Answer :

(a) Given: E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]}


As it is clear from the above electric field vector, that electric field is in the negative x direction. Therefore, the direction of propagation of the vector will be in negative y-direction i.e. –j.


(b) As we know that the general equation for electric field vector in the positive x-direction is:


..........(i)


The electric field part of an electromagnetic wave in vacuum is


E = {(3.1 N/C) cos [(1.8 rad /m) y + (5.4 × 106 rad/s)t..............(ii)


Comparing equation (i) and (ii), we get


E0 = 3.1 N


(c) Angular frequency ω = 5.4 × 106 rad/s


Wave number i.e. k = 1.8 rad/m


Wavelength of the wave can be calculated as follows:


λ = 2π/k




(d) Angular frequency of the wave can be expressed as follows:


ω = 2π v


5.4 × 106 = 2 × 3.14 × v




V = 0.86 × 106 Hz


(e) Magnetic field strength can be calculated as follows:


B0 = E0/c



B0 = 1.03 × 10-7 T


(f) Since magnetic field vector is in the negative z-direction, the general equation for magnetic field vector can be written as follows:


The expression for the magnetic field part of the wave can be written as follows:


B = B0 cos (ky + wt) k


B = {1.03 × 10-7 cos [(1.8 rad/m)y + (5.4 × 106 rad/s)t]}k


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