Evaluation of Different Quantities Through Different Surfaces.

Muhammad Kamran(k13-2495) Kamran Sadiq (k13-2489) Arslan Saeed (k13-2499)

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.

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

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.

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 areaØe=∫∫sD.dS

For Sypherical Surface

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

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.

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

Magnetic Flux

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

Magnetic Flux Through Radial Surface

Øb=∫B.ds Øb= ∫ ∫ µoI/2πp

dpdz aø. aø

Øb= ∫ ∫ µoI/2πp dpdz

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

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: 26 & z: 1 7

Integral(D.ds)=400 units

Example for div(D)=PvD=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

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

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)ayE= -(100xyz)ax -(50 (x^2)z+40y)ay

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)

Example for curl(H)=JH=(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

Thank you…. one more thing

• Think positive……..

Any question……….????