解析電磁気学・演習

回 (日付)	予習すべき事項	進んだ箇所	演習問題	演習問題略解
1 (4/11)	予習すべき事項はありません。	第1章～2.1.3節まで	第1回演習	例題: 省略 $\fbox{1}$ (a) $\bf{a}_1 = -\frac{13}{25}( 3 \hat{\bf{y}} - 4 \hat{\bf{z}} )$ $\bf{a}_2 = 2 \hat{\bf{x}} - \frac{36}{25} \hat{\bf{y}} - \frac{27}{25} \hat{\bf{z}}$ $\fbox{2}$ 省略
2 (4/18)	2.2節, 2.3節	2.2節, 2.3節	第2回演習	(1) 省略 (2) $\hat{\bf{x}} = \cos\frac{\pi}{6} \hat{\bf{\rho}} - \sin\frac{\pi}{6} \hat{\bf{\phi}}$ $\hat{\bf{y}} = \sin\frac{\pi}{6} \hat{\bf{\rho}} + \cos\frac{\pi}{6} \hat{\bf{\phi}}$ $\fbox{1} \bf{r}(x,y) = x \hat{\bf{x}} + y \hat{\bf{y}} + (2-2x-\frac{y}{2}) \hat{\bf{z}}$ $(0 \leq x \leq 1, 0 \leq y \leq 4-4x)$ $\fbox{2}$ (a) $\bf{A} = \hat{\bf{x}} + \hat{\bf{y}} + 6 \hat{\bf{z}}$ (b) $\hat{\bf{\rho}} = \frac{1}{\sqrt{2}} \hat{\bf{x}} + \frac{1}{\sqrt{2}} \hat{\bf{y}}, \hat{\bf{\phi}} = -\frac{1}{\sqrt{2}} \hat{\bf{x}} + \frac{1}{\sqrt{2}} \hat{\bf{y}}$ (c) $\bf{r}_0 = \sqrt{2} \hat{\bf{\rho}} + \sqrt{6} \hat{\bf{z}}$ (d) $\bf{A}(\bf{r}_0) = \sqrt{2} \hat{\bf{\rho}} + 6 \hat{\bf{z}}$ (e) $\bf{r}_0 = 2\sqrt{2} \hat{\bf{r}}$ (f) $\bf{A}(\bf{r}_0) = ( \frac{1}{\sqrt{2}} + 3\sqrt{3} ) \hat{\bf{r}} + ( \frac{\sqrt{6}}{2} - 3 ) \hat{\bf{\theta}}$
3 (4/25)	2.4節	2.4節	第3回演習	(1) $\frac{1}{a}$ (2) $\frac{4 \pi a^3}{3}$ $\fbox{1} \frac{1}{a}$ $\fbox{2} \frac{1}{6}$
4 (5/2)	2.5-2.7節	2.5-2.7節	第4回演習	(1) $yz$ $\fbox{1} 2y + 4yz + 3x^2 k \cos kz$ $\fbox{2} -y$
5 (5/9)	3.1-3.3節	3.1-3.3節	第5回演習	(1) $\bf{E} = \frac{\lambda_0 (x \hat{\bf{x}} + y \hat{\bf{y}} )}{2 \pi \varepsilon_0 (x^2+y^2)}$ (2) $\bf{E} = \frac{\sigma z}{2 \varepsilon_0} \hat{\bf{z}} \left ( \frac{1}{z} - \frac{1}{\sqrt{z^2+a^2}} \right )$ $\fbox{1} \bf{E} = \frac{Q}{4\pi^2\varepsilon_0(x^2+a^2)^{\frac{3}{2}}} (\pi x \hat{\bf{x}} - 2a \hat{\bf{z}} )$ $\fbox{2} \bf{E} = \frac{Q}{4\pi\varepsilon_0h} \hat{\bf{z}} \left\{ \frac{1}{\sqrt{a^2+(z-\frac{h}{2})^2}} - \frac{1}{\sqrt{a^2+(z+\frac{h}{2})^2}} \right\}$
6 (5/16)	3.4節	3.4節	第6回演習	(1) $\int\int_S \bf{E} \bullet d\bf{S} = \frac{Q}{\varepsilon_0}$ ※閉曲面の積分 (2) $E(\rho) = \frac{\lambda}{2\pi\varepsilon_0\rho}$ $\fbox{1}\int\int_S \bf{E} \bullet d\bf{S} = 0$ ※閉曲面の積分 $\fbox{2} \rho < a$ のとき $E(\rho) = 0$ $\rho \geq a$ のとき $E(\rho)=\frac{a \sigma}{\varepsilon_0 \rho}$
