發電機原理 在靠近磁鐵凹槽處, 裝上兩個繞在軟鐵芯上的線圈, 分別對準磁鐵的兩極 搖轉線圈時, 線圈導線中就感應出交流電 每當線圈轉半周, 線圈對應的磁極就改變一次, 電流方向也跟著改變 發電機 (generator): 一種將機械能轉換為電能的設備 基本上可分為直流發電機 (DC generator) 與交流發電機 (AC generator) 兩種, 交流發電機又可分為單相 (single phase) 及三相 (three phase) 兩種 大型發電機 : 一般電力公司發電廠則有水力發電機 ( 以水流推動水輪機轉動 ) 火力發電機 ( 以燃燒油 煤或瓦斯方式產生高溫高壓蒸氣, 推動蒸氣渦輪機轉動 ) 核能發電機 ( 以原子分裂產生高溫高壓使蒸氣渦輪機轉動 ) 等 發電廠的發電機通常為大容量 ( 高達數仟 MW) 的三相交流同步發電機
交流電的產生 交流電壓的產生 : 將一個矩形線圈放在磁鐵 N 極 S 極所構成的固定磁場中, 使該線圈以某一個特定方向 特定速度旋轉 線圈旋轉切割了磁通量 (magnetic flux) 使磁通量對時間發生變化, 就發生了交變的感應電勢在導體中 滑環 N S 單相交流發電機 線圈的兩個端點分別連接不同的滑環 (slip-rings), 該對滑環會隨線圈轉動而旋轉, 將滑環以電刷 (brushes) 壓住, 將電刷兩端加上負載 ( 燈泡或電阻器 ), 則會有交變的電流通過負載, 因此負載兩端會產生交變的電壓, 此種電壓即為單相交流發電機的電壓
直流發電機的基本原理 直流發電機之作用原理, 如圖 3-6 所示, 係在線圈兩端分別焊接在半圓形銅片上, 此銅片俗稱整流子, 再由電刷將電流由整流子引出 當線圈在磁場中轉動時, 雖產生交流電, 但整流子能使電流以一定的方向由 C 端電刷流出,D 端電刷流入而成直流電 整流子 直流電機的結構圖
The magnetic dipole (What is the most elementary unit of magnetism) A circular loop of a conductor carrying an electric current, which can generate a magnetic field. A circular current loop can be considered the most elementary unit of magnetism. If a current loop has area A and carries a current i, then its magnetic dipole moment is m=ia. The unit of magnetic moment: A m (amp meter ) m The torque on a magnetic dipole of moment m in a magnetic induction B is then simply In free space m B m H o (-This mean that B tries to align the dipole so that the moment m lies parallel to the induction) 4
- The energy of the dipole moment m in the presence of a magnetic induction (If no frictional forces are operating, the work done by the turning force will be conserved) E m B In free space E m H o The field produced by a current loop is identical in form to the field produced by calculation from two hypothetical magnetic poles of strength separated by a distance l. i m A S -P N l +P P: magnetic pole strength l: distance m = p.l 5
- unit system in magnetism The three unit systems Conversion factors: 1 oersted = 79.58 A/m 1 gauss = 10-4 tesla 1 emu/cm 3 = 1000 A/m 1 maxwell = 10-8 weber CGS - Gaussian MKS - Sommerfeld Kennelly 1. The Earth s magnetic field is typically H=56 A/m (0.70e) B=0.7x10-4 tesla. The saturation magnetization of iron is M 0 =1.7x10 6 A/m M r =0.8x10 6 A/m 3. Lab electromagnet H = 1.6x10 6 A/m B = tesla 6
Magnetic field calculation solenoids Magnetic field are usually produced by electromagnets (solenoids are often cylindrical in shape) A solenoid is made by winding a large number of turns of insulated Cu wire, or a similar electrical conductor, in a helical fashion on an insulated tube know as a former. The ferromagnetic core of an electromagnet generate a higher magnetic induction than a solenoid for the same magnetic field H B 1. Field at the centre of a long thin solenoid (what is the simplest way to produce a uniform magnetic field) If the solenoid has N turns wound on a former of length L and carries a current i amperes the field inside it will be H Ni L ni n N L n: the no. of turns per unit length Magnetic field lines Around a solenoid x x x x x x x x 7
A practical method of making an infinite solenoid is to make the solenoid toroidal in shape. This ensures that the field is uniform throughout the length of the solenoid. The magnetic field H N i r N: the total no. of turns r: the radius of the toroid i: the current flowing 8
. Field due to a circular coil (What is the field strength produced by the simplest form of coil geometry the single-turn) (1) Field at the center of a circular coil element of conductor By Biot-Savart, 9
() Field on the axis of a circular coil x dh = 1 idl u 4πr dl a, H H terms of C i 3 sin a ia ( a x ) i ( ) a 3 a, C 4 3/ odd order must have 4 a ( r 4 (the field at the center of 3 3 15 8a ) 4 6 ia r 6,C 6 3 This can be expressedin the formof a seriesin x (and by symmetry all H H (1 C x H axial axial o o dl u C i a x C...) zero coefficients so the form of x r = a, dh = 1 sin α 4πa (sin α)idl u dh axial dh sin r a the coil) 105 48a dh axial = 1 4πa (sin3 α)idl u x 6 the dipole field become) 10
Ex 1.3 If a coil of 100 turns and diameter 10 cm carries a current of 0.1A, calculate the magnetic field at a distance of 50 cm along the axis of the coil. x P a = 5 cm x = 50 cm i i = 0.1A a = 5 cm x = 50 cm N = 100 H =? H 100(0.05) 0.05 0.5 0.05 (0.53) 3 0.0098 A m (0.1) 3 11
