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Sir please solve the mcq,provide steps...
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Asked by subhrajayanta64 | 10 Sep, 2019, 04:37: AM
answered-by-expert Expert Answer
 
(Vectors are typed as Bold )
First, let us find out the magnetic field at test point r for the spinning charged sphere.
This is done by finding the magneteic vector potential A .
Then magnetic field is obtained from vector potential using  B = begin mathsize 12px style bold nabla end stylex A .
Left side figure shows a charged sphere of radius R, spinning with angular speed ω .
It is required to find the vector potential at test point r whose position vctor makes angle ψ with the spinning axis.
 
Let us asume the coordinate system as shown in right side of figure  so that centre of sphere coincides with origin,
position vector of test point r aligns with z-axis and spinning axis is rotated to lie in X-Z plane
 
Let us consider small area element da' = R2sinθ' dθ' dφ' . Vector potential dA at test point r is given by
 
begin mathsize 14px style d bold italic A bold left parenthesis bold italic r bold right parenthesis space equals space fraction numerator mu subscript o over denominator 4 pi end fraction space fraction numerator K left parenthesis r apostrophe right parenthesis d bold italic a bold apostrophe over denominator s end fraction end style........................(1)
where K(r') is the surface current density, s is the ditance between area element da' and test point r.
 
vector potential A(r) due to whole spinning sphere,  begin mathsize 14px style bold italic A bold left parenthesis bold italic r bold right parenthesis space equals space fraction numerator mu subscript o over denominator 4 pi end fraction space integral fraction numerator K left parenthesis bold italic r apostrophe right parenthesis d bold italic a bold apostrophe over denominator s end fraction end style.................................(2)
we have, K(r') = σ v ,   and s = ( R2 + r2 - 2Rr cosθ' )
where σ is surface charge density, v is velocity vector of area element da'
 
we have, 
begin mathsize 14px style v space equals space omega cross times r apostrophe space equals space open vertical bar table row cell x with hat on top end cell cell y with hat on top end cell cell z with hat on top end cell row cell omega sin psi end cell 0 cell omega cos psi end cell row cell R space sin theta apostrophe space cos phi apostrophe end cell cell R space sin theta apostrophe space sin phi apostrophe end cell cell R space cos theta apostrophe end cell end table close vertical bar space equals space R omega space open square brackets negative left parenthesis cos psi space sin theta apostrophe space sin ϕ apostrophe right parenthesis x with hat on top space space plus space left parenthesis space cos psi space sin theta apostrophe space cos phi apostrophe space minus space sin psi space cos theta apostrophe right parenthesis y with hat on top space plus space left parenthesis space sin psi space sin theta apostrophe space sin phi apostrophe right parenthesis z with hat on top close square brackets end style
 
It has to be noticed that each of these terms, except one, involves either sinφ' or cosφ' .
Since integration of sinφ' or cosφ' within limits 0 to 2π vanishes, with only one non-vanishing term,
vector potential relation (2) becomes
 
begin mathsize 14px style bold italic A bold left parenthesis bold italic r bold right parenthesis space equals space fraction numerator mu subscript 0 R cubed sigma omega space sin psi over denominator 2 end fraction open parentheses integral subscript 0 superscript pi fraction numerator cos theta apostrophe space sin theta apostrophe over denominator square root of R squared plus r squared minus 2 R space r space cos theta apostrophe end root end fraction space d theta apostrophe close parentheses stack space bold italic y with hat on top end style  ........................ (3)
Above integration can be worked out using u = cosθ' , then integration eqn.(3) becomes
 
begin mathsize 12px style bold italic A bold left parenthesis bold italic r bold right parenthesis space equals space minus space fraction numerator mu subscript o space R space sigma space omega space sin psi over denominator 6 space r squared end fraction open square brackets open parentheses R squared plus r squared plus R space r close parentheses open vertical bar R minus r close vertical bar space minus space open parentheses R squared plus r squared minus R space r close parentheses open parentheses R plus r close parentheses close square brackets space bold italic y with hat on top end style  .....................................(4)
Let us consider,  ω × r = - ω r sinψ begin mathsize 12px style bold italic y with hat on top end style.
Also when test point r is outside of sphere r > R, then bracketed term in eqn.(4) becomes -2R3
 
Hence begin mathsize 14px style bold italic A bold left parenthesis bold italic r bold right parenthesis space equals space fraction numerator mu subscript o R to the power of 4 sigma over denominator 3 r cubed end fraction open parentheses bold italic omega cross times bold italic r close parentheses end style  .................................(5)
Next step is to adjust the coordinate system, so that spinning axis coincides with z-axis and the point r is at (r, θ, φ)
 
Hence eqn.(5) becomes,   begin mathsize 14px style bold italic A left parenthesis r comma space theta comma space phi right parenthesis space equals space fraction numerator mu subscript o R to the power of 4 omega sigma over denominator 3 end fraction fraction numerator sin theta over denominator r squared end fraction bold phi with bold hat on top space equals space C space fraction numerator sin theta over denominator r squared end fraction bold phi with bold hat on top end style   ..............................(6)
where in eqn.(6),  C = ( μo R4 ω σ )/3   .............................(7)
 
Now to get magnetic field, we use B = begin mathsize 12px style nabla end stylex A 
 
In Spherical coordinate system, using curl of vector potential , magnetic field is written as, 
 
begin mathsize 12px style bold italic B space equals space fraction numerator 1 over denominator r space sin theta end fraction fraction numerator partial differential over denominator partial differential theta end fraction open parentheses sin theta space C space fraction numerator sin theta over denominator r squared end fraction close parentheses space bold r with bold hat on top space minus space 1 over r squared space fraction numerator partial differential over denominator partial differential r end fraction open parentheses r space C space fraction numerator sin theta over denominator r squared end fraction close parentheses bold italic theta with hat on top space equals space fraction numerator 2 C space cos theta over denominator r cubed end fraction bold r with bold hat on top space plus space fraction numerator C space sin theta over denominator r cubed end fraction bold theta with bold hat on top space end style  ......................................(8)
 
we know electric field of sphere at test point r,  
 
E(r)  = [ Q/(4πεor2) ] begin mathsize 12px style bold italic r with hat on top end style  = [ (4πR2σ) / (4πεor2) ] begin mathsize 12px style bold italic r with hat on top end style = [ (R2σ) / (εor2) ] begin mathsize 12px style bold italic r with hat on top end style ........................(9)
 
Poynting vector S  = (1/μo) [ E × B ]  =  (1/μo)  [ (R2σ) / (εor2) ] ( C sinθ / r3 ) begin mathsize 12px style bold italic phi with hat on top end style
By substituting C from eqn.(7), by considering r = R very near to spherical surface, Poynting vector becomes
 
S(θ) = [ (σ2 ω R sinθ )/εo ] begin mathsize 12px style bold italic phi with hat on top end style
Answered by Thiyagarajan K | 11 Sep, 2019, 12:27: PM

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