Chapter 19Mass Spects

ReviewCh. 17 - 19

Hint: Be able to do the homework (both theproblems to turn in AND the recommended ones)you’ll do fine on the exam!

Friday, April 9, 1999 in class

Ch. 17 - 19

You may bring one 3”X5” index card (hand-writtenon both sides), a pencil or pen, and a scientificcalculator with you.

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 x

d1

d2

d dm m2 1

2 12 2

v

Bq( )

d d2 1 2 1

2

v

Bq( )m m Mass

Spectrometer

rv

Bm

q

Let’s first determine the distance from theentrance point that the H atom hits thephotographic plate.

rv

B

m

q( . )( )

( . )( .

167 10 3 10

16 10 0 6255

27 6

19

kg m / s

C T) cm

Two singly ionized particles enter a mass spectrometer at a speed of3 X 106 m/s. The strength of the magnetic field is 0.625 T. If one of the particles is H and the other particle hits the photographic plate 110 cm further away than the H atom, what chemical element is the second particle?

Now, let’s figure out the radius of curvatureof the mystery element...

We were told that the 2nd element collides withthe photographic plate 110 cm further away fromthe entrance point than the H atom.

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 x

d1

d2

d2 - d1= 110 cm

d2 = 110 + 2r1

d2= 120 cm

r2= 60 cm

m

r B

v

q ( . )( . )( .0 6 16 10 0 6252 10

1926 m C T)

3 10 m / skg

6

Finally, let’s determine the mass ofour mystery element...

PHYSICS RULES!

Televisions rely uponelectromagnetic fieldsto produce the imageswe see.

An electron “gun” fires electronsat the television screen. Whenthey collide with the material onthe back side of the screen, coloredlight is emitted, producing one pixelin the image we see on the screen.

electromagnettelevisionscreen

Changing the strength of the magnetic fieldchanges the degree to which the electron beamis deflected.

We now make explicit a fact we’ve assumedimplicitly in dealing with magnetic fields thus far...

The superposition principle applies!

x x1 2

Btot = B1 + B2

CurrentResistanceOhm’s Law

Series & Parallel CircuitsKirchhoff’s Rules

RC CircuitsMagnetic Fields, Forces, Torques

E

A

Vd t

The charge carriers, each with charge q, move with

an average speed vd in response to the electric field. If there are n charge carriers per volume in the conductor, then the number of charge carriers passing a surface A in a time interval t is given by

Q n Av tdq

I Q

tn Avdq

V = I R

R = L / AResisitivity and Resistance are also a function of Temperature.

V R

0V

a

b c

d

Power is the changein energy per unittime.

As we move a charge q from a to b, it gains potential energy equal to PE = q V

PE

t

PE V

IV t t

q Use this form forpower supplied by

batteries!

P = I2 R

V R

0V

a

b c

d

Power is the changein energy per unittime.

As we move a charge q from c to d, it loses potential energy. The rate of energy loss is

PE

t

Use this form forpower dissipated

by resistors.

Of course, real batteries have some internal resistance, such that

V R

Rint

I

V = - I Rint

load resistor

internal resistance

terminal voltage

V R

Rint

I

What is thecurrent in thiscircuit?

= I Rint + I R

I = Rint + R

What is the powersupplied by the EMF?

What is the power supplied by the battery?

P = I V = I2 R

So what happenedto the difference?

P = I = I2 (Rint + R)

Joule Heating

V1 + V2 = V I1 = I2

due to conservation of charge!

V R2

+

_R1

I1

I2

Req = R1 + R2

Resistors in series ADD.

V R2

+

_R1VR1 = VR2

I

I1

I2

I1 = V / R1

I2 = V / R2

V VI = I1 + I2 = + R1 R2

1 1 1 = + Req R1 R2

Resistors in parallelADD INVERSELY.

1) The sum of the currents entering a junction must equal the sum of currents leaving a junction. (Conservation of Charge)

2) The sum of potential differences across all the elements on any closed loop in a circuit must be zero. (Conservation of Energy)

Junction Rule

Loop Rule

For Applying Kirchhoff’s Rules

A) assign a direction to the current in eachbranch of the circuit. Just GUESS!! If yourguess is incorrect, the current will come outas a negative number, but the magnitude willstill be correct!

B) when applying the loop rule, you mustchoose a consistent direction in which to proceed around the loop (either clockwise or counterclockwise, your choice, but stick to it).

