A Sinusoidal Electromagnetic Wave Is Propagating in Vacuum $\vec { E }$

Short Answer

A sinusoidal electromagnetic wave is propagating in vacuum.At a given point and at a particular time the $\vec { E }$ field is in the +x direction and the $\vec { B }$ field is in the -y direction,and at that point the intensity of the wave is $0.45 \mathrm {~W} / \mathrm { m } ^ { 2 }$ $\left( c = 3.0 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , \right.$ $\mu _ { 0 } = 4 \pi \times 10 ^ { - 7 } \mathrm {~T} \cdot \mathrm { m } / \mathrm { A } ,$ $\left. \varepsilon _ { 0 } = 8.85 \times 10 ^ { - 12 } \mathrm { C } ^ { 2 } / \mathrm { N } \cdot \mathrm { m } ^ { 2 } \right)$ (a)What is the direction of propagation of the wave?
(b)What is the $\vec { E }$ field amplitude at the given point?