Question 11.29: Use the duality transformation (Prob. 7.64) to construct the...

\text { With } \alpha=90^{\circ}, \text { Eq. } 7.68 \text { gives } E ^{\prime}=c B , B ^{\prime}=-\frac{1}{c} E , q_{m}^{\prime}=-c q_{6} . Use this to “translate” Eqs. 10.72, 10.73,

E ( r , t)=\frac{q}{4 \pi \epsilon_{0}} \frac{ᴫ}{( ᴫ \cdot u )^{3}}\left[\left(c^{2}-v^{2}\right) u + ᴫ \times( u \times a )\right]                                (10.72)

B ( r , t)=\frac{1}{c} \hat{ᴫ} \times E ( r , t)                       (10.73)

and 11.70:

P=\frac{\mu_{0} q^{2} a^{2}}{6 \pi c}                              (11.70)

E ^{\prime}=c\left(\frac{1}{c} \hat{ᴫ} \times E \right)=\hat{ ᴫ } \times\left(-c B ^{\prime}\right)=-c\left(\hat{ᴫ} \times B ^{\prime}\right).

=-\frac{1}{c} \frac{\left(-q_{m}^{\prime} / c\right)}{4 \pi \epsilon_{0}} \frac{ᴫ}{( ᴫ \cdot u )^{3}}\left[\left(c^{2}-v^{2}\right) u + ᴫ \times( u \times a )\right]=\frac{\mu_{0} q_{m}^{\prime}}{4 \pi} \frac{ᴫ}{( ᴫ \cdot u )^{3}}\left[\left(c^{2}-v^{2}\right) u + ᴫ \times( u \times a )\right].

P=\frac{\mu_{0} a^{2}}{6 \pi c} q_{e}^{2}=\frac{\mu_{0} a^{2}}{6 \pi c}\left(-\frac{1}{c} q_{m}^{\prime}\right)^{2}=\frac{\mu_{0} a^{2}}{6 \pi c^{3}}\left(q_{m}^{\prime}\right)^{2}.

Or, dropping the primes,

B ( r , t)=\frac{\mu_{0} q_{m}}{4 \pi} \frac{ᴫ}{( ᴫ \cdot u )^{3}}\left[\left(c^{2}-v^{2}\right) u + ᴫ \times( u \times a )\right].

E ( r , t)=-c(\hat{ ᴫ } \times B ).

P=\frac{\mu_{0} q_{m}^{2} a^{2}}{6 \pi c^{3}}.