Electricity and Magnetism Charges in Crossed E- and B-Fields

Transcription

1 Electricit and Magnetism Charges in Crossed E- and B-Fields Lana Sheridan De Anza College Nov 4, 2015

2 Last time magnetic force on a charge circular trajectories helical trajectories

3 CHECKPOINT 2 Warm Up Question: Crossed Fields The figure shows four directions for the velocit vector v : of a positivel charged particle moving The diagram through shows a uniform four electric possible field directions E : (directed for out the of the velocit page and v of represented positivel-charged with an encircled dot) particle: and a uniform whichmagnetic direction field could B :. (a) possibl Rank directions result 1, in2, and 3 according to the magnitude of the net force a a net force of zero on the particle? 1 on the particle, greatest first. (b) Of all four directions, which might result in a net force of zero? 2 B E v 1 v + v 3 v 4 (A) 1 (left) (B) 2 (up) 8-5 (C) Crossed 3 (right) Fields: The Hall Effect s we (D) just discussed, 4 (down) a beam of electrons in a vacuum can be deflected b a agnetic field. Can the drifting conduction electrons in a copper wire also be 1 Hallida, Resnick, Walker, 9th ed., page 741. F c

4 CHECKPOINT 2 Warm Up Question: Crossed Fields The figure shows four directions for the velocit vector v : of a positivel charged particle moving The diagram through shows a uniform four electric possible field directions E : (directed for out the of the velocit page and v of represented positivel-charged with an encircled dot) particle: and a uniform whichmagnetic direction field could B :. (a) possibl Rank directions result 1, in2, and 3 according to the magnitude of the net force a a net force of zero on the particle? 1 on the particle, greatest first. (b) Of all four directions, which might result in a net force of zero? 2 B E v 1 v + v 3 v 4 (A) 1 (left) (B) 2 (up) 8-5 (C) Crossed 3 (right) Fields: The Hall Effect s we (D) just discussed, 4 (down) a beam of electrons in a vacuum can be deflected b a agnetic field. Can the drifting conduction electrons in a copper wire also be 1 Hallida, Resnick, Walker, 9th ed., page 741. F c

5 Overview charged particle in E and B fields applications of crossed fields discover of the electron Hall effect

6 The Lorentz Force A charged particle can be affected b both electric and magnetic fields. This means that the total force on a charge is the sum of the electric and magnetic forces: F = qe + qv B This total force is called the Lorentz force. This can alwas be used to deduce the electromagnetic force on a charged particle in E or B fields.

7 Crossed Fields Both electric and magnetic fields interact with moving charges and produce forces on them. This can be used to stud charged particles.

8 Figure A velocit selector. Velocit Selector: Using both electric and magnetic fields Charges are accelerated with and electric field then travel down a apter 29 Magnetic Fields channel with uniform electric and magnetic fields. S B in S F B S E S v S F e Slit P r Detector arra S B in B S Source Velocit selector

9 Velocit Selector: Using both electric and magnetic fields The particles onl reach the end of the channel if F = 0. F = q E + q v B so that means qe = qv B supposing v and B are perpendicular: v = E B

10 Mass Spectrometer After selecting particles to have velocit v = E/B along the channel, the are fed into a magnetic field. S B 0, in S v S F e Slit locit selector. arged particle P Velocit selector r Detector arra S B in S E v S q Figure A mass spectrome-

11 Mass Spectrometer S F B Source S E S v S F e Slit A velocit selector. positivel charged particle g with velocit the path, S v in the r. presa magnetic field directed page and an electric field to the right, it experiences ric force q S E to the right and tic force qv S 3 S B to the left. P Velocit selector r Detector arra S B in S B 0, in S E v S q Where the collide with the detector allows us to find the radius of Mass-to-charge ratio: Figure A mass spectrometer. Positivel charged particles are sent first through a velocit selector and then into a region where the magnetic field S B0 causes the particles to move min a semicircular path and strike a detector arra at P. q = rb 0 v

12 The Discover of the Electron Orienting a magnetic field at right angles to an electric field allowed J.J. Thompson in 1897 to determine the ratio of the electron s charge to its mass: q m. This was significant because it showed that the electron was much lighter than other known particles, establishing it as a new kind of particle.

