Zadania z kolokwium

1)

Dane: A(2,−2,1)cm, B(4,2,−3)cm, q_A = 3μC, q_B = −5μC, C(−2,0,4)cm

Obliczyć: V_c, ${\overrightarrow{E}}_{C}$, $\left| {\overrightarrow{E}}_{C} \right|$

$$V = \frac{1}{4\pi\varepsilon_{0}} \bullet \frac{q}{r}$$

$$V_{\text{AC}} = \frac{1}{4\pi\varepsilon_{0}} \bullet \frac{q_{A}}{r_{\text{AC}}}$$

$$V_{\text{BC}} = \frac{1}{4\pi\varepsilon_{0}} \bullet \frac{q_{B}}{r_{\text{BC}}}$$

$$V_{C} = V_{\text{AC}} + V_{\text{BC}} = \frac{1}{4\pi\varepsilon_{0}} \bullet \frac{q_{A}}{r_{\text{AC}}} + \frac{1}{4\pi\varepsilon_{0}} \bullet \frac{q_{B}}{r_{\text{BC}}}$$

$${\overrightarrow{r}}_{\text{AC}} = \left\lbrack - 2 - 2,0 + 2,4 - 1 \right\rbrack = \left\lbrack - 4,2,3 \right\rbrack\text{cm}$$

$$r_{\text{AC}} = \left| {\overrightarrow{r}}_{\text{AC}} \right| = \sqrt{{( - 4)}^{2} + 2^{2} + 3^{2}} = \sqrt{29}\text{cm}$$

$${\overrightarrow{r}}_{\text{BC}} = \left\lbrack - 2 - 4,0 - 2,4 + 3 \right\rbrack = \left\lbrack - 6, - 2,7 \right\rbrack\text{cm}$$

$$r_{\text{BC}} = \left| {\overrightarrow{r}}_{\text{BC}} \right| = \sqrt{{( - 6)}^{2} + {( - 2)}^{2} + 7^{2}} = \sqrt{89}\text{cm}$$

$$V_{C} = \frac{1}{4\pi\varepsilon_{0}}\left( \frac{3 \bullet 10^{- 6}}{\sqrt{29} \bullet 10^{- 2}} - \frac{5 \bullet 10^{- 6}}{\sqrt{89} \bullet 10^{- 2}} \right) = \ldots$$

$$\overrightarrow{E} = \frac{1}{4\pi\varepsilon_{0}} \bullet \frac{q}{r^{3}} \bullet \overrightarrow{r}$$

$${\overrightarrow{E}}_{\text{AC}} = \frac{1}{4\pi\varepsilon_{0}} \bullet \frac{q_{A}}{{r^{3}}_{\text{AC}}} \bullet {\overrightarrow{r}}_{\text{AC}}$$

$${\overrightarrow{E}}_{\text{BC}} = \frac{1}{4\pi\varepsilon_{0}} \bullet \frac{q_{A}}{{r^{3}}_{\text{BC}}} \bullet {\overrightarrow{r}}_{\text{BC}}$$

$${\overrightarrow{E}}_{C} = {\overrightarrow{E}}_{\text{AC}} + {\overrightarrow{E}}_{\text{BC}} = \frac{1}{4\pi\varepsilon_{0}}\left( \frac{q_{A}}{{r^{3}}_{\text{AC}}} \bullet {\overrightarrow{r}}_{\text{AC}} + \frac{q_{A}}{{r^{3}}_{\text{BC}}} \bullet {\overrightarrow{r}}_{\text{BC}} \right) = \frac{1}{4\pi\varepsilon_{0}}\left( \frac{3 \bullet 10^{- 6}}{\left( \sqrt{29} \bullet 10^{- 2} \right)^{3}}\left\lbrack - 4,2,3 \right\rbrack \bullet 10^{- 2} - \frac{5 \bullet 10^{- 6}}{\left( \sqrt{89} \bullet 10^{- 2} \right)^{3}}\left\lbrack - 6, - 2,7 \right\rbrack \bullet 10^{- 2} \right) = \frac{1}{4\pi\varepsilon_{0}}\left( \left\lbrack \frac{3}{\left( \sqrt{29} \right)^{3}} \bullet \left( - 4 \right),\frac{3}{\left( \sqrt{29} \right)^{3}} \bullet 2,\frac{3}{\left( \sqrt{29} \right)^{3}} \bullet 3 \right\rbrack \bullet 10^{- 2} - \left\lbrack \frac{5}{\left( \sqrt{89} \right)^{3}} \bullet \left( - 6 \right),\frac{5}{\left( \sqrt{89} \right)^{3}} \bullet \left( - 2 \right),\frac{5}{\left( \sqrt{89} \right)^{3}} \bullet 7 \right\rbrack \bullet 10^{- 2} \right) = \frac{1}{4\pi\varepsilon_{0}}\left\lbrack - \frac{12}{\left( \sqrt{29} \right)^{3}} + \frac{30}{\left( \sqrt{89} \right)^{3}},\frac{6}{\left( \sqrt{29} \right)^{3}} + \frac{10}{\left( \sqrt{89} \right)^{3}},\frac{9}{\left( \sqrt{29} \right)^{3}} - \frac{35}{\left( \sqrt{89} \right)^{3}} \right\rbrack \bullet 10^{- 2}$$

$$\left| \overrightarrow{E} \right| = \frac{10^{- 2}}{4\pi\varepsilon_{0}}\sqrt{\left( - \frac{12}{\left( \sqrt{29} \right)^{3}} + \frac{30}{\left( \sqrt{89} \right)^{3}} \right)^{2} + \left( \frac{6}{\left( \sqrt{29} \right)^{3}} + \frac{10}{\left( \sqrt{89} \right)^{3}} \right)^{2} + \left( \frac{9}{\left( \sqrt{29} \right)^{3}} - \frac{35}{\left( \sqrt{89} \right)^{3}} \right)^{2}}$$

