Magnetism

We are probably all familiar with the facts that magnets repel and attract each other. This is similar to electric charges. This suggests that we can propose the following law (like Newton’s Law of Gravity and Coulomb’s Law for Electricity):

Fmagnetic = X p1 p2 / r122

where X would be the magnetic constant, analogous to G and k, that describes the strength of the magnetic force.

Magnetic Poles

The “p” in the previous equation stands for poles. In magnetism, we have two kinds of poles, and we call them North and South. Like poles repel and unlike poles attract, just as electric charges do.

However, unlike charges, we always have two poles! If we break a magnet (which has two poles) in half, we have not separated the two poles, rather we have two (smaller) magnets that both have two poles!

Magnetic Poles

N S

N S N S

Break one bar magnet in half, and you have two smaller bar magnets!

Magnetism

Since we cannot seem to isolate one magnetic pole like we could electric charges, the force equation that is similar to Newton’s and Coulombs Law turns out to be not very useful.

We do have a more useful alternative, though. It turns out that charges experience a force when moving near magnets!

Magnetic ForceJust like charges set up an electric field, and other charges in the

vicinity feel an electric force due to that electric field (Fel = qE), we can work with the idea that magnets set up a magnetic field in space, and charges moving through that field experience a magnetic force. There are two things to note:

1. The charges must be moving.

2. The direction of the force is “weird”:

the direction of the magnetic force is perpendicular to the velocity of the moving charge, and perpendicular to the magnetic field direction!

Magnetic Force Lawmagnitude: Fmagnetic = q v B sin(vB)

direction: right hand rule:

thumb = hand fingers

Point your right hand in the direction of v, curl you fingers in the direction of B, and the force will be in the direction of your thumb; if the charge is negative, the force direction is opposite that of your thumb (or use you left hand).

Units

Fmagnetic = q v B sin(vB)

This law effectively defines the magnetic field, B (just like Felectric=qE defined E).

The MKS units of B are:

Tesla = Nt-sec / Coul-m .

This unit turns out to be a very large one. We have a smaller unit:

10,000 Gauss = 1 Tesla .

Example

A proton is moving at a speed of 3 x 104 m/s towards the West through a magnetic field of strength 500 Gauss directed South. What is the strength and direction of the magnetic force on the proton at this instant?

qproton = +e = 1.6 x 10-19 Coul.

v = 3 x 104 m/s, West

B = 500 Gauss * 1 Tesla/10,000 Gauss

= .05 T, South

Example, cont.magnitude: Fmagnetic = q v B sin(vB)

direction: right hand rule

magnitude: F = (1.6 x 10-19 Coul) * (3 x 104 m/s) * (.05 T) * sin(90o) = 2.4 x 10-16 Nt.

direction: thumb = hand x fingers

= West x South = UP.

Note: although the force looks small, consider the acceleration: a = F/m =

2.4 x 10-16 Nt / 1.67 x 10-27 kg = 1.44 x 1011 m/s2.

Magnetic Forces

We’ll play with magnetic forces on moving charges in the Magnetic Deflection experiment in lab. At that time we’ll also discuss and experiment with the earth’s magnetic field.

Magnetic Force and Motion

Since the magnetic field is perpendicular to the velocity, and if the magnetic force is the only force acting on a moving charge, the force will cause the charge to go in a circle:

F = ma, Fmag = q v B, and a = 2r = v2/r

gives: q v B = mv2/r, or r = mv/qB .

Mass Spectrograph

We can design an instrument in which we can control the magnetic field, B. If we ionize (almost always singly) the material, we know the charge, q. We can use known voltages to get a known speed for the ions, v. We can then have the beam circle in the field and hit a target, and from that we can measure the radius, r. Hence, we can then calculate the mass using r = mv/qB .

Example

To see if this is really feasible, let’s try using realistic numbers to see what the radius for a proton should be:

q = 1.6 x 10-19 Coul; B = .05 Teslas (500 G)

m = 1.67 x 10-27 kg; Vacc = 500 volts gives:

(1/2)mv2 = qV, or v = [2qV/m]1/2 = 3.1 x 105 m/s

r = mv/qB = (1.67 x 10-27kg)*(3.1 x 105 m/s) / (1.6 x 10-19 Coul)*(0.05 T) = .065 m = 6.5 cm.

Mass Spectrometer

Heavier masses will give bigger radii, but we can shrink the radii if they become too big by using bigger magnetic fields. Note that by measuring quantities that we can easily measure (charge, radius, Voltage, magnetic field), we can determine very tiny masses! In one of our experiments in lab (Charge to Mass of the Electron), we will essentially determine the mass of an electron!

