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# Sir please solve the mcq,provide steps...

Asked by subhrajayanta64 10th September 2019, 4:37 AM

(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 = x 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

........................(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,  .................................(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,

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

........................ (3)
Above integration can be worked out using u = cosθ' , then integration eqn.(3) becomes

.....................................(4)
Let us consider,  ω × r = - ω r sinψ .
Also when test point r is outside of sphere r > R, then bracketed term in eqn.(4) becomes -2R3

Hence   .................................(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,      ..............................(6)
where in eqn.(6),  C = ( μo R4 ω σ )/3   .............................(7)

Now to get magnetic field, we use B = x

In Spherical coordinate system, using curl of vector potential , magnetic field is written as,

......................................(8)

we know electric field of sphere at test point r,

E(r)  = [ Q/(4πεor2) ]   = [ (4πR2σ) / (4πεor2) ] = [ (R2σ) / (εor2) ] ........................(9)

Poynting vector S  = (1/μo) [ E × B ]  =  (1/μo)  [ (R2σ) / (εor2) ] ( C sinθ / r3 )
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 ]
Answered by Expert 11th September 2019, 12:27 PM
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