2)

$$i = 20kA,\ q = 2mC,\ V = 50\frac{m}{s}$$

$$\overrightarrow{F} = q\overrightarrow{V} \times \overrightarrow{B}$$

$$B = \frac{\text{μi}}{2\pi r}$$

$${\overrightarrow{B}}_{1} = \left\lbrack 0, - B_{1},0 \right\rbrack$$

$${\overrightarrow{B}}_{2} = \left\lbrack 0,B_{2},0 \right\rbrack$$

$$\overrightarrow{B} = {\overrightarrow{B}}_{1} + {\overrightarrow{B}}_{2} = \left\lbrack 0 + 0, - B_{1} + B_{2},0 + 0 \right\rbrack$$

$$B_{1} = \frac{\mu_{0}i}{2\pi r_{1}} = \frac{4\pi \bullet 10^{- 7} \bullet 20 \bullet 10^{3}}{2\pi \bullet 7} = \frac{40}{7} \bullet 10^{- 4} \cong 0.58 \bullet 10^{- 3}T$$

r₁ = 5m + 2m = 7m

$$B_{2} = \frac{\mu_{0}i}{2\pi r_{2}} = \frac{4\pi \bullet 10^{- 7} \bullet 20 \bullet 10^{3}}{2\pi \bullet 2} = 2 \bullet 10^{- 3}T$$

r₂ = 2m

$$\overrightarrow{B} = \left\lbrack 0,2 \bullet 10^{- 3} - 0.58 \bullet 10^{- 3},0 \right\rbrack T = \left\lbrack 0,1.42 \bullet 10^{- 3},0 \right\rbrack T$$

$$\overrightarrow{V} = \lbrack 50,0,0\rbrack$$

$$\overrightarrow{F} = 2 \bullet 10^{- 3}\left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 50 & 0 & 0 \\ 0 & 1.42 \bullet 10^{- 3} & 0 \\ \end{matrix} \right| = 2 \bullet 10^{- 3}\left\lbrack 0,0,50 \bullet 1.42 \bullet 10^{- 3} \right\rbrack = \left\lbrack 0,0,142 \bullet 10^{- 6} \right\rbrack N = \left\lbrack 0,0,0.142 \right\rbrack\text{mN}$$

3.

$$\lambda = 100\mu\frac{C}{m},\ a = 20cm,\ b = 80cm,\ q = 2\mu C$$

$$E = \frac{1}{2\pi\epsilon} \bullet \frac{\lambda}{r},\ \overrightarrow{F} = q\overrightarrow{E},\ F = \frac{1}{2\pi\epsilon} \bullet \frac{\text{λq}}{r}$$

$$\overrightarrow{F}||\overrightarrow{\text{dl}}$$

$$\overrightarrow{F} \bullet \overrightarrow{\text{dl}} = Fdl\cos 0 = Fdl$$

$$W = \int_{L}^{}{\overrightarrow{F} \bullet \overrightarrow{\text{dl}}} = \int_{L}^{}\text{Fdl} = \int_{a}^{b}{\frac{1}{2\pi\epsilon} \bullet \frac{\text{λq}}{r} \bullet dr = \frac{\text{λq}}{2\pi\epsilon}}\int_{a}^{b}{\frac{1}{r}dr = \frac{\text{λq}}{2\pi\epsilon}\ln{\left| r \right|\left| \frac{b}{a} \right.\ = \frac{\text{λq}}{2\pi\epsilon}}}\left( \ln b - \ln a \right) = \frac{\text{λq}}{2\pi\epsilon}\ln{\left| r \right|\left| \frac{b}{a} \right.\ = \frac{\text{λq}}{2\pi\epsilon}}\ln{\frac{b}{a} = \frac{100 \bullet 10^{- 6} \bullet 2 \bullet 10^{- 6}}{2\pi \bullet \frac{1}{36\pi} \bullet 10^{- 6}}\ln{\frac{0.8}{0.2} = 3600 \bullet 10^{- 6}\ln 4 = 3.6}}\ln 4 \bullet 10^{- 3}J$$

3. Inna wersja

r_z = x − d

$$\overrightarrow{E_{1}} = \begin{bmatrix} E_{1} \\ 0 \\ 0 \\ \end{bmatrix}$$

$$\overrightarrow{E_{2}} = \begin{bmatrix} E_{2} \\ 0 \\ 0 \\ \end{bmatrix}$$

$$\overrightarrow{E} = \begin{bmatrix} E_{1} + E_{2} \\ 0 \\ 0 \\ \end{bmatrix}$$

$$E_{1} = \frac{1}{2\pi\epsilon} \bullet \frac{\lambda_{1}}{r_{1}}$$

$$E_{2} = \frac{1}{2\pi\epsilon} \bullet \frac{\lambda_{2}}{r_{2}}$$

$$E = \frac{1}{2\pi\epsilon}\left( \frac{\lambda_{1}}{r_{1}} + \frac{\lambda_{2}}{r_{2}} \right)$$

$$W = \int_{L}^{}{\overrightarrow{F} \bullet dl} = \int_{L}^{}\text{Fdl} = \int_{a}^{b}\frac{1}{2\pi\epsilon}\left( \frac{\lambda_{1}}{x} + \frac{\lambda_{2}}{x - d} \right)dx = \frac{1}{2\pi\epsilon}\left( \lambda_{1}\int_{a}^{b}\frac{1}{x}dx + \lambda_{2}\int_{a}^{b}{\frac{1}{x - d}\text{dx}} \right) = \frac{1}{2\pi\epsilon}\left( \lambda_{1}\ln{x\left| \frac{b}{a} \right.\ } + \lambda_{2}\ln{x - d\left| \frac{b}{a} \right.\ } \right) = \frac{1}{2\pi\epsilon}\left( \lambda_{1}\ln\frac{b}{a} + \lambda_{2}\ln\frac{b - d}{a - d} \right)$$