1) When you encounter a resistor in the directionof the current, the voltage drop is V = - I R

2) When you encounter a resistor opposite thecurrent, the voltage drop is V = + I R

R2

I

R1 I1

I2 - I1R1

+I2R2

3) When you encounter an emf in the directionyou’re going around the loop, the voltage change is +V

4) When you encounter an emf opposite thethe direction you’re going, the voltage change is -V

- V1

+ V2

V1 V2

5) The junction rule can only be applied n-1times in a circuit with n junctions.

6) Each new equation you write must containa current that you haven’t yet used.

7) To solve a system of equations with kunknown quantities, you need k independentequations.

Find the currents in each branch of this circuit.

I0 I1

I26

1

18 V

12

12 V+

_

+

_

1

a

f

e

c

b

d

There are two junction points, so we can applythe junction rule ONCE.

I0 = I1 + I2

Remember, each branch has it’s own current.

6

1

18 V

12

12 V+

_

+

_

1

a

f

e

c

b

d

How many possible loops are there?

b c d a e f c d a b e f c b

We have 3 unknowns, so we need 3 equationstotal. Therefore, we need use only 2 of the 3equations provided by the loop rule. (The junctionrule gave us 1 equation already!)

Loops do NOT equal currents!

q

t

Q=CV

Q=0.632CV

V R+

_ C

= the time constant = R C

q( ) ( )/t e t Q 1

V V

Q

C

c ( ) ( )

( )

/

/

t e

e

t

t

1

1

Q = final chargeon C

Charging acapacitor

Discharginga capacitor

RCI

q

t

Q0=CV0

Q=0.368CV0

= the time constant = R C

q( ) /t e t Q0

V V

Q

C

c 0

0

( ) /

/

t e

e

t

t

Q0 = initial chargeon C

Also as was the case with the other fields,the density of the magnetic field lines tellsus the magnitude of the field: the moretightly packed the lines, the stronger the field.

weaker field

stronger fieldweaker field

stronger field

What is the direction of the magneticforce in this case?

x x x x x

x x x x x

x x x x x

x x x x x

B into the page

v

+

Palm faces the directionof the magnetic force.

Thumb points in thedirection of the motionof a positive charge.

Fingers point in thedirection of the magneticfield.

F B v B q qv sin

This force acts in the direction perpendicularto the plane defined by the vectors v and Bas indicated by the right-hand rule!

WARNING: cross-product(NOT simple multiplication!)

DON’T FORGET:Forceshave

directions!

I n Aqvd

| | sin F B I L

| | sin F v BdnALq

But recall our definition of current in a wire...

Substituting, we get the simpler expression:

So what happens if we put a loop of wire carrying current I in a magnetic field?

B

I Force on top and bottom of loop are 0! But force on left and right of the loop are equal in magnitude and opposite in direction, causinga net torque on the loop!

N I A sinB

| | sin F B I L

N B I A sin

Where is the angle between the normalto the loop and the magnetic field.

Top View:

BI

normal

x

The normal is the direction perpendicularto the plane of the loop of wire.

FB

FB

FB

FB

x x x x x

x x x x x

x x x x x

x x x x x

B into the page

v +

x x x x x

x x x x x

x x x x x

x x x x x

B into the page

v

+ x x x x x

x x x x x

x x x x x

x x x x x

B into the page

v

+

x x x x x

x x x x x

x x x x x

x x x x x

B into the page

v

+

So a charged particle moving in a uniformmagnetic field will move in a circle!

The Centripetal Force!

F v Bq

Fv

C

m

r

2

rv

Bm

q

x x x x x

x x x x x

x x x x x

x x x x x

r

x x x x x

x x x x x

x x x x x

x x x x xd

We can measure dusing a photographicplate (for example) to record the location of the end of the particle’s path.

d = 2 r

mr B

v

q

Mass Spectrometer

The magnetic field around a very long,current carrying wire is given by:

B

r

0

2

I

The magnetic field is a vector quantity, so it has a direction!!!!

Where the constant of proportionality is known as the permeability of free space and is found to be

074 10 Tm / A

I

Wrap your hand aroundthe wire with your thumbpointed in the directionof the current.

Your fingers curl aroundthe wire in the directionthe magnetic field points.

I1

I2d

1

2

F

L2

0 1

I I

2 d2

If currents in same direction, wires attract.

If currents in opposite directions, wires repel.

Binside 0nI

Where n is the number of turns per unitlength of the solenoid.

The magnetic field is directed parallel to theaxis of the solenoid. The right hand rulefor currents will help you determine towardwhich end the field points.