A) \[\vec{E}=\sqrt{\frac{I}{{{\varepsilon }_{0}}C}}\cos \left[ \frac{2\pi }{\lambda }(y-ct) \right]\hat{i};\vec{B}=\frac{1}{c}E\hat{k}\]
B) \[\vec{E}=\sqrt{\frac{I}{{{\varepsilon }_{0}}C}}\cos \left[ \frac{2\pi }{\lambda }(y-ct) \right]\hat{k};\vec{B}=-\frac{1}{c}E\hat{i}\]
C) \[\vec{E}=\sqrt{\frac{2I}{{{\varepsilon }_{0}}C}}\cos \left[ \frac{2\pi }{\lambda }(y-ct) \right]\hat{k};\vec{B}=+\frac{1}{c}E\hat{i}\]
D) \[\vec{E}=\sqrt{\frac{2I}{{{\varepsilon }_{0}}C}}\cos \left[ \frac{2\pi }{\lambda }(y+ct) \right]\hat{k};\vec{B}=\frac{1}{c}E\hat{i}\]
Correct Answer: A
Solution :
| [c] E is the electric field vector, and B is the magnetic field vector of the EM wave. For electromagnetic waves E and B are always perpendicular to each other and perpendicular to the direction of propagation. The direction of propagation is the direction of E x B. So, if the wave propagates in the +Y direction then the direction of E and B should be in +X and +Z or vice versa i.e +Z and +X respectively. |
| Case 1. |
| Let us suppose\[\vec{E}\] is in \[\hat{i}\]and\[\hat{B}\] is in\[\hat{k}\] |
| Then\[\vec{E}\times \vec{B}\] will be in\[-\hat{j}\] |
| Not Possible. |
| Case 2. |
| Let us suppose \[\vec{E}\]is in \[\hat{k}\]and \[\vec{B}\]is in\[\hat{i}\] |
| Then \[\vec{E}\times \vec{B}\]will be in\[\hat{j}\] |
| This is satisfying option [c]as the electric and magnetic field also propagate in positive y direction with time so\[(y-ct)\] should be there in wave equation. |
| Also \[I=\frac{c{{\in }_{o}}}{2}|{{E}_{o}}{{|}^{2}}\] |
| \[|{{E}_{o}}|=\sqrt{\frac{2I}{c{{\in }_{o}}}}\] |
| From these, we can say that option [c] would be the best option. |
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