(1) 無限に長い直線電流の磁束密度
アンペールの法則は、定常電流について、閉曲線に沿う B \bm B B μ 0 \mu_0 μ 0 B \bm B B B r B_r B r z z z ∇ ⋅ B = 0 \nabla\cdot\bm B=0 ∇ ⋅ B = 0 2 π r h B r = 0 2\pi rh\,B_r=0 2 π r h B r = 0 0 0 0 B z B_z B z r z rz r z B z ( r 1 ) = B z ( r 2 ) B_z(r_1)=B_z(r_2) B z ( r 1 ) = B z ( r 2 ) 0 0 0 B z ≡ 0 B_z\equiv0 B z ≡ 0
別の導き方として、ビオ・サバールの法則を使い、点 ( r , 0 , 0 ) (r,0,0) ( r , 0 , 0 ) μ 0 I 4 π ∫ − ∞ ∞ r d z ′ ( r 2 + z ′ 2 ) 3 / 2 = μ 0 I 4 π ⋅ 2 r \dfrac{\mu_0I}{4\pi}\displaystyle\int_{-\infty}^{\infty}\dfrac{r\,dz'}{(r^2+z'^2)^{3/2}}=\dfrac{\mu_0I}{4\pi}\cdot\dfrac2r 4 π μ 0 I ∫ − ∞ ∞ ( r 2 + z ′2 ) 3/2 r d z ′ = 4 π μ 0 I ⋅ r 2
次元は μ 0 I / r \mu_0I/r μ 0 I / r I = 1 I=1 I = 1 r = 2 r=2 r = 2 10.0 μ 10.0\ \mu 10.0 μ μ 0 I / ( 2 r ) \mu_0I/(2r) μ 0 I / ( 2 r ) 31.4 μ 31.4\ \mu 31.4 μ μ 0 I / ( 4 π r ) \mu_0I/(4\pi r) μ 0 I / ( 4 π r ) 5.0 μ 5.0\ \mu 5.0 μ r ^ \hat{\bm r} r ^ ϕ ^ \hat{\bm\phi} ϕ ^
(2) a) 円電流の z z z
軸上では、円周上のどの素片も P から同じ距離 R R R d l ⊥ R d\bm l\perp\bm R d l ⊥ R a / R a/R a / R a 2 a^2 a 2 2 π a 2\pi a 2 π a a a a a / R a/R a / R a a a ( a 2 + z 2 ) 3 / 2 = R 3 (a^2+z^2)^{3/2}=R^3 ( a 2 + z 2 ) 3/2 = R 3 R − 2 R^{-2} R − 2 R − 1 R^{-1} R − 1
極限で確かめる。z = 0 z=0 z = 0 μ 0 I / ( 2 a ) \mu_0I/(2a) μ 0 I / ( 2 a ) ∣ z ∣ ≫ a |z|\gg a ∣ z ∣ ≫ a B z → μ 0 I a 2 2 ∣ z ∣ 3 B_z\to\dfrac{\mu_0Ia^2}{2|z|^3} B z → 2∣ z ∣ 3 μ 0 I a 2 m = π a 2 I m=\pi a^2I m = π a 2 I μ 0 m 2 π ∣ z ∣ 3 \dfrac{\mu_0m}{2\pi|z|^3} 2 π ∣ z ∣ 3 μ 0 m a = 5 a=5 a = 5 I = 1 I=1 I = 1 B z ( 0 ) = 12.57 μ B_z(0)=12.57\ \mu B z ( 0 ) = 12.57 μ z = 15 z=15 z = 15 0.397 μ 0.397\ \mu 0.397 μ 0.465 μ 0.465\ \mu 0.465 μ 17 % 17\% 17% z = 3 a z=3a z = 3 a ∣ z ∣ ≫ a |z|\gg a ∣ z ∣ ≫ a 1 / 2 1/2 1/2 a 2 a^2 a 2 a a a − z -z − z
(2) b) 円電流の磁気双極子モーメント
