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Announcements Exam 2 is one week from tomorrow. Contact me by the end of the Wednesday’s lecture if you have special circumstances different than for exam 1. Today’s agenda: Magnetic Fields. You must understand the similarities and differences between electric fields and field lines, and magnetic fields and field lines. Magnetic Force on Moving Charged Particles. You must be able to calculate the magnetic force on moving charged particles. Motion of a Charged Particle in a Uniform Magnetic Field. You must be able to calculate the trajectory and energy of a charged particle moving in a uniform magnetic field. Magnetic Forces on Currents and Current-Carrying Wires. You must be able to calculate the magnetic force on currents. Velocity Selector and Mass Spectrometer. You must be able to work these detailed examples. Magnetism Recall how there are two kinds of electrical charge (+ and -), and likes repel, opposites attract. Similarly, there are two kinds of magnetic poles (north and south), and like poles repel, opposites attract. S S N S N Repel S N N S N Repel S Attract N S N N S Attract There is an important difference between magnetism and electricity: it is possible to have isolated + or – electric charges, but isolated N and S poles have “never” been observed.* - + N S I.E., every magnet has BOTH a N and a S pole, no matter how many times you “chop it up.” SS N SN N *But see here, here, here, and here. Magnetic Fields The earth has associated with it a magnetic field, with poles near the geographic poles. The pole of a magnet attracted to the earth’s north geographic pole is the magnet’s north pole. The pole of a magnet attracted to the earth’s south geographic pole is the magnet’s south pole. N S http://hyperphysics.phyastr.gsu.edu/hbase/magnetic/ magearth.html Just as we used the electric field to help us “explain” and visualize electric forces in space, we use the magnetic field to help us “explain” and visualize magnetic forces in space. Magnetic field lines point in the same direction that the north pole of a compass would point. Later I’ll give a better definition for magnetic field direction. Magnetic field lines are tangent to the magnetic field. The more magnetic field lines in a region in space, the stronger the magnetic field. Outside the magnet, magnetic field lines point away from N poles (*why?). Huh? *The N pole of a compass would “want to get to” the S pole of the magnet. For those of you who aren’t going to pay attention until you have been told the secret behind the naming of the earth’s magnetic poles… The north pole of a compass needle is defined as the end that points towards the geographic “Santa Claus” north pole, which experts in the field of geomagnetism call “the geomagnetic north” or “the Earth’s North Magnetic Pole.” If you think about it, the people to whom compasses meant the most— sailors—defined magnetic north as the direction the north poles of their compass needles pointed. Their lives depended on knowing where they were, so I guess it is appropriate that we acknowledge their precedence. Thus, by convention, the thing we call Earth’s North Magnetic Pole is actually the south magnetic pole of Earth’s magnetic field. For those of you who are normal humans, just ignore the above. Here’s a “picture” of the magnetic field of a bar magnet, using iron filings to map out the field. The magnetic field ought to “remind” you of the earth’s field. We use the symbol B for magnetic field. Remember: magnetic field lines point away from north poles, and towards south poles. S N The SI unit* for magnetic field is the Tesla. 1 kg 1 T= Cs These units come from the magnetic force equation, which appears three slides from now. In a bit, we’ll see how the units are related to other quantities we know about, and later in the course we’ll see an “official” definition of the units for the magnetic field. *Old unit, still sometimes used: 1 Gauss = 10-4 Tesla. The earth’s magnetic field has a magnitude of roughly 0.5 G, or 0.00005 T. A powerful permanent magnet, like the kind you might find in headphones, might produce a magnetic field of 1000 G, or 0.1 T. http://liftoff.msfc.nasa.gov/academy/space/mag_field.html The electromagnet in the basement of Physics that my students (used to) use in experiments can produce a field of 26000 G = 26 kG = 2.6 T. Superconducting magnets can produce a field of over 