Lect12 - Physics 212 Lecture 12 Today's Concept: Magnetic...

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Unformatted text preview: Physics 212 Lecture 12 Today's Concept: Magnetic Force on moving charges r rr F = qv × B Physics 212 Lecture 12, Slide 1 Physics Music Who is the Artist? A) B) C) D) E) Susan Tedeschi Marcia Ball Shawn Colvin Bonnie Raitt Melissa Etheridge I guess I’m iin a bluesy n mood this week mood Wonderful albums Physics 212 Lecture 12, Slide 2 Physics Your Comments “the cross product stuff and right hand rule” Will do first thing “no main problems with it. Maybe do a computation no based problem, and some stuff like the homework please” please “What are magnetic fields? Why do they exist? What Why does moving charge create this thing called magnetism? I really still don't understand what magnetism is.” magnetism Will do last thing Good Question. We will learn Good how to produce magnetic fields next time. This topic is abstract, though, and difficult to visualize though, 40 “really need to do this earlier in the night. really Maybe then one of my comments would get posted..” get 30 20 “so i really dont get the whole x marks the so spot kinda stuff. I guess i didnt pay attention in pirate school” attention 05 10 0 Confused Confident Physics 212 Lecture 12, Slide 3 Physics Key Concepts: 1) The force on moving charges due to a The magnetic field. magnetic 2) The cross product. Today’s Plan: 1) 2) 3) 05 Review of magnetism Review of cross product Example problem Physics 212 Lecture 12, Slide 4 Physics Magnetic Observations • Bar Magnets N S N S S N N S • Compass Needles N S Magnetic Charge? N 07 S cut in half N S N S Physics 212 Lecture 12, Slide 5 Physics Magnetic Observations • Compass needle deflected by electric current I • Magnetic fields created by electric currents • Magnetic fields exert forces on electric currents (charges in motion) V V I F F F I I I F V V 12 Physics 212 Lecture 12, Slide 6 Physics Magnetism & Moving Charges All observations are explained by two equations: r rr F = qv × B r r µ0 I ds × r ˆ dB = 2 4π r 14 Today Next Week Physics 212 Lecture 12, Slide 7 Physics Cross Product Review • Cross Product different from Dot Product – A●B is a scalar; A x B is a vector – A●B proportional to the component of B parallel to A – A x B proportional to the component of B perpendicular to A • Definition of A x B – Magnitude: ABsinθ – Direction: perpendicular to plane defined by A and B with sense given by right-hand-rule 16 Physics 212 Lecture 12, Slide 8 Physics Remembering Directions: The Right Hand Rule r rr F = qv × B y x F B qv 17 z Physics 212 Lecture 12, Slide 9 Physics Preflight 2 Three points are arranged in a uniform magnetic field. The B field points into the screen. 1) A positively charged particle is located at point A and is stationary. The direction of the magnetic force on the particle is: r rr F = qv × B 50 40 30 The particle’s velocity is zero. There can be no magnetic force. 20 10 0 18 Physics 212 Lecture 12, Slide 10 Physics Preflight 4 Three points are arranged in a uniform magnetic field. The B field points into the screen. A B C D E 1) A positively charged particle is located at point A and is stationary. The direction of the magnetic force on the particle is: r rr F = qv × B BB 50 40 qv 30 F X 20 B 10 0 21 Physics 212 Lecture 12, Slide 11 Physics Cross Product Practice r rr F = qv × B • protons (positive charge) coming out of screen • Magnetic field pointing down • What is direction of force on POSITIVE charge? BB A) Left -x B) Right +x C) UP D) Down +y -y E) Zero y x B 24 s n ee cr z Physics 212 Lecture 12, Slide 12 Physics Motion of Charge q in Uniform B Field • Force is perpendicular to v – Speed does not change – Uniform Circular Motion x vx x x • Solve for R: r rr F = qv × B ⇒ F = qvB x Rx x x x vx x x x x Fx xxx q F F x x xF x 2 v a= R v2 qvB = m R xxx q q xxx xv x x x x x x q mv R= qB x x x x xv x x Uniform B into page Demo 30 Physics 212 Lecture 12, Slide 