定義式 m = 1 2 ∫ r ′ × J d V \bm m=\dfrac12\int\bm r'\times\bm J\,dV m = 2 1 ∫ r ′ × J d V J d V → I d l \bm J\,dV\to I\,d\bm l J d V → I d l d l d\bm l d l 1 2 ∣ r ′ × d l ∣ \tfrac12|\bm r'\times d\bm l| 2 1 ∣ r ′ × d l ∣ 1 2 ∮ r ′ × d l \tfrac12\oint\bm r'\times d\bm l 2 1 ∮ r ′ × d l m = I S \bm m=I\bm S m = I S m \bm m m B \bm B B m × B \bm m\times\bm B m × B − m ⋅ B -\bm m\cdot\bm B − m ⋅ B ⋅ \cdot ⋅ 2 ^2 2 a = 5 a=5 a = 5 I = 1 I=1 I = 1 7.85 × 10 − 3 7.85\times10^{-3} 7.85 × 1 0 − 3 ⋅ \cdot ⋅ 2 ^2 2
(2)a) の軸上遠方の式から m m m μ 0 m / ( 2 π ∣ z ∣ 3 ) = μ 0 I a 2 / ( 2 ∣ z ∣ 3 ) \mu_0m/(2\pi|z|^3)=\mu_0Ia^2/(2|z|^3) μ 0 m / ( 2 π ∣ z ∣ 3 ) = μ 0 I a 2 / ( 2∣ z ∣ 3 ) m = π a 2 I m=\pi a^2I m = π a 2 I π a 2 \pi a^2 π a 2 2 π a 2\pi a 2 π a ⋅ \cdot ⋅ I I I
(3) a) 回転する一様帯電球の中心の磁束密度
別の導き方は、球を z z z ( r ′ , θ ) (r',\theta) ( r ′ , θ ) r ′ d r ′ d θ r'\,dr'\,d\theta r ′ d r ′ d θ R = r ′ sin θ R=r'\sin\theta R = r ′ sin θ d I = ρ ω R r ′ d r ′ d θ dI=\rho\omega R\,r'\,dr'\,d\theta d I = ρ ω R r ′ d r ′ d θ ( z = − r ′ cos θ ) (z=-r'\cos\theta) ( z = − r ′ cos θ ) d B z = μ 0 d I R 2 2 r ′ 3 dB_z=\dfrac{\mu_0\,dI\,R^2}{2r'^3} d B z = 2 r ′3 μ 0 d I R 2 μ 0 ρ ω a 2 3 \dfrac{\mu_0\rho\omega a^2}3 3 μ 0 ρ ω a 2 r ′ r' r ′ σ \sigma σ 2 3 μ 0 σ ω r ′ \dfrac23\mu_0\sigma\omega r' 3 2 μ 0 σ ω r ′ σ = ρ d r ′ \sigma=\rho\,dr' σ = ρ d r ′ 0 0 0 a a a
球殻だけなら B = μ 0 Q ω 6 π a B=\dfrac{\mu_0Q\omega}{6\pi a} B = 6 π a μ 0 Q ω Q Q Q 3 2 \dfrac32 2 3 Q = 1 μ Q=1\ \mu Q = 1 μ ω = 100 \omega=100 ω = 100 a = 5 a=5 a = 5 B = 2.0 × 10 − 10 B=2.0\times10^{-10} B = 2.0 × 1 0 − 10 μ 0 Q ω / a \mu_0Q\omega/a μ 0 Q ω / a Q ω Q\omega Q ω μ 0 I / ( 2 a ) \mu_0I/(2a) μ 0 I / ( 2 a )
誤りは、円電流の式に I = Q ω I=Q\omega I = Q ω 2 π / ω 2\pi/\omega 2 π / ω Q ω / ( 2 π ) Q\omega/(2\pi) Q ω / ( 2 π ) μ 0 Q ω / ( 2 a ) = 1.26 × 10 − 9 \mu_0Q\omega/(2a)=1.26\times10^{-9} μ 0 Q ω / ( 2 a ) = 1.26 × 1 0 − 9 2 π 2\pi 2 π 1 / ( 6 π ) 1/(6\pi) 1/ ( 6 π ) 1.33 × 10 − 10 1.33\times10^{-10} 1.33 × 1 0 − 10 2 / 3 2/3 2/3
(3) b) 回転する一様帯電球の磁気双極子モーメント