$$\overrightarrow{E} = \frac{1}{2\pi\epsilon}\left( \frac{\lambda_{1}}{x} + \frac{\lambda_{2}}{x - d} \right)$$

$$\overrightarrow{\text{dl}} = \begin{bmatrix} \text{dt} \\ \text{adt} \\ 0 \\ \end{bmatrix}$$

$$\overrightarrow{F} \bullet \overrightarrow{\text{dl}} = \begin{bmatrix} F \\ 0 \\ 0 \\ \end{bmatrix} \bullet \begin{bmatrix} \text{dt} \\ \text{adt} \\ 0 \\ \end{bmatrix} = \begin{bmatrix} \text{Fdt} \\ 0 \\ 0 \\ \end{bmatrix}$$

$$\int_{L}^{}{\begin{bmatrix} \text{qE} \\ 0 \\ 0 \\ \end{bmatrix} \bullet \begin{bmatrix} \text{dt} \\ 0 \\ 0 \\ \end{bmatrix}} = \int_{L}^{}\text{qEdt} = \int_{a}^{b}{q\frac{1}{2\pi\epsilon}\frac{\lambda}{t}\text{dt}}$$

Zadanie przykładowe kolokwium nr.2

1)

Dane: Szukane:

I_m = 20A E = ?

f = 50Hz

ω = 2πf

a = 50cm

b = 30cm

c = 10cm

μ = μ₀

$$E = - \frac{d\Phi_{B}}{\text{dt}}$$

$$\oint_{}^{}{\overrightarrow{E} \bullet \overrightarrow{\text{dl}}} = - \frac{d}{\text{dt}}\iint_{s}^{}{\overrightarrow{B} \bullet \overrightarrow{\text{ds}}}$$

$$\Phi_{B} = \iint_{s}^{}{\overrightarrow{B} \bullet \overrightarrow{\text{ds}}}$$

/* Zwroty B₁ i B₂ się dodają, bo z przewodów i₁ i i₂ zgodnie z reg. śruby prawoskrętnej linie idą w tą samą stronę */

/* wektor ds skierowany zgodnie z osią Z */

$$B = \frac{\mu_{0}i}{2\pi r}$$

$${\overrightarrow{B}}_{1} = \begin{bmatrix} 0 \\ 0 \\ - B_{1} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ - \frac{\mu_{0}i_{1}}{2\pi r_{1}} \\ \end{bmatrix}$$

/* −B₁ bo przeciwnie do osi Z, za kartkę */

$${\overrightarrow{B}}_{2} = \begin{bmatrix} 0 \\ 0 \\ - B_{2} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ - \frac{\mu_{0}i_{2}}{2\pi r_{2}} \\ \end{bmatrix}$$

$$\overrightarrow{B} = {\overrightarrow{B}}_{1} + {\overrightarrow{B}}_{2} = \begin{bmatrix} 0 \\ 0 \\ - \frac{\mu_{0}i_{1}}{2\pi r_{1}} - \frac{\mu_{0}i_{2}}{2\pi r_{2}} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ - \frac{\mu_{0}}{2\pi}\left( \frac{i_{1}}{r_{1}} + \frac{i_{2}}{r_{2}} \right) \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ - \frac{\mu_{0}}{2\pi}\left( \frac{i_{1}}{y} + \frac{i_{2}}{d - y} \right) \\ \end{bmatrix}$$

r₁ = y,  r₂ = d − y

$$\overrightarrow{\text{ds}} = \begin{bmatrix} 0 \\ 0 \\ \text{ds} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \text{dxdy} \\ \end{bmatrix}$$

ds = dxdy

$$\Phi_{B} = \iint_{s}^{}{\begin{bmatrix} 0 \\ 0 \\ - \frac{\mu_{0}}{2\pi}\left( \frac{i_{1}}{y} + \frac{i_{2}}{d - y} \right) \\ \end{bmatrix} \bullet \begin{bmatrix} 0 \\ 0 \\ \text{dxdy} \\ \end{bmatrix}} = \iint_{s}^{}{\left( - \frac{\mu_{0}}{2\pi}\left( \frac{i_{1}}{y} + \frac{i_{2}}{d - y} \right) \right)\text{dxdy}} = - \frac{\mu_{0}}{2\pi}\left( \iint_{s}^{}{\frac{i_{1}}{y}\text{dxdy}} + \iint_{s}^{}{\frac{i_{2}}{\text{dy}}\text{dxdy}} \right) = - \frac{\mu_{0}}{2\pi}\left( i_{1}\iint_{s}^{}{\frac{1}{y}\text{dxdy}} + i_{2}\iint_{s}^{}{\frac{1}{\text{dy}}\text{dxdy}} \right)\ldots$$

$$\iint_{s}^{}{\frac{1}{y}\text{dxdy}} = \int_{c}^{c + b}{\left( \int_{0}^{a}{\frac{1}{y}\text{dx}} \right)\text{dy}} = \int_{c}^{c + b}{\frac{1}{y}x\left| \frac{a}{0} \right.\ \text{dy}} = \int_{c}^{c + b}{\left( a - 0 \right)\text{dy}} = \int_{c}^{c + b}{a\frac{1}{y}\text{dy}} = a\ln\left| y \right|\left| \frac{c + b}{c} \right.\ = a\left\lbrack \ln{\left( c + b \right) - \ln c} \right\rbrack = a\ln\frac{c + b}{c}$$

$$\iint_{s}^{}{\frac{1}{d - y}\text{dxdy}} = \int_{c}^{c + b}{\left( \int_{0}^{a}{\frac{1}{d - y}\text{dx}} \right)\text{dy}} = \int_{c}^{c + b}{\frac{1}{d - y}x\left| \frac{a}{0} \right.\ \text{dy}} = \int_{c}^{c + b}{\frac{1}{d - y}\text{ady}} = - a\int_{c}^{c + b}\frac{- 1}{d - y} = - a\ln\left( d - y \right)\left| \frac{c + b}{c} \right.\ = - a\left( \ln{\left( d - c - b \right) - \ln\left( d - c \right)} \right) = - a\ln\frac{d - c - b}{d - c} = a\ln\frac{d - c}{d - c - b}$$

