東京工業大学 工学院 電気電子系 2019年8月実施 電磁気学2 Author Zero
Description 直流電流の作る磁界に関する，以下の設問 (1) および (2)に答えよ。
(1) x − y x-y x − y a a a I I I z z z P ( 0 , 0 , z 1 ) P(0,0,z_1) P ( 0 , 0 , z 1 ) B 1 \boldsymbol{B}_1 B 1 ∼ \sim ∼ μ 0 \mu_0 μ 0
「図 2.1 のように円電流の微小線素ベクトル d s d\boldsymbol{s} d s P P P d B d\boldsymbol{B} d B d B z d\boldsymbol{B}_z d B z d B ⊥ d\boldsymbol{B}_{\perp} d B ⊥ d B z = ∣ d B z ∣ dB_z = |d\boldsymbol{B}_z| d B z = ∣ d B z ∣ d B ⊥ d\boldsymbol{B}_{\perp} d B ⊥ ∣ B ⊥ ∣ = ( 1 ◯ ) |\boldsymbol{B}_{\perp}| = (\textcircled{1}) ∣ B ⊥ ∣ = ( 1 ◯ ) d B z dB_z d B z d s d\boldsymbol{s} d s P P P R \boldsymbol{R} R R R R R = ∣ R ∣ R = |\boldsymbol{R}| R = ∣ R ∣ ( 2 ◯ ) (\textcircled{2}) ( 2 ◯ ) × \times × d B = μ 0 4 π I d s ↔ R R 3 d\boldsymbol{B} = \frac{\mu_0}{4\pi}I\frac{d\boldsymbol{s} \leftrightarrow \boldsymbol{R}}{R^3} d B = 4 π μ 0 I R 3 d s ↔ R d s d\boldsymbol{s} d s R \boldsymbol{R} R d B d\boldsymbol{B} d B d B = ∣ d B ∣ dB = |d\boldsymbol{B}| d B = ∣ d B ∣ μ 0 , I , a , z 1 \mu_0,I,a,z_1 μ 0 , I , a , z 1 d s = ∣ d s ∣ ds = |d\boldsymbol{s}| d s = ∣ d s ∣ d B = ( 3 ◯ ) dB = (\textcircled{3}) d B = ( 3 ◯ ) d B d\boldsymbol{B} d B z z z φ \varphi φ a a a z 1 z_1 z 1 cos φ = ( 4 ◯ ) \cos\varphi = (\textcircled{4}) cos φ = ( 4 ◯ ) μ 0 , I , a , z 1 \mu_0,I,a,z_1 μ 0 , I , a , z 1 d s ds d s d B z = ( 5 ◯ ) dB_z = (\textcircled{5}) d B z = ( 5 ◯ ) μ 0 , I , a , z 1 \mu_0,I,a,z_1 μ 0 , I , a , z 1 B 1 \boldsymbol{B}_1 B 1 z z z B z B_z B z B z = ( 6 ◯ ) B_z = (\textcircled{6}) B z = ( 6 ◯ ) B 1 = ( 1 ◯ , 1 ◯ , 6 ◯ ) \boldsymbol{B}_1 = (\textcircled{1}, \textcircled{1}, \textcircled{6}) B 1 = ( 1 ◯ , 1 ◯ , 6 ◯ ) B 1 \boldsymbol{B}_1 B 1
(2) 図 2.2 2.2 2.2 a a a b b b h h h N N N μ \mu μ
(a) コアの内側 ( a < r < b , 0 ≤ θ < 2 π , − h / 2 < z < h / 2 ) (a < r < b,0 \le \theta <2\pi, -h/2 < z <h/2) ( a < r < b , 0 ≤ θ < 2 π , − h /2 < z < h /2 ) 2.2 2.2 2.2 ( r , θ , z ) (r,\theta,z) ( r , θ , z ) J J J
B r = B z = 0 , B θ = μ N J / ( 2 π r ) B_r = B_z = 0,B_{\theta} = \mu NJ/(2\pi r) B r = B z = 0 , B θ = μ N J / ( 2 π r ) となることを示せ。
(b) コアの外側では，コイルに流れる直流電流 J J J
(c) コアの断面 ( a < r < b , 0 ≤ θ < 2 π , − h / 2 < z < h / 2 ) (a < r < b,0 \le \theta <2\pi, -h/2 < z <h/2) ( a < r < b , 0 ≤ θ < 2 π , − h /2 < z < h /2 ) Φ \Phi Φ ln ( x ) \ln(x) ln ( x ) x x x
Φ = μ N J h 2 π ln ( b a ) \Phi = \frac{\mu NJh}{2\pi}\ln\big(\frac{b}{a}\big) Φ = 2 π μ N J h ln ( a b ) Kai (1)
① 0
② ビオ・サバールの法則より、
d B = μ 0 I 4 π r 2 d s = μ 0 I 4 π ( a 2 + z 1 2 ) d s dB = \frac{\mu_0 I}{4\pi r^2}ds = \frac{\mu_0 I}{4\pi(a^2 + z_1^2)}ds d B = 4 π r 2 μ 0 I d s = 4 π ( a 2 + z 1 2 ) μ 0 I d s ③
μ 0 I 4 π ( a 2 + z 1 2 ) d s \frac{\mu_0 I}{4\pi(a^2 + z_1^2)}ds 4 π ( a 2 + z 1 2 ) μ 0 I d s ④
a a 2 + z 1 2 \frac{a}{\sqrt{a^2 + z_1^2}} a 2 + z 1 2 a d B z = μ 0 I 4 π ( a 2 + z 1 2 ) d s ⋅ a a 2 + z 1 2 = μ 0 I a 4 π ( a 2 + z 1 2 ) 3 2 d s \begin{aligned}