リング電流に分ける方法でも求められる。(3)a) と同じリング（半径 R = r ′ sin θ R=r'\sin\theta R = r ′ sin θ d I = ρ ω R r ′ d r ′ d θ dI=\rho\omega R\,r'\,dr'\,d\theta d I = ρ ω R r ′ d r ′ d θ d m = π R 2 d I dm=\pi R^2\,dI d m = π R 2 d I m = π ρ ω ∫ 0 a r ′ 4 d r ′ ∫ 0 π sin 3 θ d θ = π ρ ω ⋅ a 5 5 ⋅ 4 3 = 4 π 15 ρ ω a 5 m=\pi\rho\omega\int_0^ar'^4dr'\int_0^\pi\sin^3\theta\,d\theta=\pi\rho\omega\cdot\dfrac{a^5}5\cdot\dfrac43=\dfrac{4\pi}{15}\rho\omega a^5 m = π ρ ω ∫ 0 a r ′4 d r ′ ∫ 0 π sin 3 θ d θ = π ρ ω ⋅ 5 a 5 ⋅ 3 4 = 15 4 π ρ ω a 5
質量分布も同じ一様な球なら、角運動量 L = 2 5 M a 2 ω L=\tfrac25Ma^2\omega L = 5 2 M a 2 ω 2 5 M a 2 \tfrac25Ma^2 5 2 M a 2 m = Q 2 M L \bm m=\dfrac Q{2M}\bm L m = 2 M Q L m = Q 2 M ⋅ 2 5 M a 2 ω = Q ω a 2 5 m=\dfrac Q{2M}\cdot\dfrac25Ma^2\omega=\dfrac{Q\omega a^2}5 m = 2 M Q ⋅ 5 2 M a 2 ω = 5 Q ω a 2 2 3 M a 2 \tfrac23Ma^2 3 2 M a 2 m = Q ω a 2 / 3 m=Q\omega a^2/3 m = Q ω a 2 /3
(3)a) の中心の磁束密度とは B = 5 μ 0 m 4 π a 3 B=\dfrac{5\mu_0m}{4\pi a^3} B = 4 π a 3 5 μ 0 m m = 5.0 × 10 − 8 m=5.0\times10^{-8} m = 5.0 × 1 0 − 8 ⋅ \cdot ⋅ 2 ^2 2 Q ω a 2 / 3 Q\omega a^2/3 Q ω a 2 /3 8.3 × 10 − 8 8.3\times10^{-8} 8.3 × 1 0 − 8 ⋅ \cdot ⋅ 2 ^2 2 5 / 3 5/3 5/3 Q ω a 2 / 2 Q\omega a^2/2 Q ω a 2 /2 1.25 × 10 − 7 1.25\times10^{-7} 1.25 × 1 0 − 7 ⋅ \cdot ⋅ 2 ^2 2
解答
(1) 無限に長い直線電流の磁束密度
z z z z z z r r r r r r z z z C C C z z z ∮ C B ⋅ d l = μ 0 I \oint_C\bm B\cdot d\bm l=\mu_0I ∮ C B ⋅ d l = μ 0 I B \bm B B ϕ \phi ϕ B ϕ B_\phi B ϕ 2 π r B ϕ = μ 0 I 2\pi r\,B_\phi=\mu_0I 2 π r B ϕ = μ 0 I
(2) a) 円電流の z z z
電流は z z z P = ( 0 , 0 , z ) \mathrm P=(0,0,z) P = ( 0 , 0 , z ) r ′ = a ( cos φ , sin φ , 0 ) \bm r'=a(\cos\varphi,\sin\varphi,0) r ′ = a ( cos φ , sin φ , 0 ) d l = a ( − sin φ , cos φ , 0 ) d φ d\bm l=a(-\sin\varphi,\cos\varphi,0)\,d\varphi d l = a ( − sin φ , cos φ , 0 ) d φ R = ( − a cos φ , − a sin φ , z ) \bm R=(-a\cos\varphi,-a\sin\varphi,z) R = ( − a cos φ , − a sin φ , z ) d B = μ 0 I 4 π d l × R R 3 , R = a 2 + z 2
