PHYS 41 - Fields and Waves
Consider a charge moving in a magnetic field B field into plane F=ma acceleration change of direction of velocity Take F as centripetal force: 0 F qvb cos90 qvb F Centripetal v m R R mv qb Since F is perpendicular to v, the force leads to a circular motion Radius of circular motion Angular velocity: v R qb m
Radiation is produced by oscillating electrons Angular frequency of electrons = radiation frequency f qb m B fm q B fm q 6 31 450 10 Hz 9.11 10 kg 1.6 10 19 C 0.0877 T Homework : Problem 7.15 and 7.1 (both 11 th and 1 th Eds.)
Summary of Section 7.4 Charge in a magnetic field move in circular path Radius: R mv qb Angular frequency: v R qb m Hmwk Sect. 7.4: Probls 7.15 and 7.1 (11 th and 1 th Eds.)
Charges pass by an area with a B and an E crossed The E field deflects + charges down The B field deflects + charges up Particles will go through undeflected if F e and F m balance each other F F qvb qe m e E v B Only those particles with v = E/B will go through undeflected Crossed B and E fields act as a velocity selector
Velocity selectors are used in analog TV screens J.J. Thomson used a velocity selector to study the electron: Accelerated electrons through a potential V Sent them through a v-selector with E and B adjusted to let them to go undeflected Calculated the ratio of the charge to the mass of the electron 1 1 KE ev mv m B e m E VB He received the Nobel prize for discovering that the electron was a particle E His son received it for discovering that it was a wave... Hmwk : Probl. 7.7 (11 th Ed.) or 7.5 (1 th Ed.)
1 1 E KE ev mv m ; v B E B B me 31 6 9.1 10 kg 6 10 N/C 19 ev 1.6 10 150 C V 0.685 0.86 T Homework : Problem 7.31 (both 11 th or 1 th Eds.)
Summary of Section 7.5 Velocity selector v E B Ratio of charge to mass of electron e m E VB Hmwk Sect. 7.5: Problems 7.7 and 7.31 (11 th Ed.) or 7.5 and 7.31 (1 th Ed.)
Consider charges moving inside a conductor placed in a magnetic field B Each charge feels a force Magnitude: Direction: to the left F qvd B F qv B d V d = drift velocity Force is to the left Now obtain the total force on the whole length of wire Number of charges in length of wire n = density of charges = nla A = cross sectional area Total force in F nal qv length of wire db nqvda lb IlB
For cases in which B is not to perpendicular to wire F Il B F IlB sin Some examples For a segment of a wire df Idl B
Force on wire? F IlB sin 50 A 1 m 1. T sin 45 4.4 N Again using vectors: Force pointing out of the plane Hmwk: Probl. 7.34 (11 th Ed.) or 7.36 (1 th Ed.) Solution: a) b) and c): 7.06x 10-3 N 0 ˆ 0ˆ 0 ˆ F Il B I li B cos45 i Bsin 45 j iˆ ˆj kˆ 0 I l 0 0 IlB sin 45 k 4.4 Nk B 0 0 cos 45 B sin 45 0 ˆ ˆ
Find the force on the three segments of wire Segment 3: length L into the plane Segment 1 Segment Segment 1: Segment 3: F ILB sin90 0 ˆj ILBj ˆ 0 F IlBsin180 0 Segment is more interesting...
Segment 3: x components of force cancel net force will be in the y direction 0 0 180 180 F df IB dl cos x IBR 0 0 0 0 180 0 0 x 0 d cos 0 0 0 180 180 F df IB dl sin y 0 0 0 0 180 0 0 y 0 IBR d sin IRB Total force on all three segments: F ILBj ˆ IRBj ˆ IB L R ˆj Homework : Problem 7.35 and 7.39 or 7.37 and 7.39 (1 th Ed.)
Summary of Section 7.6 Forces on current carrying conductors F Il B df Idl B Add forces Hmwk Sect. 7.6: Probls 7.34, 7.35 and 7.39 (11 th Ed.) or 7.36, 7.37 and 7.39 (1 th Ed.)
Consider a current loop in a B field perpendicular to current x components of force cancel y components of force cancel net force on loop will be zero Now consider a loop tilted respect to the B field Again x and y components of force cancel net force on loop will be zero but there will be a torque on loop
Loop tilted respect to the B field Forces in x and y cancel Forces in x form a couple Segments in y direction feel forces: F Il B F IaB in ± x Separation between these forces is and torque is then: r F bsin Fbsin IaB bsin IAB sin Bsin Or in vector form: Magnetic dipole moment is: B IA Valid for all geometries
Some examples What are the directions of IA and B? A current loop in a B field has potential energy: U B Bcos
Magnetic moment: NIA Total 30 5 A 0.05 m 1.18 Am Torque, method 1: Torque, method : Bsin 0 1.18 1. sin 90 1.41 Am T Nm 0 NIBAsin 30 5 A 1. T 0.05 m sin 90 1.41 Nm Torque tends to rotate
Homework: Probls 7.4 and 7.44 (both 11 th and 1 th Eds.) Soln. 7.4: a), b) and c): 7.06x 10-3 N
U Bcos90 0 Initial potential energy: 0 Initial Final potential energy: 0 B cos 0 1.18 1. 1.41 UFinal Am T J Change in potential energy: UFinal UInitial 1.41 J Homework : Problem 7.47 (both 11 th and 1 th eds.)
How do magnets work? If for any loop in a uniform B field The forces in x and y cancel Forces form a couple that produces only a torque Then, how do magnets attract or repel? Magnets affect atomic loops in materials B field of magnets is not uniform Forces produced are not zero
Summary of Section 7.7 In loops : Forces in x and y cancel Forces in x form a couple Loops feel a torque Magnetic dipole moment is: Potential energy: B IA U B Bcos Magnets work by exerting forces on atomic loops in materials thanks to their not uniform B field Hmwk Sect. 7.7: Probls 7.4, 7.44 and 7.47 (both 11 th Ed and 1 th Eds.)
PHYS 41 - Fields and Waves
PHYS 41 - Fields and Waves