1. Evaluation of Different Quantities
Through Different Surfaces.
Muhammad Kamran(k13-2495)
Kamran Sadiq (k13-2489)
Arslan Saeed (k13-2499)

2. Introduction
In engineering fields especially in
electrical engineering in a number of
applications we are concerned to find
different quantities like fluxes
through different surfaces .This
problem leads us to use multivariable
calculus as a tool.

3. Topic To Discuss
• So we are going to discuss following topics
involving multi variable calculus as a tool.
– Electric flux
– Magnetic flux
– Source for E-fields
– Source of M-fields

4. Electric Flux
 In electromagnetism, electric flux is
the measure of flow of the electric
field through a given area. Electric
flux is proportional to the number of
electric field line going through a
normally perpendicular surface.

5. E-Flux due to Non-Uniform E-
Field
 For a non-uniform
electric field, the
electric
flux dΦE through a
small surface
area dS is given by
 dØe=D.dS
 For Whole area
Øe=∫∫sD.dS

6. For Sypherical Surface

7. For Cylinderical Surface
 da1=pdø aø
 da2=dz az
 dS=da1Xda2
 dS=pdø dz ap
 Øe=∫∫sD.dS
 For extensive
Volume
 Øe=∫∫∫vD.dV
 dV=p²dø dz

8. Magnetic Flux
 In physics, specifically electromagnetism,
the magnetic flux (often denoted Φ or ΦB)
through a surface is the surface integral of the
normal component of the magnetic
field B passing through that surface.

9. Equation of Magnetic Field
 For a varying
magnetic field, we
first consider the
magnetic flux
through an small
area element dS,
where we may
consider the field to
be constant
 A generic
surface, S, can then
be broken into
infinitesimal
elements and the
total magnetic flux
through the surface
is then the surface
integral

10. Magnetic Flux

11. Magnetic Flux Through Radial
Surface
 Magnetic flux Through any close Surface is
always 0.
 The formula of magnetic flux
Is only applicable for radial Surface
We know
B=µoH =µoI/2πp aø
ds=dpdz aø

12. Magnetic Flux Through Radial
Surface
 Øb=∫B.ds
 Øb= ∫ ∫ µoI/2πp
dpdz aø. aø
 Øb= ∫ ∫ µoI/2πp
dpdz

13. Sources of electric
fields
 If I am given with a electric flux density (D)
through a differential surface area (dS)
I can calculate its source as:
------Integral(D.dS)= Pv(volume
charge density)
 If I am only given with a electric flux density
(D)
I can calculate its source as:
------Div(D)=Pv

14. Example for Integral(D.dS)= Pv
 D=4x^2 ax + 10xz ay +3xyz az
c/m^2 & dS=dxdz ay m^2
 D.dS=10xzdxdz m
 “ S” is a surface from x: 26
& z: 1  7
 Integral(D.ds)=400 units

15. Example for div(D)=Pv
D=4(x^2) ax + 10xz ay +3xyz
az c/m^2
Div(
D)=d/dx(Dx)+d/dy(Dy)+d/dz
(Dz)
 =8x+0+3xy
 = 8x+3xy=pv

16.  If I am given with “E” and I have to find Pv
 I can find using same formulaes, I used behind
but first I’ll find D=E0E

17.  If I am given with potential field ”V” and
I am asked to find electric field “E”
 I’ll use E=-grad(V)
 grad(V)=d/dx(V) ax +d/dy(V) ay
+d/dz(V) az
 Example:
 V=50(x^2)yz +20(y^2)
 grad( V)=(100xyz)ax +(50
(x^2)z+40y)ay
E= -(100xyz)ax -(50 (x^2)z+40y)ay

18. Source of magnetic field
produced by DC-current
 If I am given with H(magnetic
field) produced by dc-current, I
can find the source current
producing it, by using:
 Curl(H)=J(current density)

19. Example for curl(H)=J
H=(y^2)z ax +2(x+1)yz ay –
(x+1)(z^2) az
Curl(H)=-(x+1)y ax
+{(y^2)+(z^2) }ay
Implies J=-(x+1)y ax
+{(y^2)+(z^2) }ay

20. Thank you…. 
one more thing• Think positive……..

21. Any question……….????