7 (5/23)	3.5節-3.7節	3.5節	第7回演習	(1) $-\int \bf{E} \bullet d\bf{r} = \frac{Q}{4\pi\varepsilon_0a}$ $\fbox{1} -\int \bf{E} \bullet d\bf{r} = \frac{Q}{4\pi\varepsilon_0a}$ $\fbox{2} \|\bf{r}\| > a$ のとき、 $-\int \bf{E} \bullet d\bf{r} = \frac{Q}{4\pi\varepsilon_0 r_0$ $\|\bf{r}\| \leq a$ のとき、 $-\int \bf{E} \bullet d\bf{r} = \frac{3Q}{8\pi\varepsilon_0 a} - \frac{Q r_0^2}{8\pi\varepsilon_0 a^3}$
8 (5/30)	4章	4章	第8回演習	(1) $\frac{Q^2d}{2\varepsilon_0S}$ (2) $\frac{Q^2d}{2\varepsilon_0S}$ $\fbox{1}$ (a) $\frac{Q^2}{8\pi\varepsilon_0} \left ( \frac 1{a_1} - \frac 1{a_2} \right )$ (b) $\frac{Q^2}{8\pi\varepsilon_0} \left ( \frac 1{a_1} - \frac 1{a_2} \right )$ (c) $C \approx \frac{4 \pi \varepsilon_0 a_1^2}d$ $\fbox{2} C=\frac{\pi \varepsilon_0}{\log \frac{b-a}a} \approx \frac{\pi \varepsilon_0}{\log \frac ba}$
9 (6/6)	5章	5章	第9回演習	例題 (1) $\frac 1{\varepsilon_1} \tan\theta_1 = \frac 1{\varepsilon_2} \tan\theta_2$ (2) (a) $C_0 = \frac{\varepsilon_1 \varepsilon_2}{\varepsilon_2 t + \varepsilon_1 (d-t)}$ (b) $0 \leq x \leq t, V(z)=\frac{\varepsilon_2 V_0}{\varepsilon_2 t + \varepsilon_1(d-t)}$ $t < x \leq d, V(z)=\frac{\varepsilon_2 t + \varepsilon_1(x-t)}{\varepsilon_2 t + \varepsilon_1(d-t)}V_0$ $\fbox{1}$ (a) いずれも $D = \frac{Q}{2 \pi \rho L}$ (b) $0 \leq \rho < b, E = \frac{Q}{2 \pi \varepsilon_1 \rho L}$ $b \leq \rho < c, E = \frac{Q}{2 \pi \varepsilon_2 \rho L}$ (c) $C = \frac{2\pi L}{\frac 1{\varepsilon_1} \log \frac ba + \frac 1{\varepsilon_2} \log \frac cb}$ (d) $C = \sigma_p = \frac Q{2 \pi b L} \left ( \frac{\varepsilon_0}{\varepsilon_2} - \frac{\varepsilon_0}{\varepsilon_1} \right )$ $\fbox{2}$ (a) $Q_0 = \frac{2 \pi \varepsilon_0 L}{\log \frac ba} V_0$ (b) $V = \frac L{L + (\varepsilon_r - 1)x} V_0$
10 (6/13)	6.1-6.3	6.1-6.3	第10回演習	例題 (1) $\bf{B} = \frac{\mu_0 I a^2}{2(a^2+z^2)^{3/2}} \hat{\bf{z}}$ (2) $B = \frac{\mu_0 I\rho}{2\pi a^2}, \quad (\rho \leq a)$ $B = \frac{\mu_0 I}{2\pi \rho}, \quad (a < \rho < b)$ $B = 0, \quad (\rho > b)$ $\fbox{1} \bf{B} = \frac{\mu_0 I}{2a} \hat{\bf{z}} \left ( \frac 1{\pi} + \frac{3}{4} \right )$ $\fbox{2} B = \frac{\mu_0 I\rho}{2\pi r_0^2}, \quad (\rho \leq r_0)$ $B = \frac{\mu_0 I}{2\pi \rho}, \quad (r_0 < \rho < R_i)$ $B = \frac{\mu_0 I}{2\pi \rho} \frac{R_o^2-\rho^2}{R_o^2 - R_i^2}, \quad (R_i < \rho < R_o)$ $B = 0, \quad (\rho > R_o)$
