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

b

Solution

Here

E_{y}^{2} + E_{z}^{2} = 9 A^{2}

, so the tip of the electric field traces a circle at constant magnitude: this is circular polarization. At fixed

x = 0

, it starts at

+ y

when

t = 0

and initially rotates toward

+ z

. This explicit rotation specifies the sense without relying on competing handedness conventions.

Using

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

, with

v_{p} = ω / k

, gives

B = \frac{3 A}{v _{p}} (\hat{y} sin ψ + \hat{z} cos ψ), B_{x} = 0.

v_{p}

c

. The magnetic flux density also has constant magnitude and is perpendicular to

E

at every instant.

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