Magnetic force: An approach with Bernoulli's equation.

[Ka-In Yen]

1. Abstract: In this paper, Bernoulli's effect is used to interpret
the magnetic force.

See my posting:
http://groups.google.com/groups?selm=9f181401.0312250209.55762999%40posting.google.com

All comments are welcome.

Ka-In Yen
yenkain@yahoo.com.tw
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[Ka-In Yen]

yenkain@yahoo.com.tw (Ka-In Yen) wrote in message news:<9f181401.0401140031.7709d6e3@posting.google.com>...

1. Abstract: In this paper, Bernoulli's effect is used to interpret
the magnetic force.

See my posting:
http://groups.google.com/groups?selm=9f181401.0312250209.55762999%40posting.google.com


Fm = F1+F2 = -(m/R)*v1*v2 - L1*(v1)^2 /2 - L2*(v2)^2 /2 (1)
Assuming the surrounding are neutral, static, and far away from
wire 1 and wire 2, then L1 and L2 terms can be ignored.
Fm = -(m/R)*v1*v2 = - mu_0 *q1*q2*v1*v2/ (4*pi* R^2)
Fm = - mu_0 * i1 * i2 / (4 * pi * R^2) (2)

My dear friends,

I am very sorry to disappoint you, the above derivation is
incomplete. To eliminate L1 and L2 terms, negative potential
masses have to be considered. A shematic diagram is shown as
figure 2.

(+q1,0)
(-q1,v1)
______________+______________________________ wire 1
^
|
R
(+q2,0)
| (-q2,v2)
V______________+_____________________________ wire 2


Figure 2

The whole electric wire is neutral; for every drifting electron(-),
there is a resting ion(+). (-q1,v1) is charges of drifting electrons,
and (+q1,0) is charges of resting ions. (-q2,v2) and (+q2,0) are
same definition.

1) To (-q1,v1) and (-q2,v2) pair, we have (m/R)*(v1+v2)^2 /2.
2) To (-q1,v1) and (+q2,0) pair, we have (-m/R)*v1^2 /2.
3) To (+q1,0) and (-q2,v2) pair, we have (-m/R)*v2^2 /2.

1) + 2) + 3) = (m/R)*v1*v2 = mu_0 *q1*q2*v1*v2/(4*pi* R^2)
= mu_0 *i1*i2 / (4*pi* R^2)



Ka-In Yen
yenkain@yahoo.com.tw


How to correctly measure an unknown length with a clock.
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