13 the plates. We can appl this same equation to the beam of electrons in Fig. 28-7; if need be, we can calculate the deflection b measuring the deflection of the beam on screen S and then working back to calculate the deflection at the end of the plates. (Because the direction of the deflection is set b the sign of the Discover of the Electron: Main Idea Electrons are accelerated along the ellow line. + e ran cting rmi- Filament Screen C B E Spot of light (not to b th- V To vacuum pump Screen S Glass envelope

14 the plates. We can appl this same equation to the beam of electrons in Fig. 28-7; if need be, we can calculate the deflection b measuring the deflection of the beam on screen S and then working back to calculate the deflection at the end of the plates. (Because the direction of the deflection is set b the sign of the Discover of the Electron: Main Idea Electrons are accelerated along the ellow line. + e ran cting rmi- Filament Screen C B E Spot of light (not to b th- V To vacuum pump Screen S Glass envelope The electric field deflects them upward. The magnetic field deflects them downward. Adjust the magnetic field until the deflections cancel out and the spot returns to the center.

15 roblem Wh does that tell us about q/m? icle in an electric field Consider onl the E-field from 2 parallel charged plates: Plate 0 E Plate m,q x = L x Fig An ink drop of mass m and charge magnitude Q is A charge particle follows a parabola, because the field is uniform. deflected in the electric field of an ink-jet printer. This is exactl like projectile motion. Let t represent the time required for the drop to pa 1 Figure from Hallida, Resnick, Walker, 9th ed, page 593.

16 Wh does that tell us about q/m? a charged particle in an electric field f an ink-jet An ink drop ive charge of ion between s with speed 1.6 cm. The ric field at all is downward of op at the far n the drop lines) is on the drop eld is directed static force of p. Thus, as the t speed v x,it ion a. Let t represent the time required for the drop to pass = v through the region between i, t + 1 the plates.during 2 at2 t the vertical and horizontal displacements of the drop are (22-31) respectivel. Eliminating t between these two equations and substituting Eq for a,we find QEL2 2mv x 2 0 E Plate Plate m,q x = L Fig An ink drop of mass m and charge magnitude Q is deflected in the electric field of an ink-jet printer. The displacement in the vertical () direction (same dir. as field 1 2 a t 2 and L v x t, ( C)( N/C)( m) 2 x

17 Wh does that tell us about q/m? a charged particle in an electric field f an ink-jet An ink drop ive charge of Plate ion between m,q s with speed E 1.6 cm. The x 0 x = L ric field at all Plate is downward of Fig An ink drop of mass m and charge magnitude Q is op at the Thefar displacement in the vertical () direction (same dir. as field deflected in the electric field of an ink-jet printer. n the drop lines) is on the drop Let t represent the time required for the drop to pass = v through the region between i, t + 1 the plates.during 2 at2 t the vertical and horizontal displacements of the drop are If the particle is movinghorizontall 1 2 a t 2 and onl L onv x t, entr into(22-31) the field, eld is directed v i, = 0. respectivel. Eliminating t between these two equations and static force of substituting Eq for a,we find p. Thus, Also as thea = F E /m, giving: t speed v QEL2 x,it 2 ion a. 2mv x = 1 F E 2 m t2 ( C)( N/C)( m) 2

18 Wh does that tell us about m/q? There is no acceleration in the x direction: x = L = v x t t = L v Therefore the deflection in the direction due to the electric field b the end of the plates (length L): = (qe)l2 2mv 2 This gives an expression for q/m: q m = 2 v 2 E L 2

19 Wh does that tell us about m/q? There is no acceleration in the x direction: x = L = v x t t = L v Therefore the deflection in the direction due to the electric field b the end of the plates (length L): = (qe)l2 2mv 2 This gives an expression for q/m: But what is the speed v? q m = 2 v 2 E L 2

20 Wh does that tell us about m/q? There is no acceleration in the x direction: x = L = v x t t = L v Therefore the deflection in the direction due to the electric field b the end of the plates (length L): = (qe)l2 2mv 2 This gives an expression for q/m: q m = 2 v 2 E L 2 But what is the speed v? v = E/B

21 How to determine v with Crossed Fields The deflection of a charged particle moving through the fields is 0, onl if F net = 0. Assuming v B: F E = F B qe = qvb v = E B Switch on both fields to get a measurement of v. Then switch off the magnetic field and measure the deflection (E-field onl): q m = 2 E B 2 L 2

22 Discover of the Electron For an electron, q = e: e m e = C/kg the mass of the electron m e is reall small.

23 Discover of the Electron For an electron, q = e: e m e = C/kg the mass of the electron m e is reall small. From this ratio and Millikan s oil drop experiments that determined e = C we can find m e. (Do it now!)

