Magnetism

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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)

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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.

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para and ferro magnets

do not retainany magnetismin absence of external field

retains domainsin which magneticfield remain in the absence of external fields

strawberry in a B-field

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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

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One big magnet!

Note that the geographical North pole is in fact the magnetic south pole B=0.3-0.6 x 10-4 Tesla

Why is it higher here?

demo:compass needlescompass

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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?a) It would point roughly to the geographical southb) It could point anywherec) It would rotate with constant angular speed

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.

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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

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direction of force on charged paricles in B-field

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 versionof Right Hand Rules given in the Textbook.

demo: bending the beam I

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example: electron in magnetic field

an electron with v=1x106 m/s is entering a area with B=1 T. The field is directed into the screen.

a) in which direction will the electron be bent, if at all?

b) how large is the force? what is the acceleration?

x x x x

x x x x

x x x xa) use right hand rule:thumb is velocity (initially to the right)index finger is field (in the screen)middle finger is force perpendicular to bothswitch 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

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question

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 screenb) out of the screenc) parallel to the B field lines

and the screend) perpendicular to the B field

lines and parallel to the screen

e) in the direction of motion

Mg

v

45o

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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

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Magnetic spectrometers

Beam fromcyclotrons

target chamber S800 spectrometerAt the cyclotron

Bending angle ~ 150o

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questionIn 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/qBFor 120Sn: M=1.991x10-25 kg v=12.8814x107 m/sB=1T q=50x1.6x10-19 C.RSn=3.2060 mFor 120Snb: M=1.991x10-25 kg v=13.0990x107 m/sB=1T q=51x1.6x10-19 C.RSb=3.1961 mRSn-RSb=3.206-3.1961=9.9x10-3 m = 9.9 mm thus separated

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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

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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 xproton

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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

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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?

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

I

B

top view

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question

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

I

A

D

C

B

x x x x

x x x x

x x x x

x x x x

x x x x

top view

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question

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

I

A

D

C

B

x x x x

x x x x

x x x x

x x x x

x x x x

top view

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Torque on a current loop

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.

I

a

b

Rotation axis

B

x

F

F

Top view

frontal view

If there is a net torque,the loop will rotate!

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Torque on a current loop

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

I

a

b

Rotation axis

B

x

F

F

Top view

frontal view

frontal view

sin=F/FL

FFL

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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 =IAN magnetic moment of 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

NAI

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example

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:

a) the area of the coil?

b) the magnetic moment of the coil?

c)the torque one the coil?

B

NAI

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

note for loncapa: area of an ellipse: ab with a, b radii in the two directions

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electric motor

By supplying electricity we can get some work done!

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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 experimentmagnetic field of a current

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How to quantify the field

0 = “permeability of free space” =

4 x 10-7 Tm/A

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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

a) What is the direction of the magnetic field near the electron due to the wire?

b) what is the magnitude of the magnetic field near the electron?

c) what is the direction of the force on the electron?

d) what is the magnitude of the force on the electron?

I=1A

2 cm

q=-1.6x10-19C

a) use 2nd right hand rule B-field goes into the screenb)

=4 x 10-7 x 1/(20.02)=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)

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question 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 up2) what is the direction of the

force on the proton? to the left to the right up down no force at all

I

x

proton movinginto the screen

1) use 2nd right hand rule(same as example on previousslide)

2) use 1st right hand rule. velocityis into the screen, B-field is into the screen: no Force (sin(00)=0)

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magnetic force between two parallel wires

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:

d

attractive if same directionrepulsive if opposite direction

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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?

a) B1=0I/(2r)=4x10-7x1/(20.01)=2x10-5 T B2= 4x10-7x2/(20.01)=4x10-5 T B1: into the screen B2: out of the screen Bnet=2x10-5 T out of the screen b) 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/(2d)= 4x10-7x1x2/(20.02)=2x10-5 N

1A 2A

2cm

1 2

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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…

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other cases: the current loop

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?

R

I I

X

right-handed current through loop: B-field in the screen

left-handed current through loop: B-field out of the screenI=2RBcenter/0=

2x0.01x10-5/4x10-7=0.16 A

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other cases II: magnetic field of a solenoid

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 B-field of solenoid

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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=4x10-7 x 3000/0.3 x 4 = 1x10-3 T use n=N/L

inside the coil, the field is uniform