11 (6/20)	6.4	6.4	第11回演習	$\fbox{1}$ (a) $\frac{\mu_0 I_0 I}{2\pi} \log\frac{d+2a}d \hat{\bf{z}}$ (b) $\frac{\mu_0 I_0 I a}{\pi} \hat{\bf{y}} \left ( \frac 1d - \frac 1{d+2a} \right )$ $\fbox{2}$ (a) $\bf{v}(t) = v_0 ( \cos\omega_c t \hat{\bf{y}} - \sin \omega_c t \hat{\bf{z}} )$ $\bf{r}(t) = \frac{v_0}{\omega_c} \{ \sin\omega_c t \hat{\bf{y}} + (\cos \omega_c t -1) \hat{\bf{z}} \}, \quad \omega_c = \frac{qB}m$ (b) $\frac{v_0}{\omega_c} \geq d$ (c) $v_0$ (d) $\frac 1{\omega_c} \sin^{-1} \frac{\omega_c d}{v_0}$
12 (6/27)	7章	7章	第12回演習	$\fbox{1}$ ① $I_B \approx 11I_A$ $\fbox{2}$ ⑤ $\Phi = \frac{2NI\mu S}{5l}$ $\fbox{3}$ (a) $\bf{B}_i = B_i cos\theta \hat{\bf{r}} - B_i \sin\theta \hat{\bf{\theta}}$ $\bf{B}_0 = B_0 cos\theta \hat{\bf{r}} - B_0 \sin\theta \hat{\bf{\theta}}$ $\bf{B}_m = \frac{\mu_0 a^3}{3 r^3} M (2 cos\theta \hat{\bf{r}} + \sin\theta \hat{\bf{\theta}})$ (b) $B_i - \frac{2\mu_0}3 M = B_0$ $\frac{B_i}{\mu} + \frac M3 = \frac{B_0}{\mu_0}$ (c) $\bf{B}_i = \frac{3\mu}{\mu + 2\mu_0} B_0 \hat{\bf{z}}$ $\bf{M} = \frac{3(\mu - \mu_0)}{\mu_0(\mu + 2\mu_0)} B_0 \hat{\bf{z}}$
13 (7/4)	8章	8章	第13回演習	$\fbox{1}$ ア. ① イ. ② ウ. ⑧ エ. ⑦ オ. ⑩ $\fbox{2}$ (a) $U = \omega_1 B_{0} ab \sin\omega_1 t$ (b) $U = B_1 \omega_2 ab \cos\omega_2 t$ (c) $U = \omega_2 B_2 ab \cos 2\omega_2 t$ $\fbox{3}$ (a) $\frac{\mu_0 I^2 l}{16\pi}$ (b) $\frac{\mu_0 l}{8 \pi}$ (c) $\frac I{2 \pi z} \hat{\bf{x}} + \frac I{2\pi (d-z)} \hat{\bf{x}}$ (d) $\frac{\mu_0 I}{2\pi} \left ( \frac 1z + \frac 1{d-z} \right )$ (e) $\frac{\mu_0 I l}{\pi} \log \frac{d-a}a$ (f) $\frac{\mu_0 l}{\pi} \log \frac{d-a}a$ (g) $\frac{\mu_0 l}{4\pi} + \frac{\mu_0 l}{\pi} \log \frac{d-a}a$
14 (7/11)	9章	9章	第14回演習	(1) 略 (2) $\frac 1{Z_0} H^2(c_0t - z)$ (3)(a) $\frac 1{Z_0}$ (b) いずれも $\frac t2 \sqrt{\frac{\varepsilon_0}{\mu_0}}$ (c) 略
15 (7/25)	補足資料	補足資料	第15回演習	(1) $\frac{Q}{4\pi\varepsilon_0}\left [ \frac{\rho \hat{\bf{\rho}} + (z-h)\hat{\bf{z}}}{\{ \rho^2 + (z-h)^2 \}^{\frac 32}} - \frac{\rho \hat{\bf{\rho}} + (z+h)\hat{\bf{z}}}{\{ \rho^2 + (z+h)^2 \}^{\frac 32}} \right ]$ (2) $\sigma = -\frac{Qh}{2\pi (\rho^2+h^2)^{\frac 32}}$ (3) $-Q$ (4) $-\frac{Q^2}{4\pi\varepsilon_0(2h)^2} \hat{\bf{z}}$

I科	電子情報システムコース
C科	電磁気学第二・演習(読替)