Chapter 28 Inductance and Electromagnetic Oscillations 3. Calculation of self-Induction ①假设线圈中的电流Ⅰ; ②求线圈中的磁通量Φ; ③由定义求出自感系数L L
Chapter 28 Inductance and Electromagnetic Oscillations 3. Calculation of Self-Induction ①假设线圈中的电流 I ； ②求线圈中的磁通量 ； ③由定义求出自感系数 L。 I L =
Chapter 28 Inductance and Electromagnetic Oscillations Example: Inductance of a Solenoid 长直螺线管,线圈密度为n,长度为l,横截面积为 S,插有磁导率为的磁介质,求线圈的自感系数L。 Solution:先设电流Ⅰ—根据安培环路定理求得H→B →L Magnetic field inside the solenoid of length l is: Bs√ Ponl
Chapter 28 Inductance and Electromagnetic Oscillations Example: Inductance of a Solenoid. nI l NI B 0 0 = = Magnetic field inside the solenoid of length l is: 一长直螺线管，线圈密度为 n，长度为 l，横截面积为 S，插有磁导率为 的磁介质，求线圈的自感系数 L 。 l S E Solution: 先设电流 I 根据安培环路定理求得 H B Φ L
Chapter 28 Inductance and Electromagnetic Oscillations Magnetic flux linkage: NR=nlBA=(monIn)op2/s L=N①B_∠ImR2n2l =/0n2S The inductance per unit length for a long solenoid near its center 2=10nS Inductance l like capacitance--depends only on the geometry of the device
Chapter 28 Inductance and Electromagnetic Oscillations 2 2 0 NB = nlBA = ( n lI)R Magnetic flux linkage: n lS I I R n l I L B 2 0 2 2 N 0 = = = l S E The inductance per unit length for a long solenoid near its center, n S l L 2 = 0 Inductance L —— like capacitance ——depends only on the geometry of the device
Chapter 28 Inductance and Electromagnetic Oscillations Example: calculating self-inductance of co-axis wire 有两个同轴圆筒形导体,其半径 R 分别为R和R,通过它们的电流 均为Ⅰ,但电流的流向相反设在 两圆筒间充满磁导率为/的均匀 磁介质,求其自感L Solution: B czar r
Chapter 28 Inductance and Electromagnetic Oscillations Example: Calculating self-inductance of co-axis wire. r I B 2 0 = 1 R2 R r Solution: 有两个同轴圆筒形导体 , 其半径 分别为 和 , 通过它们的电流 均为 , 但电流的流向相反.设在 两圆筒间充满磁导率为 的均匀 磁介质 , 求其自感L. R1 R2 I L R1 I R2 l I r
Chapter 28 Inductance and Electromagnetic Oscillations 如图在两圆筒间取一长 为l的面PQRS,并将其分 R 成许多小面元 d④=B·dS=Bldn ①=d￠= R2 ldr ri Zt r R R 2 2T R
Chapter 28 Inductance and Electromagnetic Oscillations 如图在两圆筒间取一长 为 的面 , 并将其分 成许多小面元. l PQRS Φ B S d = d = Bldr l r r I Φ Φ R R d 2π d 2 1 = = 1 2 ln 2π R Il R Φ = R1 I S P R Q R2 l I r dr