# Magnetism x x x x x x F B q.

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Magnetism x x x x x x F B q

Introduction to Magnetic Phenomena
Overview of Lecture Introduction to Magnetic Phenomena Bar magnets & Magnetic Field Lines Source of Fields: Monopoles? Currents? Magnetic forces: The Lorentz Force equation Motion of charged particle in a Constant Magnetic Field. Magnetic Force on a current-carrying wire Current Loops Magnetic Dipole Moment Torque (when in constant B field) Potential Energy (when in constant B field) Text Reference: Chapter

Magnetism Magnetic effects from natural magnets have been known for a long time. Recorded observations from the Greeks more than 2500 years ago. The word magnetism comes from the Greek word for a certain type of stone (lodestone) containing iron oxide found in Magnesia, a district in northern Greece. Properties of lodestones: could exert forces on similar stones and could impart this property (magnetize) to a piece of iron it touched. Small sliver of lodestone suspended with a string will always align itself in a north-south direction. ie can detect the magnetic field produced by the earth itself.

Like poles repel; Unlike poles attract.
Bar Magnet Bar magnet ... two poles: N and S Like poles repel; Unlike poles attract. Magnetic Field lines: (defined in same way as electric field lines, direction and density) Does this remind you of a similar case in electrostatics?

Electric Field Lines of an Electric Dipole
Magnetic Field Lines of a bar magnet

N S Magnetic Monopoles Try cutting a bar magnet in half:
One explanation: there exists magnetic charge, just like electric charge. An entity which carried this magnetic charge would be called a magnetic monopole (having + or - magnetic charge). How can you isolate this magnetic charge? Try cutting a bar magnet in half: N S In fact no attempt yet has been successful in finding magnetic monopoles in nature. Many searches have been made, due primarily to fact that existence of monopole could give an explanation (within framework of QM) for the quantization of electric charge (argument of Dirac)

Source of Magnetic Fields?
What is the source of magnetic fields, if not magnetic charge? Hypothesis: electric charge in motion! Maybe. eg current in wire surrounding cylinder (solenoid) produces very similar field to that of bar magnet. (demo) Therefore, understanding source of field generated by bar magnet lies in understanding currents at atomic level within bulk matter. Orbits of electrons about nuclei From Quantum Mechanics we learn that aarticles possess an intrinsic magnetism. Intrinsic “spin” of electrons (more important effect)

Definition of Magnetic Field
Magnetic field B is defined operationally by the magnetic force on a test charge. (We did this to talk about the electric field too) What is "magnetic force"? How is it distinguished from "electric" force? Start with some observations: CRT deflection Empirical facts: Charged particles in motion are deflected by a magnetic field, a) magnitude: µ to velocity of q b) direction: ^ to direction of q and also of B q F v mag

x x x x x x v B q ® ® ® ® ® v B q ­ ­ ­ ­ ­ ­ ­ ­ v B q F ´ F F = 0
Lorentz Force • The force F on a charge q moving with velocity v through a region of space with electric field E and magnetic field B is given by: x x x x x x v B q ® ® ® ® ® v B q ­ ­ ­ ­ ­ ­ ­ ­ v B q F F F = 0

(c) Both Ey & Bz can exist
Lecture 13, CQ An electron enters a region of space with speed v and exits the region as shown with the same speed (magnitude) v. From this information, what can we infer about the Ey and Bz fields in the region? x y v e E ? B ? - (a) Only Ey exists (b) Only Bz exists (c) Both Ey & Bz can exist

(c) Both Ey & Bz can exist
Lecture 13, CQ An electron enters a region of space with speed v and exits the region as shown with the same speed (magnitude) v. From this information, what can we infer about the Ey and Bz fields in the region? x y v e E ? B ? (a) Only Ey exists (b) Only Bz exists (c) Both Ey & Bz can exist If an Ey then there is a Force Fy = q Ey = may which will accelerate the charge in the y direction. The x velocity is unchanged. The y velocity is new and changing. Superposition tells us the overall velocity is increasing. So, NO Ey possible; this field would speed up the charge. What about Bz (ie, a field OUT of the plane)? It produces a force on the electron in the x-y plane, always perpendicular to the velocity. Therefore, Bz will NOT cause the magnitude of the velocity to change. But, Bz will change the direction of q. Bz exists in this example.

