Question

A spherical shell of radius R, carrying a uniform surface charge a, is set spinning at angular velocity (J). Find the vector potential it produces at point r (Fig. 5.45).

Answer 1

Explanation:

• vector r lies on z- axis
• J is tilted at angle Ψ
• Orient x-axis such that w lies in x-z plane

Given:

Vector potential                          

                                          $A (r) = \frac{u_o}{4*\pi } \int {\frac{K(r')}{r*} } \, da'$

Where,  K = б*v ; r* = sqrt (R^2 + r^2 -2R*r*cos(θ'))  ; da' = R^2*sin(θ')*dθ'dΦ'

Solution:

- Velocity of v point a point r' in a rotating rigid body is given by:

    v = w x r' =          

                         $\left[\begin{array}{ccc}x&y&z\\wsin(a)&0&wcos(a)\\Rsin(b')cos(c')&Rsin(b')sin(c')&Rcos(b')\end{array}\right]$      

- where a =  Ψ and b' = θ' and c' = Φ'

v = R*w [-(cos Ψ *sin θ' *sin Φ') x + (cos Ψ *sin θ' *cos Φ' - sin Ψ * cos θ') y

+ (cos Ψ *sin θ' *sin Φ') z ]

- Notice that terms like sin Φ' and cos Φ' contribute to zero:

                                   $\int\ {sin (c')} \, dc' = \int\ {cos (c')} \, dc' = 0$

- Hence,

         

               $A(r) = - \frac{u_o *R^3*w*sigma*sin(a)}{2}*\int\limits^p_0 {\frac{cos (b')*sin(b') }{\sqrt{R^2 + r^2 -2*r*R*cos(b')} } } \, db' y$

- Evaluate integral u = cos (b')

                $= - \frac{1}{3R^2*r^2} * [ (R^2 + r^2 +rR)*(R - r) - (R^2 + r^2 -rR)*(R + r)]$

- From we can determine two cases when r > R and r < R

Hence,

r < R          $A(r) = - \frac{u_o *R*sigma}{3} * (w x r)$

r > R          $A(r) = - \frac{u_o *R^4*sigma}{3r^3} * (w x r)$

- Reverting back to original coordinate system given in figure 5.45:

r < R              $A(r) = - \frac{u_o *R*w*sigma}{3} *r*sin(b) . c$

r > R              $A(r) = - \frac{u_o *R^4*w*sigma}{3r^2} *sin(b) . c$

Where,     b = θ and c = direction along Φ.

Hence, A ( r , θ , Φ )