dB_z &= \frac{\mu_0 I}{4\pi(a^2 + z_1^2)}ds \cdot \frac{a}{\sqrt{a^2 + z_1^2}} \\
&= \frac{\mu_0 I a}{4\pi(a^2 + z_1^2)^{\frac{3}{2}}}ds
\end{aligned} d B z = 4 π ( a 2 + z 1 2 ) μ 0 I d s ⋅ a 2 + z 1 2 a = 4 π ( a 2 + z 1 2 ) 2 3 μ 0 I a d s ⑤
μ 0 I a 4 π ( a 2 + z 1 2 ) 3 2 d s \frac{\mu_0 I a}{4\pi(a^2 + z_1^2)^{\frac{3}{2}}}ds 4 π ( a 2 + z 1 2 ) 2 3 μ 0 I a d s B z = ∫ 0 2 π μ 0 I a 4 π ( a 2 + z 1 2 ) 3 2 a d θ = μ 0 I a 2 2 ( a 2 + z 1 2 ) 3 2 \begin{aligned}
B_z &= \int_0^{2\pi} \frac{\mu_0 I a}{4\pi(a^2 + z_1^2)^{\frac{3}{2}}}ad\theta \\
&= \frac{\mu_0 I a^2}{2(a^2 + z_1^2)^{\frac{3}{2}}}
\end{aligned} B z = ∫ 0 2 π 4 π ( a 2 + z 1 2 ) 2 3 μ 0 I a a d θ = 2 ( a 2 + z 1 2 ) 2 3 μ 0 I a 2 ⑥
μ 0 I a 2 2 ( a 2 + z 1 2 ) 3 2 \frac{\mu_0 I a^2}{2(a^2 + z_1^2)^{\frac{3}{2}}} 2 ( a 2 + z 1 2 ) 2 3 μ 0 I a 2 (2)
(a)
アンペールの法則より、積分路を円周上にすると、
2 π r B θ = μ N J ⇒ B θ = μ N J 2 π r 2\pi r B_{\theta} = \mu NJ \Rightarrow B_{\theta} = \frac{\mu NJ}{2\pi r} 2 π r B θ = μ N J ⇒ B θ = 2 π r μ N J (b)
積分路 C 1 C_1 C 1
H 3 l − H 4 l = 0 ⇒ H 3 = H 4 H_3 l - H_4 l = 0 \Rightarrow H_3 = H_4 H 3 l − H 4 l = 0 ⇒ H 3 = H 4 積分路 C 2 C_2 C 2
H 1 l − H 2 l = N J l ⇒ H 1 − H 2 = N J H_1 l - H_2 l = NJl \Rightarrow H_1 - H_2 = NJ H 1 l − H 2 l = N J l ⇒ H 1 − H 2 = N J H 4 H_4 H 4 0 0 0 H 3 = 0 S H_3 = 0S H 3 = 0 S
よって、H 2 = 0 H_2 = 0 H 2 = 0 H 1 = N J H_1 = NJ H 1 = N J
コアの外側では，磁界がゼロとなる。
(c)
Φ = B S = h ∫ a b μ N J 2 π r d r = μ N h J 2 π ∫ a b 1 r d r = μ N h J 2 π log b a \begin{aligned}
\Phi &= BS \\
&= h\int_a^b \frac{\mu NJ}{2\pi r}dr \\
&= \frac{\mu NhJ}{2\pi} \int_a^b \frac{1}{r}dr \\
&= \frac{\mu NhJ}{2\pi} \log \frac{b}{a}
\end{aligned} Φ = BS = h ∫ a b 2 π r μ N J d r = 2 π μ N h J ∫ a b r 1 d r = 2 π μ N h J log a b (d)
U = ∭ u 1 2 μ B 2 d u = h ∫ 0 2 π ∫ a b 1 2 μ ⋅ μ 2 N 2 J 2 4 π 2 r 2 r d r d θ = μ h N 2 J 2 8 π 2 ⋅ ∫ 0 2 π d θ ∫ a b 1 r d r = μ h N 2 J 2 8 π 2 ⋅ 2 π [ log r ] a b = μ h N 2 J 2 4 π log b a \begin{aligned}
U &= \iiint_u \frac{1}{2\mu}B^2 du \\
&= h \int_0^{2\pi}\int_a^b \frac{1}{2\mu} \cdot \frac{\mu^2N^2J^2}{4\pi^2 r^2} rdrd\theta \\
&= \frac{\mu h N^2 J^2}{8\pi^2} \cdot \int_0^{2\pi} d\theta \int_a^b \frac{1}{r}dr \\
&= \frac{\mu hN^2 J^2}{8\pi^2} \cdot 2\pi \bigg[\log r\bigg]_a^b \\
&= \frac{\mu h N^2 J^2}{4 \pi} \log \frac{b}{a}
\end{aligned} U = ∭ u 2 μ 1 B 2 d u = h ∫ 0 2 π ∫ a b 2 μ 1 ⋅ 4 π 2 r 2 μ 2 N 2 J 2 r d r d θ = 8 π 2 μ h N 2 J 2 ⋅ ∫ 0 2 π d θ ∫ a b r 1 d r = 8 π 2 μ h N 2 J 2 ⋅ 2 π [ log r ] a b = 4 π μ h N 2 J 2 log a b (e)
1 2 L J 2 = 1 2 J 2 ⋅ μ h N 2 2 π log b a \frac{1}{2}LJ^2 = \frac{1}{2}J^2 \cdot \frac{\mu h N^2}{2\pi} \log \frac{b}{a} 2 1 L J 2 = 2 1 J 2 ⋅ 2 π μ h N 2 log a b 従って、
L = μ h N 2 2 π log b a L = \frac{\mu h N^2}{2\pi} \log\frac{b}{a} L = 2 π μ h N 2 log a b 
