a) \vec E=2E_0\cos(n\pi/2)\cos(kx-\omega t+n\pi/2)\hat z, amplitude 2E_0\cos(n\pi/2); b) n par: interferência construtiva; n ímpar: interferência destrutiva; c) \vec B=-\frac{2E_0}{c}\cos(n\pi/2)\cos(kx-\omega t+n\pi/2)\hat y, orientação -\hat y, amplitude B_0=\frac{2E_0}{c}|\cos(n\pi/2)|.
Como pensou
We need answer physics question in Portuguese. Need solve. Let's restate.
There are two coherent EM waves in vacuum along x. Superposition of electric fields, both parallel z:
E = E0 { cos[k(x+d) - ω t] + cos(kx - ω t) } \hat{z}
where k wave number, ω angular frequency. E0, d positive. Assumes distances d satisfy kd = nπ for n=0,1,2,... Answer:
a) Show E can be rewritten as a wave propagating in sense x>0, with amplitude expressed only as function of E0 and n.
b) Indicate relation between n and type of interference generated by superposition.
c) Determine resultant magnetic field B, indicating orientation, considering amplitude modulus B0.
Need solve step by step.
Using trig identity: cos A + cos B = 2 cos((A+B)/2) cos((A-B)/2). Let A = k(x+d) - ωt = kx + kd - ωt. B = kx - ωt. Then (A+B)/2 = kx - ωt + kd/2. (A-B)/2 = kd/2 = nπ/2.
So E = E0 * 2 cos(kx - ωt + kd/2) cos(kd/2) \hat{z}. Since kd=nπ, cos(kd/2)=cos(nπ/2). For n=0,1,2,... cos(nπ/2): n=0 ->1; n=1 ->0; n=2 -> -1; n=3 ->0; n=4 ->1 etc. Amplitude = 2 E0 cos(nπ/2). But for n odd amplitude zero, no wave? But question says "pode ser reescrito como uma onda que se propaga no sentido x > 0, com amplitude expressa apenas em função de E0 e n." If amplitude zero for odd n, still can be written as amplitude 0. Maybe they expect amplitude = 2E0 cos(nπ/2). However perhaps they want absolute value? Since amplitude positive? "amplitude expressa apenas em função de E0 e n" maybe 2E0 cos(nπ/2) can be negative. Amplitude usually nonnegative, so |2E0 cos(nπ/2)| = 2E0 |cos(nπ/2)|. But propagation direction? Let's examine.
Original superposition: cos[k(x+d)-ωt] + cos(kx-ωt). Each term is wave propagating +x? cos(kx - ωt) is +x; cos[k(x+d)-ωt] = cos(kx - ωt + kd) also +x. Both propagate +x. Superposition = 2 cos(kd/2) cos(kx - ωt + kd/2). This is a wave propagating +x (phase kx - ωt + φ0, phase velocity ω/k >0). Amplitude = 2E0 cos(kd/2) = 2E0 cos(nπ/2). But if negative, can absorb sign into phase: 2E0 cos(nπ/2) cos(kx - ωt + kd/2) = |2E0 cos(nπ/2)| cos(kx - ωt + kd/2 + π if negative). Since question says amplitude expressa apenas em função de E0 e n, maybe they allow signed amplitude? In physics amplitude often positive; but "amplitude" maybe magnitude. Let's see item b relation between n and type of interference. If n even: cos(nπ/2)=±1, constructive? Interference type: constructive when amplitude maximum 2E0; destructive when zero. n even (0,2,4,...) constructive? Let's check: n=0: phase difference kd=0 => in phase? Original waves? cos(k(x+d)-ωt) and cos(kx-ωt) difference = kd = 0 mod 2π for n=0? n=0 => in phase, constructive, amplitude 2E0. n=1: kd=π, difference π => destructive (out of phase), amplitude 0. n=2: kd=2π => constructive, amplitude 2E0? But cos(nπ/2)=cos π=-1, amplitude magnitude 2E0, but with phase shift π. Still constructive (max amplitude). n=3: kd=3π => destructive? Difference 3π => cos(3π/2)=0, amplitude 0. So relation: n par constructive, n ímpar destructive. Let's formulate: n even -> constructive interference (amplitude 2E0); n odd -> destructive interference (amplitude 0). Wait n=0 included even.