10 T. Never get near an operating superconducting magnet while wearing a watch or belt buckle with iron in it! Today’s agenda: Magnetic Fields. You must understand the similarities and differences between electric fields and field lines, and magnetic fields and field lines. Magnetic Force on Moving Charged Particles. You must be able to calculate the magnetic force on moving charged particles. Motion of a Charged Particle in a Uniform Magnetic Field. You must be able to calculate the trajectory and energy of a charged particle moving in a uniform magnetic field. Magnetic Forces on Currents and Current-Carrying Wires. You must be able to calculate the magnetic force on currents. Velocity Selector and Mass Spectrometer. You must be able to work these detailed examples. Magnetic Fields and Moving Charges A charged particle moving in a magnetic field experiences a force. The following equation (part of the Lorentz Force Law) predicts the effect of a magnetic field on a moving charged particle: F = qv B force on particle velocity of charged particle Oh nooo! The little voices are back. magnetic field vector What is the magnetic force if the charged particle is at rest? What is the magnetic force if v is (anti-)parallel to B? Vector notation conventions: is a vector pointing out of the paper/board/screen (looks like an arrow coming straight for your eye). is a vector pointing into the paper/board/screen (looks like the feathers of an arrow going away from eye). Direction of magnetic force--Use right hand rule: fingers out in direction of v, thumb perpendicular to them rotate your hand until your palm points in the direction of B (or bend fingers through smallest angle from v to B) thumb points in direction of F on + charge Your text presents two alternative variations (curl your fingers, imagine turning a right-handed screw). There is one other variation on the right hand rule. I’ll demonstrate all variations in lecture sooner or later. still more variations: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html Complication: your Physics 23 text uses the right hand rule shown in the figure, so we’ll use the same rule in Physics 2135. There are a number of variations of this rule. Unfortunately, most of the Youtube videos I find say to use your palm for A , your thumb for B , and your outstretched fingers for A B . This includes the MIT Open Courseware site. I’ll show you later that both ways are correct. I am going to use our textbook’s technique. You can use whatever works for you! F = q vB sinθ Direction of magnetic force: F + Fingers of right hand out in direction of velocity, thumb perpendicular to them. Bend your fingers until they point in the direction of magnetic field. Thumb points in direction of magnetic force on + charge. B - v F B v Force is “down” because charge is -. v F? - B I often do this: fingers out in direction of velocity, thumb perpendicular to them. Rotate your hand until your palm points in the direction of magnetic field. Thumb points in direction of magnetic force on + charge. “Foolproof” technique for calculating both magnitude and direction of magnetic force. F = qv B ˆi F = q det v x B x ˆj vy By kˆ v z B z http://www.youtube.com/watch?v=21LWuY8i6Hw All of the right-hand rules are just techniques for determining the direction of vectors in the cross product without having to do any actual math. ^ Example: a proton is moving with a velocity v = v0j in a region ^ of uniform magnetic field. The resulting force is F = F0i. What is the magnetic field (magnitude and direction)? To be worked at the blackboard. Blue is +, red is -. Image from http://www.mathsisfun.com/algebra/matrix-determinant.htm ^ Example: a proton is moving with a velocity v = v0j in a region ^ of uniform magnetic field. The resulting force is F = F0i. What is the magnetic field (magnitude and direction)? v = v 0 ˆj F = F0 ˆi F = qv B ˆi ˆ F = F0 i = q det v x B x ˆj vy By kˆ v z = e B z ˆi det 0 B x ˆj v0 By kˆ 0 B z F0ˆi = e ˆi v 0 Bz B y 0 ˆj 0 B z B x 0 kˆ 0 B y B x v 0 ^ Example: a proton is moving with a velocity v = v0j in a region ^ of uniform magnetic field. The resulting force is F = F0i. What is the magnetic field (magnitude and direction)? F0 ˆi = ev 0B z ˆi 0jˆ eB x v 0kˆ = ev 0B z ˆi + 0 ˆj+ eB x v 0 kˆ F0 = ev 0B z F0 Bz = ev 0 and and 0 = eB x v 0 Bx = 0 What is By???? It could be anything. Not enough information is provided to find By. We can’t find the