maxwell’s equations, poynting vector, and energy...

Maxwell’s Equations, Poynting Vector, and Energy Flow

W15D1

1

Announcements

Gradebook closes Wednesday at 5 pm. Sunday Tutoring 1-5 pm in 26-152 Final Exam Monday May 20 from 9 am – 12 noon in Johnson Athletic Center Second Floor (Indoor Track)

2

Online Evaluations Are Now Open •  http://web.mit.edu/subjectevaluation/ •  You have until Monday, May 20 at 9 AM. •  Please evaluate all subjects in your list. •  Don’t forget your Tas. •  Please write comments.

Your feedback is read and valued!

Maxwell’s Equations

!E ⋅ ndA

S"∫∫ = 1

ε0

ρ dVV∫∫∫ (Gauss's Law)

!B ⋅ ndA

S"∫∫ = 0 (Magnetic Gauss's Law)

!E ⋅d !s

C#∫ = − d

dt!B ⋅ ndA

S∫∫ (Faraday's Law)

!B ⋅d!s

C#∫ = µ0

!J ⋅ ndA

S∫∫ + µ0ε0

ddt

!E ⋅ ndA

S∫∫ (Maxwell - Ampere's Law)

4

Poynting Vector and Power

S =E ×B

µ0

Power per unit area: Poynting vector

P =S ⋅ n da

opensurface

∫∫

Power through a surface

5

Energy Flow in Resistors, Capacitors and Inductors

6

Energy Flow: Resistor

S =E ×B

µ0

On surface of resistor direction is INWARD

7

Group Problem: Resistor

Power Consider a cylindrical resistor of length l, radius a, resistance R and current I through the resistor. Recall that there is an electric field in the wire given by a)  What are vector expressions for the electric and magnetic fields on the surface of the resistor? b)  Calculate the flux of the Poynting vector through the surface of the resistor in terms of the electric and magnetic fields. c)  Express your answer to part b) in terms of the current I and resistance R. Does your answer make sense?

8

I

al

Ik

r

ˆr

E l = ΔV = I R

Displacement Current

dQdt

= ε0

dΦE

dt≡ Idis

E = Q

ε0 A⇒Q = ε0EA = ε0ΦE

!B ⋅d!s

C"∫ = µ0

!J ⋅ nda

S∫∫ + µ0ε0

ddt

!E ⋅ nda

S∫∫

= µ0(Icon + Idis )

So we had to modify Ampere’s Law:

9

Group Problem: Capacitor A circular capacitor of spacing d and radius R is in a circuit carrying the steady current I. Neglect edge effects. At time t = 0 it is uncharged. The point P lies a distance R from the central axis.

a)  Find the electric field E(t) at P (mag. & dir.) b)  Find the magnetic field B(t) at P (mag. & dir.) c)  Find the Poynting vector S(t) at P (mag. & dir.) d)  What is the flux of the Poynting vector into/out of the

capacitor? e)  How does this compare to the time derivative of the energy

stored in the electric field?

I I

d

P

Rk

r

ˆ

unit vectors at point P

+Q(t) Q(t)

10

CQ: Capacitor

For the capacitor shown in the figure, the direction of the Poynting vector at the point P is in the

11

I I

d

P

Rk

r

ˆ

unit vectors at point P

+Q(t) Q(t)

1. + r − direction 2. +k − direction

3. +θ − direction 4. − r − direction

5. − k − direction 6. −θ − direction

Inductors

I L

LI = ΦSelf

ε = −

dΦB

dt= −L dI

dt

12

1.  Find the magnetic field B(t) at P (dir. and mag.) 2.  Find the electric field E(t) at P (dir. and mag.) 3.  Find the Poynting vector S(t) at P (dir. and mag.) 4.  What is the flux of the Poynting vector into/out of the inductor? 5.  How does this compare to the time derivative of the energy

stored in the magnetic field?

Group Problem: Inductor A solenoid with N turns, radius a and length h, has a current I(t) that is decreasing in time. Consider a point P located at a radial distance a from the center axis of the solenoid at the inner edge of the wires. Neglect edge effects.

