Modern Physics 7 Question 3

3. An electromagnetic wave is represented by the electric field

E=E0n^sin[ωt+(6y8z)]. Taking unit vectors in x,y and z-directions to be i^,j^,k^, the direction of propagation s^, is

(a) s^=3i^4j^5

(b) s^=4k^+3j^5

(c) s^=3j^+4k^5

(d) s^=4j^3k^5

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Solution:

  1. Standard expression of electromagnetic wave is given by

E=E0n^[sin(ωtkr^)]

Here, k is the propagation vector. Direction of propagation in this case is k^.

Given expression of electromagnetic wave,

E=E0n^sin[ωt+(6y8z)]E=E0n^sin[ωt(8z6y)]

Comparing Eq. (ii) with Eq. (i), we get

kr^=8z6y Here, r^=xi^+yj^+zk^ and k=kxi^+kyj^+kzk^kr^=xkx+yky+zkz

From Eqs. (iii) and (iv), we get

xkx= zero kx=0yky=6yky=6zkz=8zkz=8

Hence, k=6j^+8k^

So, direction of propagation,

s^=k^=k|k|=6j^+8k^62+82=6j^+8k^10=3j^+4k^5



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