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PHY 204: Elementary Physics II -- Slides PHY 204: Elementary Physics II (2021)

2020

21. Maxwell's equations. Electromagnetic waves

Gerhard Müller University of Rhode Island, [email protected]

Robert Coyne University of Rhode Island, [email protected]

Follow this and additional works at: https://digitalcommons.uri.edu/phy204-slides

Recommended Citation Müller, Gerhard and Coyne, Robert, "21. Maxwell's equations. Electromagnetic waves" (2020). PHY 204: Elementary Physics II -- Slides. Paper 46. https://digitalcommons.uri.edu/phy204-slides/46https://digitalcommons.uri.edu/phy204-slides/46

This Course Material is brought to you for free and open access by the PHY 204: Elementary Physics II (2021) at DigitalCommons@URI. It has been accepted for inclusion in PHY 204: Elementary Physics II -- Slides by an authorized administrator of DigitalCommons@URI. For more information, please contact [email protected]. Dynamics of Particles and Fields

Dynamics of Charged Particle:

• Newton’s equation of motion: ~F = m~a. • : ~F = q(~E +~v ×~B).

Dynamics of Electric and Magnetic Fields:

I q • Gauss’ law for electric field: ~E · d~A = . e0 I • Gauss’ law for magnetic field: ~B · d~A = 0. I dΦ Z • Faraday’s law: ~E · d~` = − B , where Φ = ~B · d~A. dt B I dΦ Z • Ampere’s` law: ~B · d~` = µ I + µ e E , where Φ = ~E · d~A. 0 0 0 dt E

Maxwell’s equations: 4 relations between fields (~E,~B) and sources (q, I).

tsl314 Gauss’s Law for

The net electric flux ΦE through any closed is equal to the net charge Qin inside divided by the constant e0: I ~ ~ Qin Qin −12 2 −1 −2 E · dA = 4πkQin = i.e. ΦE = with e0 = 8.854 × 10 C N m e0 e0 The closed surface can be real or fictitious. It is called “”. The symbol H denotes an over a closed surface in this context.

• Gauss’s law is a general relation between and electric field. • In : Gauss’s law is equivalent to ’s law. • Gauss’s law is one of four Maxwell’s equations that govern cause and effect in and .

tsl44 Gauss’s Law for

The net magnetic flux ΦB through any closed surface is equal to zero: I ~B · d~A = 0.

There are no magnetic charges. Magnetic field lines always close in themselves. No matter how the (closed) Gaussian surface is chosen, the net magnetic flux through it always vanishes.

The figures below illustrate Gauss’s laws for the electric and magnetic fields in the context of an electric dipole (left) and a magnetic dipole (right).

tsl236 Gauss’ Law for Electric and Magnetic Fields

I I q ~B · d~A = 0 ~E · d~A = in e0

tsl508 Magnetic flux and Faraday’s law

• Magnetic field ~B (given) • Surface S with perimeter loop (given) • Surface A (given) • Area vector ~A = Anˆ (my choice) • Positive direction around perimeter: ccw (consequence of my choice) Z Z ~ ~ ~ • Magnetic flux: ΦB = B · dA = B · ndAˆ

d~B • Consider situation with 6= 0 dt • Induced electric field: ~E I • Induced EMF: E = ~E · d~` (integral ccw around perimeter) dΦ • Faraday’s law: E = − B dt

tsl411 Ampere’s` Law (Restricted Version)

The circulation integral of the magnetic field ~B around any closed curve (loop) C is equal to the net IC flowing through the loop: I ~ ~ −7 B · d` = µ0IC, with µ0 = 4π × 10 Tm/A

The symbol H denotes an integral over a closed curve in this context. Note: Only the component of ~B tangential to the loop contributes to the integral.

The positive current direction through the loop is determined by the right-hand rule.

tsl237 Ampere’s` law (Full Version)

• Conduction current: I. dΦ • : I = e E . D 0 dt I dΦ • Ampere’s` law: ~B · d~` = µ (I + I ) = µ I + µ e E . 0 D 0 0 0 dt

tsl315 Faraday’s law and Ampere’s` law

I dΦ I dΦ ~E · d~s = − B ~B · d~s = µ I + µ e E dt 0 0 0 dt

tsl509 Traveling Waves

Mechanical waves travel in some medium. Examples: sound wave, violin string, surface water wave. While the wave propagates, the medium undergoes periodic motion.

Distinguish:

(1) direction of wave propagation, (2) direction in which medium moves.

Transverse wave: (1) and (2) are to each other.

Longitudinal wave: (1) and (2) are parallel to each other.

Electromagnetic waves are transversely oscillating electric and magnetic fields. Electromagnetic waves travel in the vacuum. There is no medium.

