Question:
An electromagnetic wave is represented by the electric field
\(\vec{E}=E_0\hat{n}\sin\left[\omega t+(6y-8z)\right]\)
. Taking unit vectors in x, y and z directions to be \(\hat{i},\hat{j},\hat{k}\), the direction of propagation \(\hat{s}\), is :
An electromagnetic wave is represented by the electric field
\(\vec{E}=E_0\hat{n}\sin\left[\omega t+(6y-8z)\right]\)
. Taking unit vectors in x, y and z directions to be \(\hat{i},\hat{j},\hat{k}\), the direction of propagation \(\hat{s}\), is :
\(\vec{E}=E_0\hat{n}\sin\left[\omega t+(6y-8z)\right]\)
. Taking unit vectors in x, y and z directions to be \(\hat{i},\hat{j},\hat{k}\), the direction of propagation \(\hat{s}\), is :
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A wave \(\sin(\omega t+\vec k\cdot\vec r)\) travels opposite to \(\vec k\).
Updated On: Nov 5, 2026
- \(\hat{s}=\frac{3\hat{i}-4\hat{j}}{5}\)
- \(\hat{s}=\frac{-4\hat{k}+3\hat{j}}{5}\)
- \(\hat{s}=\left(\frac{-3\hat{j}+4\hat{k}}{5}\right)\)
- \(\hat{s}=\frac{4\hat{j}-3\hat{k}}{5}\)
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The Correct Option is C
Solution and Explanation
The argument is \(\omega t+(6y-8z)=\omega t+\vec k\cdot\vec r\) with \(\vec k=6\hat j-8\hat k\), \(|\vec k|=10\).
A wave of the form \(\sin(\omega t-\vec k\cdot\vec r)\) travels along \(+\vec k\); with the plus sign the wave travels along \(-\vec k\).
\(\hat s=-\frac{6\hat j-8\hat k}{10}=\frac{-3\hat j+4\hat k}{5}\).
Answer: (3) \(\hat{s}=\left(\frac{-3\hat{j}+4\hat{k}}{5}\right)\)
A wave of the form \(\sin(\omega t-\vec k\cdot\vec r)\) travels along \(+\vec k\); with the plus sign the wave travels along \(-\vec k\).
\(\hat s=-\frac{6\hat j-8\hat k}{10}=\frac{-3\hat j+4\hat k}{5}\).
Answer: (3) \(\hat{s}=\left(\frac{-3\hat{j}+4\hat{k}}{5}\right)\)
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