# Magnetism The Magnetic Force x x x v F B q  v F B q   v F = 0 B q.

## Presentation on theme: "Magnetism The Magnetic Force x x x v F B q  v F B q   v F = 0 B q."— Presentation transcript:

Magnetism The Magnetic Force x x x v F B q  v F B q   v F = 0 B q

Today... Magnetic forces: The Lorentz Force equation Motion of charged particle in a Constant Magnetic Field.

Magnetic Fields We saw last lecture that some substances, particularly iron, possess a property we call magnetism that exerts forces on other magnetic materials We also saw that single magnetic charges (magnetic monopoles) did not exist We saw that magnetic fields, shown up by iron filings look similar to electric dipole fields Also that magnetic fields seem to be associated with moving charges What is this "magnetic force"? How is it related to and distinguished from the "electric" force?

Consider a positive charge q moving in the field of a magnet with velocity, experimentally we find: 1.If q moves in the +z direction and the field points in the +y direction then the force is in the –x direction. The force is proportional to the velocity and the field 2.If q moves in the +x direction the force is in the +z direction, again proportional to and Magnetic Forces

3.If q moves in the +y direction there is no force 4.If q is at rest there is no force 5.The force is proportional to 6.The force is proportional to the sign and magnitude of q The magnetic force on a moving charge is proportional to q, v p and B, where v p is the velocity component perpendicular to the field, while the direction of is perpendicular to both and and depends on the sign of q

Lorentz Force We can add the effect of an Electric Field and get the “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: F x x x v B q  v B q F = 0  v B q F     BvqEqF 

Reminder: The Cross Product The cross (vector) product of two vectors is a third vector –Remember the dot (scalar) product multiplied two vectors to produce a scalar ABC  A B CA X B = CA B CA X B = C CThe magnitude of C is given by: C = AB sin  C ABThe direction of C is perpendicular to the plane defined by A and B, and in the direction defined by the right hand rule, rotating from A to B. UIUC

Cartesian components of the cross product: C A B C = A X B C X = A Y B Z - B Y A Z C Y = A Z B X - B Z A X C Z = A X B Y - B X A Y ABC Reminder: The Cross Product Note: B X A = - A X B Drawing 3-dimensional vectors, conventionally – a vector going into the slide – a vector coming out of the slide

Three points are arranged in a uniform magnetic field. The B field points into the screen. Consider the force on a positively charged particle in the following conditions 1) It is located at point A and is stationary. v=0 The magnetic force is zero 2) The positive charge moves from point A toward B. v in x direction, B in z, RH rule says F in –y The direction of the magnetic force on the particle is to the left 3) The positive charge moves from point A toward C. Rotate our x axis to be along the direction A-C F will be perpendicular to that line and upwards Motion in a magnetic field x y z

Question 1 Two protons each move at speed v in the x-y plane (as shown in the diagram) in a region of space which contains a constant B field in the -z -direction. Ignore the interaction between the two protons. –What is the relation between the magnitudes of the forces on the two protons? (a) F 1 < F 2 (b) F 1 = F 2 (c) F 1 > F 2 B x y z 1 2 v v

Question 1 1.a 2.b 3.c 15 0 of 300

Question 1 Two protons each move at speed v in the x-y plane (as shown in the diagram) in a region of space which contains a constant B field in the -z -direction. Ignore the interaction between the two protons. –What is the relation between the magnitudes of the forces on the two protons? (a) F 1 < F 2 (b) F 1 = F 2 (c) F 1 > F 2 B x y z 1 2 v v The magnetic force is given by: In both cases the angle between v and B is 90° Therefore F 1 = F 2.

Question 2 Two protons each move at speed v in the x-y plane (as shown in the diagram) in a region of space which contains a constant B field in the -z-direction. Ignore the interaction between the two protons. –What is F 2x,the x-component of the force on the second proton? B x y z 1 2 v v (a) F 2x < 0 (b) F 2x = 0 (c) F 2x > 0

Question 2 1.a 2.b 3.c 15 0 of 300

Question 2 Two protons each move at speed v in the x-y plane (as shown in the diagram) in a region of space which contains a constant B field in the -z-direction. Ignore the interaction between the two protons. –What is F 2x,the x-component of the force on the second proton? B x y z 1 2 v v (a) F 2x < 0 (b) F 2x = 0 (c) F 2x > 0 To determine the direction of the force, we use the right-hand rule. The directions of the forces are shown in the diagram F 2x < 0