24 Discover of the Electron For an electron, q = e: e m e = C/kg the mass of the electron m e is reall small. From this ratio and Millikan s oil drop experiments that determined e = C we can find m e. (Do it now!) m e = kg

25 The Hall effect Or, how to use a current and a field to create a potential difference.

26 he particles that were lighting The Hall effect that the Or, two how deflecting to use forces a current and a field to create a potential difference. vb Electrons flowing in a conductor can also be deflected b a magnetic field! (28-7) d of the charged particles pass- 8-6 and rearranging ield (28-8) hus, the crossed fields allow us ugh Thomson s apparatus. in all matter. He also claimed hdrogen) b a factor of more 6.15.) His m/ q measurement, idered to be the discover of B d i v d F B + B + i + v d High + F E + E Low F B i (a) i (b)

27 rticles that were lighting The Hall effect the two deflecting forces Electrons are pushed to the right until so much negative charge has built up on the right side that the electrostatic force balances the magnetic force. (28-7) i i e charged particles passd rearranging ield (28-8) he crossed fields allow us homson s apparatus. l matter. He also claimed gen) b a factor of more His m/ q measurement, d to be the discover of B d v d F B + B + + v d High + F E + E Low F B i i f a positivel charged parut of the page and repre- (a) At this point we have crossed fields and the potential difference between the left and the right side stabilizes. i (b)

28 The Hall effect The Hall effect allows us to learn man things about the charge carriers in a conductor: their charge their volume densit their drift velocit (for a given current)

29 The Hall effect i Suppose the charge carriers (a) in a conductor (b) were positivel charged: positivel charged parof the page and reprea) Rank directions 1, 2, le, greatest first. (b) Of B Low + + F E F B + v d i E High i i an be deflected b a copper wire also be We would get the opposite polarit for the potential difference! (c) Fig A strip of copper carring a current i is immersed in a magnetic field B :.

30 The Hall effect The constant potential difference that appears across the conductor once the current has stabilized is called the Hall potential difference. V = Ed where d is the width of the conductor.

31 The Hall effect The constant potential difference that appears across the conductor once the current has stabilized is called the Hall potential difference. V = Ed where d is the width of the conductor. V is eas to measure, as is d. This means we can determine the horizontal E-field also.

32 The Hall effect The constant potential difference that appears across the conductor once the current has stabilized is called the Hall potential difference. V = Ed where d is the width of the conductor. V is eas to measure, as is d. This means we can determine the horizontal E-field also. Since the electric force and magnetic force balance: F E = F B ee = ev d B v d = E B We can use our knowledge to estimate v d.

33 The Hall effect Alternativel, we can estimate the densit of charge carriers, n. Remember: v d = I n e A Equating this with our expression for v d on the previous slide: E B = I n e A Rearranging, and using V = Ed and letting t = A/d be the conductor thickness: n = B I e( V )t

34 The Hall effect Remembering V = Ed and t = A/d is the conductor thickness: n = B I e( V )t V is called the Hall Potential Difference: V = B I net

35 The Hall effect - example v : question tron has charge q and is moving through a magnetic field with velocit, the magnetic force acting on the electron is given b Eq Because q is negative, the direction of of edge F : is length opposite d = the 1.5 cross cm, product moving in B : B v : the, which is in A solid metal cube, positive direction at a constant velocit v of magnitude 4.0 m/s. The cube moves through a uniform magnetic field B of magnitude T in the positive z direction. v F : B This is the crossproduct result. z B d (a) d d x v B (b) x This is the resulting electric field. The weak electric field creates a weak electric force. Which cube face is at a lower electric potential and which is at a higher electric potential because of the motion through the field? 1 Hallida, Resnick, Walker, 9th ed, page 743.