$$\ldots = - \frac{\mu_{0}}{2\pi}\left( i_{1}a\ln\frac{c + b}{c} + i_{2}a\ln\frac{d - c}{d - c - b} \right) = - \frac{\mu_{0}a}{2\pi}\left( I_{m}\sin\text{ωt} \bullet \ln{\frac{c + b}{c} +}I_{m}\cos\text{ωt} \bullet \ln\frac{d - c}{d - c - b} \right)$$

$$E = - \frac{d\Phi_{B}}{\text{dt}} = \frac{\mu_{0}}{2\pi}\left( I_{m}\omega\ln\frac{c + b}{c} \bullet \cos\text{ωt} - I_{m}\omega\ln\frac{d - c}{d - c - b} \bullet \sin\text{ωt} \right)\lbrack V\rbrack$$

1.Inny przykład

$$d = - \frac{d\Phi_{B}}{\text{dt}}$$

$$\Phi_{B} = \iint_{s}^{}{\overrightarrow{B} \bullet \overrightarrow{\text{ds}}}$$

$$\overrightarrow{B} = \frac{\mu_{0}i}{2\pi r}$$

$$B_{1} = \frac{\mu_{0}i_{1}}{2\pi r_{1}}$$

$$B_{2} = \frac{\mu_{0}i_{2}}{2\pi r_{2}}$$

$$\overrightarrow{B} = {\overrightarrow{B}}_{1} \times {\overrightarrow{B}}_{2} = \left\lbrack 0,0, - B_{1} + B_{2} \right\rbrack = \left\lbrack 0,0, - \frac{\mu_{0}i_{1}}{2\pi r_{1}} + \frac{\mu_{0}i_{2}}{2\pi r_{2}} \right\rbrack = \left\lbrack 0,0,\frac{\mu_{0}}{2\pi}\left( - \frac{i_{1}}{x} + \frac{i_{2}}{y} \right) \right\rbrack$$

$$s:\left\{ \begin{matrix} 5cm \leq x \leq 25cm \\ 5cm \leq y \leq 15cm \\ \end{matrix} \right.\ $$

$\overrightarrow{\text{ds}}\bot s$ds prostopadłe do powierzchni całkowania

$$\overrightarrow{\text{ds}} = \left\lbrack 0,0,ds \right\rbrack = \left\lbrack 0,0,dxdy \right\rbrack$$

$$\overrightarrow{B} \bullet \overrightarrow{\text{ds}} = \frac{\mu_{0}}{2\pi}\left( - \frac{i_{1}}{x} + \frac{i_{2}}{y} \right)\text{ds}$$

$$\Phi_{B} = \iint_{s}^{}{\frac{\mu_{0}}{2\pi}\left( - \frac{i_{1}}{x} + \frac{i_{2}}{y} \right)\text{dxdy}} = \frac{\mu_{0}}{2\pi}\int_{d}^{b + d}{\int_{c}^{a + c}{\left( - \frac{i_{1}}{x} + \frac{i_{2}}{y} \right)\text{dxdy}}} = \frac{\mu_{0}}{2\pi}\int_{d}^{b + d}\left( - i_{1}\ln x \right) + i\frac{i_{2}}{y}{\lbrack x\rbrack}_{c}^{a + c}dy = \frac{\mu_{0}}{2\pi}\int_{d}^{b + d}{\left( i_{1}\ln\frac{c}{a + c} + \frac{i_{2}a}{y} \right)\text{dy}} = \frac{\mu_{0}}{2\pi}\left( i_{1}\ln\frac{c}{a + c} \right)y + i_{2}a\ln y\left\| \frac{b + d}{d} \right.\ = \frac{\mu_{0}}{2\pi}\left( i_{1}\left( b + d \right)\ln\frac{c}{a + c} + i_{2}a\ln\left( b + d \right) - i_{1}d\ln\frac{c}{a + c} - i_{2}a\ln d \right) = \frac{\mu_{0}}{2\pi}\left( i_{1}b\ln{\frac{c}{a + c} + i_{2}a\ln\frac{b + d}{d}} \right)$$

$$\Phi_{B} = \frac{\mu_{0}}{2\pi}\left( i_{1}b\ln{\frac{c}{a + c} + i_{2}a\ln\frac{b + d}{d}} \right)$$

$$E = - \frac{d\Phi_{B}}{\text{dt}} = \frac{\mu_{0}}{2\pi}\left( I_{m1}\omega_{1}\cos\omega_{1}t \bullet b\ln\frac{c}{a + c} - I_{m2}\omega_{2}\sin{\omega_{2}t} \bullet b\ln\frac{b + d}{d} \right) = \frac{4\pi \bullet 10^{- 7}}{2}\left( 10 \bullet 2\pi \bullet 50 \bullet 0.1\ln\frac{0.05}{0.25} \bullet \cos\left( 25 \bullet 50 \bullet t \right) - 20 \bullet 100 \bullet 0.2 \bullet \ln\frac{0.15}{0.05}\sin{} \right)$$

2)

R = 10cm = 0.1

y = ax² − 0.2

y = 0, x = 0.1

ax² = 0.2

a(0.1)² = 0.2

a = 20

y = 20x² − 0.2

$$\overrightarrow{B} = \frac{\mu_{i}}{4\pi}\int_{L}^{}\frac{\overrightarrow{\text{dl}} \times \overrightarrow{r}}{r^{3}}$$

$$L_{1}:\left\{ \begin{matrix} x^{'} = R\cos\varphi \\ y^{'} = R\sin\varphi \\ z^{'} = 0 \\ \end{matrix} \right.\ $$

