Q. 11

# Suppose tha

(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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