(a) F1 < F2 (b) F1 = F2 (c) F1 > F2 (a) F2x < 0 (b) F2x = 0
Lecture 13, CQ Two protons each move at speed v (as shown in the diagram) toward a region of space which contains a constant B field in the -z-direction. What is the relation between the magnitudes of the forces on the two protons? (a) F1 < F2 (b) F1 = F2 (c) F1 > F2 What is F2x, the x-component of the force on the second proton? (a) F2x < 0 (b) F2x = 0 (c) F2x > 0

(a) F1 < F2 (b) F1 = F2 (c) F1 > F2
Lecture 13, CQ Two protons each move at speed v (as shown in the diagram) toward a region of space which contains a constant B field in the -z-direction. What is the relation between the magnitudes of the forces on the two protons? (a) F1 < F2 (b) F1 = F2 (c) F1 > F2 The magnetic force is given by: In both cases the angle between v and B is 90°!! Therefore F1 = F2.

(a) F1 < F2 (b) F1 = F2 (c) F1 > F2 (a) F2x < 0 (b) F2x = 0
Lecture 13, CQ F1 Two protons each move at speed v (as shown in the diagram) toward a region of space which contains a constant B field in the -z-direction. What is the relation between the magnitudes of the forces on the two protons? F2 (a) F1 < F2 (b) F1 = F2 (c) F1 > F2 What is F2x, the x-component of the force on the second proton? (a) F2x < 0 (b) F2x = 0 (c) F2x > 0 To determine the direction of the force, we use the right-hand rule. As shown in the diagram, F2x < 0.

B-Fields and Circular Motion
" R cancels R !! " © Copyright, Department of Physics, University of Illinois at Urbana-Champaign.

Trajectory in Constant B Field
x x x x x x v B q Suppose charge q enters B field with velocity v as shown below. What will be the path q follows? R F v F Force is always ^ to velocity and B. What is path? Path will be circle. F will be the centripetal force needed to keep the charge in its circular orbit. Calculate R:

x x x x x x v F B q R Lorentz force: centripetal acc: Newton's 2nd Law: Þ Þ This is an important result, with useful experimental consequences !

Ratio of charge to mass for an electron
1) Turn on electron ‘gun’ DV ‘gun’ 2) Turn on magnetic field B R 3) Calculate B … next week; for now consider it a measurement 4) Rearrange in terms of measured values, V, R and B & Þ See problem 29-38

(a) W1 < W (b) W1 = W (c) W1 > W
Lecture 13, CQ L B v A proton, moving at speed v, enters a region of space which contains a constant B field in the -z-direction and is deflected as shown. B v1 Another proton, moving at speed v1 = 2v, enters the same region of space and is deflected as shown. Compare the work done by the magnetic field (W for v, W1 for v1) to deflect the protons? (a) W1 < W (b) W1 = W (c) W1 > W

(a) W1 < W (b) W1 = W (c) W1 > W
Lecture 13, CQ L B v A proton, moving at speed v, enters a region of space which contains a constant B field in the -z-direction and is deflected as shown. B v1 Another proton, moving at speed v1 = 2v, enters the same region of space and is deflected as shown. Compare the work done by the magnetic field (W for v, W1 for v1) to deflect the protons? (a) W1 < W (b) W1 = W (c) W1 > W Remember that the work done W is defined as: Also remember that the magnetic force is always perpendicular to the velocity: Therefore, the work done is ZERO in each case:

Lawrence's Insight "R cancels R"
We have derived the radius of curvature of the trajectory of a charged particle in a constant magnetic field. E.O. Lawrence realized in 1929 an important feature of this equation which became the basis for his invention of the cyclotron. Þ Þ Þ Rewrite in terms of angular velocity w ! R does indeed cancel R in above eqn. So What?? The angular velocity is independent of R!! Therefore the time for one revolution is independent of the particle's energy! This fact can be exploited to provide the mechanism for acceleration of charge.

