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[The naive explanation for the pressure of light offered in Section 9.2.3 has its flaws, as you discovered if you worked Problem 9.11. Here’s another account, due originally to Planck.] A plane wave traveling through vacuum in the z direction encounters a perfect conductor occupying the region z0, and reflects back:

E(z,t)=E0[cos(kz-ωt)-cos(kz+ωt)]x^,(z>0),

(a) Find the accompanying magnetic field (in the region role="math" localid="1657454664985" (z>0).

(b) Assuming inside the conductor, find the current K on the surface z=0, by invoking the appropriate boundary condition.

(c) Find the magnetic force per unit area on the surface, and compare its time average with the expected radiation pressure (Eq. 9.64).

Short Answer

Expert verified

(a) The magnetic field is B=E0c[cos(kz-ωt)+cos(kz+ωt)]y^

(b) The current K on the surface is K=2E0μ0ccos(ωt)x^

(c) The magnetic force per unit area is f=ε0E02, and it is twice the pressure in Eq. 9.64.

Step by step solution

01

Expression for the electric field for : (z>0)

Write the expression for electric field for (z>0)

E(z,t)=E0[cos(kz-ωt)-cos(kz+ωt)]x^ …… (1)

Here, k is the wave number, ωis the angular frequency and t is the time.

02

Determine the accompanying magnetic field:

(a)

Since, (E×B)points in the direction of propagation, write the equation for the magnetic field.

B=E0c[cos(kz-ωt)+cos(kz+ωt)]y^

Therefore, the magnetic field is B=E0c[cos(kz-ωt)+cos(kz+ωt)]y^

03

Determine the current K on the surfacez=0 :

(b)

It is known that K×(-Z^)=1μ0B

Substitute B=E0c[cos(kz-ωt)+cos(kz+ωt)]y^in the above equation.

K×(-z^)=1μ0E0c[cos(kz-ωt)+cos(kz+ωt)]y^

=E0μ0c[2cos(ωt)]y^K=2E0μ0ccos(ωt)x^

Therefore, the current K on the surface is K=2E0μ0ccos(ωt)x^
04

Determine the magnetic force per unit area on the surface:

(c)

Write the expression for the force per unit area.

f=K×Bavg

Substitute K=2E0μ0ccos(ωt)x^expression and Bavg=(cosωt)y^in the above expression.

f=2E02μ0c2cos(ωt)x^×[cos(ωt)y^]

f=2ε0E02cos2(ωt)z^

It is known that the time average of cos2(ωt)is12Hence, the above equation becomes,

f=2ε0E0212z^

f=ε0E02

This is twice the pressure in Eq. 9.64, but that was for a perfect absorber, whereas this is a perfect reflector.

Therefore, the magnetic force per unit area isf=ε0E02,and it is twice the pressure in Eq. 9.64.

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Most popular questions from this chapter

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[The naive explanation for the pressure of light offered in section 9.2.3 has its flaws, as you discovered if you worked Problem 9.11. Here's another account, due originally to Planck.] A plane wave travelling through vaccum in the z direction encounters a perfect conductor occupying the region z0, and reflects back:

E(z,t)=E0[coskz-ωt-coskz+ωt]x^,(z>0)

  1. Find the accompanying magnetic field (in the region (z>0))
  2. Assuming B=0inside the conductor find the current K on the surface z=0, by invoking the appropriate boundary condition.
  3. Find the magnetic force per unit area on the surface, and compare its time average with the expected radiation pressure (Eq.9.64).

Calculate the reflection coefficient for light at an air-to-silver interface (μ1=μ2=μ0,ε=ε0,σ=6×107(Ωm)-1)at optical frequencies(ω=4×1015/s).

Consider a particle of charge q and mass m, free to move in the xyplane in response to an electromagnetic wave propagating in the z direction (Eq. 9.48—might as well set δ=0)).

(a) Ignoring the magnetic force, find the velocity of the particle, as a function of time. (Assume the average velocity is zero.)

(b) Now calculate the resulting magnetic force on the particle.

(c) Show that the (time) average magnetic force is zero.

The problem with this naive model for the pressure of light is that the velocity is 90°out of phase with the fields. For energy to be absorbed there’s got to be some resistance to the motion of the charges. Suppose we include a force of the form ymv, for some damping constant y.

(d) Repeat part (a) (ignore the exponentially damped transient). Repeat part (b), and find the average magnetic force on the particle.

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