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
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"
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