13

N turns; each turn carries a current I

k

I(t)

a

I(t)

z

hPa

rˆ

CQ: Inductor

14

N turns; each turn carries a current I

k

I(t)

a

I(t)

z

hPa

rˆ

A solenoid has a current I(t) that is increasing in time. The direction of the Poynting vector at the point P is in the

1. + r − direction 2. +k − direction

3. +θ − direction 4. − r − direction

5. − k − direction 6. −θ − direction

Energy Flow: Inductor Direction on surface of inductor with increasing current is INWARD

S =E ×B

µ0

15

Energy Flow: Inductor Direction on surface of inductor with decreasing current is OUTWARD

S =E ×B

µ0

16

Power & Energy in Circuit Elements

P =

S ⋅ dA

Surface∫∫

uE = 12 ε0 E2

uB = 1

2µ0B2

Dissipates energy

Power flows in or out resulting in dissipating, storing or releasing energy

17

Resistor: Capacitor: Inductor

Maxwell’s Equations in Differential Form

18

Div, Grad, and Curl

divergence:

nabla:

curl:

Laplacian:

gradient:

19

!∇ = ∂

∂xi + ∂

∂yj+ ∂

∂zk

grad f =

!∇f = ∂ f

∂xi + ∂ f

∂yj+ ∂ f

∂zk

div!A =!∇⋅!A =

∂Ax

∂x+∂Ay

∂y+∂Az

∂z

curl

!A =!∇×!A =

∂Az

∂y−∂Ay

∂z

⎛

⎝⎜⎞

⎠⎟i +

∂Ax

∂z−∂Az

∂x⎛⎝⎜

⎞⎠⎟

j+∂Ay

∂x−∂Ax

∂y

⎛

⎝⎜⎞

⎠⎟k

∇2 ≡

!∇⋅!∇ = ∂2

∂x2 +∂2

∂y2 +∂2

∂z2

18.02 Divergence Theorem

!∇⋅!C

volumeV∫∫∫ dV =

!C ⋅ n dA

closed surfacecontainingV

"∫∫ Divergence Theorem

20

Meaning of the divergence of a vector field at a point P. Consider a sphere of radius ε centered about the point P shown below right. Then

!∇⋅!C

!∇⋅!C (P) ≡ lim

ε→0

!C ⋅ n dA

sphere"∫∫

Volume of sphere

P

C

The divergence of a vector field is a function, and it’s value at a point is a scalar quantity

Group Problem: Maxwell’s Eqs. in Differential Form via the Divergence Theorem

Use the Divergence Theorem and Maxwell’s Equations in integral form to find expressions for:

21

!E ⋅ ndA

S"∫∫ = 1

ε0

ρ dVV∫∫∫ (Gauss's Law)

!B ⋅ ndA

S"∫∫ = 0 (Magnetic Gauss's Law)

!E ⋅d !s

C#∫ = − d

dt!B ⋅ ndA

S∫∫ (Faraday's Law)

!B ⋅d!s

C#∫ = µ0

!J ⋅ ndA

S∫∫ + µ0ε0

ddt

!E ⋅ ndA

S∫∫ (Maxwell - Ampere's Law)

!∇⋅!C

volumeV∫∫∫ dV =

!C ⋅ n dA

closed surfacecontainingV

"∫∫ Divergence Theorem

1)!∇⋅!E = ?

2)!∇⋅!B = ?

Group Problem Ans.: Maxwell’s Eqs. in Differential Form via the Divergence Theorem

Gauss’s Law for Electric Fields: Gauss’s Law for Magnetic Fields:

22

!B ⋅ ndA

S"∫∫ =

math !∇⋅!B

volumeV∫∫∫ dV =

physics

0⇒

!∇⋅!B = 0

!∇⋅!E

volumeV∫∫∫ dV =

math !E ⋅ ndA

S"∫∫ =

physics 1ε0

ρ dVV∫∫∫ ⇒

!∇⋅!E = ρ

ε0

18.02 Stokes’ Theorem

(!∇×!C) ⋅ n dA

open surface S∫∫ =

!C ⋅d!s

closer pathboundary of S

"∫ Stokes' Theorem

23

Meaning of a component of the curl of a vector field at a point P: Consider a circle of radius ε centered about the point P shown below right. Then

!∇×!C

!∇×!C (P) ≡ lim

ε→0

!C ⋅d!s

circle"∫

Area of circle

P

C

n

The component of the curl of the vector perpendicular to the plane of the circle is

Group Problem: Maxwell’s Eqs. in Differential Form via Stokes Theorem

Use the Stokes Theorem and Maxwell’s Equations in integral form to find expressions for:

