A) \[\frac{1}{c}(6\widehat{k}+8\widehat{i})\,cos[(6x+8z-10ct)]\]
B) \[\frac{1}{c}(6\widehat{k}-8\widehat{i})\,cos[(6x+8z-10ct)]\]
C) \[\frac{1}{c}(6\widehat{k}+8\widehat{i})\,cos[(6x-8z+10ct)]\]
D) \[\frac{1}{c}(6\widehat{k}-8\widehat{i})\,cos[(6x+8z+10ct)]\]
Correct Answer: B
Solution :
\[E=10\widehat{j}\text{ }cos\left( 6x+8z \right)\] Phase angle of E at\[t=0\text{ }is\text{ }6x+8z\]. As E and B oscillate in same phase \[\therefore \text{ }at,\text{ }t=0\] Phase angle of B must be \[6x+8z\] Direction of wave propagation \[=\,\,\frac{6\widehat{j}+8\widehat{k}}{\sqrt{36+64}}\] \[=\,\,\frac{6\widehat{j}+8\widehat{k}}{10}\] Let \[\overrightarrow{B}=a\widehat{i}+b\widehat{j}+d\widehat{k}\] and unit vector in direction of propagation of EM wave is \[\frac{\overrightarrow{E}\times \overrightarrow{B}}{\left| \overrightarrow{E}\parallel \overrightarrow{B} \right|}\] \[\frac{\overrightarrow{E}\times \overrightarrow{B}}{\left| \overrightarrow{E}\parallel \overrightarrow{B} \right|}\,\,=\,\,\frac{6\widehat{i}\,+\,8\widehat{k}}{10}\] \[\frac{10d\,\widehat{i}+\widehat{k}(-10a)}{10\,\times \,\sqrt{{{a}^{2}}+{{b}^{2}}+{{a}^{2}}}}=\,\frac{6\,\widehat{i}+8\widehat{k}}{10}\] \[\left| \overrightarrow{B} \right|\,=\,\frac{E}{c}\,=\,\frac{10}{c}\] \[c\frac{[10\,d\widehat{i}-10a\widehat{k}]}{10\times 10}\,\,=\,\,\frac{6\widehat{i}+8\widehat{i}}{10}\] \[c\left| d\,\widehat{i}-a\,\widehat{k} \right|\,\,=\,\,6\,\widehat{i}+8\,\widehat{k}\] \[d=\,\frac{6}{c}\] \[a=-\frac{8}{c}\,\,\,\,\,\therefore \,\,\,\,\overrightarrow{B}=-\,\frac{8}{c}\,\widehat{i}+\frac{6}{c}\,\widehat{k}\] \[\therefore \,\,\,B=\,\frac{6\widehat{k}-8\widehat{i}}{c}\,\cos (6x+8z-10ct)\]
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