學院物理/磁場

定義

電量為 q 的電荷，以速度 v 射入，受到與運動方向垂直的作用力 F，則區域內有磁場。

${\vec {F}}=q\,{\vec {v}}\times {\vec {B}}=i\,{\vec {l}}\times {\vec {B}}$

應用

質譜儀

${\begin{aligned}&F=qvB={\frac {mv^{2}}{r}}\\&r={\frac {mv}{qB}}\end{aligned}}$

速度選擇器

${\begin{aligned}&qE=qvB\\&v={\frac {E}{B}}\end{aligned}}$

電荷的螺線運動

${\begin{aligned}&r={\frac {mv}{qB}}\\&T={\frac {2\pi r}{v}}={\frac {2\pi m}{qB}}\\&\omega ={\frac {qB}{m}}\\&P=v_{x}T\end{aligned}}$

例題

三

電子受電位差 ΔV' 從靜止加速進入均勻磁場，電子受垂直行進方向磁場作用，行進路徑彎曲為半徑 r，求磁場與角頻率？

${\begin{aligned}{\frac {mv^{2}}{2}}&-0=-q\Delta V\\v=&{\sqrt {\frac {-2q\Delta V}{m}}}\\&B={\frac {mv}{er}}\\&\omega ={\frac {v}{r}}\end{aligned}}$

四

圖28.20

半徑 R 電流 I 的封閉半圓導線在 xy 平面上受 y 軸方向的均勻磁場，求導線受力？

由於直線導線和磁場垂直，

${\begin{aligned}&F_{1}=2iaB{\hat {z}}\\&F_{2}=-2iaB{\hat {z}}\end{aligned}}$

五

Torque

${\begin{aligned}&\end{aligned}}$

霍爾效應

例題

六

${\begin{aligned}&\end{aligned}}$

畢歐沙伐

$d{\vec {B}}={\frac {\mu i}{4\pi }}{\frac {d{\vec {l}}\times {\hat {r}}}{r^{2}}}$

例題

一

θ₁

θ₂

{\begin{matrix}\longrightarrow \\i\quad {\vec {l}}\quad x\end{matrix}}

↖

{\vec {r}}

圖29.4

${\begin{aligned}&d{\vec {l}}\times {\hat {r}}=\sin \left(\pi -{\frac {\pi }{2}}-\theta \right)\,dx\,{\hat {\odot }}=\cos \theta \,dx\,{\hat {\odot }}\\&{\vec {B}}={\frac {\mu i}{4\pi }}\int {\frac {d{\vec {l}}\times {\hat {r}}}{r^{2}}}={\frac {\mu i}{4\pi }}\int {\frac {\cos \theta \,dx}{r^{2}}}\,{\hat {\odot }}\\&B={\frac {\mu i}{4\pi }}\int {\frac {\left|d{\vec {l}}\times {\hat {r}}\right|}{r^{2}}}={\frac {\mu i}{4\pi }}\int {\frac {\cos \theta \,dx}{r^{2}}}={\frac {\mu i}{4\pi }}\int {\frac {y\,dx}{r^{3}}}\\&\qquad x=y\tan \theta \qquad dx=y\sec ^{2}\theta \,d\theta \qquad r=y\sec \theta \\&={\frac {\mu i}{4\pi }}\int {\frac {y^{2}\sec ^{2}\theta \,d\theta }{y^{3}\sec ^{3}\theta }}={\frac {\mu i}{4\pi y}}\int \cos \theta \,d\theta ={\frac {\mu i}{4\pi y}}{\Big [}\sin \theta {\Big ]}_{\theta _{2}}^{\theta _{1}}\\&={\frac {\mu i}{4\pi y}}{\big (}\sin \theta _{1}-\sin \theta _{2}{\big )}\\&\\&\theta _{1}={\frac {\pi }{2}}\qquad \theta _{2}=-{\frac {\pi }{2}}\qquad B={\frac {\mu i}{2\pi y}}\end{aligned}}$

