Magnetism

1. Magnetism PHY232 – Spring 2007 Jon Pumplin http://www.pa.msu.edu/~pumplin/PHY232 (Ppt courtesy of Remco Zegers)

2. magnetism Magnetic fields are produced by moving electrical charges – i.e., currents) • macroscopic (e.g. currents in a wire) • microscopic (electrons in atomic orbit and rotating around their own axis) PHY232 - Pumplin - magnetism

3. magnets • the magnetic field produced by electrons tend to cancel each other, so most materials are not magnetic • in certain ‘ferromagnetic’ materials (iron) neighboring electrons can couple and form domains (< 1mm) that are magnetic. Since there are many domains that have different orientation, the material is overall not magnetized • when an external magnetic field is applied the fields in the different domains align and the whole object becomes magnetic • after the external field is removed, a material like iron becomes unmagnetized quickly, but some remain magnetized and can be used as ‘permanent’ magnets. PHY232 - Pumplin - magnetism

4. para and ferro magnets strawberry in a B-field do not retain any magnetism in absence of external field retains domains in which magnetic field remain in the absence of external fields PHY232 - Pumplin - magnetism

5. magnetic poles and fields • magnets have ‘north’ and ‘south’ poles and field lines point in the direction of force on a North magnetic pole. • unlike the case of electrical fields, where positive charges can exists separate from negative charges, north and south poles always come together. There are no monopoles discovered so far. demo: magnetic field lines (ohp) broken magnet PHY232 - Pumplin - magnetism

6. demo: compass needles compass One big magnet! Why is it higher here? Note that the geographical North pole is in fact the magnetic south pole B=0.3-0.6 x 10-4 Tesla PHY232 - Pumplin - magnetism

7. The compass needle in fact wants to point into the earth (along the direction of the field line). But if hold parallel to earth, it can’t do that and will point wherever. There is no reason for it to rotate though. question • If you are standing exactly at the (magnetic) south • Pole (I.e. near the geographical north pole), and are • holding a compass parallel with the earth’s surface, • in which direction would the needle point? • It would point roughly to the geographical south • It could point anywhere • It would rotate with constant angular speed PHY232 - Pumplin - magnetism

8. charged particles moving in a magnetic field • A charged particle q that is moving with a velocity v in a magnetic field B will feel a force where q: charge of particle v: velocity of paticle B: magnetic field : angle between velocity vector and field direction PHY232 - Pumplin - magnetism

9. direction of force on charged paricles in B-field demo: bending the beam I • magnitude of the force • you can find the direction of the force using the right hand rule. It holds for positive charges. For negative charges switch the direction of the force In the 3pm lecture (Section 2), we will use the version of Right Hand Rules given in the Textbook. PHY232 - Pumplin - magnetism

10. an electron with v=1x106 m/s is entering a area with B=1 T. The field is directed into the screen. in which direction will the electron be bent, if at all? how large is the force? what is the acceleration? • use right hand rule: • thumb is velocity (initially to the right) • index finger is field (in the screen) • middle finger is force perpendicular to both • switch direction because negative charge • b) F=|q|vBsin=1.6x10-19 x 1x106 x 1=1.6x10-13 N • a=F/m=1.6x10-13 N/9.11x10-31 kg =1.76x1017 m/s2 example: electron in magnetic field x x x x x x x x x x x x PHY232 - Pumplin - magnetism

11. A Magnesium ion (Z=12) with all its electrons removed is moving in a field of 0.1 T as shown. What direction will the force act? a) into the screen b) out of the screen c) parallel to the B field lines and the screen d) perpendicular to the B field lines and parallel to the screen e) in the direction of motion question Mg 45o v PHY232 - Pumplin - magnetism

12. Charged particle in a magnetic field • Let’s assume a charged particle is moving in a uniform magnetic field so that the velocity is perpendicular to the field. • The particle will follow a curved path and is directed towards the center • Use Newton’s second law and the equation for centripetal acceleration demo: bending the beam II PHY232 - Pumplin - magnetism

13. Magnetic spectrometers Beam from cyclotrons target chamber S800 spectrometer At the cyclotron Bending angle ~ 150o PHY232 - Pumplin - magnetism

14. question In a nuclear reaction two types of fully ionized particles are created. 120Sn with Z=50 and v=12.8814x107 m/s (Tin) 120Sb with Z=51 and v=13.099x107 m/s (Antimony) Both have a mass of 1.991x10-25 kg and pass through a 180o magnetic spectrometer with B=1T. If the detector used to locate the particles can separate events that are 2 mm away from each other, are 120Sn and 120Sb separated? r= mv/qB For 120Sn: M=1.991x10-25 kg v=12.8814x107 m/s B=1T q=50x1.6x10-19 C. RSn=3.2060 m For 120Snb: M=1.991x10-25 kg v=13.0990x107 m/s B=1T q=51x1.6x10-19 C. RSb=3.1961 m RSn-RSb=3.206-3.1961=9.9x10-3 m = 9.9 mm thus separated PHY232 - Pumplin - magnetism