But item a "Mostre que E pode ser reescrito como uma onda que se propaga no sentido x > 0, com amplitude expressa apenas em função de E0 e n." Using identity, yes. They may expect amplitude = 2E0 cos(nπ/2). But maybe they want "2E0 (-1)^{n/2}" for even n? But n odd amplitude zero. Let's see they explicitly assume kd = nπ for n=0,1,2,... So n includes odd; then amplitude zero for odd, wave not propagating? Actually E=0 for odd n; is that "onda"? It is zero field, can be considered wave with zero amplitude. So yes.
For part c: Determine resultant magnetic field B, indicating orientation, considering amplitude B0. Need use relation between E and B for EM wave in vacuum. For each plane wave propagating in +x, E parallel z, B parallel y? Need orientation: For EM wave propagating in +x, E × B direction = propagation direction. With E in z direction, if E = E0 cos(...) \hat{z}, propagation +x, then B = (E/c) \hat{y}? Let's verify cross product: E × B should point +x. If E = z, B = y, z × y = -x, not +x. Wait right hand: \hat{z} × \hat{y} = -\hat{x}. So to get +x, B must be along -y? Let's check: E = z, B = ? For plane wave, B = (1/c) \hat{k} × E where \hat{k} propagation direction. If \hat{k}=+\hat{x}, E = E_z \hat{z}, then \hat{x} × \hat{z} = -\hat{y}. So B = (E/c)(-\hat{y}). Alternatively from Faraday: ∇×E = -∂B/∂t; for E_z(x - ct), B_y = -E_z/c? Let's derive sign.
Let E = E0 cos(kx - ωt) \hat{z}. Plane wave solution: E = E0 f(x - ct) \hat{z}. Maxwell: ∇×E = (∂E_z/∂x) \hat{y}? Compute: ∇×E = (0, ∂E_z/∂x, 0)? Curl components: (∂Ez/∂y - ∂Ey/∂z)=0, (∂Ex/∂z - ∂Ez/∂x)? Actually formula: (curl E)_y = ∂Ex/∂z - ∂Ez/∂x = -∂Ez/∂x. So ∇×E = -∂Ez/∂x \hat{y}. Faraday: ∇×E = -∂B/∂t => -∂Ez/∂x \hat{y} = -∂B/∂t => ∂B/∂t = ∂Ez/∂x \hat{y}. For f(x - ct), ∂Ez/∂x = f'. ∂B/∂t = f' \hat{y}; integrate t: B = -(1/c) f \hat{y}? Wait ∫ f'(x - ct) dt = -(1/c) f(x - ct). Because d/dt f(x - ct) = -c f'. So B = -(1/c) f \hat{y} = -(1/c) E_z \hat{y}. Yes B = - (E/c) \hat{y}. So B along -y. Equivalently B = (1/c) \hat{k} × E? \hat{k} = x, E = z; \hat{x}×\hat{z} = -\hat{y}; B = (1/c)(-\hat{y}) E. Correct. Some textbooks B = (1/c) k × E = (1/c) \hat{k} × E; yes.
Alternatively if E = cos(kx - ωt + φ), B = (1/c) cos(kx - ωt + φ) (-\hat{y}) = -(E/c) \hat{y}. So resultant B = (1/c) (- \hat{y}) E. Need amplitude B0 = |E_amplitude|/c = (2E0 |cos(nπ/2)|)/c? Or maybe signed? Since B field expression: B = - (2E0/c) cos(nπ/2) cos(kx - ωt + kd/2) \hat{y}. But if n odd, zero. If they define B0 modulus, then B0 = |2E0 cos(nπ/2)|/c = (2E0/c)|cos(nπ/2)|. But part says "considerando que possui amplitude de módulo B0" maybe means they want express B in terms of B0. So B = ± B0 cos(...) \hat{y}, with amplitude B0. Need determine orientation: -y direction (or +y depending sign of cos). Let's be precise.
Let's solve thoroughly.
Given superposition:
E = E0 [ cos(kx + kd - ωt) + cos(kx - ωt) ] \hat{z}.
Use trig: cos A + cos B = 2 cos((A+B)/2) cos((A-B)/2).
A = kx + kd - ωt; B = kx - ωt.
(A+B)/2 = (2kx + kd - 2ωt)/2 = kx + kd/2 - ωt = kx - ωt + kd/2.
(A-B)/2 = (kd)/2 = nπ/2.
Thus E = 2 E0 cos(nπ/2) cos(kx - ωt + nπ/2) \hat{z}. Since kd = nπ, kd/2 = nπ/2.
