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磁矩

2023-06-17 14:20 作者:sn42  | 我要投稿

磁矩和電偶極矩類似,是電磁學中的重點。將一個帶有電流的回路放在勻強磁場(為了簡化問題我們暫且只研究勻強),會有繞質心的力矩產(chǎn)生,使回路轉動。由于整體沒有電流,所以回路受合外力為0。

我們研究如下問題

  1. 回路(設它沒有支路)所受安培力的力矩:

    (下述環(huán)路積分均在回路上且沿電流方向)

    由力矩定義和安培力公式寫出:

    %5Cmathbf%7BM%7D%20%3D%5Coint%20%5Cmathbf%7Br%7D%20%5Ctimes%20%5Cmathrm%7Bd%7D%5Cmathbf%7BF%7D%3D%0A%5Coint%20%5Cmathbf%7Br%7D%20%5Ctimes%20(I%5Cmathrm%7Bd%7D%5Cmathbf%7Br%7D%5Ctimes%5Cmathbf%7BB%7D)%3D%0AI%5Coint(%5Cmathbf%20r%20%5Ccdot%20%5Cmathbf%20B)%5Ccdot%20%5Cmathrm%7Bd%7D%5Cmathbf%7Br%7D-I%5Coint%20(%5Cmathbf%20r%20%5Ccdot%20%5Cmathrm%7Bd%7D%5Cmathbf%7Br%7D)%5Ccdot%5Cmathbf%7BB%7D%3D%0AI%5Coint(%5Cmathbf%20r%20%5Ccdot%20%5Cmathbf%20B)%5Ccdot%20%5Cmathrm%7Bd%7D%5Cmathbf%7Br%7D

    第三個等號用到了矢量分析公式:%5Cmathbf%20A%20%5Ctimes%20(%5Cmathbf%20B%20%5Ctimes%20%5Cmathbf%20C)%3D(%5Cmathbf%20A%20%5Ccdot%5Cmathbf%20C)%5Cmathbf%20B-(%5Cmathbf%20A%20%5Ccdot%5Cmathbf%20B)%5Cmathbf%20C

    第四個等號是因為左式第二項為0,不能理解的話可以展開成分量形式證明

    接下來有一個突破口:%5Coint%20(%5Cmathbf%20r%20%5Ccdot%20%5Cmathbf%20B)%5Cmathrm%20d%20%5Cmathbf%20r%2B%0A%5Coint%20(%5Cmathrm%20d%20%5Cmathbf%20r%20%5Ccdot%20%5Cmathbf%20B)%5Cmathbf%20r%3D%5Cmathbf%200

    展開為分量形式可證。

    另外有:2%5Cmathbf%20S%5Ctimes%5Cmathbf%20B%3D%5Coint%20(%5Cmathbf%20r%5Ctimes%5Cmathrm%20d%5Cmathbf%20r)%5Ctimes%5Cmathbf%20B%3D%0A%5Coint(%5Cmathbf%20r%5Ccdot%20%5Cmathbf%20B)%5Cmathrm%20d%5Cmathbf%20r-%5Coint(%5Cmathrm%20d%5Cmathbf%20r%5Ccdot%20%5Cmathbf%20B)%5Cmathbf%20r

    兩式聯(lián)立得到%5Cmathbf%20S%5Ctimes%5Cmathbf%20B%3D%5Coint%20(%5Cmathbf%20r%20%5Ccdot%20%5Cmathbf%20B)%5Cmathrm%20d%20%5Cmathbf%20r

    帶回力矩的表達式得到%5Cmathbf%20M%3DI%5Cmathbf%20S%5Ctimes%5Cmathbf%20B

    這里的S是廣義的面積,帶有方向,計算式為%5Cmathbf%20S%3D%5Cfrac%7B1%7D%7B2%7D%20%5Coint%20%5Cmathbf%20r%5Ctimes%5Cmathrm%20d%5Cmathbf%20r

    如果回路在一個平面上則可以簡單的用回路沿電流方向的右手螺旋定則確定S的方向。

    磁矩在磁場中具有的能量E%3D%5Cint_%7B0%7D%5E%7B%5Ctheta%7D%20-ISB%5Csin%20%5Ctheta%5Cmathrm%20d%20%5Ctheta%3D-ISB%5Cmathrm%20cos%5Ctheta%3D-I%5Cmathbf%20S%5Ccdot%5Cmathbf%20B

    積分后沒有帶常數(shù)是因為沒有必要。


  2. 圓環(huán)形線圈在空間中產(chǎn)生的磁場:

    一個帶有很小的半徑R的,帶有電流I的圓環(huán)在空間直角坐標系Oxyz中,O為其圓心,其面積指向z軸,空間中磁導率為%5Cmu_0

    我們計算到O距離為r_0,與水平面夾角為%5Cvarphi的點A(r_0%5Ccos%20%5Cvarphi%2C0%2Cr_0%5Csin%20%5Cvarphi)處的磁場:

    圓環(huán)上位置為(R%5Ccos%20%5Ctheta%2CR%5Csin%20%5Ctheta%2C0)的大小為R%5Cmathrm%20d%20%5Ctheta(-%5Csin%20%5Ctheta%2C%5Ccos%20%5Ctheta%2C0)的微元指向A的向量:%5Cmathbf%20r%3D(r_0%5Ccos%20%5Cvarphi%20-R%5Ccos%5Ctheta%2C-R%5Csin%5Ctheta%2Cr_0%5Csin%5Cvarphi),

