 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2019/5/solution

E

E_{0}

its peak amplitude;

x

the position along propagation;

t

the time;

k = 2 π / λ

the wavenumber; and

ω = 2 π f

λ

f

the frequency, and

\hat{y}

the unit vector in the direction of the electric field. Constant phase

k x - ω t

moves along

+ x

ω / k

Put

θ = k x - ω t

\nabla \times E = - k E_{0} sin θ \hat{z} = - \frac{\partial B}{\partial t} .

The accompanying oscillating magnetic field, excluding any independent static background, is therefore

B = (k / ω) E_{0} cos θ \hat{z}

. To determine the propagation speed, also use the source-free Ampere-Maxwell law:

\nabla \times B = μ_{0} ϵ_{0} \frac{\partial E}{\partial t}, k \frac{k}{ω} E_{0} sin θ = μ_{0} ϵ_{0} ω E_{0} sin θ .

Thus

v = \frac{ω}{k} = \frac{1}{μ _{0} ϵ _{0}} = c, B (x, t) = \frac{E _{0}}{c} cos (k x - ω t) \hat{z} .

The electric field and magnetic field oscillate in phase, perpendicular to one another and to propagation;

E \times B

points along

+ x

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