This is a wave propagating +x because phase = kx - ωt + const. If amplitude 2E0 cos(nπ/2) nonzero; if zero, trivial. We can write amplitude A_n = 2 E0 cos(nπ/2). Maybe for n odd, A_n=0; for n even, A_n=2E0(-1)^{n/2}. If they require amplitude positive, A_n = 2E0 |cos(nπ/2)| and maybe phase adjusted.
b) Type of interference: n even -> constructive interference (maximum amplitude 2E0); n odd -> destructive interference (zero amplitude). Because phase difference between original waves is Δφ = (kx+kd-ωt) - (kx-ωt)=kd = nπ. For n even, Δφ = 2mπ -> in phase -> constructive. For n odd, Δφ = (2m+1)π -> opposite phase -> destructive. That is clearer. Let's state: n par -> interferência construtiva; n ímpar -> destrutiva.
c) Magnetic field. For each plane wave, relation B = (1/c) \hat{u} × E, where \hat{u} = \hat{x} propagation direction. Since E along z, \hat{x} × \hat{z} = -\hat{y}. Therefore B = - (1/c) E \hat{y}. Resultant E = A_n cos(kx - ωt + nπ/2) \hat{z}. Then B = - (A_n/c) cos(kx - ωt + nπ/2) \hat{y}. If A_n = 2E0 cos(nπ/2). So B = - (2E0/c) cos(nπ/2) cos(kx - ωt + nπ/2) \hat{y}. Orientation: negative y direction (sentido -y) for the phase when amplitude positive? But amplitude modulus B0 = |2E0 cos(nπ/2)|/c. The direction may be ±y depending on phase; but orientation (polarization direction) is along y axis, with sign negative in expression? Let's phrase: o campo magnético é paralelo ao eixo y, no sentido -y (ou +y se incorporarmos o sinal da amplitude); sua amplitude é B0 = |2E0 cos(nπ/2)|/c. Since question asks "indicando sua orientação, considerando que possui amplitude de módulo B0." Maybe answer: B = - (A_n/c) cos(...) \hat{y}, ou seja, ao longo de -y; amplitude B0 = |A_n|/c = 2E0|cos(nπ/2)|/c. Could also write B = B0 ...? Need be careful.
Let's inspect original: "Determine o campo magnético \vec{B} resultante dessa superposição, indicando sua orientação, considerando que possui amplitude de módulo B0." They might want B vector with amplitude B0. They don't ask relation between B0 and E0? "considerando que possui amplitude de módulo B0" maybe they want in terms of B0. But B0 not given? It says amplitude of modulus B0, but we can determine B0 from E0 and c? Maybe they want expression B = B0 ... with orientation. But maybe "considerando que possui amplitude de módulo B0" means assume amplitude is B0 (positive), so express B in terms of B0 and cos? However B0 depends on n. Let's parse: "Determine o campo magnético \vec{B} resultante dessa superposição, indicando sua orientação, considerando que possui amplitude de módulo \(B_0\)." Maybe they want derive B from Faraday/Maxwell and state orientation; amplitude modulus \(B_0\) perhaps given symbol, not necessarily numerically? But no relation given. In EM waves, B0 = E0/c generally? But here resultant amplitude maybe 2E0 cos(...)/c, so B0 = 2E0|cos(nπ/2)|/c. They might expect B = (E/c) in perpendicular direction, amplitude B0. Let's see.
Could also derive using Faraday law: ∇×E = -∂B/∂t. Let's do that to show orientation and amplitude. Since E = E_z(x,t) \hat{z}. Compute ∇×E = -∂E_z/∂x \hat{y}. Then ∂B/∂t = ∂E_z/∂x \hat{y}. Integrating w.r.t. t: B = - (1/c) E_z \hat{y} + const. For wave propagating +x, constant 0. So B = - (E/c)\hat{y}. Then amplitude modulus B0 = |E_amp|/c = |2E0 cos(nπ/2)|/c. But "considerando que possui amplitude de módulo B0" maybe just use B0 = E0/c? Hmm.
Let's examine possible issue: The superposition of two waves with same amplitude and same polarization but different phase yields a standing wave? Wait no, both propagate same direction, so no standing wave; it's a traveling wave with amplitude modulation? Actually if both same frequency/direction, superposition is another traveling wave with amplitude depending on phase difference. Yes.
Let's consider if original expression maybe has one wave traveling +x and another? cos[k(x+d)-ωt] = cos(kx - ωt + kd), same direction. So no standing wave.
Need answer in exam style: item by item.
Let's craft final with derivation.