d\bm B=\frac{\mu_0I}{4\pi}\frac{d\bm l\times\bm R}{R^3},\qquad R=\sqrt{a^2+z^2}
d B = 4 π μ 0 I R 3 d l × R , R = a 2 + z 2 d l × R = a d φ ( z cos φ , z sin φ , a ) d\bm l\times\bm R=a\,d\varphi\,\bigl(z\cos\varphi,\ z\sin\varphi,\ a\bigr) d l × R = a d φ ( z cos φ , z sin φ , a ) x , y x,y x , y φ \varphi φ 0 0 0 2 π 2\pi 2 π 0 0 0 z z z φ \varphi φ B z = μ 0 I 4 π ⋅ a 2 ( a 2 + z 2 ) 3 / 2 ⋅ 2 π = μ 0 I a 2 2 ( a 2 + z 2 ) 3 / 2
B_z=\frac{\mu_0I}{4\pi}\cdot\frac{a^2}{(a^2+z^2)^{3/2}}\cdot2\pi=\frac{\mu_0Ia^2}{2(a^2+z^2)^{3/2}}
B z = 4 π μ 0 I ⋅ ( a 2 + z 2 ) 3/2 a 2 ⋅ 2 π = 2 ( a 2 + z 2 ) 3/2 μ 0 I a 2 + z +z + z
(2) b) 円電流の磁気双極子モーメント
磁気双極子モーメントは m = 1 2 ∮ r ′ × I d l \bm m=\dfrac12\oint\bm r'\times I\,d\bm l m = 2 1 ∮ r ′ × I d l r ′ × d l = a 2 d φ z ^ \bm r'\times d\bm l=a^2d\varphi\,\hat{\bm z} r ′ × d l = a 2 d φ z ^ m = I 2 ⋅ a 2 ⋅ 2 π z ^ = π a 2 I z ^ \bm m=\dfrac I2\cdot a^2\cdot2\pi\,\hat{\bm z}=\pi a^2I\,\hat{\bm z} m = 2 I ⋅ a 2 ⋅ 2 π z ^ = π a 2 I z ^ m = I S \bm m=I\bm S m = I S S = π a 2 S=\pi a^2 S = π a 2
(3) a) 回転する一様帯電球の中心の磁束密度
体積電荷密度は ρ = 3 Q 4 π a 3 \rho=\dfrac{3Q}{4\pi a^3} ρ = 4 π a 3 3 Q r ′ \bm r' r ′ v = ω z ^ × r ′ \bm v=\omega\hat{\bm z}\times\bm r' v = ω z ^ × r ′ J = ρ v \bm J=\rho\bm v J = ρ v B = μ 0 4 π ∫ J × ( 0 − r ′ ) r ′ 3 d V \bm B=\dfrac{\mu_0}{4\pi}\displaystyle\int\dfrac{\bm J\times(\bm 0-\bm r')}{r'^3}\,dV B = 4 π μ 0 ∫ r ′3 J × ( 0 − r ′ ) d V v × ( − r ′ ) = r ′ × v \bm v\times(-\bm r')=\bm r'\times\bm v v × ( − r ′ ) = r ′ × v z z z x ′ v y − y ′ v x = ω ( x ′ 2 + y ′ 2 ) = ω r ′ 2 sin 2 θ x'v_y-y'v_x=\omega(x'^2+y'^2)=\omega r'^2\sin^2\theta x ′ v y − y ′ v x = ω ( x ′2 + y ′2 ) = ω r ′2 sin 2 θ x , y x,y x , y φ \varphi φ 0 0 0 ( r ′ , θ , φ ) (r',\theta,\varphi) ( r ′ , θ , φ ) B z = μ 0 ρ ω 4 π ∫ 0 2 π d φ ∫ 0 π sin 3 θ d θ ∫ 0 a r ′ d r ′ = μ 0 ρ ω 4 π ⋅ 2 π ⋅ 4 3 ⋅ a 2 2 = μ 0 ρ ω a 2 3