Question 3 Two protons each move at speed v in the x-y plane (as shown in the diagram) in a region of space which contains a constant B field in the -z-direction. Ignore the interaction between the two protons. –Inside the B field, the speed of each proton: B x y z 1 2 v v (a) decreases (b) increases (c) stays the same

Question 3 1.a 2.b 3.c 15 0 of 300

Question 3 Two protons each move at speed v in the x-y plane (as shown in the diagram) in a region of space which contains a constant B field in the -z-direction. Ignore the interaction between the two protons. –Inside the B field, the speed of each proton: B x y z 1 2 v v (a) decreases (b) increases (c) stays the same Although the proton does experience a force (which deflects it), this is always  to. Therefore, there is no possibility to do work So kinetic energy is constant and is constant

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

Radius of Circular Orbit Lorentz force: qvBF  centripetal acc: R v a 2  x x x v F B q F v R Newton's 2nd Law: maF   R v mqvB 2   qB mv R  This is an important result, with useful experimental consequences !

Ratio of charge to mass for an electron In 1897 J.J.Thomson used a cathode ray tube to measure e/m for an electron Used an electric and magnetic field in opposition to cancel force and thus deflection of the electron Electron accelerated through a voltage V by the “electron gun” giving it kinetic energy Velocity when it enters the fields

e/m for an electron In an electric field alone the spot is deflected Then apply a magnetic field at right angles (Force in opposite direction) until the deflection is reduced to zero Force due to electric field Force due to magnetic field When the fields cancel

Measurement of particle energies Many experiments using particles measure their velocity (energy or momentum) by measuring their curvature in a magnetic field Cloud chambers Bubble chambers Magnetic spectrometers

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. 5) What is the direction of the magnetic field in chamber 1? a) Up b) Down c) Left d) Right e) Into page f) Out of page Preflight 12:

6) What is the direction of the magnetic field in chamber 2? a) Up b) Down c) Left d) Right e) Into page f) Out of page Preflight 12:

In chamber 1, the velocity is initially up. Since the particle’s path curves to the right, the force is to the right as the particle enters the chamber. Three ways to figure out the direction of B from this: 1) Put your thumb in the direction of the F (right) and your fingers in the direction of v (up) The way that your fingers curl is the direction of B. 2) Put your palm in the direction of F (right), and your thumb in the direction of v (up), your fingers (keep them straight) point in the direction of B. 3) Keep your thumb, index and middle fingers at right angles from each other. Your thumb points in the direction of v (up), middle finger points towards F (right), then the index finger gives the the direction of B (out of page)

8) Compare the magnitude of the magnetic field in chamber 1 to the magnitude of the magnetic field in chamber 2. Preflight 12: a) B 1 > B 2 b) B 1 = B 2 c) B 1 < B 2

The magnetic force is always perpendicular to v. The force doesn’t change the magnitude of v, it only changes the particle’s direction of motion. The force gives rise to a centripetal acceleration. The radius of curvature is given by:

Lecture 12, Act 3 (a) W 1 < W (b) W 1 = W (c) W 1 > W L B B v v B B v1v1 v1v1 – Compare the work done by the magnetic field ( W for v, W 1 for v 1 ) to deflect the protons. 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. Another proton, moving at speed v 1 = 2v, enters the same region of space and is deflected as shown.

Lecture 12, Act 3 (a) W 1 < W (b) W 1 = W (c) W 1 > 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: – Compare the work done by the magnetic field ( W for v, W 1 for v 1 ) to deflect the protons. 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. Another proton, moving at speed v 1 = 2v, enters the same region of space and is deflected as shown. L B B v v B B v1v1 v1v1

Summary Lorentz force equation: –Static B-field does no work –Velocity-dependent force given by right hand rule formula Next time: magnetic forces and dipoles Reading assignment: Chapter 28.1 and 28.3 Examples: 28.2, 28.3, and 28.6 through 28.10

The Hall Effect Which charges carry current? Positive charges moving counterclockwise experience upward force Upper plate at higher potential Negative charges moving clockwise experience upward force Upper plate at lower potential Equilibrium between electrostatic & magnetic forces: This type of experiment led to the discovery (E. Hall, 1879) that current in conductors is carried by negative charges (not always so in semiconductors). Can be used as a B-sensor; used in some ABS to detect shaft rotation speed – ferromagnetic rotating blades interupt the magnetic field  oscillating voltage

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