Potential detail: In part a, use identity:
\[
\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\]
with \(A=k(x+d)-\omega t\), \(B=kx-\omega t\). Then:
\[
\vec E = 2E_0 \cos\left(\frac{kd}{2}\right)\cos\left(kx-\omega t + \frac{kd}{2}\right)\hat z.
\]
Since \(kd=n\pi\), \(\cos(kd/2)=\cos(n\pi/2)\), so:
\[
\vec E = 2E_0\cos\frac{n\pi}{2}\cos\left(kx-\omega t+\frac{n\pi}{2}\right)\hat z.
\]
Thus wave propagating +x with amplitude \(2E_0\cos(n\pi/2)\). If we define amplitude as nonnegative, \(A_n = 2E_0 |\cos(nπ/2)|\) and maybe phase constant. But they said amplitude expressa apenas em função de E0 e n; I can write \(A_n=2E_0\cos(nπ/2)\), acentuando que para n ímpar é nula. Or \(A_n = 2E_0(-1)^{n/2}\) for even n? Let's see for n=0: 2E0; n=2: -2E0; n=4: 2E0. So \(A_n = 2E_0 (-1)^{n/2}\) for n even; zero for odd. But not necessary.
Part b: "n par: interferência construtiva; n ímpar: interferência destrutiva." Maybe mention phase difference \(Δφ = kd = nπ\). For n even = 2mπ, cosΔφ=1 => construtiva; n odd = (2m+1)π, cosΔφ=-1 => destrutiva. Actually if phase difference = 2mπ, in phase; if (2m+1)π, opposite phase. Yes.
Part c:
Magnetic field relation: For EM wave in vacuum:
\[
\vec B = \frac{1}{c}\hat u \times \vec E
\]
where \(\hat u = \hat x\). So:
\[
\vec B = \frac{1}{c}\hat x \times \left[A_n \cos(kx-\omega t+nπ/2)\hat z\right] = -\frac{A_n}{c}\cos(kx-\omega t+nπ/2)\hat y
\]
with \(A_n=2E_0\cos(nπ/2)\). Thus orientation: eixo y, sentido \(-\hat y\) (quando \(A_n\cos(...)>0\); o vetor pode ter sinal global dependente da fase, mas a direção de oscilação é y). Amplitude: \(B_0 = \frac{|A_n|}{c} = \frac{2E_0}{c}|\cos(nπ/2)|\). Since they say "amplitude de módulo B0", maybe write:
\[
\vec B = -B_0\cos(kx-\omega t+nπ/2)\hat y?
\]
But if B0 = |A_n|/c, then coefficient = -B0 sign(A_n). Better write:
\[
\vec B = -\frac{2E_0}{c}\cos\frac{nπ}{2}\cos(kx-\omega t+\frac{nπ}{2})\hat y
\]
and \(B_0 = \left|\frac{2E_0}{c}\cos\frac{nπ}{2}\right|\). Or if using B0:
\[
\vec B = -B_0 \operatorname{sgn}(\cos(nπ/2)) \cos(...)\hat y
\]
ugly. Maybe just leave in terms of E0 and n. But item says "considerando que possui amplitude de módulo B0" maybe maybe they want "\(\vec B = - \frac{E}{c}\hat y\), com \(B_0 = E_0/c\)" Wait if n odd resultant zero, no amplitude; B0 would be zero. If n even, each original wave has B amplitude E0/c; superposition of two magnetic fields? Wait each original wave has electric field amplitude E0 and magnetic field amplitude E0/c in same direction (both -y). Superposition of B fields should be same linear combination as E: B_total = - (1/c) E_total \hat y. So amplitude = A_n/c = 2E0 cos(nπ/2)/c. Thus B0 = 2E0|cos(nπ/2)|/c. That's it.
But perhaps because they asked "considerando que possui amplitude de módulo B0" maybe they want \(B_0 = E_0/c\) for each wave? Let's read: "Determine o campo magnético \vec{B} resultante dessa superposição, indicando sua orientação, considerando que possui amplitude de módulo \(B_0\)." It might mean "Determine the resultant magnetic field \vec{B} of this superposition, indicating its orientation, considering that it has amplitude of modulus \(B_0\)." Maybe they expect \(B_0 = 2E_0\cos(nπ/2)/c\) and orientation. "considerando que possui amplitude de módulo B0" might just tell "call its amplitude B0" so final answer should have B0. But since B0 not previously defined, we define it. I'll write both.