36 Reasoning: When the cube first begins to move through the magnetic field, its electrons do also. Because each electron has charge q and is moving through a magnetic field with velocit v :, the magnetic force F : B acting on the electron is given b Eq Because q is negative, the direction of F : is opposite the cross product B : B v :, which is in Free charges in the conductor will feel a force as the move along with the entire conductor through the field. The Hall effect - example question The free charges are electrons. We have to find the direction of the force on them. v This is the crossproduct result. z B d (a) d d x v B (b) x This is the resulting electric field. The weak electric field creates a weak electric force.

37 er and lower electric potential? The Hall effect - example question he electric Free charges field in thecreated conductor b will the feel charge a forceseparation as the move along roduces with the an entire electric conductor force F through : theon field. E qe : each electron E : The free charges are electrons. We have to find the direction of the force on them. is the magnetic e on an electron. KEY IDEAS Electrons are forced to the left face, leaving the right face positive. A v B F B (c) x (d) x

38 The Hall effect - example v : question tron has charge q and is moving through a magnetic field with velocit, the magnetic force acting on the electron is given b Eq Because q is negative, the direction of of edge F : is length opposite d = the 1.5 cross cm, product moving in B : B v : the, which is in A solid metal cube, positive direction at a constant velocit v of magnitude 4.0 m/s. The cube moves through a uniform magnetic field B of magnitude T in the positive z direction. v F : B This is the crossproduct result. z B d (a) d d x v B (b) x This is the resulting electric field. The weak electric field creates a weak electric force. What is the potential difference between the faces of higher and lower electric potential? 1 Hallida, Resnick, Walker, 9th ed, page 743.

39 The Hall effect - example question When does the potential difference between the faces stabilize?

40 v B The Hall effect - example question x When does the potential (c) difference between (d) the faces stabilize? (b) v B F B x x The weak electric field creates a weak electric force. More migration creates a greater electric field. The forces now balance. No more electrons move to the left face. E E F B F E x x (f ) (g) F B F E x l cube moves at constant velocit through a uniform magnetic field. (b) F agnetic force acting on an electron forces the electron E = F to the B left face, d leaving the opposite face positive. (e) (f) The resulting weak electric force on the next electron, but it too is forced to the left face. Now (g) the (h) the electric force matches the magnetic force. (h)

41 v B The Hall effect - example question x When does the potential (c) difference between (d) the faces stabilize? (b) v B F B x x The weak electric field creates a weak electric force. More migration creates a greater electric field. The forces now balance. No more electrons move to the left face. E E F B F E x x (f ) (g) F B F E x l cube moves at constant velocit through a uniform magnetic field. (b) F agnetic force acting on an electron forces the electron E = F to the B left face, d leaving the opposite face positive. (e) (f) The resulting weak electric force on the next electron, but it too is forced ( ee to ) = evb the left face. Now (g) the (h) the electric force matches the magnetic V force. d = vb V = vbd (h)

42 v B The Hall effect - example question x When does the potential (c) difference between (d) the faces stabilize? (b) v B F B x x The weak electric field creates a weak electric force. More migration creates a greater electric field. The forces now balance. No more electrons move to the left face. E E F B F E x x (f ) (g) F B F E x l cube moves at constant velocit through a uniform magnetic field. (b) F agnetic force acting on an electron forces the electron E = F to the B left face, d leaving the opposite face positive. (e) (f) The resulting weak electric force on the next electron, but it too is forced ( ee to ) = evb the left face. Now (g) the (h) the electric force matches the magnetic V force. d = vb V = vbd V = 3.0 mv (h)

43 Related Effects the Hall effect in semiconductors - can be more complex! Depends on the material. the quantum Hall effect - can observe quantization of the Hall potential difference. Can be used to measure the charge of the electron.

44 Summar charged particles in crossed-fields charge and mass of the electron Hall effect Homework Serwa & Jewett: PREVIOUS: Ch 29, Obj Qs: 7; Problems: 13, 15, 23, 73, 80 Ch 29, Problems: 25, 29, 55, (27)