing physics lecturers. i
noticed in my notes, for a similar shaped circuit as the one i have got,
i
have writen beside it " series/paralell circuit, total S= centre+1/2 side
reluctances(assuming symetrical)." i did not understand what this meant
and
have looked throught books but cannot seem to find an explanation.
if that means what i think it does then the total reluctance would be
(S=flux length/mu*A) + (2*S=flux length/mu*A)
=(2.5^-3/1200*mu0*16*10^-3) + (2*( 5*10^-3/ 1200*mu0*10*10^-3))
i am not sure however and genuinely cannot find anything in any books. i
am
not lazy, looking for someone to do it for me. i am just confused as to
what
to do.
today however, i had an idea after seeing a different question in a
book.
what i did was :
S1= core length/Ur*U0*A = 0.025/(1200 *U0 * 0.16) =103.6 at/wb
S2= 2*(flux path lenght / Ur*U0*A) = 2* (0.05/ 1200 *U0 *0.1) =2* 331.57
at/wb
= 663.15 at/wb
therefore total reluctance = 663.15+766.75 at/wb
for S2, i multiplied this by 2 because i treated them as two seperate
shapes. you may not be able to see what i mean as you do not have a
diagram,
which is with the question, and i am not allowed to send attachments.
then i worked out the self inductance by using the formula L= N^2/S,
however
my answer for this depends on the answer for reluctance being correct.
paul


The central difficulty you expressed was in visualizing reluctance,
analogous to working out circuit resistance but of a magnetic path. If
you had a resistor R1, whose ends were connected by two parallel side
resistors R2 each, you would not have much trouble in visualizing the
total
circuit resistance through the center limb as R1 + R2/2

But I should apologize to you. You sent me a diagram so that I could place
it in a web accessible position.
Unfortunately, this was a diagram embedded in a Microsoft Word 97 product,
which guaranteed I could not extract it to the web (unwilling as I am to
keep up the race to next years' products of the Evil Kingdom) It was by
all accounts a conventional shell transformer (i.e. one with a central
limb
and two side limbs) and the only dimension not mentioned elsewhere was the
side limb cross section, which you gave above.

To summarize - refreshing on number representation, so you can quickly
check that
1.23E-3 = 1.23*10^-3 = 0.00123 might be helful, and the idea of
reluctances
being treated like series and parallel resistances would not be hard to
keep in mind.

Do you want to persevere with the remaining parts of the problem? I am
finding your
attack helpful.

Brian W



Brian Whatcott
Altus OK Eureka!