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એક વિદ્યુતચુંબકીય તરંગ $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$ દિશામાં પ્રવર્તે છે જ્યાં તેનું પોલારાઈજેશન $\hat{\mathrm{k}}$ દિશામાં છે.તો ચુંબકીયક્ષેત્રનું સાચું સ્વરૂપ નીચે પૈકી કયું હશે?
$\mathrm{B}_{0} \frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}} \cos \left(\omega \mathrm{t}-\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)$
$\mathrm{B}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}} \cos \left(\omega \mathrm{t}-\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)$
$\mathrm{B}_{0} \hat{\mathrm{k}} \cos \left(\omega \mathrm{t}-\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)$
$\mathrm{B}_{0} \frac{\hat{\mathrm{j}}-\hat{\mathrm{i}}}{\sqrt{2}} \cos \left(\omega \mathrm{t}+\mathrm{k} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)$
Solution
Direction of polarisation $=\hat{\mathrm{E}}=\hat{\mathrm{k}}$
Direction of propagation $=\hat{\mathrm{E}} \times \hat{\mathrm{B}}=\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}$
$\therefore \hat{\mathrm{E}} \times \hat{\mathrm{B}}=\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}$
$\hat{\mathrm{B}}=\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}}$