B x x x x x x + V - Cyclotron "Magnetic Resonance Accelerator":
"Dees" in constant magnetic field B Alternating voltage V is applied between the Dees at the orbital frequency f: + - V Particle will acquire an additional kinetic energy DT = qV each time it crosses the gap (ie twice per revolution.. E=0 in Dees!). Radius of D-Rings The particle will escape the D-rings when the velocity obtains a value of:

(a) fp < fa (b) fp = fa (c) fp > fa (a) E1 = 1/2 E0 (b) E1 = E0
Lecture 14, CQ A cyclotron is used to accelerate protons and alpha particles (He nucleus = 2 protons + 2 neutrons). What is the relation between fp, the frequency of the applied voltage for protons, and fa , the frequency of the applied voltage for alphas? (a) fp < fa (b) fp = fa (c) fp > fa If protons are accelerated to a final energy E0 when the applied voltage is V0, to what final energy E1 would they be accelerated if the applied voltage were V1 = 1/2 V0? (a) E1 = 1/2 E0 (b) E1 = E0 (c) E1 = 2E0

(a) fp < fa (b) fp = fa (c) fp > fa
Lecture 14, CQ A cyclotron is used to accelerate protons and alpha particles (He nucleus = 2 protons + 2 neutrons). What is the relation between fp, the frequency of the applied voltage for protons, and fa , the frequency of the applied voltage for alphas? (a) fp < fa (b) fp = fa (c) fp > fa The applied frequency must match the natural orbital frequency. Alpha particles have twice the charge, but about four times the mass as protons. The orbital frequency is given by: Therefore, the orbital frequency for alphas must be about half that of protons.

(a) fp < fa (b) fp = fa (c) fp > fa (a) E1 = 1/2 E0 (b) E1 = E0
Lecture 14, CQ A cyclotron is used to accelerate protons and alpha particles (He nucleus = 2 protons + 2 neutrons). What is the relation between fp, the frequency of the applied voltage for protons, and fa , the frequency of the applied voltage for alphas? (a) fp < fa (b) fp = fa (c) fp > fa If protons are accelerated to a final energy E0 when the applied voltage is V0, to what final energy E1 would they be accelerated if the applied voltage were V1 = 1/2 V0? (a) E1 = 1/2 E0 (b) E1 = E0 (c) E1 = 2E0 Decreasing the accelerating voltage will decrease the energy per turn. However, the protons will just make more turns! ie the final energy for a proton is just determined by the strength of the field and the radius of the cyclotron! However, the final energy of alphas will not be same as the final energy of protons..stay tuned for our next calculation..

Þ K.E. = T = 1/2 mv2 Þ 1 MeV = 106 eV = 106 (1.6 ´ 10-19) J
Example 1935 cyclotron at UI MeV deuterons Cyclotron was placed in a magnetic field of 1.24 Tesla and was able to accelerate deuterons from rest to an energy of 1 MeV. Calculate required radius for the D-rings. Þ K.E. = T = 1/2 mv2 Þ 1 MeV = 106 eV = 106 (1.6 ´ 10-19) J Þ R = 16 cm m » 2mp = 3.34 ´ kg q = 1.6 ´ C Accelerating voltage V = 50kV. Therefore, deuterons must make 10 orbits to achieve 1 MeV.

Yet another example Measuring curvature of charged particle in magnetic field is usual method for determining momentum of particle in modern experiments: eg B e+ e- End view: B into screen

Magnetic Forces on a Current Wire

Magnetic Force on a Current
Consider a current-carrying wire in the presence of a magnetic field B. There will be a force on each of the charges moving in the wire. What will be the total force dF on a length dl of the wire? Suppose current is made up of n charges/volume each carrying charge q and moving with velocity v through a wire of cross-section A. N S Number density Force on each charge = Total force = Þ Current = Yikes! Simpler: For a straight length of wire L carrying a current I, the force on it is:

Magnetic Force on a Current Loop
Force on top path cancels force on bottom path (F = IBL) 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 x x x x x x x x x x B i Consider loop in magnetic field as on right: If field is ^ to plane of loop, the net force on loop is 0! F Force on right path cancels force on left path. (F = IBL) B x . F If plane of loop is not ^ to field, there will be a non-zero torque on the loop!

