Biot-Savart if current loop shape is a parameterized function - doesn't that define dl vector?



 Science > Physics > Biot-Savart if current loop shape is a parameterized function - doesn't that define dl vector?

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Topic: Science > Physics
User: "Monty Hall"
Date: 10 Apr 2005 11:06:57 PM
Object: Biot-Savart if current loop shape is a parameterized function - doesn't that define dl vector?
I have a quick question about calculating mag field @ r using Biot-Savart.
Given:
current loop shape in x-y plane:
rp(m) = ( rpx(m), rpy(m), 0 ),
biot-savart (k=constant scalar, X=cross product):
dB(r) = k dl X (r - rp) / |r - rp|^3
I'm not sure how to obtain dl only given rp(m). It would seem that dl in
this case would be dl = d/dm rp(m). If this is so, why do books use dl and
not d rp????
Regards,
Monty
.

User: "Schoenfeld"

Title: Re: Biot-Savart if current loop shape is a parameterized function - doesn't that define dl vector? 11 Apr 2005 06:00:05 PM
Monty Hall wrote:

I have a quick question about calculating mag field @ r using

Biot-Savart.


Given:

current loop shape in x-y plane:
rp(m) = ( rpx(m), rpy(m), 0 ),

biot-savart (k=constant scalar, X=cross product):
dB(r) = k dl X (r - rp) / |r - rp|^3

I'm not sure how to obtain dl only given rp(m). It would seem that

dl in

this case would be dl = d/dm rp(m). If this is so, why do books use

dl and

not d rp????

dl = (D(rp)/D(rpx)) i + (D(rp)/D(rpy)) j + (D(rp)/D(rpz)) k)
where
D = partial derivative
i,j,k = basis vectors


Regards,

Monty

.


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