α = −φ → φ = −α

/* φ ma przeciwny kierunek od kierunku prądu */

$$L_{1}:\left\{ \begin{matrix} x^{'} = R\cos{( - \alpha)} \\ y^{'} = R\sin{( - \alpha)} \\ z^{'} = 0 \\ \end{matrix} \right.\ $$

$$L_{1}:\left\{ \begin{matrix} x^{'} = R\cos\alpha \\ y^{'} = - R\sin\alpha \\ z^{'} = 0 \\ \end{matrix} \right.\ $$

−π ≤ α ≤ 0

$$L_{2}:\left\{ \begin{matrix} x^{'} = - t \\ y^{'} = 20t^{2} - 0.2 \\ z^{'} = 0 \\ \end{matrix} \right.\ $$

−0.1m ≤ t ≤ 0.1m

$$\overrightarrow{B} = \frac{\mu_{i}}{4\pi}\int_{L_{1}}^{}\frac{{\overrightarrow{\text{dl}}}_{1} \times {\overrightarrow{r}}_{1}}{{r^{3}}_{1}} + \frac{\mu_{i}}{4\pi}\int_{L_{2}}^{}\frac{{\overrightarrow{\text{dl}}}_{2} \times {\overrightarrow{r}}_{2}}{{r^{3}}_{2}}$$

$${\overrightarrow{\text{dl}}}_{1} = \begin{bmatrix} dx' \\ dy' \\ dz' \\ \end{bmatrix} = \begin{bmatrix} - R\sin\alpha\text{dα} \\ - R\cos\alpha\text{dα} \\ 0 \\ \end{bmatrix}$$

$${\overrightarrow{r}}_{1} = \left\lbrack x - x^{'},y - y^{'},z - z' \right\rbrack = \left\lbrack x - R\cos\alpha,y + R\sin\alpha,z \right\rbrack$$

$$r_{1} = \sqrt{\left( x - R\cos\alpha \right)^{2} + \left( y + R\sin\alpha \right)^{2} + z^{2}}$$

$${\overrightarrow{\text{dl}}}_{1} \times {\overrightarrow{r}}_{1} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - R\sin\alpha\text{dα} & - R\cos\alpha\text{dα} & 0 \\ x - R\cos\alpha & y + R\sin\alpha & z \\ \end{matrix} \right| = \begin{bmatrix} - R_{z}\cos\text{αdα} \\ R_{z}\sin\text{αdα} \\ - R\sin\text{αdα}\left( y + R\sin\alpha \right) + R\cos\alpha\left( x - R\cos\alpha \right)\text{dα} \\ \end{bmatrix} = \begin{bmatrix} - R_{z}\cos\text{αdα} \\ R_{z}\sin\text{αdα} \\ \left( - R_{y}\sin\alpha - R^{2}\operatorname{}\alpha + R_{x}\cos\alpha - R^{2}\operatorname{}\alpha \right)\text{dα} \\ \end{bmatrix} = \begin{bmatrix} - R_{z}\cos\text{αdα} \\ R_{z}\sin\text{αdα} \\ \left( R_{x}\cos\alpha - R_{y}\sin\alpha - R^{2} \right)\text{dα} \\ \end{bmatrix}$$

$${\overrightarrow{B}}_{1} = \frac{\mu_{i}}{4\pi}\int_{L_{1}}^{}\frac{\begin{bmatrix} - R_{z}\cos\text{αdα} \\ R_{z}\sin\text{αdα} \\ \left( R_{x}\cos\alpha - R_{y}\sin\alpha - R^{2} \right)\text{dα} \\ \end{bmatrix}}{\left( \sqrt{\left( x - R\cos\alpha \right)^{2} + \left( y + R\sin\alpha \right)^{2} + z^{2}} \right)^{3}}$$

$$B_{1x} = \frac{\mu_{i}}{4\pi}\int_{- \pi}^{0}\frac{- R_{z}\cos\text{αdα}}{\left( \sqrt{\left( x - R\cos\alpha \right)^{2} + \left( y + R\sin\alpha \right)^{2} + z^{2}} \right)^{3}}$$

$$B_{1y} = \frac{\mu_{i}}{4\pi}\int_{- \pi}^{0}\frac{R_{z}\sin\text{αdα}}{\left( \sqrt{\left( x - R\cos\alpha \right)^{2} + \left( y + R\sin\alpha \right)^{2} + z^{2}} \right)^{3}}$$

$$B_{1z} = \frac{\mu_{i}}{4\pi}\int_{- \pi}^{0}\frac{\left( R_{x}\cos\alpha - R_{y}\sin\alpha - R^{2} \right)\text{dα}}{\left( \sqrt{\left( x - R\cos\alpha \right)^{2} + \left( y + R\sin\alpha \right)^{2} + z^{2}} \right)^{3}}$$

${\overrightarrow{\text{dl}}}_{2} = \begin{bmatrix} - dt \\ 40tdt \\ 0 \\ \end{bmatrix}$, ${\overrightarrow{r}}_{2} = \begin{bmatrix} x + t \\ y - 20t^{2} + 0.2 \\ z \\ \end{bmatrix}$

$${\overrightarrow{\text{dl}}}_{2} \times {\overrightarrow{r}}_{2} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - dt & 40tdt & 0 \\ x + t & y - 20t^{2} + 0.2 & z \\ \end{matrix} \right| = \begin{bmatrix} 40ztdt \\ \text{zdt} \\ \left( - \left( y - 20t^{2} + 0.2 \right) - 40t\left( x + t \right) \right)\text{dt} \\ \end{bmatrix}$$

$${\overrightarrow{B}}_{2} = \frac{\mu_{i}}{4\pi}\int_{L_{2}}^{}\frac{{\overrightarrow{\text{dl}}}_{2} \times {\overrightarrow{r}}_{2}}{{r^{3}}_{2}}$$