13 Physics R= LHC mv p = qB qB 17 miles diameter Physics 212 Lecture 12, Slide 14 Physics Preflight 6 The drawing below shows the top view of two interconnected chambers. Each chamber has a unique magnetic field. A positively charged particle is fired into chamber 1, and observed to follow the dashed path shown in the figure. BB r rr F = qv × B 50 40 qv 30 . B 20 F 10 0 34 Physics 212 Lecture 12, Slide 15 Physics Preflight 8 The drawing below shows the top view of two interconnected chambers. Each chamber has a unique magnetic field. A positively charged particle is fired into chamber 1, and observed to follow the dashed path shown in the figure. Observation: R2 > R1 R= 36 mv qB 60 50 40 B1 > B2 30 20 10 0 Physics 212 Lecture 12, Slide 16 Physics Calculation A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region q,m containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. x0/2 exits here XXXXXXXXX X X X X X X X X X x0 XXXXXXXXX d XXXXXXXXX enters here B B What is B? E • Conceptual Analysis – What do we need to know to solve this problem? (A) Lorentz ForcerLaw r rr ( F = qv × B + qE ) – (C) V definition (D) Conservation of Energy/Newton’s Laws – (B) E field definition BB (E) All of the above Absolutely ! We need to use the definitions of V and E and either conservation of energy or Newton’s Laws to understand the motion of the particle before it enters the B field. We need to use the Lorentz Force Law (and Newton’s Laws) to determine what happens in the magnetic field. Physics 212 Lecture 12, Slide 17 Physics Calculation A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region q,m containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. x0/2 exits here XXXXXXXXX X X X X X X X X X x0 XXXXXXXXX d XXXXXXXXX enters here B B What is B? E • Strategic Analysis – – – Calculate v, the velocity of the particle as it enters the magnetic field Use Lorentz Force equation to determine the path in the field as a function of B Apply the entrance-exit information to determine B Let’s Do It !! Physics 212 Lecture 12, Slide 18 Physics Calculation A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region q,m containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. exits here x0/2 XXXXXXXXX X X X X X X X X X x0 XXXXXXXXX d XXXXXXXXX enters here B B What is B? E • What is v0, the speed of the particle as it enters the magnetic field ? 2E m (A) (A) vo = 2qEd vo = m (B) vo = 2ad (C) 2qE md (D) (D) vo = vo = qEd m (E) BB • Why?? – Conservation of Energy • • – Initial: Energy = U = qV = qEd Final: Energy = KE = ½ mv02 Energy Newton’s Laws • • a = F/m = qE/m v02 = 2ad 2 vo = 2 1 mv 2 = qEd o 2 qE d m vo = vo = 2qEd m 2qEd m Physics 212 Lecture 12, Slide 19 Physics Calculation A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region q,m containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? vo = x0/2 exits here XXXXXXXXX X X X X X X X X X x0 XXXXXXXXX d XXXXXXXXX enters here B B 2qEd m E • What is the path of the particle as it moves through the magnetic field? XXXXXXXXX XXXXXXXXX XXXXXXXXX (A) X X X X X X X X X (A) (B) XXXXXXXXX XXXXXXXXX XXXXXXXXX XXXXXXXXX XXXXXXXXX XXXXXXXXX XXXXXXXXX (C) X X X X X X X X X (C) • Why?? – Path is circle ! • • • Force is perpendicular to the velocity Force produces centripetal acceleration Particle moves with uniform circular motion BB Physics 212 Lecture 12, Slide 20 Physics Calculation A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region q,m containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. exits here XXXXXXXXX X X X X X X X X X x0 XXXXXXXXX d XXXXXXXXX enters here B B 2qEd m vo = What is B? x0/2 E • What is the radius of path of particle? R = xo R = 2 xo 1 R = xo 2 (A) (A) (B) (C) mvo R= qB (D) (D) 2 vo R= a BB (E) • Why?? R xo/2 XXXXXXXXX XXXXXXXXX XXXXXXXXX XXXXXXXXX Physics 212 Lecture 12, Slide 21 Physics Calculation A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region q,m containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? 