[no subject]

["Steven T. Ratliff" <STR@NWC.EDU>]

You can derive the equation from either Ampere's law or the law of Biot
and Savart.
It is not either law, although it can be derived from either one.

The equation given is just the magnetic field due to a long wire at a
distance r from the wire.




Steven T. Ratliff
Professor of Physics
Northwestern College
3003 Snelling Ave. N.
St. Paul,  MN     55113-1598
U. S. A.

Internet:  stratliff@nwc.edu





giustifamily <giustifamily@BLUEYONDER.CO.UK>
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04/12/02 12:49 PM
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hi all. is B=mu noughttimes current/2pi r not biot-savarts law. one of my
parents friends, a physics teacher, says it is ampere's law. who is
correct?

----- Original Message -----
From: "Brian Whatcott" <inet@INTELLISYS.NET>
To: <PHYS-L@lists.nau.edu>
Sent: Wednesday, April 10, 2002 1:35 AM


> At 04:28 PM 4/9/02, Paul  Giusti wrote:
> > ... a small question i cannot do. it is
> > on analogue electronics...
>
> > an amplifier has a second order gain/frequency response with a lower
cut-off
> > frequency of 350Hz, an upper cut-off frequency of 4kHz and a mid-band
> > voltage of 50 dB.
> > sketch the graph of gain against frequency.
> >
> > estimate the voltage gain in dB at the following frequencies.
> > i)    20Hz
> > ii)    350Hz
> > iii)    2kHz
> > iv)    10kHz
> >
> >
> > thanks if any of you can help.
> > paul
>
> An RC circuit has the property that the output voltage halves when the
> frequency doubles.
> As you will know, the R element holds a constant value, but the C
element
> has a reactance proportional to 1/fC.
>   A doubling is labeled as an octave, and a voltage halving is described
as
> -6 dB on this logarithmic scale. So a single pole filter is said to fall
> off at -6dB/octave.
> A two pole filter falls at  -12 dB/octave. This is also called a second
> order response.
>
> If I sketch a pass band gain of 50 dB, with a roll off at -12 dB to 0 dB
at
> 4kHz, and a slope of 12 dB from 0 dB at 350 Hz, I find I am sketching
the
> impossible.
>
> It could just be that the corner frequencies specified are meant to be
the
> pole frequencies at which the slopes begin.
> If I suppose the lower pole of  350 Hz represents 50 dB  voltage gain
then
> I might expect 0 dB at 22 Hz, and the high cutoff of 0dB at 64 kHz.
>
> If this were the case, the gain at  20 Hz would be 0dB, at 350 Hz = 50
> dB,   2 kHz = 50 dB,   10 kHz = 32dB
>
> I am however doubtful that I have correctly interpreted the intent of
this
> question.
>
>
> Brian Whatcott
>    Altus OK                      Eureka!