(3) off-axis field a circular coil In the vast majority of cases there is no closed- form analytic solution for the field generated by a current - carrying conductor In the case of the off-axis field of the single circular loop the analysis leads to an elliptic integral which has no exact solution. By Biot -Savart law idlu dh r : the distance from the coil 4r idlsin dh 4 ( x a ) where now a is also a function of instead of being a constant. In the case of the off - axis field, the fieldstrength can be calculated from this eq. by a computer using numerical techniques. 1
Chap Magnetization and Magnetic Moment (When going on to consider magnetic materials it is first necessary to define quantities which represent the response of these materials to the field. These quantities are magnetic moment and magnetization. Once that has been done, we can consider another property. The susceptibility, which is closely related to the permeability.) Consider the effect that a magnetic materials has on the magnetic induction B when a field pass throught it. Magnetization lager (paramagnets, ferromagnets) Materials can alter the magnetic induction (B) Smaller ( diamagnet) Magnetic moment (m) (can we use the torque on a specimen in a field of know strength to define its magnetic properties?) m B The torque on the dipole in the presence of a magnetic field in free space is given by The m can be expressed as the maximum torque on a magnetic dipole τ max divided by B m max B max 0 H (in freespace) Unit of m: A m (ampere meter ) This formula applies equally to a current loop or a bar magnet N S Pole strength 13
The unit of magnetic moment The magnetic moment of 1 ampere meter experiences a maximum torque of 1 newton meter when oriented perpendicular to magnetic induction of 1 tesla. i M A S -P N l +P P: magnetic pole strength l: distance In the case of current loop m=ia i: the current flowing (A) A: the cross-sectional area of the loop (m²) In the case of a bar magnet m=pl p: the pole strength (A m) l:the dipole length (m) 14
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In a current loop E m B 當 m // B時, E min 會產生 The torque turns the loop until its plane is normal to the field direction. m = ia In a bar magnet The magnetic moment vector in a bar magnet tends to align itself with B under the action at the torque. E m B 當 m // B時, E 會產生 min m A torque aligns the magnet parallel to the local direction B. m = p l p: amp meter l : dipole length M m v Magnetic moment Volume of a solid In the Sommerfeld convention this is given p 0 m pl l 0 The pole strength is arising from the more traditional CGS Treatment of magnetism, in which pole strength was defined in terms of the magnetic flux Φ emanating from a single magnetic pole 16
Magnetization (M) How are the magnetic properties of the material and the magnetic induction B related? the magnetic moment m M, magnetization, M per unit volume V A bar magnet with flux density ϕ at the center, dipole length l and with cross-sectional area A has a magnetic moment m, given by l The magnetization M is given by M m Al A 0 B 0 ( m l ) 0 ( B A) m - In this case there are no conventional external electric currents present to generate on external magnetic field (H), so B=μ 0 M M m Al A 0 B 0 0 M B H contribution If both M and H are present then their contributions can be summed. 17
Relation between H, M, B (Can we define a universal equation relating these three magnetic quantities H, M, B?) Magnetic induction B consists of two contribution B ( H M) 0 H M In free space B=μ 0 H (The magnetic induction in free space) In a material B=μ 0 M (The contribution to the induction from the magnetization of a material) Unit: B tesla H M A m A m solenoid H is generated from electromagnet outside the material permanent magnet (by electrical currents) spin M is generated by orbital angular momentum of electrons within the solid A related quantity, the magnetic polarization or intensity of magnetization I is used in the Kennelly convention. This is defined by I=μ 0 M unit of I : tesla 18