- #1
Somali_Physicist
- 117
- 13
I am sure that the vector potential of a toroid isn't 0 even though its magnetic field is , does anyone have a derivation for the its the vector potential at a point P(x,y,z) outside the toroid ? i expect that since its curl is 0 we have a general form of :
A = f(!x)ex + g(!y)ey+m(!z)ez
Such that a general function L(!k) implies that L is not the parameter k dependant.
intuitive feel:
If we rotate the torous on its back facing a zy plane we get an infinite stack of circles with a hole.Integrating over these stacks from 0 to the thickness r allows us to add up all the vector potentials.This sum is the total vector potential.An infinite stack of circles must be a cylinder therefore if we found the vector potential of a cylinder 1 - vector potential of an emptying cylinder.
A(c1) - A(c2) = A(c)
uNI/4π ( ∫1/r1 dl - ∫1/r 2dl) =A(c)
if we imagine a line to the peak of each cylinder we have two triangles which we can use to relate dl tp x,y,z from there i would assume we use cylindrical coordinates to get solvable integral.However this seems long winded and prone to error.
Does anyone know how to do this?
A = f(!x)ex + g(!y)ey+m(!z)ez
Such that a general function L(!k) implies that L is not the parameter k dependant.
intuitive feel:
If we rotate the torous on its back facing a zy plane we get an infinite stack of circles with a hole.Integrating over these stacks from 0 to the thickness r allows us to add up all the vector potentials.This sum is the total vector potential.An infinite stack of circles must be a cylinder therefore if we found the vector potential of a cylinder 1 - vector potential of an emptying cylinder.
A(c1) - A(c2) = A(c)
uNI/4π ( ∫1/r1 dl - ∫1/r 2dl) =A(c)
if we imagine a line to the peak of each cylinder we have two triangles which we can use to relate dl tp x,y,z from there i would assume we use cylindrical coordinates to get solvable integral.However this seems long winded and prone to error.
Does anyone know how to do this?