Physics 169. Luis anchordoqui. Kitt Peak National Observatory. Monday, March 13, 17

Transcription

1 Physics 169 Kitt Peak National Observatory Luis anchordoqui 1

2 6.1 Magnetic Field Stationary charges experienced an electric force in an electric field Moving charges experienced a magnetic force in a magnetic field More explicitly ~F E = q ~ E (electric force) ~F B = q~v ~ B (magnetic force) Magnetic field (T) : Unit = Tesla 1T=1Cmoving at 1m/sexperiencing 1N Common Unit 1 Gauss (G) = 10 4 T magnetic field on Earth s surface Direction of magnetic force determined from right hand rule 2

3 Right Hand Rule 3

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5 Example 0.1 C What s the force on charge moving at velocity ~v = (10ˆ 20ˆk) m/s in a magnetic field B ~ =( 3î +4ˆk) 10 4 T ~F = q~v B ~ =+0.1 (10ˆ 20ˆk) ( 3î +4ˆk) 10 4 N = 10 5 (40î + 60ˆ + 30ˆk) N N Note that ~F = q~v ~ B F ~ = qvb sin ) Magnetic force is maximum when Magnetic force is minimum (0) is angle between when ~v and = 90, (i.e.~v? B) ~ =0 or 180 (i.e. ~vk B) ~ 5

6 6.2 Motion of A Point Charge in Magnetic Field Since ~ FB? ~v -field only changes direction of velocity but Generally ~F B = q~v ~ B = qv? B Only need to consider motion? component to -field We have circular motion Magnetic force provides centripetal force on moving charge particles not its magnitude ) F B = m v2 r q vb = m v2 r ) r = mv q B r radius of circular motion 6

7 Time for moving around one orbit Independent of v Use it to measure T = 2 r v = 2 m qb (non-relativistic) m/q Cyclotron Period Generally charged particles with constant velocity moves in helix in presence of constant -field 7

8 Note -field does NO work on particles -field does NOT change K.E. of particles Particle Motion in Presence of E-field & B-field Special Case E ~? B ~ ~F = q ~ E + q~v ~ B When Lorentz Force F ~ E = F ~ B qe = qvb ) v = E/B ) v = E B Charged particles moving at will pass through i.e. crossed ~E and fields without vertical displacement velocity selector Applications Cyclotron (Lawrence & Livingston 1934) Measuring e/m for electrons (Thomson 1897) Mass Spectrometer (Aston 1919) 8

9 6.3 Magnetic Force on Currents Current = many charges moving together Consider a wire segment of length carrying current i L in a magnetic field Total magnetic force Recall i = nqv d A ) Magnetic force on current F ~ = i L ~ B ~ ~L = vector with ~ L = For infinitesimal wire segment d ~ l =(q~v d B {z ~ } ) nal {z } force on one charge carrier { d ~ F = id ~ l ~ B Total number of charge carrier length of current segment direction = direction of current 9

10 Example Force on a semicircle current loop d ~ l = infinitesimal arc length element? ~ B ) dl = Rd ) df = irb d By symmetry argument we only need to consider vertical forces ) Net force F = Z 0 = irb F =2iRB df sin Z 0 sin d (downward) df sin 10

11 Method 2 Write d ~ l in î, ˆ components d ~ l = dl sin î + dl cos ˆ = Rd ( sin î + cos ˆ ) = Bˆk (into the page) ) d ~ F = id ~ l ~ B ) ~ F = Z = irb sin d ˆ irb cos î 0 d ~ F = irb = 2iRBˆ h Z 0 sin d ˆ + Z 0 i cos d î 11

12 Example 2 Current loop in B-field For segment2 F 2 = ibb sin(90 For segment4 + ) =ibb cos (pointing downward) F 2 = ibb sin(90 4 ) =ibb cos (pointing upward) 12

13 For segment 1 For segment 3 F 1 = iab F 3 = iab ) Net force on current loop =0 But net torque on loop about O = = iab b 2 = i ab {z} B sin sin + iab b 2 sin A = area of loop Suppose loop is a coil with N turns of wires Total torque = NiAB sin 13

14 Define Unit vector ˆn to represent area-vector (using right hand rule) Then we can rewrite torque equation as ~ = NiAˆn ~ B Define NiAˆn = ~µ = Magnetic dipole moment of loop ~ = ~µ ~ B 14