magnitude and direction of the magnetic field using only the information given! ^ Example: a proton is moving with a velocity v = v0j in a region ^ of uniform magnetic field. The resulting force is F = F0i. What is the minimum magnitude magnetic field that can be present? Bz = F0 ev 0 By = 0 B min = and Bx = 0 for a minimum magnitude magnetic field, so... F0 ev 0 Homework Hint F = q vB sinθ A charged particle moving along or opposite to the direction of a magnetic field will experience no magnetic force. Conversely, the fact that there is no magnetic force along some direction does not mean there is no magnetic field along or opposite to that direction. It’s OK to use F = q vB if you know that v and B are perpendicular, or you are calculating a minimum field to produce a given force (understand why). Alternative* view of magnetic field units. F = q vB sinθ F B= q v sinθ N N = B = T = C m/s A m *“Official” definition of units coming later. Remember, units of field are force per “something.” Today’s agenda: Magnetic Fields. You must understand the similarities and differences between electric fields and field lines, and magnetic fields and field lines. Magnetic Force on Moving Charged Particles. You must be able to calculate the magnetic force on moving charged particles. Motion of a Charged Particle in a Uniform Magnetic Field. You must be able to calculate the trajectory and energy of a charged particle moving in a uniform magnetic field. Magnetic Forces on Currents and Current-Carrying Wires. You must be able to calculate the magnetic force on currents. Velocity Selector and Mass Spectrometer. You must be able to work these detailed examples. Motion of a charged particle in a uniform magnetic field Example: an electron travels at 2x107 m/s in a plane perpendicular to a 0.01 T magnetic field. Describe its path. Example: an electron travels at 2x107 m/s in a plane perpendicular to a 0.01 T magnetic field. Describe its path. The force on the electron (remember, its charge is -) is always perpendicular to the velocity. If v and B are constant, then F remains constant (in magnitude). The above paragraph is a description of uniform circular motion. B v F F - v The electron will move in a circular path with a constant speed and acceleration = v2/r, where r is the radius of the circle. Motion of a proton in a uniform magnetic field v Bout FB r v + + FB FB v + The force is always in the radial direction and has a magnitude qvB. For circular motion, a = v2/r so mv 2 F = q vB = r q rB mv v= r= m qB Thanks to Dr. Waddill for the use of the picture and following examples. The period T is 2πr 2πm T= = v qB The rotational frequency f is called the cyclotron frequency qB 1 f= = T 2πm Remember: you can do the directions “by hand” and calculate using magnitudes only. Helical motion in a uniform magnetic field If v and B are perpendicular, a charged particle travels in a circular path. v remains constant but the direction of v constantly changes. If v has a component parallel to B, then v remains constant, and the charged particle moves in a helical path. There won’t be any test problems on helical motion. B v + v Apply B-field perpendicular Electrons confined to move to in aplane plane Thanks to Dr. Yew San Hor for this and the next slide. Apply B-field perpendicular to plane Lorentz Force Law If both electric and magnetic fields are present, F = q E + v B . Applications See your textbook for numerical calculations related the next two slides. If I have time, I will show the mass spectrometer today. The energy calculation in the mass spectrometer example is often useful in homework. Velocity Selector Bout ---------------- v + q E + FE = qE FB = qv B Thanks to Dr. Waddill for the use of the picture. When the electric and magnetic forces balance then the charge will pass straight through. This occurs when FE = FB or qE = qvB or E v= B Mass Spectrometers Mass spectrometers separate charges of different mass. When ions of fixed energy enter a region of constant magnetic field, they follow a circular path. V The radius of the path depends on the mass/charge ratio and speed of the ion, and the magnitude of the magnetic field. x = 2r and mv r= qB Worked example with numbers at the end of this lecture. B S +q x r Thanks to Dr. Waddill for the use of the picture. Today’s agenda: Magnetic Fields. You must understand the similarities and differences between electric fields and field lines, and