2 B= xo 2mEd q (A) (A) vo = 2qEd m exits here x0/2 XXXXXXXXX X X X X X X X X X x0 XXXXXXXXX d XXXXXXXXX enters here B B R = 1 x0 2 E B= v m B=E 2qEd (B) 1 B= xo (C) E 2mEd q B= (D) (D) mvo qxo BB (E) • Why?? r r F = ma 2 vo qvo B = m R m vo B= qR B= m2 q xo 2qEd m 2 xo 2mEd q B= Physics 212 Lecture 12, Slide 22 Physics Follow-Up A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region q,m containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? B= 2 xo x0/2 exits here XXXXXXXXX X X X X X X X X X x0 XXXXXXXXX d XXXXXXXXX enters here B B 2mEd q E • Suppose the charge of the particle is doubled (Q = 2q),while keeping the mass constant. How does the path of the particle change? X X X X X X X X X XX X X X X X X X X X X X X X X X X X X X X X X X X X XX X X X X X X X X X X X X X X X X X X X X X X X X X XX X X X X X X X X X X X X X X X X BB X X X X X X X X X XX X X X X X X X X X X X X X X X X (A) (A) (B) (C) (C) 1. q changes fl v changes No slam dunk.. As Expected ! Several things going on here 2. q & v change fl F changes 3. v & F change fl R changes Physics 212 Lecture 12, Slide 23 Physics Follow-Up A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region q,m containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? B= 2 xo x0/2 exits here XXXXXXXXX X X X X X X X X X x0 XXXXXXXXX d XXXXXXXXX enters here B B 2mEd q E • Suppose the charge of the particle is doubled (Q = 2q),while keeping the mass constant. How does the path of the particle change? BB How does v, the new velocity at the entrance, compare to the original velocity v0? vo (A) v = (A) 2 • Why?? 1 mv 2 2 v (B) v = o (C) v = vo 2 2 = QEd = 2qEd = 2 1 mvo 2 (D) v = 2vo (D) 2 v 2 = 2vo (E) v = 2vo v = 2vo Physics 212 Lecture 12, Slide 24 Physics Follow-Up A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region q,m containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? B= 2 xo x0/2 exits here XXXXXXXXX X X X X X X X X X x0 XXXXXXXXX d XXXXXXXXX enters here B B 2mEd q E • Suppose the charge of the particle is doubled (Q = 2q),while keeping the mass constant. How does the path of the particle change? v = 2vo How does F, the magnitude of the new force at the entrance, compare to F0, the magnitude of the original force? (A) F = (A) Fo 2 (B) F = Fo (C) F = 2 Fo (D) F = 2 Fo (D) BB (E) F = 2 2 Fo • Why?? F = QvB = 2q ⋅ 2vo ⋅ B F = 2 2 Fo Physics 212 Lecture 12, Slide 25 Physics Follow-Up A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region q,m containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? B= 2 xo x0/2 exits here XXXXXXXXX X X X X X X X X X x0 XXXXXXXXX d XXXXXXXXX enters here B B 2mEd q E • Suppose the charge of the particle is doubled (Q = 2q),while keeping the mass constant. How does the path of the particle change? v = 2vo F = 2 2 Fo How does R, the radius of curvature of the path, compare to R0, the radius of curvature of the original path? (A) R = (A) Ro 2 (B) R = Ro 2 (C) R = Ro (D) R = 2 Ro (D) (E) R = 2 Ro • Why?? v2 F =m R BB v2 R=m F 2 2 2vo vo R R=m =m =o 2 2 Fo 2 Fo 2 Physics 212 Lecture 12, Slide 26 Physics Follow-Up A particle of charge q and mass m is accelerated from rest by an electric field E through a distance d and enters and exits a region q,m containing a constant magnetic field B at the points shown. Assume q,m,E,d, and x0 are known. What is B? B= 2 xo x0/2 exits here XXXXXXXXX X X X X X X X X X x0 XXXXXXXXX d XXXXXXXXX enters here B B 2mEd q E • Suppose the charge of the particle is doubled (Q = 2q),while keeping the mass constant. How does the path of the particle change? X X X X X X X X X XX X X X X X X X X X X X X X X X X X X X X X X X X X XX X X X X X X X X X X X X X X X X X X X X X X X X X XX X X X X X X X X X X X X X X X X X X X X X X X X X XX X X X X X X X X X X X X X X X X (A) (A) (B) Ro R= 2 (C) (C) A Check: (Exercise for Student) Given our result for B (above), can you show: R= 1 2mEd B Q R R= o 2 Physics 212 Lecture 12, Slide 27 Physics ...
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