Question:
The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression
\(\vec E(x,y)=10\hat j\cos[(6x+8z)]\).
The magnetic field \(\vec B(x,z,t)\) is given by - (c is the velocity of light)
The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression
\(\vec E(x,y)=10\hat j\cos[(6x+8z)]\).
The magnetic field \(\vec B(x,z,t)\) is given by - (c is the velocity of light)
\(\vec E(x,y)=10\hat j\cos[(6x+8z)]\).
The magnetic field \(\vec B(x,z,t)\) is given by - (c is the velocity of light)
Show Hint
Find the propagation direction from the phase 6x + 8z, then use B = (1/c)(n-hat x E).
Updated On: Nov 5, 2026
- \(\frac1c(6\hat k+8\hat i)\cos[(6x+8z-10ct)]\)
- \(\frac1c(6\hat k-8\hat i)\cos[(6x+8z-10ct)]\)
- \(\frac1c(6\hat k+8\hat i)\cos[(6x-8z+10ct)]\)
- \(\frac1c(6\hat k-8\hat i)\cos[(6x-8z+10ct)]\)
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The Correct Option is B
Solution and Explanation
At \(t=0\) the phase is \(6x+8z\), so the wave vector is \(\vec k=6\hat i+8\hat k\) with \(|k|=10\) and \(\omega=10c\). The wave travels along \(\hat n=\frac{6\hat i+8\hat k}{10}\) and the phase is \(6x+8z-10ct\).
For an EM wave \(\vec B=\frac1c\,\hat n\times\vec E\).
\(\hat n\times\hat j=\frac{6\,\hat i\times\hat j+8\,\hat k\times\hat j}{10}=\frac{6\hat k-8\hat i}{10}\).
\(\vec B=\frac1c\cdot\frac{6\hat k-8\hat i}{10}\times10\cos(6x+8z-10ct)=\frac1c(6\hat k-8\hat i)\cos(6x+8z-10ct)\).
Answer: (2) \(\frac1c(6\hat k-8\hat i)\cos[(6x+8z-10ct)]\)
For an EM wave \(\vec B=\frac1c\,\hat n\times\vec E\).
\(\hat n\times\hat j=\frac{6\,\hat i\times\hat j+8\,\hat k\times\hat j}{10}=\frac{6\hat k-8\hat i}{10}\).
\(\vec B=\frac1c\cdot\frac{6\hat k-8\hat i}{10}\times10\cos(6x+8z-10ct)=\frac1c(6\hat k-8\hat i)\cos(6x+8z-10ct)\).
Answer: (2) \(\frac1c(6\hat k-8\hat i)\cos[(6x+8z-10ct)]\)
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