Other uses

A cyclotron is an instrument used to accelerate charged particles to very high speeds. It uses magnetic fields to bend the charges around in circles to keep them in one place while they’re being accelerated.

Magnetic bottles can be used to contain high energy plasmas in fusion research.

Magnetic Forces

The Computer Homework Vol. 4 #1, Magnetic Deflection, deals with magnetic forces on charged particles. We’ll discuss strategies for completing this assignment in class.

Creating Magnetic Fields

Gravitational fields acted on masses, and masses set up gravitational fields:

Fgravity = mg where g = GM/r2

Electric fields acted on charges, and charges set up electric fields:

Felectric = qE where E = kQ/r2

Magnetic fields acted on moving charges; do moving charges set up magnetic fields?

Creating Magnetic Fields

Magnetic fields acted on moving charges; do moving charges set up magnetic fields?

YES! From the gravitational and electric cases, we can guess that we will need:

a constant that describes the strength similar to G and k;

an inverse square relationship with distance (1/r2); and

a dependence on what is acted upon (like m and q) - in this case, qv.

Creating Magnetic Fields

But we also have the right hand rule in the magnetic force equation, and we’ll need a right hand rule in the field generating equation also.

Magnetic Fields

B = (o/4) q v sin(vr) / r2

direction: right hand rule

thumb = hand x fingers

Point your hand in the direction of v, curl you fingers in the direction of r, and the field will be in the direction of your thumb; if the charge is negative, the field direction is opposite that of your thumb.

Magnetic ConstantThe constant (o/4) is a seemingly strange way of

writing a constant that serves the same purpose as G and k, but that is exactly what it does.

The value: (o/4) = 1 x 10-7 T*m*sec/Coul (or 1 x 10-7 T*m/Amp).

In fact, the constant k is sometimes written as:

k = 1/(4o).

[In both cases, the sub zero on and indicates the field is in vacuum.]

Creating Magnetic Fields

A single moving charge does create a magnetic field in the space around it, but since both the charge and the constant are very small, we usually don’t have to worry about these effects - except in two cases:

1. magnetic materials (permanent magnets) and

2. currents (electromagnets).

Creating Magnetic Fields:1. Permanent Magnets

Materials are made up of atoms that have electrons moving around the central nucleus (more on this in part 4). These moving electrons create magnetic fields. However, in almost all materials the magnetic fields created by the orbiting electrons tend to cancel - except in some materials like iron. This is the basis of making magnets out of iron bars.

Creating Magnetic Fields2. Electromagnets

If we have a series of charges moving (which means a current), then we can generate an appreciable magnetic field.

A moving charge (qv) actually can be thought of as a current over a small length:

qv = q(L/t) = I L

so that we have for each small length:

B = (o/4) I L sin(Ir) / r2

with direction: thumb(field) = hand(current) x fingers(radius).

Special Cases

For the current over a complete circuit, the field produced at any particular point in space will depend on where the point is and the shape of the circuit.

Loop of Current

For the field at the center of a current loop we have the special equation:

Bat center of loop = o N I / 2R

where N is the number of turns in the wire,

I is the current in each loop, and

R is the radius of the loop.

ExampleWhat is the field strength at the center of a loop

that has 0.2 Amps running through 3400 turns, where the radius of the loop is 6 cm?

Bat center of loop = o N I / 2R

B = (4 x 10-7 T*m/A) * 3400 * 0.2A / (2*.06m)

= .0071 T = 71 Gauss.(As you move away from this center, the field will

decrease.)

Review

Basic equation for calculating magnetic fields:

B = (o/4) I L sin(Ir) / r2

with direction: thumb(field) = hand(current) x fingers(radius).

(where the IL is interchangeable with qv).

Review - continuedBasic equation for calculating magnetic fields:

B = (o/4) I L sin(Ir) / r2

with direction: thumb(field) = hand(current) x fingers(radius).

UNITS: Note all equations for B should have a o, an I, and a (distance/distance2) or (1/distance).

• Bat center of loop = o N I / 2R

• Bhelmholtz coil = 8oNI/[125]1/2 R

• Blong straight wire = o I / 2a

Forces, Fields and Currents

If currents can create magnetic fields (as we just saw), then currents should be acted upon by fields. Another way of saying the same thing is: fields act on moving charges [F = qvB sin(vB)], and currents consist of moving charges [qv = IL] , so we have:

Magnetic Force on Current

Fon current = I L B sin(IB)

with direction: thumb(force) = hand(current) x fingers(field).

