 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2014-resit/4/a/solution

Write

ψ = k x - ω t

and

v_{p} = ω / k

E = A sin ψ (\hat{y} + 6 \hat{z}) .

It has linear polarization, along the fixed line

E_{z} = 6 E_{y}

tan^{- 1} 6 ≃ 80.5 4^{\circ}

from

+ y

toward

+ z

B = v_{p}^{- 1} \hat{x} \times E

. Hence

B = \frac{A}{v _{p}} sin ψ (- 6 \hat{y} + \hat{z}), B_{x} = 0.

v_{p} = c

. Both fields are transverse, and

E \times B

points along

+ x

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