- #1
Mr-R
- 123
- 23
Dear all,
I am trying use Ampere's law and came across a problem which is a result of my lack of understanding.
\oint B.dl=uI
My lecturer derived the above equation by defining a point in the middle of a potato shaped loop. Where he stated that dl in this case is just rdθ. (r is the line joining dl to the point defined in the middle). And using B=\frac{uI}{2πr} he integrated it with respect to rdθ from θ=0 to 2π to get \oint B.dl=uI.
Now, I am trying to do the same for a circular loop. Obviously I got the same answer uI. But its wrong as it should be \frac{uI}{2r}. The problem is that I am using the same approch he used for the general shape but not getting the same answer. I am using B=\frac{uI}{2πr} where I should actually use B=\frac{uI}{4πr^{2}}
Could someone explain why can't I do that?
P.s sorry if this is messy. It is my first time.
Rashid
I am trying use Ampere's law and came across a problem which is a result of my lack of understanding.
\oint B.dl=uI
My lecturer derived the above equation by defining a point in the middle of a potato shaped loop. Where he stated that dl in this case is just rdθ. (r is the line joining dl to the point defined in the middle). And using B=\frac{uI}{2πr} he integrated it with respect to rdθ from θ=0 to 2π to get \oint B.dl=uI.
Now, I am trying to do the same for a circular loop. Obviously I got the same answer uI. But its wrong as it should be \frac{uI}{2r}. The problem is that I am using the same approch he used for the general shape but not getting the same answer. I am using B=\frac{uI}{2πr} where I should actually use B=\frac{uI}{4πr^{2}}
Could someone explain why can't I do that?
P.s sorry if this is messy. It is my first time.
Rashid