24

!E ⋅ ndA

S"∫∫ = 1

ε0

ρ dVV∫∫∫ (Gauss's Law)

!B ⋅ ndA

S"∫∫ = 0 (Magnetic Gauss's Law)

!E ⋅d !s

C#∫ = − d

dt!B ⋅ ndA

S∫∫ (Faraday's Law)

!B ⋅d!s

C#∫ = µ0

!J ⋅ ndA

S∫∫ + µ0ε0

ddt

!E ⋅ ndA

S∫∫ (Maxwell - Ampere's Law)

(!∇×!C)

open surface S∫∫ ⋅ n dA =

!C ⋅d!s

closer pathboundary of S

"∫ Stokes's Theorem

1)!∇×!E =!?

2)!∇×!B =!?

Faraday’s Law: Maxwell-Ampere Law

Group Problem Ans.: Maxwell’s Eq’s in Differential Form via Stoke’s Theorem

!E ⋅d!s

closer pathboundary of S

"∫ =math

(!∇×!E)

open surfaceS

∫∫ ⋅ n dA =physics

− ddt

!B ⋅ ndA

S∫∫ ⇒

!∇×!E = − ∂

!B∂t

25

(!∇×!B)

open surfaceS

∫∫ ⋅ n dA =math !

B ⋅d!sC"∫ =

physics

µ0

!J ⋅ ndA

S∫∫ + µ0ε0

ddt

!E ⋅ ndA

S∫∫ ⇒

!∇×!B = µ0

!J + µ0ε0

d!E

dt

Differential Version Maxwell’s Equations

!∇⋅!E = ρ

ε0

(Gauss's Law)!∇⋅!B = 0 (Magnetic Gauss's Law)

!∇×!E = − ∂

!B∂t

(Faraday's Law)

!∇×!B = µ0

!J + µ0ε0

∂!E∂t

(Maxwell - Ampere's Law)

26

Maxwell’s Equations in Free Space

In free space where Maxwell’s Equations simplify (notice the symmetry)

27

!∇⋅!E = 0 (Gauss's Law)

!∇⋅!B = 0 (Magnetic Gauss's Law)

!∇×!E = − ∂

!B∂t

(Faraday's Law)

!∇×!B = µ0ε0

∂!E∂t

(Maxwell - Ampere's Law)

ρ = 0,!J =!0

Appendix 1

Divergence and Stokes Theorem

28

Consider a vector field and a small cube of volume ΔV=ΔxΔyΔz. Flux of vector field is the sum over all six faces Consider two planar faces 1 and 2 located at z and z+Δz. Flux through just those two surfaces is where we used the Taylor formula

Divergence Theorem Proof (Sketch):

29

!C ⋅ n dA

closed surfacecontainingV

"∫∫ =!C j ⋅ n jdA

j=1

j=6

∑

!C(xp , yP , z + Δz) ⋅ n1ΔxΔy +

!C(xp , yP , z) ⋅ n2ΔxΔy

= (Cz (xp , yP , z + Δz)−Cz (xp , yP , z))ΔxΔy =∂Cz

∂zΔxΔyΔz

!E(x, y, z)

(x, y, z) ..

(xP , yP , z)

(xP , yP , z + z)n1

n2

.

+x

+ y

+z

O

z

C(xp , yP , z + z)

C(xp , yP , z)

Cz (xp , yP , z + Δz)−Cz (xp , yP , z) =

∂Ez

∂zΔz

Repeat the same argument for the two other pairs of faces:

Divergence Theorem Proof

(Sketch):

30

!C j ⋅ n jdA

j=1

j=6

∑ =∂Cx

∂x+∂Cy

∂y+∂Cz

∂z

⎛

⎝⎜⎞

⎠⎟ΔxΔyΔz =

!∇⋅!C ΔV

(x, y, z) ..

(xP , yP , z)

(xP , yP , z + z)n1

n2

.