Let’s now consider how this force will work on complete current circuits, not just individual lengths.

Forces on rectangular current loop

Consider the situation in the figure below:

A current loop (with current direction going

counter-clockwise) is situated in a Magnetic

Field going from North to South poles.

N B S

I

Forces on rectangular current loop

We need to consider the forces on each of the four sides of the current loop.

The force on the top and bottom of the loop are zero, since the field and current are either parallel or anti-parallel (IB = 0o or 180o).

I Ftop=0

N B S

I Fbottom=0

Forces on rectangular current loop

The current on the left side is going down while the field is directed to the right, so that means the force is directed out of the screen, and the magnitude is: Fleft = I L B sin(90o) = I L B.

F

N B S L

Forces on rectangular current loop

The current on the right side is going up while the field is directed to the right, so that means the force is directed into the screen, and the magnitude is: Fright = I L B sin(90o) = I L B.

F

N B S L

Net Force

Ftop = 0 Fbottom = 0

Fleft = ILB out Fright = ILB in

As we can see, the NET FORCE (F) is zero.

However, since the force is pushing out on the left and in on the right, there is a Torque! The loop will tend to rotate about an axis through the center.

Torque on rectangular current loop

Recall that torque is: = r F sin(rF). In the figure below we can see that r = w/2. Thus the Fleft gives a torque of (w/2)ILB, and the Fright also gives a torque of (w/2)ILB.

r

F F

N B S L

w

Torque on rectangular current loop

Since both torques are trying to rotate the loop in the same direction, the total torque is:

= wILB. We note that wL = A (width times length = Area). Also, we can have several loops (N) that will each give a torque.

F F

N B S L

w r

Torque on a loop

The final result for this loop is:

= N A I B sin(IB) sin(rF) .

In this orientation, IB = 90o and rF = 90o .

If the loop does rotate, we see that IB remains at 90o (the current still goes up and down, the field still goes to the right), but rF changes as the loop rotates!

= N A I B sin(rF) .

Electric Meter

One of the early types of current meters is one where we have such a loop of current in a magnetic field, and we have a restraining spring. As the current increases, the torque increases, but we have a restraining spring to keep it from rotating completely around. The bigger the current, the bigger the torque, and the loop will turn through a bigger angle. Attach a pointer onto the loop, and we have the (analogue) meter.

Electric Motor

To create an electric motor that will continue to spin when a current is applied, we need to keep the current going up the right side, even when the side originally on the left becomes the right side due to the spinning. We can accomplish this by using a set of brushes as indicated in the next slide.

Electric Motor + -

N S

This diagram will be explained in class, since it involves three dimensions. The two green “C’s” are actually one ring that is split. The ring is in and out of the screen.

Electric Motor

= N A I B sin(rF) .

Since we switch the current to make it always run up whatever wire is on the left side, we make sin(rF) always positive.

To find the average torque, we need to determine the average of sin() from 0o through 180o. From the calculus, we find that its average value is 2/.

Average Torque

Thus we get for the average torque:

average = (2/) N A I B .

But we usually describe motors by their power, not by their torque. Recall that Power is energy per time, and energy is force through a distance. For rotations, this becomes: energy is torque through an angle, and power is torque through an angle per time (but recall = t).

Average Power

Putting all of this together gives:

Pavg = N A I B (2/), and with f, we have:

Pavg = 4 f N A I B.

Note that the power depends not only on the details of the motor (N, A, B) and the current (I), but also on how fast the motor spins (f)!

ExampleDesign an electric motor that has a power of 1/2

hp when it rotates at a frequency of 120 Hz (120 cycles/sec * 60 sec/min = 7200 rpm).

Pavg = 0.5 hp* (746 Watts/1 hp) = 373 Watts.

f = 120 Hz.

Design = ? (This means we specify N, A, I and B such that P = 373 Watts when f=120 Hz. We have some “free” choices!)

Example, cont.

Pavg = 4 f N A I B.

Pavg = 373 Watts, f = 120 Hz.

Let’s start by choosing some reasonable values: choose an area (A) of 10 cm x 10 cm = .01 m2; choose a magnetic field (B) of 1000 Gauss (0.1 T). We can also choose a current (I) of 5 amps. This means that we can now solve for the number of turns, N.

Example, cont.

Pavg = 4 f N A I B.

Pavg = 373 Watts, f = 120 Hz.

A = .01 m2 B = 0.1 T I = 5 amps.

N = P / [4 f A I B]

= 373 Watts / [4 * 120 Hz * .01 m2 * 5 Amps * .1 T] = 155 turns.