    其長度可近似看作r_0-R%5Ccos%5Ctheta%5Ccos%5Cvarphi

    于是這個微元產(chǎn)生的磁場:%5Cmathrm%20d%5Cmathbf%20B%3D%5Cfrac%7B%5Cmu_0I%5Cmathrm%20d%5Cmathbf%20l%5Ctimes%20%5Cmathbf%20r%20%7D%7B4%5Cpi%20r%5E3%7D%20%0A%3D%5Cfrac%7B%5Cmu_0IR%5Cmathrm%20d%20%5Ctheta(-%5Csin%20%5Ctheta%2C%5Ccos%20%5Ctheta%2C0)%5Ctimes(r_0%5Ccos%20%5Cvarphi%20-R%5Ccos%5Ctheta%2C-R%5Csin%5Ctheta%2Cr_0%5Csin%5Cvarphi)%20%7D%7B4%5Cpi%20(r_0-R%5Ccos%5Ctheta%5Ccos%5Cvarphi)%5E3%7D%20

    %0A%3D%5Cfrac%7B%5Cmu_0IR%5Cmathrm%20d%20%5Ctheta(r_0%5Csin%5Cvarphi%5Ccos%5Ctheta%2Cr_0%5Csin%5Cvarphi%5Csin%5Ctheta%2CR-r_0%5Ccos%5Cvarphi%5Ccos%5Ctheta)%20%7D%7B4%5Cpi%20r_0%5E3%20%7D(1%2B3%5Cfrac%7BR%7D%7Br_0%7Dcos%5Cvarphi%5Ccos%5Ctheta)%20

    對θ在0到2%5Cpi積分得到

    %5Cmathbf%20B%3D%5Cfrac%7B%5Cmu_0I%5Cpi%20R%5E2%7D%7B4%5Cpi%20r_0%5E3%7D(3%5Csin%5Cvarphi%5Ccos%5Cvarphi%2C0%2C2-3%5Ccos%5E2%5Cvarphi)

  3. 求圓環(huán)形線圈在空間中產(chǎn)生的磁感線

    不妨只分析軸截面上的磁感線,建立平面直角坐標系xOy

    由2中B的表達式知(%5Cmathrm%20d%20x%2C%5Cmathrm%20d%20y)%E4%B8%8E(3%5Csin%5Cvarphi%5Ccos%5Cvarphi%2C2sin%5E2%5Cvarphi-cos%5E2%5Cvarphi)%E5%85%B1%E7%BA%BF

    %5Cfrac%7B%5Cmathrm%20d%20y%7D%7B%5Cmathrm%20d%20x%7D%20%3D%5Cfrac%7B2y%7D%7B3x%7D%20-%5Cfrac%7Bx%7D%7B3y%7D%20,設%5Cfrac%7By%7D%7Bx%7D%20%3Dk%0A,有%5Cfrac%7B%5Cmathrm%20d%20kx%7D%7B%5Cmathrm%20d%20x%7D%20%3Dk%2Bx%5Cfrac%7B%5Cmathrm%20d%20k%7D%7B%5Cmathrm%20d%20x%7D%20%3D%5Cfrac%7B2%7D%7B3%7D%20k-%5Cfrac%7B1%7D%7B3k%7D%20

    整理得到%5Cfrac%7B3k%5Cmathrm%20d%20k%7D%7Bk%5E2%2B1%7D%20%3D-%5Cfrac%7B%5Cmathrm%20d%20x%7D%7Bx%7D%20

    左右積分得%5Cfrac%7B3%7D%7B2%7D%20%5Cln(k%5E2%2B1)%3D-%5Cln%20x%2BC_1

    %5Cfrac%7By%5E2%7D%7Bx%5E2%7D%2B1%3DC_2x%5E%7B-%5Cfrac%7B2%7D%7B3%7D%7D,這就是單條磁感線的表達式,C2是常數(shù)

    如果磁感線的密度正比于磁場強度,那么則需要對C2進行一些控制,讓磁感線簇能夠體現(xiàn)強度。

    磁感線于x軸右交點橫坐標:C_2%5E%5Cfrac%7B3%7D%7B2%7D,設n來表示第n根磁感線,在n足夠大時,有:

    %5Cfrac%7B1%7D%7B%5Crho%7D%20%3D%5Cfrac%7B%5Cmathrm%20d%20x%7D%7B%5Cmathrm%20d%20n%7D%20%3D%5Cfrac%7B%5Cmathrm%20d%20C_2%5E%5Cfrac%7B3%7D%7B2%7D%7D%7B%5Cmathrm%20d%20n%7D%3D%0A%5Cfrac%7B3C_2%5E%5Cfrac%7B1%7D%7B2%7D%5Cmathrm%20d%20C_2%7D%7B2%5Cmathrm%20d%20n%7D%5Cpropto%20%0Ax%5E3%3DC_2%5E%5Cfrac%7B9%7D%7B2%7D

    %5Cfrac%7B%5Cmathrm%20d%20C_2%7D%7BC_2%5E4%7D%5Cpropto%20%5Cmathrm%20d%20n

    C_2%3DC_3n%5E%7B-%5Cfrac%7B1%7D%7B3%7D%7D

    所以%5Cfrac%7By%5E2%7D%7Bx%5E2%7D%2B1%3DC_3n%5E%7B-%5Cfrac%7B1%7D%7B3%7D%7Dx%5E%7B-%5Cfrac%7B2%7D%7B3%7D%7D,這是磁感線簇的表達式,n是正整數(shù)。

    最后附上一張磁感線簇的圖:




磁感線簇

中間是空的只是因為再向內(nèi)需要很多線,電腦計算力有限。實際上,他不可能像這樣無限向中間延伸,小到一定程度時,R<<r0將失效,我們求得的磁感線也就失真了。


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