2.InductancesIn a linear medium, the magnetic flux @through the closed circuitis alsoproportionaltothe currentI.The magnetic flux linked with the current I is called the magneticflux linkage with the current I, and it is denoted as Y. The ratio of ytoIis denoted byL,henceyL.1It is called the inductance of the circuit, with the unit henry (H)and the inductance can be also considered as the magnetic flux linkageper unit current.In linear media, the inductance of a circuit depends only on theshape and the sizes,but not on the current.The magnetic flux linkage is different from the magnetic flux, anditisassociatedwithacurrentU
2. Inductances In a linear medium, the magnetic flux through the closed circuit is also proportional to the current I. I L = It is called the inductance of the circuit, with the unit henry (H), and the inductance can be also considered as the magnetic flux linkage per unit current. In linear media, the inductance of a circuit depends only on the shape and the sizes, but not on the current. The magnetic flux linked with the current I is called the magnetic flux linkage with the current I, and it is denoted as . The ratio of to I is denoted by L, hence The magnetic flux linkage is different from the magnetic flux, and it is associated with a current
If the magnetic flux is linked with a current N times, then themagnetic flux linkage will be increased by N times. If only a part of themagnetic flux is linked to a current, the magnetic flux linkage must bereducedproportionatelyA loop coil with N turns the magnetic flux linkage with the currentis y =N@,and theinductanceof theloopcoil withNturnsisNd下111Ii1212Supposewe havetwo loopcurrents.athe magnetic flux linkage P, linked withr2-r1dl2currentI,consistsoftwoparts:oneisr2ygenerated by the magnetic flux caused byOcurrent I, itself, andit is denoted as Y,Another Y 2 is produced by the magnetic flux at loopl, by current I,uK
If the magnetic flux is linked with a current N times, then the magnetic flux linkage will be increased by N times. If only a part of the magnetic flux is linked to a current, the magnetic flux linkage must be reduced proportionately. I N I L = = Suppose we have two loop currents, the magnetic flux linkage 1 linked with current I 1 consists of two parts: one is generated by the magnetic flux caused by current I 1 itself, and it is denoted as 11 . dl1 O z y x dl2 l2 l1 I2 I1 r2 - r1 r2 r1 A loop coil with N turns the magnetic flux linkage with the current is = N , and the inductance of the loop coil with N turns is Another 12 is produced by the magnetic flux at loopl 1 by current I 2
Hence, the magnetic flux linkage Y linked with current I isYl =4 +4l2The magnetic flux linkage P, linked with current I, is4, =421 + 422n4nIf the surrounding medium is linear, then all the ratios.422andare independent of the currents since all the magnetic7Pflux linkages are proportional to the current generating them.Yi2PLetM12 =12where Ln is called the self-inductance of loop li, and M, is called themutualinductancefromloopl,to loop l42Similarly,wedefineM.011,1where L22 is called the self-inductance of loop l2, and M2, is called themutualinductance fromloop l, to loopl2u7
The magnetic flux linkage 2 linked with current I 2 is 2 =21 +22 If the surrounding medium is linear, then all the ratios, , , and are independent of the currents since all the magnetic flux linkages are proportional to the current generating them. 1 11 I 2 12 I 2 22 I 1 21 I where L11 is called the self-inductance of loop l 1 , and M12 is called the mutual inductance from loop l 2 to loop l 1 . Similarly, we define 2 22 22 I L = 1 21 21 I M = where L22 is called the self-inductance of loop l 2 , and M21 is called the mutual inductance from loop l 1 to loop l 2 . Hence, the magnetic flux linkage 1 linked with current I 1 is 1 =11 +12 1 11 11 I L = 2 12 12 I M Let =
Substitute the above parameters Lu, L22 , M2 , and M2, intothe above equation, we haveY, = M21l, + L2212Y = L.I, + M1212Forlinearhomogeneous media, we can prove thatM12 = M21Since we can find the mutualinductances between any two loopcircuits as follows:dl, :dldl,.dlMa-M12 =r-r1J124元Consideringdl, dl, = dl, dl, 2 -rl=|r -rl , we haveM12 = M21u√
Substitute the above parameters L11，L22，M12 ，and M21 into the above equation, we have 1 11 1 12 2 = L I + M I 2 21 1 22 2 = M I + L I For linear homogeneous media, we can prove that M12 = M21 Since we can find the mutual inductances between any two loop circuits as follows: − = 2 1 2 1 1 2 21 d d 4π l l M r r l l − = 1 2 1 2 2 1 12 d d 4π l l M r r l l Considering ，we have 1 2 2 1 2 1 1 2 dl dl = dl dl , r −r = r −r M12 = M21
dl, dldl, -dlM-IJMo-司If dl, I dl, everywhere, the mutual inductances be zeroIf dl, /l dl, everywhere, the mutual inductances will be maximumIn electronic devices,if we need to increase the magnetic couplingbetween two coils,the two coils should be placed parallelto each other.If the magnetic coupling needs to be eliminated, they should beperpendiculartoeach other.The mutualinductance could be positive or negative, while theself-inductanceis always positiveU7
− = 2 1 2 1 1 2 21 d d 4π l l M r r l l − = 1 2 1 2 2 1 12 d d 4π l l M r r l l In electronic devices, if we need to increase the magnetic coupling between two coils, the two coils should be placed parallel to each other. If the magnetic coupling needs to be eliminated, they should be perpendicular to each other. The mutual inductance could be positive or negative, while the self-inductance is always positive. If everywhere, the mutual inductances be zero. d 1 d 2 l ⊥ l If everywhere, the mutual inductances will be maximum. 1 d 2 dl // l