/* B_2x, B_2y, B_2z analogicznie */

B_x = B_1x + B_2x

B_y = B_1y + B_2y

B_z = B_1z + B_2z

2. Inny przykład

$$L_{1}:\left\{ \begin{matrix} 0 \leq x' \leq 20 \\ y^{'} = 0 \\ z^{'} = 0 \\ \end{matrix} \right.\ \rightarrow L_{1}:\left\{ \begin{matrix} x^{'} = t \\ y^{'} = 0 \\ \frac{z^{'} = 0}{0 \leq t \leq 20cm} \\ \end{matrix} \right.\ $$

L₂:

y = ax² + bx + c

y = a(x − x₁)(x − x₂)

y = a(x − 0)(x − 20)

y = ax(x − 20)

40 = 10a(10 − 20)

40 = −100a

$$a = - \frac{40}{100} = - \frac{2}{5}$$

$$y = \left( - \frac{2}{5} \right)x^{2} - 20\left( - \frac{2}{5} \right)x$$

$$y = - \frac{2}{5}x^{2} + 8x$$

$$L_{2}:\left\{ \begin{matrix} x^{'} = s \\ y^{'} = - \frac{2}{5}s^{2} + 0.5 \\ \frac{z^{'} = 0}{- 20cm \leq 5 \leq 0} \\ \end{matrix} \right.\ $$

$$\overrightarrow{A} = \frac{\mu_{i}}{4\pi}\int_{L_{1}}^{}\frac{{\overrightarrow{\text{dl}}}_{1}}{r_{1}} + \frac{\mu_{i}}{4\pi}\int_{L_{2}}^{}\frac{{\overrightarrow{\text{dl}}}_{2}}{{\overrightarrow{r}}_{2}}$$

$${\overrightarrow{\text{dl}}}_{1} = \left\lbrack dx^{'},dy^{'},dz' \right\rbrack$$

dx^′ = 1dt

dy^′ = 0

dz^′ = 0

$${\overrightarrow{\text{dl}}}_{1} = \left\lbrack dt,0,0 \right\rbrack$$

$${\overrightarrow{r}}_{1} = \left\lbrack x - t,y,z \right\rbrack$$

$$r_{1} = \sqrt{\left( x - t \right)^{2} + y^{2} + z^{2}}$$

$${\overrightarrow{A}}_{1} = \frac{\mu_{i}}{4\pi}\int_{L_{1}}^{}\frac{\left\lbrack dt,0,0 \right\rbrack}{\sqrt{\left( x - t \right)^{2} + y^{2} + z^{2}}}$$

$$A_{1x} = \frac{\mu_{i}}{4\pi}\int_{0}^{0.2}\frac{\text{dt}}{\sqrt{\left( x - t \right)^{2} + y^{2} + z^{2}}}$$

A_1y = 0

A_1z = 0

$${\overrightarrow{\text{dl}}}_{2} = \left\lbrack dx^{'},dy^{'},dz' \right\rbrack$$

$${\overrightarrow{\text{dl}}}_{2} = \left\lbrack - ds,\left( - \frac{2}{5} \bullet 2s - 8 \right)ds,0 \right\rbrack$$

$${\overrightarrow{\text{dl}}}_{2} = \left\lbrack - ds, - \left( \frac{4}{5}s + 8 \right)ds,0 \right\rbrack$$

$$r_{2} = \sqrt{\left( x + s \right)^{2} + \left( y - \left( - \frac{2}{5}s^{2} - 8s \right) \right)^{2} + z^{2}}$$

$$r_{2} = \sqrt{\left( x + s \right)^{2} + \left( y + \frac{2}{5}s^{2} + 8s \right)^{2} + z^{2}}$$

$$A_{2} = \frac{\mu_{i}}{4\pi}\int_{L_{2}}^{}\frac{\left\lbrack - ds, - \left( \frac{4}{5}s + 8ds \right),0 \right\rbrack}{\sqrt{\left( x + s \right)^{2} + \left( y + \frac{2}{5}s^{2} + 8s \right)^{2} + z^{2}}}$$

2. Inny przykład

y = ax²

x = −1,  y = 1

1 = a • 1²

a = 1

y = ax + b

$$a = \frac{3}{3} = 1$$

b = 2

$$L_{1}:\left\{ \begin{matrix} x^{'} = t \\ y^{'} = t^{2} \\ \frac{z^{'} = 0}{- 1 \leq t \leq 2} \\ \end{matrix} \right.\ $$

$$L_{2}:\left\{ \begin{matrix} x^{'} = t - 5 \\ y^{'} = - 5 + 2 \\ z^{'} = 0 \\ \end{matrix} \right.\ $$

$$\overrightarrow{B} = \frac{\mu_{0}i}{4\pi}\int_{L}^{}\frac{\overrightarrow{\text{dl}} \times \overrightarrow{r}}{r^{3}}$$

$$\overrightarrow{B} = \frac{\mu_{0}i}{4\pi}\int_{L_{1}}^{}\frac{\overrightarrow{\text{dl}} \times {\overrightarrow{r}}_{1}}{{r_{1}}^{3}} + \frac{\mu_{0}i}{4\pi}\int_{L_{2}}^{}\frac{\overrightarrow{\text{dl}} \times {\overrightarrow{r}}_{2}}{{r_{2}}^{3}}$$

$${\overrightarrow{\text{dl}}}_{1} = \left\lbrack dx^{'},dy^{'},dz' \right\rbrack = \left\lbrack dt,2tdt,0 \right\rbrack$$

$${\overrightarrow{\text{dl}}}_{2} = \left\lbrack - ds, - ds,0 \right\rbrack = \left\lbrack - 1, - 1,0 \right\rbrack$$

$${\overrightarrow{r}}_{1} = \left\lbrack x - x^{'},y - y^{'},\ z - z' \right\rbrack = \left\lbrack x - t,y - t^{2},z \right\rbrack$$

