An electric field of a plane radio wave travelling in the positive x direction i

Question

An electric field of a plane radio wave travelling in the positive x direction is polarized along the y-axis and has angular frequency a The electric field magnitude is Eo. Write down a general form of the wave function describing the field E(x, t). Use the differential form of Faraday's law to show that the magnetic field B must be directed along the z-axis and its magnitude is Eo/c (c is the speed of light). 3 marks] Now the field E is in the direction i 3j and the field B is in the [1 mark] direction 3i -j 5k. What is the direction K of wave propagation?

Explanation / Answer

Answer:- The equation for electric field, for a plane wave travelling in x-direction and electric field polarized in y-direction can be written as-

E(x, t) = E0 cos(kx - wt) y , the direction is in y-axis.

From Faraday's law, curl of above equation is equal to negative time derivative of magnetic field i.e-

curl of E(x, t) = - d(B)/dt. So taking curl of above equation we get-

d(E(x,t)/dx = -d(B)/dt

=> kE0 sin(kx - wt) z = -d(B)/dt ,  z denotes z-cap i.e direction.

=> kE0 sin(kx - wt) dt z = -dB, on integrating both the sides we get-

B(x, t) = (kE0/w) cos(kx - wt) z, note k = wave number and w = angular frequency and w/k = speed of light. Thus-

B(x, t) = E0/c cos(kx - wt) z, hence B(x, t) is polarized in z-direction.

Direction of wave propagation is always in the direction of cross product of E and B. So cross product of i + 3j and 3i - j + 5k will give direcyion of wave propagation. So cross product of these two is-

Direction of wave propagation = (i + 3j ) X (3i - j + 5k ) = 15i - 5j - 10k .

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