Waves transport energy and, in some cases, information, but not the medium itself (if there is a medium). tsl316 Sinusoidal Transverse Traveling Wave

Wave function: y(x, t) = A sin(kx − ωt)

2π • k = (wave number) λ • λ (wavelength) 2π • ω = = 2πf (angular frequency) T ω 1 • f = = (frequency) 2π T • T (period) λ ω • c = = λf = (wave speed) T k

tsl317 Wave Equation

• y(x, t) = A sin(kx − ωt) (displacement) ∂y • v(x, t) = = −ωA cos(kx − ωt) (velocity) ∂t ∂2y • a(x, t) = = −ω2A sin(kx − ωt) (acceleration) ∂t2 ∂y • = kA cos(kx − ωt) (slope of wave form) ∂x ∂2y • = −k2A sin(kx − ωt) (curvature of wave form) ∂x2 ∂2y/∂t2 ω2 • = = c2 (ratio of second derivatives) ∂2y/∂x2 k2 ∂2y ∂2y • Wave equation: = c2 ∂t2 ∂x2

tsl318 Electromagnetic Wave (1)

Maxwell’s equations for electric and magnetic fields in free space (no sources):

I I • Gauss’ laws: ~E · d~A = 0, ~B · d~A = 0. I dΦ I dΦ • Faraday’s and Ampere’s` laws: ~E · d~` = − B , ~B · d~` = µ e E . dt 0 0 dt

Consider fields of particular directions and dependence on space:

~E = Ey(x, t)ˆj, ~B = Bz(x, t)kˆ. y Gauss’ laws are then automatically satisfied. E

Use the cubic Gaussian surface to show that

• the net electric flux Φ is zero, E B • the net magnetic flux ΦB is zero. x

z tsl319 Electromagnetic Plane Wave (2)

I dΦ • Faraday’s law, ~E · d~` = − B , dt applied to loop in (x, y)-plane becomes ∂ [E (x + dx, t) − E (x, t)]dy = − B (x, t)dxdy y y ∂t z ∂ ∂ ⇒ E (x, t) = − B (x, t)(F) ∂x y ∂t z y I dΦ • Ampere’s` law, ~B · d~` = µ e E , 0 0 dt applied to loop in (x, z)-plane becomes dy ∂ B [−B (x + dx, t) + B (x, t)]dz = µ e E (x, t)dxdz z z 0 0 ∂t y dx ∂ ∂ E ⇒ − B (x, t) = µ e E (x, t)(A) ∂x z 0 0 ∂t y x dz z dx

tsl320 Electromagnetic Plane Wave (3)

2 2 2 2 ∂ ∂ ∂ Ey ∂ B ∂ B ∂ Ey Take partial derivatives (F) and (A): = − z , − z = µ e . ∂x ∂t ∂x2 ∂t∂x ∂t∂x 0 0 ∂t2

2 2 ∂ Ey ∂ Ey ⇒ = c2 (E) (wave equation for electric field). ∂t2 ∂x2

2 2 2 2 ∂ ∂ ∂ Ey ∂ B ∂ B ∂ Ey Take partial derivatives (F) and (A): = − z , − z = µ e . ∂t ∂x ∂t∂x ∂t2 ∂x2 0 0 ∂t∂x

∂2B ∂2B ⇒ z = c2 z (B) (wave equation for magnetic field). ∂t2 ∂x2

1 c = √ (speed of light). e0µ0

Sinusoidal solution:

• Ey(x, t) = Emax sin(kx − ωt)

• Bz(x, t) = Bmax sin(kx − ωt) tsl321 Electromagnetic Plane Wave (4)

For given wave number k the angular frequency ω is determined, for example by substitution of Emax sin(kx − ωt) into (E).

For given amplitude Emax the amplitude Bmax is determined, for example, by substituting Emax sin(kx − ωt) and Bmax sin(kx − ωt) into (A) or (F).

ω E ⇒ = max = c. k Bmax

The direction of wave propagation is determind by the :

1 ~S = ~E ×~B. µ0

tsl322 Energy Transport in Electromagnetic Plane Wave

Fields: Ey(x, t) = Emax sin(kx − ωt), Bz(x, t) = Bmax sin(kx − ωt).

1 2 1 2 3 Energy density: u(x, t) = e0Ey(x, t) + Bz (x, t). [J/m ] 2 2µ0

2 2 2 1 2 Use the amplitude relations e0Emax = e0c Bmax = Bmax. µ0

2 2 1 2 2 EmaxBmax 2 u(x, t) = e0Emax sin (kx − ωt) = Bmax sin (kx − ωt) = sin (kx − ωt). µ0 cµ0

Energy transported across area A in time dt: dU(x, t) = u(x, t)Acdt. [J]

1 dU Power transported per unit area: = u(x, t)c = S(x, t). [W/m2] A dt

Intensity (average power transported per unit area):

¯ EmaxBmax e0c 2 c 2 2 I = S = = Emax = Bmax. [W/m ] 2µ0 2 2µ0 tsl323 Momentum Transport in Electromagnetic Plane Wave

The momentum transported by an electromagnetic wave is proportional to the energy transported.

~p ~S 1 Momentum density: = , where ~ = ~ ×~ is the Poynting vector. 2 S E B V c µ0

When the wave is absorbed by a material surface it exerts an impulse ~Fdt = ∆~p.

The resulting radiation pressure is the average force per unit area:

F¯ p p dx p S¯ I P = = = = c = = . abs A Adt Adx dt V c c

2S¯ 2I The radiation pressure exerted by a reflected wave is twice as large: P = = . ref c c

tsl324