Question

Given vector potential at a distance 7cm of a solenoide axis which has a length of infinite with radius is 2cm and have magnetic field 0,3T ?

Answer 1

Answer

Relationship between potential vector and magnetic field can given as

$\vec{B}=\vec{\nabla}\times \vec{A}$

Multiplying by $\vec{ds}$ Both side

$\int\vec{B}.\vec{ds}=\int\vec{\nabla}\times (\vec{A}.\vec{ds})$

By stoke's thoerm

$\int\vec{\nabla}\times (\vec{A}.\vec{ds}) =\int \vec{A} . \vec{dl}$

So

$\int \vec{A}.\vec{dl}=\int \vec{B}.\vec{ds}$

Consider a solenoid of radius "a"

Potential vector and dl vector are in same direction and magnetic field and ds are in same direction

So

$\int {A}{dl}=\int {B}{ds}$

Potential vector and magnetic field are constant so

$A\int dl=B\int {ds}$

$A(2\pi r) =B(\pi a^2)$

$A=\frac{Ba^2}{2r}\\$

So by putting the value in above equation

$A=\frac{Ba^2}{2r}\\$

$A=\frac{0.3(0.02)^2}{2(0.07)}=8.57\times10^{-4}Tm$