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The magnetic field of a plane electromagnetic wave is given by B = 3 * 10^(-8) * sin[200*pi*(y + c*t)] * i_hat T, where c = 3 * 10⁸ m/s. What is the corresponding electric field?

1. E = -10^(-6) * sin[200*pi*(y + c*t)] * k_hat V/m
2. E = -9 * sin[200*pi*(y + c*t)] * k_hat V/m
3. E = 9 * sin[200*pi*(y + c*t)] * k_hat V/m
4. E = 3 * 10^(-8) * sin[200*pi*(y + c*t)] * k_hat V/m

Correct answer: E = -9 * sin[200*pi*(y + c*t)] * k_hat V/m

Solution

The argument y + c*t = constant implies y decreases as t increases, so the wave propagates in the -y direction. Magnitude: E0 = c * B0 = 3*10⁸ * 3*10⁻⁸ = 9 V/m. Direction: for propagation in -y and B in +x (i_hat), the Poynting vector S = (1/mu0)*(E x B) must be in -y. Testing E = -k_hat (i.e., -z direction): E x B = (-k_hat) x (i_hat) = -(k x i) = -j_hat = -y_hat. This matches propagation in -y. So E = -9*sin[200*pi*(y+ct)] * k_hat V/m.

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