15 6.4 Sources of Magnetic Field experiences magnetic force in A moving charge can generate -field ( ~ B -field Magnetic field due to moving point charge = µ 0 4 q~v ˆr r 2 = µ 0 4 q~v ~r r 3 Permeability of free space (magnetic constant) µ 0 = T m/a (or equivalently N/A 2 ) 15

16 = µ 0 4 qv sin r 2 ( maximum minimum when when = 90 =0/180 at P 0 =0= B ~ at P 1 at P 2 < ~ B at P 3 A single moving charge will NOT generate a steady magnetic field stationary charges generate steady -field steady currents generate steady -field ~E 16

17 Principle of Superposition Magnetic field at point P can be obtained by integrating contribution from individual current segments d B ~ = µ 0 4 Z = ) d ~ B = µ 0 4 Note For current around a whole circuit Biot-Savart Law Coulomb s Law Basic element of Basic element of ~E entire circuit dq ~v ˆr r 2 dq ~v = dq d~s dt = id~s id~s ˆr r 2 d ~ B = Z entire circuit is to magnetic field as is to electric field -field Electric charges -field Current element Biot-Savart Law µ 0 4 dq id~s id~s ˆr r 2 17

18 Example 1 Magnetic field due to straight current segment ) d~s ˆr = dz sin = dz sin( ) = dz d r = d dz p d2 + z 2 (Trigonometric identity) 18

19 db = µ 0 4 idz r 2 d r = µ 0i 4 d dz (d 2 + z 2 ) 3/2 ) B = Z L/2 L/2 db = µ 0id 4 Z +L/2 L/2 dz (d 2 + z 2 ) 3/2 B = µ 0i 4 d z (z 2 + d 2 ) 1/2 +L/2 L/2 Limiting Cases ) B = µ 0i 2 d Recall E = 2 0 d B = µ 0i 4 d When L L 2 L 1/2 L d2 d (B-field due to long wire) 1/2 2 = L 4 + d2 1/2 L 2 4 direction of B-field determine from right-hand screw rule for an infinite long line of charge 19

20 Example 2 A circular current loop For every current element id~s 1 generating a magnetic field d B ~ 1 at P there is an opposite current element id~s 2 generating d B ~ 2 so that d ~ B 1 sin = d ~ B 2 sin 20

21 Only vertical component of -field needs to be considered at point P ) -field at point P B = db = µ 0 4 Z B = ) B = µ 0 ir 2 2(R 2 + z 2 ) 3/2 around circuit Z 2 0 * d~s? ˆr z} { ids sin 90 db = µ 0i 4 R r 3 ) B = µ 0iR 2 2r 3 r 2 cos {z } vertical component µ 0 i cos 4 r 2 2Z 2 0 d ds {z} Rd direction of B-field determined from right-hand screw rule 21

22 Limiting Cases B-field at center of loop For ) z R z =0 ) B = µ 0i 2R B = µ 0 ir 2 2z 3 1+ R2 z 2 3/2 Recall E-field for an electric dipole E,= A circular current loop is also called a magnetic dipole Z Arc current B = db cos {z } B = µ 0i 4 R around circuit = µ 0i 4 when µ 0 ir 2 2z 3 / 1 z 3 p 4 0 x 3 z =0) =0here R z = } length { of arc R {z} r 3 R = r =0 Z 0 ds {z} Rd 22

23 Example 3 Magnetic field of a solenoid Solenoid can produce a strong and uniform magnetic field inside its coils ils. Consider a solenoid of length L consisting of N turns of wire 23

24 Define n = Number of turns per unit length = N L Consider -field at distance from center of solenoid For segment of length dz number of current turns d = ndz ) Total current = nidz Using result from one coil in Example 2 we get -field from coils of length at distance from center dz db = µ 0(ni dz)r 2 2r 3 z 24

25 However r = p R 2 +(z d) 2 ) B = Z +L/2 B = µ 0ni 2 L/2 = µ 0niR 2 2 " db L (Integrating over entire solenoid) Z +L/2 L/2 2 r + d + 2 R 2 L d dz [R 2 +(z d) 2 ] 3/2 r R 2 + L 2 d L 2 d 2 # along negative z direction Ideal Solenoid then L R B = µ 0ni [1 + 1] 2 ) B = µ 0 ni direction of B-field determined from right-hand screw rule 25

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