The electric field associated with an electromagnetic wave travelling in vacuum is given by E₀ (3y + 4z + t) , i , where is the angular frequency. All quantities are in SI units. The correct statement(s) about this wave is/are: [Given: speed of light in vacuum c = 3 10^8 ms ⁻¹ .]

Step-by-step solution

The given electric field is E = E₀ (3y + 4z + t) , i . Comparing the phase = 3y + 4z + t with the standard wave equation phase k r - t , we can rewrite it as -(-3y - 4z - t) . Thus, the wave vector is k = -3 j - 4 k . The direction of wave propagation is given by the unit vector n : n = k | k | = -3 j - 4 k (-3)^2 + (-4)^2 = - 1 5 (3 j + 4 k ) The magnitude of the wave vector is | k | = 5 m ⁻¹ . The angular frequency is: = c| k | = (3 10^8) 5 = 1.5 10^9 rad s ⁻¹ The magnetic field B is given by: B = 1 c ( n E ) = 1