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A plane electromagnetic wave of frequency 500 MHz travels through vacuum along the +y direction. At a certain point and instant the magnetic field is B = 8.0*10⁻⁸ (z-hat) T. Taking c = 3*10⁸ m/s, find the electric field vector at that point. (x-hat, y-hat, z-hat are unit vectors along x, y, z.)

1. -24 (x-hat) V/m
2. 2.6 (x-hat) V/m
3. 24 (x-hat) V/m
4. -2.6 (x-hat) V/m

Correct answer: -24 (x-hat) V/m

Solution

Magnitude: E0 = c*B0 = (3*10⁸)*(8.0*10⁻⁸) = 24 V/m. For the direction, the wave moves along +y (y-hat), so E x B must point along +y. With B along +z, we need E direction (E-hat) such that E-hat x (z-hat) = y-hat. Since (-x-hat) x (z-hat) = -(x-hat x z-hat) = -(-y-hat) = +y-hat, E must be along -x. Therefore E = -24 (x-hat) V/m.

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