二

圖29.5

${\begin{aligned}&B={\frac {\mu i}{4\pi r^{2}}}\int dl\end{aligned}}$

三

x

y

{\begin{aligned}&\to i\\&\;d{\vec {l}}\end{aligned}}

↗

{\vec {r}}

↘

{\begin{matrix}d{\vec {B}}\\\propto \\d{\vec {l}}\times {\hat {r}}\end{matrix}}

圖29.6

${\begin{aligned}&d{\vec {l}}\times {\hat {r}}=dl\,\sin {90}^{\circ }\;{\hat {x}}=dl\,{\hat {x}}\\&B={\frac {\mu i}{4\pi }}\int {\frac {\left|d{\vec {l}}\times {\hat {r}}\right|}{r^{2}}}={\frac {\mu i}{4\pi }}\int {\frac {dl}{r^{2}}}={\frac {\mu iy}{2r^{2}}}={\frac {\mu iy}{2\left(x^{2}+y^{2}\right)}}\\&\qquad \int dl=\int y\,d\theta =y\int d\theta =y2\pi \qquad r^{2}=x^{2}+y^{2}\\&dB_{x}={\frac {\mu iy\,dl}{4\pi r^{3}}}\qquad \quad \;\;r^{3}=\left(x^{2}+y^{2}\right)^{\frac {3}{2}}\\&B_{x}={\frac {Ba}{r}}={\frac {\mu iy^{2}}{2r^{3}}}={\frac {\mu iy^{2}}{2\left(x^{2}+y^{2}\right)^{\frac {3}{2}}}}\\&\mu _{A}=i{\vec {A}}\qquad \mu _{A}=iy2\pi \\&x\to 0\qquad B_{x}={\frac {\mu i}{2y}}={\frac {\mu \mu _{A}}{4\pi y^{2}}}\end{aligned}}$

μ_A 磁矩，μ = μ₀ = 磁導系數

電流磁力

$F={\frac {\mu ii_{0}l}{y2\pi }}$

例題

四

電流磁力

安培定律

$\int {\vec {B}}\,d{\vec {l}}=\mu i$

例題

五

a

圖29.13

${\begin{aligned}&B(r\geq a)={\frac {\mu i}{r2\pi }}\\&B(r<a)={\frac {\mu i_{0}}{r2\pi }}={\frac {\mu ir^{2}}{r2\pi a^{2}}}={\frac {\mu ir}{2\pi a^{2}}}\\&\qquad i_{0}={\frac {ir^{2}}{a^{2}}}\end{aligned}}$

六

線圈磁場

高斯定律

維基百科中的相關條目：

高斯定律

例題

七

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

圖29.21

${\begin{aligned}&\Phi =\int B\,dA\\&B={\frac {\mu i}{4\pi r^{2}}}\int {\frac {\left|dl\times r\right|}{r^{2}}}={\frac {\mu i}{2\pi y}}\end{aligned}}$

法拉第定律

維基百科中的相關條目：

法拉第定律

$\epsilon =-{\frac {d\Phi }{dt}}=-{\frac {d}{dt}}\int B\,dA$

例題

一

線圈

${\begin{aligned}&\epsilon =-N{\frac {d\Phi }{dt}}\\&i={\frac {\left|\epsilon \right|}{R}}={\frac {Blv}{R}}\end{aligned}}$

二

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

----WW----

R

圖30.9

${\begin{aligned}&i={\frac {Blv}{R}}\\&mv{\frac {dv}{dt}}=-\left({\frac {Blv}{R}}^{2}\right)R\\&v=v_{0}e^{-t\,/{\tfrac {mR}{B^{2}l^{2}}}}=v_{0}e^{\tfrac {-B^{2}l^{2}t}{mR}}={\frac {dx}{dt}}\\&x=v_{0}{\frac {mR}{B^{2}l^{2}}}\end{aligned}}$

三

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

圖30.10

${\begin{aligned}&\epsilon ={\frac {-d\Phi }{d}}t={\frac {Bl^{2}\,d\theta }{2\,dt}}={\frac {Bl^{2}\omega }{2}}\\&d\Phi =B\,dA={\frac {Bl^{2}\pi \,d\theta }{2\pi }}={\frac {Bl^{2}\,d\theta }{2}}\end{aligned}}$

四

冷次

六

General 螺線管半徑 a 求 r 處 E

圖30.16

${\begin{aligned}&B=\mu ni\\&Ea2\pi ={\frac {-d\Phi }{dt}}={\frac {-\pi a^{2}dB}{dt}}={\frac {-\pi a^{2}\mu ndi}{dt}}\end{aligned}}$

七

發電

${\begin{aligned}&\epsilon =NBA\omega \\&\omega =2\pi fi={\frac {\epsilon }{R}}\end{aligned}}$

八

馬達的感應電流

${\begin{aligned}&i={\frac {\epsilon }{R}}\end{aligned}}$

電感

維基百科中的相關條目：

電感

$\epsilon ={\frac {-L\,di}{dt}}$

例題

一

螺線管

${\begin{aligned}&L={\frac {N\Phi }{i}}={\frac {\mu N^{2}A}{l}}\end{aligned}}$

二

RL 電路

${\begin{aligned}&{\frac {\epsilon }{R}}-i={\frac {\epsilon }{R}}e^{\frac {-Rt}{L}}\\&i={\frac {\epsilon }{R}}\left(1-e^{\frac {-Rt}{L}}\right)\end{aligned}}$