$${\overrightarrow{r}}_{2} = \left\lbrack x + 5,y + 5 - 2,z \right\rbrack$$

$$r_{1} = \sqrt{\left( x - t \right)^{2} + \left( y - t^{2} \right)^{2} + z^{2}}$$

$$r_{2} = \sqrt{\left( x + 5 \right)^{2} + \left( y + 5 - 2 \right)^{2} + z^{2}}$$

$$B_{1} = \frac{\mu_{0}i}{4\pi}\int_{L_{1}}^{}\frac{\left\lbrack 2ztdt, - zdt,\left( y^{2} - 2txt\text{dt}^{2} \right)\text{dt} \right\rbrack}{\sqrt{\left( x - t \right)^{2} + \left( y - t^{2} \right)^{2} + z^{2}}}$$

$${\overrightarrow{\text{dl}}}_{1} \times {\overrightarrow{r}}_{1} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \text{dt} & 2tdt & 0 \\ x - t & y - t^{2} & z \\ \end{matrix} \right| = \begin{bmatrix} 2ztdt \\ - zdt \\ \left( y - t^{2} \right)dt - 2t\left( x - t \right)\text{dt} \\ \end{bmatrix}$$

$$B_{1x} = \frac{\mu_{0}i}{4\pi}\int_{- 1}^{2}\frac{2ztdt}{\left( \sqrt{\left( x - t \right)^{2} + \left( y - t^{2} \right)^{2} + z^{2}} \right)^{3}}$$

itd…

3. W polu elektromagnetycznym porusza się ładunek q równolegle do osi 0Z z prędkością V. Oblicz się działającą na ten ładunek

$$\left\{ \begin{matrix} \frac{\overrightarrow{E} = \left\lbrack E_{x}\left( z,t \right),\ 0,\ 0 \right\rbrack}{E_{x}\left( z,t \right) = 1500\sin\left( 6 \bullet 10^{- 6} - 1800t \right)\lbrack T\rbrack} \\ \frac{\overrightarrow{B} = \left\lbrack 0,B_{y}\left( z,y \right),0 \right\rbrack}{B_{y}\left( z,y \right) = S\sin\left( 6 \bullet 10^{- 6} - 1800t \right)\lbrack T\rbrack} \\ \end{matrix} \right.\ $$

$\overrightarrow{E}$ – pole elektryczne

$\overrightarrow{B}$ – pole magnetyczne

$\overrightarrow{F} = q\left( \overrightarrow{E} + \overrightarrow{V} \times \overrightarrow{B} \right)$ Siła działająca

$$\overrightarrow{V} = \left\lbrack 0,0,V \right\rbrack$$

$$\overrightarrow{B} = \left\lbrack 0,B_{y}\left( z,t \right),0 \right\rbrack$$

$$\overrightarrow{V} \times \overrightarrow{B} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & V \\ 0 & B_{y}\left( z,t \right) & 0 \\ \end{matrix} \right|\frac{= \hat{i}\left( 0 - VB_{y}\left( z,t \right) \right) + \hat{j}\left( 0 + 0 \right) + \hat{k}(0 - 0)}{= \left\lbrack - VB_{y}\left( z,t \right),0,0 \right\rbrack}$$

$$\overrightarrow{F} = q\left( \left\lbrack E_{x}\left( z,t \right),\ 0,\ 0 \right\rbrack + \left\lbrack - VB_{y}\left( z,t \right),0,0 \right\rbrack \right)$$

3. Sprawdzenie praw Maxwella dla fal elektromagnetycznych

$\overrightarrow{\text{rot}}\overrightarrow{E} = - \frac{d\overrightarrow{B}}{\text{dt}}$Prawo Maxwella-Faraday’a

$$\left\{ \begin{matrix} \overrightarrow{E} = \left\lbrack E_{x},0,0 \right\rbrack \\ E_{x}\left( z,t \right) = 1500\ \sin\left( 62.82z - 18.84 \bullet 10^{9}t \right)\lbrack V/m\rbrack \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} \overrightarrow{B} = \left\lbrack 0,B_{y},0 \right\rbrack \\ B_{y}\left( z,t \right) = S\sin\left( 62.82z - 18.84 \bullet 10^{9}t \right)\lbrack\mu T\rbrack \\ \end{matrix} \right.\ $$

$$1\ \overrightarrow{\text{rot}}\overrightarrow{E} = \overrightarrow{\nabla} \times \overrightarrow{E} = \left\lbrack \frac{d}{\text{dx}},\frac{d}{\text{dy}},\frac{d}{\text{dz}} \right\rbrack \times \left\lbrack E_{x},0,0 \right\rbrack = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{d}{\text{dx}} & \frac{d}{\text{dy}} & \frac{d}{\text{dz}} \\ E_{x} & 0 & 0 \\ \end{matrix} \right| = \hat{i}\left( 0 - 0 \right) + \hat{j}\left( \frac{dE_{x}}{\text{dz}} - 0 \right) + \hat{k}\left( 0 - \frac{d - E_{x}}{\text{dy}} \right)$$

$$\overrightarrow{\text{rot}}\overrightarrow{E} = \left\lbrack 0,\frac{dE_{x}(z,t)}{\text{dz}},\frac{dE_{x}(z,t)}{\text{dy}} \right\rbrack = \left\lbrack 0,\frac{dE_{x}(z,t)}{\text{dz}},0 \right\rbrack$$

$$\frac{d}{\text{dz}}\left( 1500\sin\left( k_{z} - \omega t \right) \right) = 1500\cos\left( k_{z} - \omega t \right) \bullet K = 1500 \bullet 62.8\cos\left( k_{z} - \omega t \right)$$

$$\overrightarrow{\text{rot}}\overrightarrow{E} = \left\lbrack 0,94200\cos\left( k_{z} - \omega t \right),0 \right\rbrack$$

