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Suppose that the electric field part of an electromagnetic wave in vacuum is

`vec"E" = {(3.1"N"/"C") cos[(1.8 ("rad")/"m")"y" + (5.4 xx 10^8 ("rad")/"s")"t"]}hat"i"`

**(a) **What is the direction of propagation?

**(b) **What is the wavelength λ?

**(c) **What is the frequency v?

**(d) **What is the amplitude of the magnetic field part of the wave?

**(e) **Write an expression for the magnetic field part of the wave.

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#### Solution

**(a)** From the given electric field vector, it can be inferred that the electric field is directed along the negative x-direction. Hence, the direction of motion is along the negative y-direction i.e., `-hat"j"`.

**(b)** It is given that,

`vec"E" = 3.1"N"/"C" cos[(1.8 ("rad")/"m")"y" + (5.4 xx 10^8 ("rad")/"s")"t"]hat"i"` ...........(1)

The general equation for the electric field vector in the positive x direction can be written as:

`vec"E" = "E"_0 sin ("kx" - ω"t")hat"i"` ........(2)

On comparing equations (1) and (2), we get

Electric field amplitude, E_{0} = 3.1 N/C

Angular frequency, ω = 5.4 × 10^{8} rad/s

Wave number, k = 1.8 rad/m

Wavelength, `lambda = (2pi)/1.8` = 3.490 m

**(c)** Frequency of wave is given as:

`"v" = ω/(2pi)`

= `(5.4 xx 10^8)/(2pi)`

= 8.6 × 10^{7} Hz

**(d)** Magnetic field strength is given as:

`"B"_0 = "E"_0/"c"`

Where,

c = Speed of light = 3 × 10^{8} m/s

∴ `"B"_0 = 3.1/(3 xx 10^8)`

= 1.03 × 10^{−7} T

**(e)** On observing the given vector field, it can be observed that the magnetic field vector is directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as:

`vec"B" = "B"_0 cos ("ky" + ω"t")hat"k"`

= `{(1.03 xx 10^-7 "T") cos [(1.8 "rad"/"m")"y" + (5.4 xx 10^6 "rad"/"s")"t"]}hat"k"`

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