NCERT Exemplar · Class 12 Physics Electromagnetic Waves

(d) \(\displaystyle E_x\), \(\displaystyle E_y\) are in phase, so \(\displaystyle \mathbf{E}\) keeps one direction: plane polarised, not circular. (a) and (b) are identical and differ from the \(\displaystyle \mathbf{B}\) below.\[\mathbf{B}=\frac{1}{c}\,\hat{\mathbf{k}}\times\mathbf{E},\qquad \hat{\mathbf{k}}\times\hat{\mathbf{i}}=\hat{\mathbf{j}},\quad \hat{\mathbf{k}}\times\hat{\mathbf{j}}=-\hat{\mathbf{i}} \] \[\mathbf{B}=\frac{1}{c}\left(E_1\hat{\mathbf{j}}-E_2\hat{\mathbf{i}}\right)\cos(kz-\omega t) \] \[\mathbf{E}\cdot\mathbf{B}=\frac{1}{c}\left(-E_1E_2+E_2E_1\right)\cos^{2}(kz-\omega t)=0 \] \[\text{(a), (b): } B_x=\frac{E_1}{c}\cos(kz-\omega t),\ \ B_y=-\frac{E_2}{c}\cos(kz-\omega t)\ \neq\ \text{derived }\mathbf{B}\ \text{unless } E_1=-E_2 \] \[\tan\theta=\frac{E_y}{E_x}=\frac{E_2}{E_1}=\text{constant}\ \Rightarrow\ \text{plane polarised} \] \[\text{circular: } E_x=E_0\cos(kz-\omega t),\ E_y=E_0\sin(kz-\omega t)\ \text{(not the case here)} \]

(b), (c) Both fields are transverse, and \(\displaystyle \mathbf{E}=c(\mathbf{B}\times\hat{\mathbf{k}})\) follows from \(\displaystyle \mathbf{B}\). (a) fails: \(\displaystyle 1/\omega\) is not \(\displaystyle 1/c\). (d) fails: \(\displaystyle \mathbf{E},\mathbf{B}\) are not parallel to \(\displaystyle \hat{\mathbf{k}}\).\[\mathbf{B}=\frac{1}{c}\,\hat{\mathbf{k}}\times\mathbf{E} \] \[\mathbf{B}\times\hat{\mathbf{k}}=\frac{1}{c}(\hat{\mathbf{k}}\times\mathbf{E})\times\hat{\mathbf{k}}=\frac{1}{c}\left[\mathbf{E}(\hat{\mathbf{k}}\cdot\hat{\mathbf{k}})-\hat{\mathbf{k}}(\mathbf{E}\cdot\hat{\mathbf{k}})\right]=\frac{\mathbf{E}}{c}\ \Rightarrow\ \mathbf{E}=c\,(\mathbf{B}\times\hat{\mathbf{k}}) \] \[\nabla\cdot\mathbf{E}=0\ \Rightarrow\ \hat{\mathbf{k}}\cdot\mathbf{E}=0,\qquad \hat{\mathbf{k}}\cdot\mathbf{B}=\frac{1}{c}\,\hat{\mathbf{k}}\cdot(\hat{\mathbf{k}}\times\mathbf{E})=0 \] \[\frac{1}{\omega}\neq\frac{1}{c}\ \ (\omega=ck);\qquad \frac{1}{\omega}(\mathbf{k}\times\mathbf{E})=\frac{k}{\omega}(\hat{\mathbf{k}}\times\mathbf{E})=\frac{1}{c}(\hat{\mathbf{k}}\times\mathbf{E})\ \ \text{only for the wave vector } \mathbf{k}=k\hat{\mathbf{k}} \]

(b), (d) Both fields are transverse (no \(\displaystyle x\) component) and \(\displaystyle \mathbf{E}\times\mathbf{B}\) points along \(\displaystyle x\). (a), (c) have a field along \(\displaystyle x\).\[E_x=0,\quad B_x=0\ \Rightarrow\ \text{(a), (c) rejected} \] \[\mathbf{B}=\frac{1}{c}\,\hat{\mathbf{i}}\times\mathbf{E}\quad(\text{wave along }+x) \] \[\text{(b): } \hat{\mathbf{i}}\times E_y\hat{\mathbf{j}}=E_y\hat{\mathbf{k}}\ \Rightarrow\ B_z=+\frac{E_y}{c} \] \[\text{(d): } \hat{\mathbf{i}}\times E_z\hat{\mathbf{k}}=-E_z\hat{\mathbf{j}}\ \Rightarrow\ B_y=-\frac{E_z}{c} \]

(a), (c), (d) The wave has the oscillation frequency of the charge. \(\displaystyle 10^{9}\ \mathrm{Hz}=1000\ \mathrm{MHz}\) is the top of the radio-wave band.\[f_{\text{wave}}=f_{\text{charge}}=10^{9}\ \mathrm{Hz} \] \[\lambda=\frac{c}{f}=\frac{3\times10^{8}\ \mathrm{m\,s^{-1}}}{10^{9}\ \mathrm{Hz}}=0.3\ \mathrm{m} \]

(b), (d) Only an accelerated charge radiates. Rest and constant velocity give no acceleration; a circular orbit and a fall in a field do.\[\text{(a), (c): } a=0\ \Rightarrow\ \text{no radiation} \] \[\text{(b): } a=\frac{v^{2}}{r}\neq0\ \Rightarrow\ \text{radiates} \] \[\text{(d): } a=\frac{qE}{m}\neq0\ \Rightarrow\ \text{radiates} \]

(a), (c), (d) Energy \(\displaystyle U\) carries momentum \(\displaystyle U/c\): absorbed, it transfers \(\displaystyle U/c\); reversed, \(\displaystyle 2U/c\). A real surface reflects partly.\[p=\frac{F}{A},\qquad F=\frac{\Delta(\text{momentum})}{\Delta t},\qquad I=\frac{U}{A\,\Delta t} \] \[\text{absorbed: } \Delta(\text{momentum})=\frac{U}{c}\ \Rightarrow\ p=\frac{U}{c\,A\,\Delta t}=\frac{I}{c} \] \[\text{reflected: } \Delta(\text{momentum})=\frac{U}{c}-\left(-\frac{U}{c}\right)=\frac{2U}{c}\ \Rightarrow\ p=\frac{2I}{c} \] \[\text{real surface: } \frac{I}{c}<p<\frac{2I}{c} \]