b field conducting sphere
DESCRIPTION
University Electromagnetism: Magnetic field (induction) of a rotating conducting charged sphere (with surface charges)TRANSCRIPT
B-field of a rotating charged conducting sphere 1
Magnetic Field of a Rotating Charged Conducting Sphere
© Frits F.M. de Mul
B-field of a rotating charged conducting sphere 2
B-field of a rotating charged conducting sphere
Question:
Calculate B-field in arbitrary points on the axis of rotation inside and outside the sphere
Available:
A charged conducting sphere (charge Q, radius R), rotating with ω rad/sec
ω
B-field of a rotating charged conducting sphere 3
Calculate B-field in point P inside or outside the sphere
P
P
O
ω
Analysis and Symmetry (1)
Assume Z-axis through O and P.
zP
Z
Y
XCoordinate systems:
- X,Y, Z
θ
ϕ
r
- r, θ, ϕ
B-field of a rotating charged conducting sphere 4
Analysis and Symmetry (2)
Conducting sphere,
all charges at surface:
surface density: σ = Q/(4πR2) [C/m2]
ω
P
P
zP
Y
X
Z
θ
ϕ
r
ORotating charges will establish a “surface current”
Surface current density j’ [A/m] will be a function of θ
j’
B-field of a rotating charged conducting sphere 5
Analysis and Symmetry (3)
ω
P
zP
Y
X
Z
θ
ϕ
r
O
T
Cylinder- symmetry around Z-axis:
dBz
Z-components only !!
Direction of contributions dB:
P
O
dB
T
θ
r
er
dl
20 .
4 Pr
I redldB
×=π
µBiot & Savart :
rPdB dB, dl and er mutual. perpendic.
B-field of a rotating charged conducting sphere 6
Approach (1): a long wire
dB
20 .
4 Pr
I redldB
×=π
µBiot & Savart :
note:
r and vector er !!
dB ⊥ dl and er
dB ⊥ ∆ AOP
Z
YX
P
z
I.dl in dz at z
dl
er rP
yP
α
A
O
B-field of a rotating charged conducting sphere 7
Approach (2): a volume current
dB
dvrP
20
4rej
dB×=
πµ
Biot & Savart :
dB ⊥ dl and er
dB ⊥ ∆ AOP
j: current density [A/m2]
Z
Y
P
j.dA.dl = j.dvdl
er
yP
dA
jA
OrP
B-field of a rotating charged conducting sphere 8
Approach (3): a surface current
dB
dArP
20
4rej
dB×′
=π
µ
Biot & Savart :
dB ⊥ dl and er
dB ⊥ ∆ AOP
Z
Y
P
dl
er
yP
dl
j’A
OrP
Current strip at surface: j’: current density[A/m]
j’.db.dl = j’.dA
dldb
B-field of a rotating charged conducting sphere 9
Approach (4)
Z
dϕ
R
ϕ
θ
dθ
R sinθ
Conducting sphere,
surface density: σ = Q/(4πR2)
surface element:
dA = (R.dθ).(R.sinθ. dϕ)
R.dθ.R.sinθ. dϕ
Surface element:
B-field of a rotating charged conducting sphere 10
Conducting sphere (1)
dA = db.dl
Surface charge σ.dA on dA will rotate with ω
dl = R.sinθ.dϕ ; db= R dθ Needed:
• j , er , rP
dArP
20
4rej
dB×′=
πµ with j’ in [A/m]
R.sinθ.dϕ
Z
R
ϕ
dθ
dϕ
R sinθ R.dθ
ω
θ
B-field of a rotating charged conducting sphere 11
Conducting sphere (2)
Z
R
ϕ
dθ
dϕ
R sinθ
R.dθ R.sinθ.dϕ
ω
dA = db.dl
dl = R.sinθ.dϕ db= Rdθ
Full rotation over 2π.Rsinθ in 2π/ω s.
