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A plane electromagnetic wave travels in the -z direction toward a sheet of paper. At time t = 0, the wave strikes the paper at normal incidence. At point O (at the origin), the magnetic field vector points in the -y direction with magnitude 4.0 * 10⁻⁸ T (this is the maximum). The wave frequency is 1.0 * 10¹⁶ Hz. The +x direction is to the right and +y is upward. Which of the following options are correct?

1. The electric field E at t = 0 at point O is 12 i-hat V/m
2. The electric field E at t = 2.5 * 10⁻¹⁷ s at z = 0.75 * 10⁻⁸ m is 6*sqrt(2) i-hat V/m
3. The electric field E at t = 5 * 10⁻¹⁷ s at z = 1.5 * 10⁻⁸ m is -6*sqrt(2) i-hat V/m
4. The electric field E at t = 7.5 * 10⁻¹⁷ s at z = 2.25 * 10⁻⁸ m is -12 i-hat V/m

Correct answer: The electric field E at t = 0 at point O is 12 i-hat V/m

Solution

Wave travels in -z direction. E0 = c*B0 = 3*10⁸ * 4*10⁻⁸ = 12 V/m. Direction: k-hat = -z-hat, B = -y-hat*B0, so E = (k-hat x B-hat)*E0... use: S = E x B must point in -z. E in +x, B in -y: E x B = x-hat x (-y-hat) = -(x x y) = -z-hat. Correct! E(z,t) = 12 cos(kz + omega*t) i-hat where k = omega/c. omega = 2*pi*10¹⁶, k = omega/c = 2*pi*10¹⁶/(3*10⁸) = 2*pi/3 * 10⁸ rad/m. At t=0, z=0: E=12 i-hat. Option 1: TRUE. For option 2: t=2.5*10⁻¹⁷, z=0.75*10⁻⁸. kz = (2pi/3)*10⁸ * 0.75*10⁻⁸ = (2pi/3)*0.75 = pi/2. omega*t = 2pi*10¹⁶ * 2.5*10⁻¹⁷ = 2pi*0.25 = pi/2. Phase = pi/2+pi/2 = pi. cos(pi)=-1. E = -12 i-hat. Option 2 says 6*sqrt(2) — FALSE. Option 3: t=5*10⁻¹⁷, z=1.5*10⁻⁸. kz=pi*10⁸*1.5*10⁻⁸=... wait k=omega/c=2pi*10¹⁶/3*10⁸. kz=2pi*10¹⁶/(3*10⁸)*1.5*10⁻⁸=2pi*1.5/(3*10⁻⁸*10⁸) — let me compute carefully. k=2pi*f/c=2pi*10¹⁶/(3*10⁸). kz=2pi*10¹⁶/(3*10⁸)*1.5*10⁻⁸=2pi*1.5*10⁸/(3*10⁸)=2pi*0.5=pi. omega*t=2pi*10¹⁶*5*10⁻¹⁷=2pi*0.5=pi. Total phase=2pi. cos(2pi)=1. E=12 i-hat, not -6*sqrt(2). Option 3 FALSE. Option 4: t=7.5*10⁻¹⁷, z=2.25*10⁻⁸. kz=2pi*10¹⁶/(3*10⁸)*2.25*10⁻⁸=2pi*2.25/(3*10⁰)... =2pi*2.25*10⁸/(3*10⁸)=2pi*0.75=3pi/2. omega*t=2pi*10¹⁶*7.5*10⁻¹⁷=2pi*0.75=3pi/2. Phase=3pi. cos(3pi)=-1. E=-12 i-hat. Option 4 TRUE.

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