Let's ensure no sign mistakes. Let's verify with explicit n=0:
E = E0[cos(kx -ωt)+cos(kx -ωt)] = 2E0 cos(kx-ωt) \hat z. Propagating +x. B = - (2E0/c) cos(kx-ωt) \hat y. Since wave propagates +x, E along +z, Poynting vector E×B = (2E0 cos)( - (2E0/c) cos) (z×y) = - (4E0^2/c cos^2)(-x) = +x. Yes.
For n=2:
E = E0[cos(kx+2π -ωt)+cos(kx-ωt)] = E0[cos(kx-ωt)+cos(kx-ωt)] = 2E0 cos(kx-ωt) \hat z? Wait cos(kx+2π -ωt)=cos(kx-ωt), so same as n=0. Our formula: 2E0 cos(π) cos(kx -ωt + π) \hat z = 2E0 (-1)(-cos(kx-ωt)) = 2E0 cos(kx-ωt). Yes.
For n=1:
E= E0[cos(kx+π -ωt)+cos(kx-ωt)] = E0[-cos(kx-ωt)+cos(kx-ωt)] =0. Formula gives 2E0 cos(π/2)=0. Good.
Part c for n=1: B=0.
Now, possible issue: The two waves are coherent and propagate in vacuum. The magnetic field of each wave: For a wave with electric field \(E_i = E_0 \cos(k(x+d)-ωt)\hat z\), the magnetic field is \(B_i = \frac{1}{c} \hat x × E_i = -\frac{E_0}{c}\cos(k(x+d)-ωt)\hat y\). Similarly second \(B_2 = -\frac{E_0}{c}\cos(kx-ωt)\hat y\). Sum \(B = -\frac{E_0}{c}[\cos(k(x+d)-ωt)+\cos(kx-ωt)]\hat y = -\frac{1}{c}\vec E \times? Actually B = -E_total/c \hat y\). Yes.
Thus final expression:
\[
\vec B(x,t)= -\frac{2E_0}{c}\cos\left(\frac{n\pi}{2}\right)\cos\left(kx-\omega t+\frac{n\pi}{2}\right)\hat y.
\]
Amplitude modulus:
\[
B_0 = \frac{2E_0}{c}\left|\cos\frac{n\pi}{2}\right|.
\]
Orientation: \(-y\) direction (ou seja, perpendicular a E e à direção de propagação, formando com \hat x e \hat z o triedro; vetor no sentido negativo do eixo y).
Need be mindful: If amplitude modulus \(B_0\) requested, maybe they don't want E0? But no relation given; maybe in terms of E0 and n. I'll state \(B_0 = \frac{2E_0}{c}|\cos(nπ/2)|\).
Let's craft final response in Portuguese.
Potential formatting: Use equations and paragraphs.
Let's write:
a) Usando \(\cos A+\cos B=2\cos((A+B)/2)\cos((A-B)/2)\):
...
Portanto \(\vec E = ...\). Logo é onda progressiva no sentido +x, com amplitude \(A_n=2E_0\cos(nπ/2)\) (ou, em módulo, \(2E_0|\cos(nπ/2)|\)).
b) A diferença de fase entre as duas ondas é \(\Delta\phi = kd = nπ\). Se n par, \(\Delta\phi=2mπ\), interferência construtiva; se n ímpar, \(\Delta\phi=(2m+1)π\), interferência destrutiva.
c) Para onda EM no vácuo, \(\vec B = \frac{1}{c}\hat{u}\times\vec E\), com \(\hat{u}=\hat x\). Como \(\hat x\times\hat z=-\hat y\):
\[
\vec B = -\frac{1}{c}\vec E = -\frac{2E_0}{c}\cos\frac{nπ}{2}\cos\left(kx-\omega t+\frac{nπ}{2}\right)\hat y.
\]
Assim, orientação: ao longo do eixo y, sentido \(-\hat y\) (a amplitude algébrica pode mudar de sinal conforme n, mas a direção de oscilação é y). Módulo da amplitude:
\[
B_0 = \frac{2E_0}{c}|\cos(nπ/2)|.
\]
Se preferir em termos de B0, \(B = -B_0 \operatorname{sgn}(\cos(nπ/2))\cos(...)\hat y\)? Maybe not.
Let's ensure "considerando que possui amplitude de módulo B0" maybe they expect "B = -B0 cos(...) \hat y" maybe with B0 = E0/c? But no. Let's maybe phrase: "Chamando de \(B_0\) o módulo da amplitude, \(B_0 = \frac{2E_0}{c}|\cos(nπ/2)|\)." Then vector.