Charge on ring with radius R.sinθ and width db is: σ. 2πR.sinθ . db
current: dI = σ.2πR.sinθ.db / (2π/ω) = σ ω R sinθ .db
current density: j’ =σωR sinθ [A/m]
θ
B-field of a rotating charged conducting sphere 12
Conducting sphere (3)
R
ϕ
dθ
dϕ
R sinθ
R.dθ R.sinθ.dϕ
dArP
20
4rej
dB×′
=π
µP
zP
j’
errP
dA = R.dθ. R.sinθ. dϕ
j’ ⊥ er :
=> | j’ x er | = j’.er = j’
j’ = σωR sin θ
ω
θ
B-field of a rotating charged conducting sphere 13
Conducting sphere (4)
R
ϕ
dθ
dϕ
R sinθ
P
zP
j’
errP
dArP
20
4rej
dB×′
=π
µ
dA = Rdθ R.sinθ. dϕ
Z-components only !!
dBz α
Pr
R θα sincos =
Cylinder- symmetry: P
O
dB
θ R
rPzP α er
j’ = σωR sin θ
ω
θ
B-field of a rotating charged conducting sphere 14
Conducting sphere (5)
R
ϕ
dθ
dϕ
R sinθ
P
zP
j’
errP
dArP
20
4rej
dB×′
=π
µ
dA = Rdθ.R.sinθ. dϕ
P
O
dBdBz
θ
α
R
rPzP α
Pr
R θα sincos =
rP2= (R.sinθ)2 +
(zP - R.cosθ)2
PP
z r
RdRdR
r
RdB
θϕθθθσωπ
µ sinsin...
sin
4 20=
j’ = σωR sin θ
ω
θ
B-field of a rotating charged conducting sphere 15
Conducting sphere (6)
R
ϕ
dθ
dϕ
R sinθ
P
zP
j’
errP
with rP2= (R.sinθ)2 + (zP - R.cosθ)2
ϕθθσωπ
µdd
r
RdB
P
z .sin
4 3
340=
Integration: 0<θ<π ; 0<ϕ< 2π
PP
z r
RdRdR
r
RdB
θϕθθθσωπ
µ sinsin...
sin
4 20=
ω
θ
B-field of a rotating charged conducting sphere 16
Conducting sphere (7)
ω
P
P
zP
Y
X
Z
θ
ϕ
R
O
this result holds for zP>R ;
for -R<zP<R the result is:
zeB Rσωµ03
2 =
and for zP<-R: zeB 3
40
.3
2
pz
R
−= σωµ
zeB 3
40
.3
2 :result
pz
Rσωµ=
B-field of a rotating charged conducting sphere 17
Conducting sphere (8)
inside sphere: constant field !!ω
P
P
zP
Y
X
Z
θ
ϕ
r
O
result for |zP|>R :
result for |zP|<R :
zeB 3
40
.3
2
pz
Rσωµ=
zeB Rσωµ03
2 =
B directed along +ez for all points everywhere on Z-axis !!
B-field of a rotating charged conducting sphere 18
Conducting sphere (9)
With surface density: σ = Q/(4πR2) :
result for |zP| > R : zz eeB 3
20
3
40
6.3
2
pp z
RQ
z
R
πωµσωµ ==
result for |zP| < R : zz eeB
R
QR
πωµσωµ
63
2 0
0 ==
B-field of a rotating charged conducting sphere 19
Conducting sphere (10)
Plot of B for:
Q = 1
µ0 = 1
ω = 1
(in SI-units)
zP / R
zeB 3
20
6
pz
RQ
πωµ=
zeBR
Q
πωµ
6 0=
B-field of a rotating charged conducting sphere 20
Conclusions (1)
Homogeneously charged sphere
(see other presentation)
zeB3
20
10
Pz
RQ
πωµ= ( ) zeB 22
30 35
20 PzR
R
Q −=π
ωµ|zP| < R|zP| > R
Conducting sphere
|zP| > R |zP| < R
zeB3
20
6
pz
RQ
πωµ= zeB
R
Q
πωµ
6 0=
B-field of a rotating charged conducting sphere 21
Conclusions (2)
Plot of B for:
Q = 1
µ0 = 1
ω = 1
(in SI-units)
zP / R
Homogeneously charged sphere
Conducting sphere
The end !!