+x

+ y

+z

O

z

C(xp , yP , z + z)

C(xp , yP , z)

For any arbitrary volume, divide the volume into infinitesimal Cubes and add up the contributions yielding

!∇⋅!C

volumeV∫∫∫ dV =

!C ⋅ n dA

closed surfacecontainingV

"∫∫ Divergence Theorem

Consider a vector field and a small rectangle in the x-y plan of area ΔA=ΔxΔy. The line integral of the vector field around the closed boundary is the sum of four contributions Consider two opposite paths at y and y+Δy. Each side contributes to the line integral where we used the Taylor Theorem :

Stokes’ Theorem Proof (Sketch):

31

!C ⋅d!s

closer pathboundary of S

"∫ =!C j ⋅d

!s jj=1

j=4

∑

!C(xP , y, zP ) ⋅ Δxi +

!C(xP , y + Δy, zP ) ⋅(−Δx i)

= (Cx (xP , y, z)−Cx (xP , y + Δy, zP ))Δx = −∂Cx

∂yΔxΔy

!C(x, y, z)

Cx (xP , y + Δy, zP )−Cx (xP , y, zP ) =

∂Cx

∂yΔy

x x + x

iy

y + yd s3 = x i

d s1 = x i

j

xP

C(xP , y, zP )

C(xP , y + y, zP )

k

Added these two pieces then yields

where

This is the infinitesimal version of Stokes’ Theorem. The integral version for a closed curve and any surface with that path as a boundary resulting from dividing the surface into small rectangular patches and added up all the contributions.

Stokes’s Theorem Proof (Sketch):

32

!C j ⋅d

!s jj=1

j=4

∑ =∂Cy

∂x−∂Cx

∂y

⎛

⎝⎜⎞

⎠⎟ΔxΔy = (

!∇×!C)zΔxΔy = (

!∇×!C) ⋅ n dA

n = k

(!∇×!C) ⋅ n dA

open surface S∫∫ =

!C ⋅d!s

closer pathboundary of S

"∫ Stokes' Theorem

Appendix 2

Derivation of Three

Dimensional Wave Equations

33

Vector Identity (try to prove this in Cartesian Coordinates): Derivation of Wave Equation for Electric Field in Vacuum:

Wave Equation for Electric Field

34

!∇× (

!∇×!E) = −∇2

!E +!∇(!∇⋅!E)

Start with Faraday's Law: !∇×!E = − ∂

!B∂t

take curl of both sides: !∇× (

!∇×!E) = − ∂

∂t(!∇×!B)

apply vector identity:−∇2!E +!∇(!∇⋅!E) = − ∂

∂t(!∇×!B)

apply!∇⋅!E = 0 : ∇2

!E = ∂

∂t(!∇×!B)

apply generalized Ampere'e Law !∇×!B = µ0ε0

∂!E∂t

: ∇2!E = µ0ε0

∂2!E

∂t2

use ε0µ0 = 1/ c2 : ∇2!E = 1

c2

∂2!E

∂t2

Group Prob.: Wave Equation for Magnetic Field

35

Use!∇×!B = µ0ε0

∂!E∂t

and the vector identity !∇× (

!∇×!B) = −∇2

!B +!∇(!∇⋅!B)

along with !∇⋅!B = 0,

!∇×!E = − ∂

!B∂t

, and ε0µ0 = 1/ c2

to derive the wave equation for the magnetic field in vacuum

∇2!B = 1

c2

∂2!B

∂t2 = ∇2!B

Maxwell’s Equations can be rewritten as “three dimensional wave equations” for the electric and magnetic fields where (in Cartesian coordinates)

Summary: Maxwell’s Equations and Three Dimensional Wave Equations

36

∇2!E = 1

c2

∂2!E

∂t2

∇2!B = 1

c2

∂2!B

∂t2

∇2 ≡

!∇⋅!∇ = ∂2

∂x2 +∂2

∂y2 +∂2

∂z2

Special Case: Plane Wave

37

Consider a an electric field, , in vacuum that only varies with respect to x and t, but is independent of y and z. Then the zero divergence of the electric field requires that Because the components of the fields are independent of y and z There are two solutions to this equation: Ex = 0 or Ex is uniform in space. We have ruled out uniform fields and therefore the electric field only has non-zero y- and z- components which are each independent of y and z. We shall consider the special case (linear polarization) in which there is only a non-zero y-component:

∂Ey

∂y= 0 and

∂Ez

∂z= 0 ⇒

∂Ex

∂x= 0

!E(x,t)

!∇⋅!E =

∂Ex

∂x+∂Ey

∂y+∂Ez

∂z= 0

!E(x,t) = Ey (x,t) j+ E z (x,t)k

!E(x,t) = Ey (x,t) j