Need final line: "Resposta final: ..." Include item results short. For a: \(\vec E = 2E_0\cos(nπ/2)\cos(kx-ωt+nπ/2)\hat z\), amplitude \(2E_0\cos(nπ/2)\). b: n par construtiva, n ímpar destrutiva. c: \(\vec B = -\frac{2E_0}{c}\cos(nπ/2)\cos(kx-ωt+nπ/2)\hat y\), amplitude \(B_0=\frac{2E_0}{c}|\cos(nπ/2)|\), orientação -y. But need align with format: "Resposta final: a) ...; b) ...; c) ...".
Let's also discuss possible "n=0,1,2,..." In item a if n odd amplitude zero. But if amplitude expressed only as function of E0 and n, maybe \(\cos(nπ/2)\) is function; if they want positive, \(|...|\).
Let's consider if using \(kd = nπ\) with \(n=0,1,2,\ldots\) and \(d\) positive; for n=0, d=0? They said d positive constant but assume distances d that satisfy kd=nπ; n=0 implies d=0 not positive, contradiction? Maybe they allow d→0? The text says "Assuma distâncias d que satisfazem a relação kd = nπ para n = 0,1,2,...". It may include zero mathematically, but d positive constant; maybe n=0 not physical but included for math. We need not worry. For n=0, two waves coincide, constructive.
Let's also consider if there is any issue with "duas ondas eletromagnéticas coerentes entre si" and superposition of fields, in vacuum. Yes.
Let's formulate final elegantly.
Potential extra: In part a, show "pode ser reescrito como uma onda que se propaga no sentido x>0" means identify phase \(kx - ωt + nπ/2\) and \(k>0, ω>0\). Since both original waves propagate +x. Good.
Let's write final with trig identity and definitions.
Now, final answer in Portuguese. Use notation \(\hat{z}\), \(\hat{y}\), \(\hat{x}\). Use \(c\) for speed of light. Not given in problem but expected.
Let's produce final.
a) Usando a identidade trigonométrica
\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2},
com
A=k(x+d)-\omega t,\qquad B=kx-\omega t,
temos
\frac{A+B}{2}=kx-\omega t+\frac{kd}{2},\qquad
\frac{A-B}{2}=\frac{kd}{2}.
Como kd=n\pi, resulta
\vec E=2E_0\cos\!\left(\frac{n\pi}{2}\right)
\cos\!\left(kx-\omega t+\frac{n\pi}{2}\right)\hat z.
Essa expressão representa uma onda que se propaga no sentido x>0, pois a fase é do tipo kx-\omega t+\text{constante}. Sua amplitude é
A_n=2E_0\cos\!\left(\frac{n\pi}{2}\right)
(ou, em módulo, 2E_0\left|\cos\frac{n\pi}{2}\right|).
b) A diferença de fase entre as duas ondas é
\Delta\phi=kd=n\pi.
- Se n é par, \Delta\phi=2m\pi, então as ondas chegam em fase: interferência construtiva, com amplitude máxima 2E_0.
- Se n é ímpar, \Delta\phi=(2m+1)\pi, então as ondas chegam em oposição de fase: interferência destrutiva, com amplitude nula.
c) Para uma onda eletromagnética plana no vácuo,
\vec B=\frac{1}{c}\,\hat u\times\vec E,
sendo \hat u=\hat x a direção de propagação. Como \hat x\times\hat z=-\hat y,
\vec B=-\frac{1}{c}\vec E
=-\frac{2E_0}{c}\cos\!\left(\frac{n\pi}{2}\right)
\cos\!\left(kx-\omega t+\frac{n\pi}{2}\right)\hat y.
Portanto, o campo magnético é paralelo ao eixo y, no sentido -\hat y, e o módulo de sua amplitude é
B_0=\frac{2E_0}{c}\left|\cos\!\left(\frac{n\pi}{2}\right)\right|.
Resposta final: a) \vec E=2E_0\cos(n\pi/2)\cos(kx-\omega t+n\pi/2)\hat z, amplitude 2E_0\cos(n\pi/2); b) n par: interferência construtiva; n ímpar: interferência destrutiva; c) \vec B=-\frac{2E_0}{c}\cos(n\pi/2)\cos(kx-\omega t+n\pi/2)\hat y, orientação -\hat y, amplitude B_0=\frac{2E_0}{c}|\cos(n\pi/2)|.