Every magnet has a north pole and south pole.

Magnets - Intro The lodestone is a naturally occurring mineral called magnetite. It was found to attract certain pieces of metal. o one knew why. ome early Greek philosophers thought the lodestone had a soul. Today, we have a better idea of what's going on. All magnetism is found to come from moving charges. ometimes it is obvious how the charges are moving; other times, it s more subtle. Magnetism from Magnetic Materials: Bar magnets, refrigerator magnets, compasses Magnetism form moving charges: currents, electromagnets, motors The fundamental mechanisms needed to understand magnetic interactions involves quantum mechanics. Thus, for our purposes, we'll have to be content with providing a quantitative description of the phenomenon, without making efforts to explain the deeper why questions. ome basics about magnets Every magnet has a north pole and south pole. The force between two of the same poles is repulsive while the force between two opposite poles is attractive. These are not the same thing as positive and negative charges, although they may sound similar. opposite poles similar poles (the magnetic field is given the letter B) Cutting a magnet will never create a separate north and south pole. You ll always just end up with two smaller magnets. Page 1 of 27

Just like electrostatic fields, we can use field lines to portray the magnetic fields. These are the magnetic field lines in the vicinity of a bar magnet. They point from orth to outh. Field Lines & Field Vectors These are two ways of showing the same information. The field vectors can be used to show the strength and direction at a particular point in space. They are tangent to the field lines If we surround a bar magnet with many little magnets, they will line up along the field lines. Moving charges create fields magnetic field lines Current in a wire will create a magnetic field. The shape of that magnetic field is circular with the wire at the center of the circle. The strength of the field decreases as the distance from the wire is increased. Page 2 of 27

Right-hand rule Let's bend the wire ow we have a loop with a current in it. The same Right Hand Rule can be used to figure out the B field near this loop. Just let your thumb point in the direction of the conventional current direction. Then curl your fingers as shown. The magnetic field points in the direction of your fingers. Inside the loop, the B field vectors are all pointing the same direction. Page 3 of 27

And what if we have many loops, all parallel to each other? This is like a coil, or solenoid. Again, the RHR will guide us in figuring out the direction of the magnetic field lines. The magnetic field A magnetic field is usually given the symbol B. The I units for the magnetic field strength are Teslas. 1Tesla = ewton Coulomb meter/second A Tesla is a rather large magnetic field, so sometimes we use the unit of Gauss, which is given by: 1 Tesla = 10 4 Gauss The directions are often perpendicular to the plane of the page (or screen) into the screen out of the screen First let's agree on a coordinate system This one has x and y on the horizontal plane. z point upwards. Page 4 of 27

ow, we'll consider a uniform Magnetic field B, which in this case points only in the +y direction. B = B0 j We can add a charged particle in the B field. ince this particle is electrically charged, and there is no electric field, there will be no force on the particle We can also impart a velocity v to the particle. Let's say that v is only pointing the +x direction. v = v0 i ow, if we have a charged particle moving in a magnetic field, that particle will experience a magnetic force given by: F B = q v B F B in this case. This is a force directed 90 to both the magnetic field and the velocity vectors: k Page 5 of 27

The right hand rule 1. Your thumb points in the direction of the moving particle. 2. Index finger in direction of magnetic field. 3. Middle finger shows the force Question: Which way will the magnetic force be directed on the particle as it enters the field? v a) Up b) Down c) Left d) Right e) Into the screen f) Out of the screen Question: Which way will the magnetic force be directed on the particle as it enters the field? v a) Up b) Down c) Left d) Right e) Into the screen f) Out of the screen Page 6 of 27

Calculating the magnetic force: Let's not forget about the little q in the equation: F B = q v B If the q is negative, then the sign (and therefore the direction) of the force vector will flip. The Cross product (or vector product) is one way of multiplying two vectors: It produces a third vector in a direction perpendicular to both the multiplicands. i = j k j = k i k = i j k j = i i k = j j i = k Magnitude of the cross product: A = a b = a b sin θ. o, for the strength of the magnetic force on a charged particle: F = qvb sin θ This tells us that if the particle is moving in the direction of the magnetic field ( θ = 0), there will be no force. Page 7 of 27