B_z=\frac{\mu_0\rho\omega}{4\pi}\int_0^{2\pi}\!d\varphi\int_0^\pi\!\sin^3\theta\,d\theta\int_0^ar'\,dr'=\frac{\mu_0\rho\omega}{4\pi}\cdot2\pi\cdot\frac43\cdot\frac{a^2}2=\frac{\mu_0\rho\omega a^2}3
B z = 4 π μ 0 ρ ω ∫ 0 2 π d φ ∫ 0 π sin 3 θ d θ ∫ 0 a r ′ d r ′ = 4 π μ 0 ρ ω ⋅ 2 π ⋅ 3 4 ⋅ 2 a 2 = 3 μ 0 ρ ω a 2 ρ \rho ρ
(3) b) 回転する一様帯電球の磁気双極子モーメント
m = 1 2 ∫ r ′ × J d V \bm m=\dfrac12\displaystyle\int\bm r'\times\bm J\,dV m = 2 1 ∫ r ′ × J d V J = ρ ω z ^ × r ′ \bm J=\rho\,\omega\hat{\bm z}\times\bm r' J = ρ ω z ^ × r ′ r ′ × v \bm r'\times\bm v r ′ × v z z z ω ( x ′ 2 + y ′ 2 ) = ω r ′ 2 sin 2 θ \omega(x'^2+y'^2)=\omega r'^2\sin^2\theta ω ( x ′2 + y ′2 ) = ω r ′2 sin 2 θ m z = 1 2 ρ ω ∫ r ′ 2 sin 2 θ d V = 1 2 ρ ω ⋅ 2 π ∫ 0 π sin 3 θ d θ ∫ 0 a r ′ 4 d r ′ = 1 2 ρ ω ⋅ 2 π ⋅ 4 3 ⋅ a 5 5 = 4 π 15 ρ ω a 5
m_z=\frac12\rho\omega\int r'^2\sin^2\theta\,dV=\frac12\rho\omega\cdot2\pi\int_0^\pi\sin^3\theta\,d\theta\int_0^ar'^4dr'=\frac12\rho\omega\cdot2\pi\cdot\frac43\cdot\frac{a^5}5=\frac{4\pi}{15}\rho\omega a^5
m z = 2 1 ρ ω ∫ r ′2 sin 2 θ d V = 2 1 ρ ω ⋅ 2 π ∫ 0 π sin 3 θ d θ ∫ 0 a r ′4 d r ′ = 2 1 ρ ω ⋅ 2 π ⋅ 3 4 ⋅ 5 a 5 = 15 4 π ρ ω a 5 ρ = 3 Q / ( 4 π a 3 ) \rho=3Q/(4\pi a^3) ρ = 3 Q / ( 4 π a 3 ) x , y x,y x , y 0 0 0
最終答
(1) 無限に長い直線電流の磁束密度
B = B ϕ ϕ ^ , B ϕ = μ 0 I 2 π r , B r = B z = 0 \bm B=B_\phi\,\hat{\bm\phi},\quad B_\phi=\dfrac{\mu_0I}{2\pi r},\quad B_r=B_z=0 B = B ϕ ϕ ^ , B ϕ = 2 π r μ 0 I , B r = B z = 0
(2) a) 円電流の z z z
B z = μ 0 I a 2 2 ( a 2 + z 2 ) 3 / 2 B_z=\dfrac{\mu_0I\,a^2}{2\,(a^2+z^2)^{3/2}} B z = 2 ( a 2 + z 2 ) 3/2 μ 0 I a 2 + z +z + z z = 0 z=0 z = 0 μ 0 I 2 a \dfrac{\mu_0I}{2a} 2 a μ 0 I
(2) b) 円電流の磁気双極子モーメント
∣ m ∣ = π a 2 I |\bm m|=\pi a^2I ∣ m ∣ = π a 2 I + z +z + z
(3) a) 回転する一様帯電球の中心の磁束密度
B = μ 0 Q ω 4 π a B=\dfrac{\mu_0Q\omega}{4\pi a} B = 4 π a μ 0 Q ω + z +z + z ω \omega ω
(3) b) 回転する一様帯電球の磁気双極子モーメント
∣ m ∣ = Q ω a 2 5 |\bm m|=\dfrac{Q\omega a^2}5 ∣ m ∣ = 5 Q ω a 2 + z +z + z