(a) Fy < 0 (b) Fy = 0 (c) Fy > 0
Lecture 14, CQ x y A current I flows in a wire which is formed in the shape of an isosceles triangle as shown. A constant magnetic field exists in the -z direction. What is Fy, net force on the wire in the y-direction? (a) Fy < 0 (b) Fy = 0 (c) Fy > 0

(a) Fy < 0 (b) Fy = 0 (c) Fy > 0
Lecture 14, CQ y A current I flows in a wire which is formed in the shape of an isosceles triangle as shown. A constant magnetic field exists in the -z direction. What is Fy, net force on the wire in the y-direction? (a) Fy < 0 (b) Fy = 0 (c) Fy > 0 x The forces on each segment are determined by: L F1 F2 F3 q Lsinq From symmetry, Fx = 0 For the y-component: Therefore:

B x q F . Þ rxF Þ t = AIB sinq F = IBL r A = wL = area of loop F
Calculation of Torque Suppose the coil has width w (the side we see) and length L (into the screen). The torque is given by: B x . F q w Þ r F rxF F = IBL Þ t = AIB sinq where A = wL = area of loop Note: if loop ^ B, sinq = 0 Þ t = 0 maximum t occurs when loop parallel to B

Magnetic Dipole Moment
We can define the magnetic dipole moment of a current loop as follows: magnitude: m = AI B x . F q direction: ^ to plane of the loop in the direction the thumb of right hand points if fingers curl in direction of current. q m Torque on loop can then be rewritten as: Þ t = AIB sinq Note: if loop consists of N turns, m = NAi

Electric Dipole Analogy
B x . F q m E +q -q p (per turn)

Potential Energy of Magnetic Dipole in B-Field

Potential Energy of Dipole
B x . F q m Work must be done to change the orientation of a dipole (current loop) in the presence of a magnetic field. Define a potential energy U (with zero at position of max torque) corresponding to this work. Þ \ Þ Þ

Potential Energy of Dipole
Stable Equilibrium negative work positive work Unstable Equilibrium B x m B x m B x m t = 0 t = 0 U = 0

(b) (a) (c) It will not rotate (a) U0 is minimum (b) U0 is maximum
Lecture 14, CQ x y I R B a b A circular loop of radius R carries current I as shown in the diagram. A constant magnetic field B exists in the +x direction. Initially the loop is in the x-y plane. The coil will rotate to which of the following positions? (a) (b) (c) It will not rotate What is the potential energy U0 of the loop in its initial position? (a) U0 is minimum (b) U0 is maximum (c) neither

(b) (a) (c) It will not rotate
Lecture 14, CQ x y I R B a b A circular loop of radius R carries current I as shown in the diagram. A constant magnetic field B exists in the +x direction. Initially the loop is in the x-y plane. The coil will rotate to which of the following positions? (b) (a) (c) It will not rotate The coil will rotate if the torque on it is non-zero: The magnetic moment m is in +z direction. Therefore the torque t is in the +y direction. Therefore the loop will rotate as shown in (b).

(a) U0 is minimum (b) U0 is maximum (c) neither
Lecture 14, CQ x y I R B a b A circular loop of radius R carries current I as shown in the diagram. A constant magnetic field B exists in the +x direction. Initially the loop is in the x-y plane. What is the potential energy U0 of the loop in its initial position? (a) U0 is minimum (b) U0 is maximum (c) neither The potential energy of the loop is given by: In its initial position, the loop has its magnetic moment vector pointing in the positive z direction. Therefore the initial potential energy is ZERO. However, this does NOT mean that the potential energy is a minimum. When the loop is in the y-z plane and has its magnetic moment pointing in the same direction as the field, its potential energy is NEGATIVE and is in fact the minimum. Since U0 is not minimum, the coil will rotate and convert potential energy to kinetic energy!!

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