Question: A negatively-charged particle travels parallel to magnetic field lines within a region of space. Which one of the following statements concerning the particle is true? a) There is a force directed perpendicular to the magnetic field. b) There is a force perpendicular to the direction in which the particle is moving. c) The force slows the particle. d) The force accelerates the particle. e) The force has a magnitude of zero newtons. Question: A positively-charged particle is stationary in a constant magnetic field within a region of space. Which one of the following statements concerning the particle is true? a) The particle will not move. b) The particle will accelerate in the direction perpendicular to the field. c) The particle will accelerate in the direction parallel to the field. d) The particle will accelerate in the direction opposite to the field. e) The particle will move with constant velocity in the direction of the field. uniform B Field (into screen) charged particle uniform B Field (into screen) force vector Page 8 of 27

uniform B Field (into screen) Circular Motion Play/Pause Circular Motion For the case of uniform circular motion, we had from ewton's 2nd law: ow, we know the force due to the magnetic field: Thus, together we get: F = m v2 r F B = q v B F = q vb q vb = mv2 r Page 9 of 27

Question: A charged particle is moving through a constant magnetic field. Does the magnetic field do work on the charged particle? a) yes, because the force is acting as the particle is moving through some distance b) no, because the magnetic force is always perpendicular to the velocity of the particle c) no, because the magnetic field is a vector and work is a scalar quantity d) no, because the magnetic field is conservative e) no, because the magnetic force is a velocity-dependent force Usually, we'll be concerned with the radius of motion: r = mv q B Let's not forget our friends period and frequency too: 2πr T = = v 2πm q B 1 f = = T q B 2πm Example Problem: An electron is accelerated from rest through a potential difference of 500V, then injected into a Uniform magentic field. Once in the field, is completes half a revolution in 2.0ns. What is the radius of the orbit. What will happen if the is a component of the initial velocity not perpendicular to the B-field? Helical Motion! Page 10 of 27

Example Problem: An electron has a kinetic energy of 22.5 ev. It moves into a region with a uniform magnetic field given by 455 microtesla. The velocity of the particle is such that is makes angle of 50 degrees with the magnetic field. What is the pitch of the helical path? What happens if there is an electric field, E, and a magnetic field, B, in the same region of space? v motion from magnetic field motion from electric field Answer: The fields will both affect the particle as the would normally. Two forces will exist and we can just add them. Offically: The Lorentz Force law: This sum of the forces on the particle will account for its dynamical properties. The Hall Effect What happens when the is equal to the? F E F = q (E + v B) F E = F B F B Page 11 of 27

The Hall Effect - setup The charge separation will lead to a potential difference across the width of the device, d.: V Hall. This potential difference will increase, until the force due to the electric field it creates is equal to the force from the magnetic field: F = qe E = V/w For the force due to the magentic field, we have F B = qvb What is v? It's the drift velocity of the charges: v d = I nqa ( n is the density of the charges, A is the cross sectional area). and so... Let's say the forces are equal: The force from the electric field is: The force due to the magnetic field is Thus: F E F E = = q F B F B V w = q IB nqa V d = BI nqa Page 12 of 27

I : current we can measure Which things do we know? V: that's a measureable quantity (use a voltmeter) B: The magnetic field is something we control (and know) A: cross sectional area = w h q: charge on an electron, we know that n: density of charges: we don't always know that. V w = BI nqa Question: A charged particle enters a uniform magnetic field. What happens to the kinetic energy of the particle? a) it increases b) it decreases c) it stays the same d) it depends on the direction of the velocity e) it depends on the direction of the magnetic field ince currents are really nothing more than moving charges, there should be a force on a current carrying wire if a magnetic field is present. Indeed, it should follow the same format as before. In this case, we have a wire of length L with a current traveling in the direction of L. ince q = It = i L v, d F B becomes the force on a current carrying wire. F = I B L B v d = q V Page 13 of 27

The force: F = I B L B The magnitude of this force is just: F B = IlB sin θ figure it will be π/2). l is the length of the wire and I is the conventional current. θ is the angle between the wire and the field. (In this And, just to be sure, the force on the electron current is also pointing in the same direction. (Otherwise, we would have problems) Example Problem: 15 V 10 cm 3 Ω This circuit is partially exposed to a 50 mt magnetic field, as shown. What is the net force on the circuit. +y +x Question: +y V +x +z wire between the poles of the magnet? A portion of a loop of wire passes between the poles of a magnet as shown. When the switch is closed and a current passes through the circuit, what is the movement, if any, of the a) The wire moves toward the north pole of the magnet (-x). b) The wire moves toward the south pole of the magnet (+x). c) The wire moves out of the screen (toward us, +z). d) The wire moves into the screen (away from us, -z). e) The wire doesn t move. Page 14 of 27