$$2 - \frac{d\overrightarrow{B}}{\text{dt}} = - \frac{d}{\text{dt}}\left\lbrack 0,B_{y},0 \right\rbrack = \left\lbrack 0, - \frac{dB_{y}}{\text{dt}},0 \right\rbrack$$

$$- \frac{d}{\text{dt}}5\sin\left( k_{z} - \omega t \right) \bullet 10^{- 6} = - 5\cos\left( k_{z} - \omega t \bullet \left( - \omega \right) \right) \bullet 10^{- 6} = 5 \bullet 18.84 \bullet 10^{9} \bullet 10^{- 6}\cos\left( k_{z} - \omega t \right) = 94.2 \bullet 10^{2}\cos\left( k_{z} - \omega t \right)$$

$$- \frac{d\overrightarrow{B}}{\text{dt}} = \left\lbrack 0,94200\cos\left( k_{z} - \omega t \right),0 \right\rbrack$$

3 czy 0, 94200cos(k_z−ωt) = 0, 94200cos(k_z−ωt)? 

Tak, prawo jest poprawne i fala spełnia prawo

$\overrightarrow{\text{rot}}\overrightarrow{B} = \mu E\frac{d\overrightarrow{E}}{\text{dt}}$Prawo Maxwella-Ampere’a

$$\left\{ \begin{matrix} \overrightarrow{E} = \left\lbrack E_{x},0,0 \right\rbrack \\ E_{x}\left( z,t \right) - 1500\sin\left( {62.8}_{z} - 18.84 \bullet 10^{9}t \right)\lbrack V/m\rbrack \\ \end{matrix} \right.\ $$

$$\overrightarrow{B} = \left\lbrack 0,B_{y},0 \right\rbrack$$

B_y(z,t) = 5sin(62.8_z−18.84•10⁹t) • 10⁻⁶[T]

$$1\ \overrightarrow{\text{rot}}\overrightarrow{B} = \left\lbrack - \frac{dB_{y}\left( z,t \right)}{\text{dz}},0,\frac{dB_{y}\left( z,t \right)}{\text{dx}} \right\rbrack = \left\lbrack - \frac{dB_{y}\left( z,t \right)}{\text{dz}},0,0 \right\rbrack$$

$$- \frac{dB_{y}\left( z,t \right)}{\text{dz}} = - \frac{d}{\text{dz}}5\sin\left( k_{z} - \omega t \right) \bullet 10^{- 6} = - 5\cos\left( k_{z} - \omega t \right) \bullet K \bullet 10^{- 6} = - 5 \bullet 62.8 \bullet \cos\left( k_{z} - \omega t \right) \bullet 10^{- 6} = - 314\cos\left( k_{z} - \omega t \right) \bullet 10^{- 6}$$

$$\overrightarrow{\text{rot}}\overrightarrow{B} = \left\lbrack - 314\cos\left( k_{z} - \omega t \right) \bullet 10^{- 6},0,0 \right\rbrack$$

$$2\ \mu E\frac{d\overrightarrow{E}}{\text{dt}},\ c = \frac{1}{\sqrt{\text{μE}}} \rightarrow \mu E = \frac{1}{c^{2}}$$

$$\text{μE}\frac{d\overrightarrow{E}}{\text{dt}} = \frac{1}{c^{2}} \bullet \frac{d\overrightarrow{E}}{\text{dt}} = \frac{1}{c^{2}} \bullet \frac{d}{\text{dt}}\left\lbrack E_{x},0,0 \right\rbrack = \left\lbrack \frac{1}{c^{2}} \bullet \frac{d}{\text{dt}}E_{x},0,0 \right\rbrack$$

$$\frac{1}{c^{2}}\frac{d}{\text{dt}}E_{x} = \frac{1}{\left( 9 \bullet 10^{8} \right)^{2}} \bullet \frac{d}{\text{dt}}1500\sin\left( k_{z} - \omega t \right) = \frac{1}{9} \bullet 10^{- 16} \bullet 1500\cos\left( k_{z} - \omega t \right) \bullet \left( - \omega \right) = - \frac{1500 \bullet 18.84 \bullet 10^{9} \bullet 10^{- 16}}{9} \bullet \cos\left( k_{z} - \omega t \right) = - \frac{9140 \bullet 10^{- 7}}{9} \bullet \cos\left( k_{z} - \omega t \right) = - 314 \bullet \cos\left( k_{z} - \omega t \right) \bullet 10^{- 6}$$

3 czy −314 • cos(k_z−ωt) • 10⁻⁶ =   − 314 • cos(k_z−ωt) • 10⁻⁶?

Tak, prawo jest poprawne i fala spełnia prawo

$\text{div}\overrightarrow{B} = 0$Prawo Maxwella-Gauss’a dla magnetyzmu

$$\left\{ \begin{matrix} {\overrightarrow{B}}_{y} = \lbrack 0,B_{y},0\rbrack \\ B_{y}\left( z,t \right) = 5\sin\left( {62.8}_{z} - 18.84 \bullet 10^{7}t \right) \bullet 10^{- 6}\lbrack T\rbrack \\ \end{matrix} \right.\ $$

$$\text{div}\overrightarrow{B} = \overrightarrow{\nabla} \bullet \overrightarrow{B} = \left\lbrack \frac{d}{\text{dx}},\frac{d}{\text{dy}},\frac{d}{\text{dz}} \right\rbrack \bullet \left\lbrack 0,B_{y}\left( z,t \right),0 \right\rbrack = \frac{d_{0}}{\text{dx}} + \frac{dB_{y}\left( z,t \right)}{\text{dy}} + \frac{d_{0}}{\text{dz}} = 0 + \frac{d}{\text{dy}}\left( \sin\left( k_{z} - \omega t \right) \bullet 10^{- 6} \right) + 0 = 0 + 0 = 0$$

czy 0 = 0?

Tak, prawo jest poprawne i fala spełnia prawo