magnetic fields and field lines. Magnetic Force on Moving Charged Particles. You must be able to calculate the magnetic force on moving charged particles. Motion of a Charged Particle in a Uniform Magnetic Field. You must be able to calculate the trajectory and energy of a charged particle moving in a uniform magnetic field. Magnetic Forces on Currents and Current-Carrying Wires. You must be able to calculate the magnetic force on currents. Velocity Selector and Mass Spectrometer. You must be able to work these detailed examples. Magnetic Forces on Currents So far, I’ve lectured about magnetic forces on moving charged particles. F = qv B Actually, magnetic forces were observed on current-carrying wires long before we discovered what the fundamental charged particles are. Experiment, followed by theoretical understanding, gives F = IL B. For reading clarity, I’ll use L instead of the l your text uses. If you know about charged particles, you can derive this from the equation for the force on a moving charged particle. It is valid for a straight wire in a uniform magnetic field. Here is a picture to help you visualize. It came from http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html. Homework Hint F = ILB sin F max = ILB For fixed L, B: this is the biggest force you can get from a given current I. For bigger F, you need bigger I. F B I L F = IL B Valid for straight wire, length L inside region of magnetic field, constant magnetic field, constant current I, direction of L is direction of conventional current I. You could apply this equation to a beam of charged particles moving through space, even if the charged particles are not confined to a wire. What if the wire is not straight? B I ds dF dF = I ds B F = dF F = I ds B Integrate over the part of the wire that is in the magnetic field region. Homework Hint: if you have a tiny piece of a wire, just calculate dF; no need to integrate. Note: I generally use ds for an infinitesimal piece of wire, instead of dl. Font choice may make “l” look like “1.” It’s a pain to search the fonts for a script lowercase l: l. Example: a wire carrying current I consists of a semicircle of radius R and two horizontal straight portions each of length L. It is in a region of constant magnetic field as shown. What is the net magnetic force on the wire? I B R L L y x There is no magnetic force on the portions of the wire outside the magnetic field region. First look at the two straight sections. F = IL B L B, so F1 =F2 = ILB I y F1 R L F2 B L x Next look at the semicircular section. Calculate the infinitesimal force dF on an infinitesimal ds of current-carrying wire. I y F1 dF ds d R L F2 B L x ds subtends the angle from to +d. Why did I call that angle instead of ? The infinitesimal force is dF =I ds B. Because we usually use for the angle in the cross product. ds B, so dF = I ds B. Arc length ds =R d. Finally, dF =I R d B. Calculate the ycomponent of F. dFy = I R d B sin I Fy = dFy y 0 Fy = I R d B sin 0 dFy d F1 dF ds R L L x Fy = I R B sind 0 Fy = -I R B cos 0 Fy = 2 I R B F2 B Interesting—just the force on a straight horizontal wire of length 2R. Does symmetry give you Fx immediately? Or, you can calculate the x component of F. I y dFx = -I R d B cos Fx = - I R d B cos F1 dF ds d dFx R L F2 B L x 0 Fx = -I R B cosd 0 Fx = - I R B sin 0 Fx = 0 Sometimes-Useful Homework Hint Symmetry is your friend. Total force: F = F1 + F2 + Fy F = ILB + ILB + 2IRB I y Fy F1 dF ds R L F2 B L F = 2IB L + R F = 2IB L + R ˆj x We probably should write the force in vector form. Possible homework hint: how would the result differ if the magnetic field were directed along the +x direction? If you have difficulty visualizing the direction of the force using the right hand rule, pick a ds along each different segment of the wire, express it in unit vector notation, and calculate the cross product. Example: a semicircular closed loop of radius R carries current I. It is in a region of constant magnetic field as shown. What is the net magnetic force on the loop of wire? FC B R y x I We calculated the force on the semicircular part in the previous example (current is flowing in the same direction there as before). F =2 I R B C Next look at the straight section. FS = IL B L B, and L=2R so y FS = 2IRB FC B R I FS x Fs is directed in the –y direction (right hand rule). Fnet =FS +Fc = -2IRB ˆj+2IRB ˆj = 0 The net force on the closed loop is zero! This is