15. What we did so far • Moving charged particles make magnetic field • North and South poles cannot exist independently • The magnitude of a force on a charge particle in a magnetic field: F=qvBsin where  is the angle between v and B. • The direction of the force is given by the (first) right-hand rule • for + particles: use directly • for – particles: after using the right hand-rule, reverse the direction of the force • For a particle moving in a direction perpendicular to a magnetic field PHY232 - Pumplin - magnetism

16. Question • a proton is moving from left to right into a field of which the field lines point into the screen. As a result, the proton will a) continue along its original trajectory b) bend upwards c) bend downwards d) bend into the screen e) bend out of the screen x x x x x x x x x x x x proton PHY232 - Pumplin - magnetism

17. magnetic force on a conducting wire • consider positive charges moving through a wire. Each particle feels a force, hence there is a net force on the wire • N: total number of charges • n: charges per unit volume • Use: see earlier • To get • More general: • where : angle between I and B vectors I I PHY232 - Pumplin - magnetism

18. top view I B electrons are moving left to right, so force due to B is up (out of the screen). When floating Fgravity=FB-field mg=Bil so m=Bil/g=1x1x1/9.81=0.102 kg question: a floating wire • a 1 m long copper wire of unknown mass is held horizontally with a current of 1 A going through it. It is placed in a horizontal magnetic field whose field lines are perpendicular to the wire. When the magnetic field is 1 T, one can let go of the wire without if falling down. What is its mass? PHY232 - Pumplin - magnetism

19. a rectangular looped copper wire carrying a current is placed horizontally in a B-field pointing down. Disregarding any other forces, it will move a) in direction of vector A b) in direction of vector B c) in direction of vector C d) in direction of vector D e) none of the above question x x x x x x x x x x x x x x x x x x x x A top view D B I C PHY232 - Pumplin - magnetism

20. a rectangular looped copper wire carrying a current is placed horizontally in a B-field pointing down. Disregarding any other forces, it will move a) in direction of vector A b) in direction of vector B c) in direction of vector C d) in direction of vector D e) none of the above correct answer: e) It will not move at all. Forces on left and right sides will cancel and likewise for top and bottom sides question x x x x x x x x x x x x x x x x x x x x A top view D B I C PHY232 - Pumplin - magnetism

21. Consider a current loop with dimension a x b in a B-field parallel to the loop. The force F on the right side (length b): F=BIb (pointing into the screen in the top view or downward in the frontal view) The force F on the left side (length b): F=BIb (pointing out of the screen in the top view or upward in the frontal view) force on up/down side (length a) is zero With the given rotation axis: Torque: =Fd=(BIb x a/2) + (BIb x a/2) =BIba=BIA with A=axb: surface of loop. Torque on a current loop Top view Rotation axis b I B a F x frontal view F If there is a net torque, the loop will rotate! PHY232 - Pumplin - magnetism

22. Torque on a current loop Top view Rotation axis • Now the loop makes an angle  with the B-field as shown right • To calculate the torque we only need the force perpendicular to the rotating loop: FL=Fsin =FLd=(BIb x a/2)sin + (BIb x a/2) sin =BIbasin=BIAsin • If there would be N loops: =BIANsin b I B a F  frontal view x F F frontal view  FL sin=F/FL  PHY232 - Pumplin - magnetism

23. So… • The general equation for a torque on a loop of N windings of wire is: • with • B: magnetic field strength • I: Current through the loop • A: area of the loop (also holds for non-rectangular loops) • N: number of windings • : angle between B and line perpendicular to loop • =IANmagnetic momentof the coil: it is a vector perpendicular to the coil.  is also the angle between  and B. Note that  is independent of B and , so it describes the properties of the coil when placed in a field. Unit: Am2  B  A N I PHY232 - Pumplin - magnetism

24. A circular coil of 5 windings is placed in a B-field of 2 T that makes and angle =60o with the line perpendicular to the coil. The radius of the coil is 3 cm, and the current through the coil is 0.5 A. What are: the area of the coil? b) the magnetic moment of the coil? c)the torque one the coil? note for loncapa: area of an ellipse: ab with a, b radii in the two directions example  B  A N I A=r2= (0.03)2=2.82x10-3 m2 =IAN=0.5 x 2.82x10-3 x 5=7.1x10-3Am2 =Bsin= 7.1x10-3 x 2 T x 0.866=1.23x10-2 Nm PHY232 - Pumplin - magnetism

