Notes 32 Magnetic Force and Torque

Transcription

1 ECE 3318 Appied Eectricity and Magnetism Spring 18 Prof. David R. Jackson Dept. of ECE Notes 3 Magnetic orce and Torque 1

2 orce on Wire q Singe charge: = q( v ) v (derivation omitted) Wire: = d C d

3 orce on Wire (cont.) orce on wire: = d C The contour C is in the direction of the reference direction for the current. C Note: There is no tota force on a wire due to the magnetic fied produced by itsef. Therefore, the magnetic fied in the formua is taken as that due to a other currents. d 3

4 Eampe ind the force on wire = = L h y 1 1 Assume that the wires are ong compared to the separation. Note: The force on wire comes from the magnetic fied due to wire 1. 4

5 Eampe (cont.) The contour C runs from point A to point (eft to right), defined by the current reference direction on wire. = C = L A h y 1 1 C C ( ˆ ) = d = d y ( ˆ ) µ ( ˆ ) 1 1 π h 5

6 Eampe (cont.) = d y ( ˆ ) µ ( ˆ ) C π h ( ˆ ) 1 = µ + L 1 π h d 1 = ˆ µ L [ N] π h (The force is a repusive force, assuming that both currents are positive.) = = L 1 = 1 6

7 Definition of Amp Assume 1 = = = ˆ µ L π h [ ] N Note: = 1 Definition of Amp: [A]: 1 [N/m] when 1.[m] = = h = = force per unit ength on wire 7

8 Definition of Amp (cont.) orce per unit ength: = µ [N/m] π h so ( 1.) π ( ) 7 = µ 1 1. µ = π (an eact constant) [H/m] 8

9 Eampe n a power substation, the current in two parae buses is 1 [ka]. The buses are 1 [m] apart. What is the force per unit ength between the two buses? 1 = = 1 1 = ˆ µ π h [N/m] (attractive force) 9

10 Eampe (cont.) ( 7 ) = ˆ 4π 1 ( ) ( ) π 1. ˆ ( ) =. [N/m] or ( ) ˆ =.45 [bs/m] Note: 1 [kg] 9.8 [N] =. [bs] 1

11 Eampe During a ighting strike, the current in two parae wires inside the wa of a house reaches 1 [ka]. The wires are 1 [cm] apart. What is the force per unit ength between the two wires? 1 1 = ˆ µ π h [N/m] (attractive force) 11

12 Eampe (cont.) ( 7 ) = ˆ 4π 1 ( ) ( ) π.1 ˆ ( ) = [N/m] or ( ) ˆ = 449. [bs/m] Note: 1 [kg] 9.8 [N] =. [bs] 1

13 Torque on Current Loop A panar current oop of wire with an arbitrary shape carries a current. Define magnetic dipoe moment of oop: m ˆn ( A ) ˆn y A = area of oop ˆn = unit norma (defined by RH rue) Note: There is no net force on the oop if the magnetic fied is a constant. ut there is a torque. = magnetic fu density vector (assumed constant here). Torque vector on oop: T = m (Pease see the tetbooks for a derivation.) 13

14 DC Motor A oop rotating in a DC magnetic fied is shown beow. y S N N ε r ˆn φ S N = ˆ ( na ˆ ) ( ˆcosφ ˆsinφ) ( ˆ ) T = m = = A + y ( ) T = ˆ A sinφ A commutator is needed to reverse the current every 18 o and make the torque in same direction. 14

15 DC Motor (cont.) rushes Commutator (rings) Note: The rings are positioned so that the torque changes sign (the current in the oop reverses) at the point where the torque is ero. The commutator reverses the oop current every 18 o of rotation. (t keeps the current fowing cockwise in the picture above.) ( ) T = ˆ A sinφ 15

16 DC Motor (cont.) n practice, there are mutipe oops and commutator segments. The torque is thus more constant as the armature turns. 16

17 Eampe n a DC motor, the armature consists of N = 1, turns (oops) of wire, with each turn being L =.1 [m] in ength (in the direction parae to the ais). The magnetic fu density produced by the stator is =.5 [T]. The radius of the armature is R =.5 [m]. ind the maimum torque on the armature. The current through the motor is 3 [A]. Assume that the magnetic fied is constant and perpendicuar to the oop ais (i.e., we are at the point of maimum torque in the rotation cyce). T = N m ˆ m = ŷ ( A ) A= L( R) ( ) [ ] [ ] N = = T = 4 1,.5, 3 A R = ˆ Ais of rotation L 17

18 Eampe (cont.) We then have T = ˆ T where or T = 15 [Nm] R L T = 11.5 [foot b] Ais of rotation 18

19 orce from a Magnet y = ˆ ( < ) A (base area) S N N ε r Large bock of iron µ r >> 1 g Assume that most of the stored energy is inside the gap region (where there is air, and H = / µ is the strongest) A U H = ( ) H dv dv Ag g µ = µ µ V V gap The magnetic fied is amost constant inside the air gap region, since the gap is sma. 19

20 orce from a Magnet (cont.) y = ˆ A (base area) S N N ε r Large bock of iron µ r >> 1 g Principe of virtua work : so ( ) du = dg H du = dg H Note: The force is the force we woud eert on the bock of iron to keep it fied in position. We then have: A = µ