Question: What will happen to this metal wire in the uniform B Field? a) It will not move b) It will accelerate to the right c) It will accelerate to the left d) It will accelerate up e) It will accelerate down Rotations in a B Field Play/Pause Here's a simple (oversimplified) current loop in a uniform magnetic field. Perspective view Top view: Let's label each of the sides of the loop Page 15 of 27

If we go through and use the rules about magnetic force on a current carrying wire, we'll see that the forces will be given by: 1. no force ( θ = 0) 2. pointed up( θ = 90) 3. no force ( θ = 180) 4. pointed down ( θ = 270) And thus, we'll expect to see some rotation of the loop (i.e. a torque must be present). The loop will rotate until it reaches this position, at which point there will no longer be torque on the loop. What happens if the loop rotates past the vertical position? (we would expect to since it should be accelerating until that point and would then rotate past vertical) Page 16 of 27

Here is a loop of wire, with a current I, in a uniform magnetic field. We can go through and use our force on a wire equation to find the direction of the force at each point along the wire. We ll see that the total net force ends up being zero, but that doesn t mean there isn t a torque! L/2 loop The the total torque will be the sum of the two torques from the top and the bottom of the loop. (the ones on the side = 0, since sin(0) = 0) Adding the torques from the top and bottom sides: But, we know the torque since we know the force and distance away from the axis. Which reduces to: If there were multiple loops, say, in a coil, we could just add the torque from each one ( ) Magnetic Dipole τ = τ top + τ bottom 1 1 τ = ILB L sin θ + ILB L sin θ 2 2 τ = I L 2 B sin θ = IAB sin θ µ We can make that expression a little more compact by defining the magnetic dipole moment: μ loop μ = IA uniform B field where I is the current through the loop, A is the area of the loop, and is the number of turns. The dipole moment is a vector. It point parallel to the surface normal of the loop. The direction (up or down) is given by the rotation of the current. Page 17 of 27

Another right hand rule The magnetic dipole moment will point in the direction of the thumb! loop µ Thus, we can simplify the expression a little bit more: τ = μ uniform B field which should be somewhat reminiscent of the torque on an electric dipole in an electric field. B Question: Consider the relationships between the directions of the torque acting on a magnetic dipole in a magnetic field, the magnetic field, and the magnetic dipole moment. Which one of the following statements regarding these directions is true? a) The torque is parallel to both the magnetic field and the dipole moment. b) The torque is perpendicular to the magnetic field, but parallel to the dipole moment. c) The torque is parallel to the magnetic field, but perpendicular to the dipole moment. d) The torque is perpendicular to both the magnetic field and the dipole moment. Question: What will this current loop do? a) Rotate Clockwise b) Rotate Counterclockwise c) othing d) Move Up e) Move Down f) Move Left g) Move Right Page 18 of 27

Magnetic field around a wire magnetic field lin We saw that there was a magnetic force created by a moving charge in a magentic field. ince physics is usually symmetric, we would expect there to be some magnetic effects from a moving charge. Indeed, if we set charges in motion through a wire, then a magentic field is created around the wire. Another representation The electric field strength from a point charge was given by: E = 1 4πϵ 0 q r 2 r Let's just naivly write a similar law for magnetic fields: B = 1 some constants something to do with moving charges something to do with the square of the distance a vector for direction Let's ask about the B field at a point near a wire. We can quantify the moving charges by qv For the direction of things, we'll use the radially directed vector r. Using the cross product, we can establish the direction. v r Page 19 of 27

The Biot and avart Law In a manner similar to electrostatics, we can figure out the magnetic field at a distance r away from a moving charge (i.e. current) by the following: db = μ 0 4π i ds r 2 This μ 0 is the permeability constant and is given by: μ 0 = 4π 10 7 T m/a = 1.26 10 6 T m/a r Example Problem: how that the field a distance d away from an infinitely long wire with current I is given by: B = μ 0 I 2πd +y P +x 0 Example Problem: Find the magnetic field at the center of the circular arc. Page 20 of 27