25. electric motor By supplying electricity we can get some work done! PHY232 - Pumplin - magnetism

26. creating magnetic field with current • So far, we have seen that magnetic field can affect the motion of charged particles. • However, the reverse is also true: moving charge can create magnetic fields. • First seen by Hans Oersted who noted that a current through a wire creates a magnetic field. • A second right-hand rule can be used to find the direction of the magnetic field demo: Oersted experiment magnetic field of a current PHY232 - Pumplin - magnetism

27. How to quantify the field • 0 = “permeability of free space” = 4 x 10-7 Tm/A PHY232 - Pumplin - magnetism

28. a) use 2nd right hand rule • B-field goes into the screen • =4 x 10-7 x 1/(20.02)=1x10-5 T • c) use 1st right-hand rule and notice that the electron is negative. Force points to the right. • d) F=qvBsin=1.6E-19x1E6x1E-5x1= • = 1.6E-18 N (note sin(90)=1) an electron passing a wire • an electron with v=1x106 m/s is moving parallel to a wire carrying a current I=1A at a distance of 2 cm, in the same direction as the current • What is the direction of the magnetic field near the electron due to the wire? • what is the magnitude of the magnetic field near the electron? • what is the direction of the force on the electron? • what is the magnitude of the force on the electron? I=1A q=-1.6x10-19C 2 cm PHY232 - Pumplin - magnetism

29. a proton is passing by a wire carrying current and is moving perpendicular to the wire, into the screen 1) what is the direction of the B-field near the proton? into the screen out of the screen to the left to the right up 2) what is the direction of the force on the proton? to the left to the right up down no force at all I x • use 2nd right hand rule • (same as example on previous • slide) 2) use 1st right hand rule. velocity is into the screen, B-field is into the screen: no Force (sin(00)=0) question proton moving into the screen PHY232 - Pumplin - magnetism

30. if we place two parallel wires next to each other, the current in wire 2 creates a field near wire 2, at distance d from wire 1: The force on wire 1 due to wire 2 is then: Note so that the force per unit length is: magnetic force between two parallel wires d attractive if same direction repulsive if opposite direction PHY232 - Pumplin - magnetism

31. question • two wires are placed parallel, one carrying a current of 1A and the other of 2A, in the same direction. The distance between the two wires is 2 cm • a) what is the magnitude of the B-field exactly in between the two wires? • b) if a proton moves parallel to the two wires with v=1x105 m/s, exactly in between the two and in the same direction as the current, what is the magnitude of the force on the proton? • c) what is the force per unit length between the two wires? • B1=0I/(2r)=4x10-7x1/(20.01)=2x10-5 T • B2= 4x10-7x2/(20.01)=4x10-5 T • B1: into the screen B2: out of the screen • Bnet=2x10-5 T out of the screen • F=qvBsin=1.6x10-19x105 x 2x10-5 x sin(90)=3.2x10-19 N • (directed to the right, use 1st right-hand rule) • c) F/l= 0I1I2/(2d)= 4x10-7x1x2/(20.02)=2x10-5 N 1A 2A 2 1 2cm PHY232 - Pumplin - magnetism

32. note • the procedure of the previous slide can be used for any number of wires. In case of 4 wires (see lon-capa), one can calculate the force of one on the wires by adding the forces of each of the other three wires on that wire… PHY232 - Pumplin - magnetism

33. magnetic field inside a current loop example: A person wants to find the current in a superconducting coil with diameter of 2 cm. She measures the magnetic field at the center to be 1x10-5 T. What is the current? other cases: the current loop I I R X right-handed current through loop: B-field in the screen left-handed current through loop: B-field out of the screen I=2RBcenter/0= 2x0.01x10-5/4x10-7= 0.16 A PHY232 - Pumplin - magnetism

34. a solenoid is a collection of coils stacked on top of each other Inside a perfect solenoid, the field lines are parallel and the field uniform outside the solenoid, the field pattern looks like that of a bar magnet. For the field inside of a solenoid: where I is the current and n is the number of turns (n) per unit length l of the solenoid note that the field at the center does not depend on the radius of the turns other cases II: magnetic field of a solenoid B-field of solenoid PHY232 - Pumplin - magnetism

35. inside the coil, the field is uniform example • A perfect coil is 30 cm long and has 3000 windings. Its radius is 2cm. What is the field strength along the central line inside the coil if the current is 4 A? • The field strength along a line parallel to the central line but 5mm away from the center is … along the central line? a) lower than b) the same as c) higher than B=0nI=4x10-7 x 3000/0.3 x 4 = 1x10-3 T use n=N/L PHY232 - Pumplin - magnetism