true in general for closed loops in a uniform magnetic field. Today’s agenda: Magnetic Fields. You must understand the similarities and differences between electric fields and field lines, and magnetic fields and field lines. Magnetic Force on Moving Charged Particles. You must be able to calculate the magnetic force on moving charged particles. Motion of a Charged Particle in a Uniform Magnetic Field. You must be able to calculate the trajectory and energy of a charged particle moving in a uniform magnetic field. Magnetic Forces on Currents and Current-Carrying Wires. You must be able to calculate the magnetic force on currents. Velocity Selector and Mass Spectrometer. You must be able to work these detailed examples. Velocity Selector Bout ---------------- v + E The magnetic field is uniform inside the dashed-line region. What speed will allow the proton to pass undeflected through region of uniform electric field? Velocity Selector Bout FE = qE ---------------- v + E + v FB = qv B When the electric and magnetic forces balance then the charge will pass straight through. FE = FB q E = q vB or E v= B Velocity Selector Bout FE = qE ---------------- v + E + v FB = qv B This only works if the electric and magnetic fields are perpendicular and oriented so that qv B is antiparallel to E. Also, we simplified the calculation by making v perpendicular to B. Mass Spectrometers Mass spectrometers separate charges of different mass. When ions of fixed energy enter a region of constant magnetic field, they follow a circular path. The radius of the path depends on the mass/charge ratio and speed of the ion, and the magnitude of the magnetic field. x = 2r and mv r= qB Worked example with numbers to follow. V S +q x B R Thanks to Dr. Waddill for the use of the picture. Example: ions from source S enter a region of uniform magnetic field B that is perpendicular to the ion path. The ions follow a semicircle and strike the detector plate at x = 1.7558 m from the point where they entered the field. If the ions have a charge of 1.6022 x 10-19 C, the magnetic field has a magnitude of B = 80.0 mT, and the accelerating potential is V = 1000.0 V, what is the mass of the ion? V S B +q x R Step 1: find the final speed of the ions if they have a charge of 1.6022 x 10-19 C and the accelerating potential is V = 1000.0 V? V Let’s focus on the part of the diagram where the ions from the source are being accelerated. S B +q If the ions are “boiling” off a source filament, their initial velocity directions will be almost random, so their average speed will be almost zero. Let’s assume the ions start at rest (corresponding to zero average speed). x R Step 1: find the final speed of the ions if they have a charge of 1.6022 x 10-19 C and the accelerating potential is V = 1000.0 V? V Ef Ei Wother if 0 0 Kf Uf Ki Ui Wother if S +q vi = 0 +q v = ? Kf Uf +Ui Uf Ui U 1 mv 2 U qV 2 2qV v m Notes: we don’t know m, so let’s keep this symbolic for now. Besides, we want to solve for m later. Also, V must be negative, which makes sense (a + charge would be accelerated away from a higher potential). Don’t get your v’s and V’s mixed up! Use a script v for speed! Step 2: find the radius of the circular path followed by the ions in the region of uniform magnetic field of magnitude B. V S B +q x R Step 2: find the radius of the circular path followed by the ions in the region of uniform magnetic field of magnitude B. Let’s focus on the magnetic field region of the diagram. +q v R x B Step 2: find the radius of the circular path followed by the ions in the region of uniform magnetic field of magnitude B. For motion in a circular path... B +q v2 FB ma m R FB qv B qvBsin 90 2 v q vB m R x v FB R mv R qB On an exam, we’ll probably let you get away without using the absolute value signs, even though it makes our brains ache when we do that. Step 3: solve for m, using q=1.6022x10-19 C, V=1000.0 V, B = 80.0 mT, and R=x/2=1.7558/2=0.8779 m. V Now we can put it all together. S 2qV v m mv R qB B +q We know q, R, B, and V. We have two unknowns, v and m, and two equations, so we can solve for m. x R 2qV m 2qV m 2V m 2 2 m v m R2 2 2 2 2 2 2 2 q B q B q B q B 2 Step 3: solve for m, using q=1.6022x10-19 C, V=-1000.0 V, B = 80.0 mT, and R=x/2=1.7558/2=0.8779 m. R 2 V 2V m q B2 S B +q q B2 R 2 m 2V x 1.6022 10 80 10 0.8779 m 19 3 2 2 R 2 1000 m 3.9515 1025 kg uranium-238! 1 atomic mass unit (u) equals 1.66x10-27 kg, so m=238.04 u.