The force between two wires. wire 1 wire 2 wire 1 wire 2 attractive force between the wires repulsive force between the wires Ampere's Law The magnetic analogue to Gauss' Law. B ds = μ 0I enc The loop integral of the product B ds around any closed loop is equal to the currents enclosed Field around a wire Let's take a simple case of the current carrying wire. We construct and Amperian surface around the wire as shown. wire Amperian Loop B ds = B cos θ ds = B ds The last expression is just B times the circumference of the Amperian loop. B 2πr o, according to Ampere's law, this should be equal to the current enclosed by the loop (times a constant) μ 0 I B = 2πr Page 21 of 27

Field Inside a wire We can use this to also find the magnetic field inside a current carrying wire. First, we'll assume that the current is uniform within the wire. Amperian Loop wire B ds = B(2πr) ince the current is uniform within the wire, the current enclosed will just be given by the ratio of the loop to the total wire: πr 2 I enc = I πr 2 Which yields for B: μ 0 I B = ( ) r 2πR 2 Question: The drawing shows two long, straight wires that are parallel to each other and carry a current of magnitude I toward you. The wires are separated by a distance 2d and are equidistant from the origin. Which one of the following expressions correctly gives the total magnetic field at the origin of the x, y coordinate system? -x d +y -y d +x a) B = + b) B = c) B = + d) B = e) B = 0 μ 0 I j 2d μ 0 I j 2πd μ 0 I j πd μ 0 I j πd Page 22 of 27

Question: The drawing shows two long, straight wires that are parallel to each other and carry a current of magnitude I in the directions shown. The wires are separated by a distance 2d and are equidistant from the origin. Which one of the following expressions correctly gives the total magnetic field at the origin of the x, y coordinate system? -x d +y -y d +x a) B = + b) B = c) B = + d) B = e) B = 0 μ 0 I j 2d μ 0 I j 2πd μ 0 I j πd μ 0 I j πd Rings of current olenoid The solenoid is essentially a stack of current carrying rings. olenoid Upon applying an electrical current through the solenoid, magentic fields will be set up inside and out. Page 23 of 27

olenoid olenoid: Representations The idealized solenoid will be considered just a stack of current loops. Inside the field is effectively uniform. Outside the solenoid, the field is much weaker, and non uniform, so we'll call it negligeable. Ampere's Law on the solenoid d h Amperian loop c Let's integrate around the loop: a b b B ds = B ds + B ds + B ds + B ds a b c c d d a Page 24 of 27

h d Amperian loop c B ds = Bh a b The current enclosed will be given by the number of wires,, times the current in each one: I enc = I = Inh (where n is the number of turns per unit length. Ampere's Law then given us: Bh = μ 0 Inh = ni B sol μ o Question: A wire, connected to a battery and switch, passes through the center of a long current-carrying solenoid as shown in the drawing. When the switch is closed and there is a current in the wire, what happens to the portion of the wire that runs inside of the solenoid? a) There is no effect on the wire. b) The wire is pushed downward. c) The wire is pushed upward. d) The wire is pushed toward the left. e) The wire is pushed toward the right. Example Problem: A 1 meter long MRI solenoid generates a 1.2 T magnetic field. To create this field, it sends 100 A through how many turns? Page 25 of 27

Toroids Ampere's Law and the Toroid Amperian Loop We'll make an Amperian loop inside To evaluate B ds as we go around the loop (clockwise), we can rely on the symmetry of the toroid to tell us that the angle between B and ds will be 0. Thus, B(2πr) = μi where is the number of windings around the circle. B tor = I μ 0 2πr Loops as dipole I Biot-avard Law can predict a value for the magentic field strenth along the axis of the loop: B(z) = μ 0 IR 2 2( R 2 + z 2 ) (3/2) Which, through a little rearrangement can be written as μ 0 IA μ 0 B(z) = = 2πz 3 2π μ z 3 Page 26 of 27

Example Problem: A particle with charge +1 C moves with a velocity equal to v = 1 i + 2 j + 1k. If a magnetic field given by: B = 1 i + 2 j + 1k is applied, what will the force be on the particle? (Express the answer in unit vector notation) Page 27 of 27