Magnetic and electric fields

1. Unit 7 - Lesson 10Unit 7 - Lesson 10
Magnetic Force
Nelson Reference:Nelson Reference:
Page 382 – 389, 392 – 394Page 382 – 389, 392 – 394 (Review of Gr 11)
Page 397 – 400Page 397 – 400 (New)
Magnetic Force on a ChargeMagnetic Force on a Charge
Velocity Selector andVelocity Selector and
Mass SpectrometerMass Spectrometer
{McGraw-Hill Reference Pages: 355-359}{McGraw-Hill Reference Pages: 355-359}

2. In Grade 11, you wereIn Grade 11, you were
taught thetaught the Right Hand
Rules (RHR’s) for(RHR’s) for
magnetism. The drawingmagnetism. The drawing
on the right, shows aon the right, shows a
conductor going into theconductor going into the
screen (or page) andscreen (or page) and
surrounded by ansurrounded by an
external magnetic fieldexternal magnetic field
from the horseshoefrom the horseshoe
magnetic. The directionmagnetic. The direction
ofof FMagnetic is as shownis as shown
usingusing RHR # _____. TheThe
green dot indicatesindicates
current is moving ______current is moving ______
the screen (or page).the screen (or page).
N
S
F M a g n e t ic
T h e d ir e c t io n o f
F M a g n e t ic
is f o u n d
b y u s in g R H R # _ _ _ _

3.  If the current was moving into the page an ___If the current was moving into the page an ___
would have been used to show this direction.would have been used to show this direction.
Magnetism Review
 A magnetic force will be exerted on a currentA magnetic force will be exerted on a current
carrying conductor if the conductor iscarrying conductor if the conductor is ⊥⊥ to theto the
magnetic field. No force is exerted if themagnetic field. No force is exerted if the
conductor is _________ to the magnetic field.conductor is _________ to the magnetic field.
 FMagnetic = B ⊥IL, B is the magnetic field strength inis the magnetic field strength in
Tesla (T),Tesla (T), I is current in Amperes (A),is current in Amperes (A), L is theis the
length of the conductor in metres (M) in thelength of the conductor in metres (M) in the
magnetic field.magnetic field.
 Only moving charge(s) create a magnetic field,Only moving charge(s) create a magnetic field,
stationary charges do not.stationary charges do not.

4.  A single charged particleA single charged particle
will be deflected by awill be deflected by a
magnetic field if the chargemagnetic field if the charge
movesmoves ⊥ to the field.to the field.
 In which direction will theIn which direction will the e-
be deflected in the topbe deflected in the top
diagram and how will itdiagram and how will it
move in the magnetic field?move in the magnetic field?
 By using RHR #3 for aBy using RHR #3 for a
negative charge, the, the e-
isis
initially deflectedinitially deflected
downward. However, thedownward. However, the
magnetic force is alwaysmagnetic force is always ⊥
to the (changing) DOM, soto the (changing) DOM, so
it moves in a semicircle.it moves in a semicircle.
UP
M a g n e tic F ie ld
e -
e -
e -
e -
e -
D O M
UP
M a g n e tic F ie ld

5. The magnetic force is alwaysThe magnetic force is always
directeddirected ⊥ the DOM of thethe DOM of the
charge socharge so circular motion
occurs. We can see that. We can see that
________ = FM ..
To calculateTo calculate FM we use:we use:
FM = qv B ⊥⊥,,, for a magnetic field, for a magnetic field
that isthat is ⊥ the DOM of thethe DOM of the
charge wherecharge where q is theis the
charge in Coulombs,charge in Coulombs, v is theis the
(tangential) speed (in m/s)(tangential) speed (in m/s)
andand B ⊥ is the strength of theis the strength of the
field in Tesla (T).field in Tesla (T).
e -
e -
e -
e -
FM
FM
F
M
F M
UP
M a g n e t ic f ie ld n o t
s h o w n , o n ly F M
a c t in g o n e -

6.  If the DOM of the charge isIf the DOM of the charge is not
⊥ to (but alsoto (but also not parallel to) theparallel to) the
magnetic field, you must findmagnetic field, you must find
the component of the fieldthe component of the field ⊥ toto
the DOM of the charge; this isthe DOM of the charge; this is
BsinBsinθθ in the diagram. Thein the diagram. The
force acting on the chargeforce acting on the charge
would now be:would now be:
FM = qv Bsinθ
 Instead, if a current carryingInstead, if a current carrying
conductor was in a field thatconductor was in a field that
was notwas not ⊥ to the conductor, theto the conductor, the
force on the conductor could beforce on the conductor could be
found from:found from:
FMagnetic = IL Bsinθ
DOM
e -
θ
M a g n e t ic F ie ld

7. Velocity Selector: Allows only charged particles of aAllows only charged particles of a
specific speed to pass through; all other particlesspeed to pass through; all other particles
are filtered out. The 1are filtered out. The 1stst
electric field accelerateselectric field accelerates
the charge and the final speed will depend on thethe charge and the final speed will depend on the
particle’s q/m ratio. The 2particle’s q/m ratio. The 2ndnd
electric field will applyelectric field will apply
a _____ ward force and the magnetic field willa _____ ward force and the magnetic field will
apply a ______ward force to the +ve charge.apply a ______ward force to the +ve charge.
Δ d
D O M+ q
+ + + + +
- - - - -
C
UP
M a g n e t ic f ie ld
g o e s in t o t h e " p a g e "
C h a r g e
a c c e le r a t e d t o
it s f in a l s p e e d

8. If a charge is to pass the through, the electricIf a charge is to pass the through, the electric
and magnetic forces must be ______ . So:and magnetic forces must be ______ . So:
FM = Fq
qv B ⊥ =q E , use only magnitudes for, use only magnitudes for B & E
v = E/B ⊥⊥
The speed (m/s) of the charge passingThe speed (m/s) of the charge passing
through the velocity selector equals thethrough the velocity selector equals the
magnitude of the electric field intensitymagnitude of the electric field intensity
(N/C) divided by the magnitude of the(N/C) divided by the magnitude of the
magnetic field intensity.magnetic field intensity.

9. Mass Spectrometer:Mass Spectrometer: Adds a 2Adds a 2ndnd
magnetic field to themagnetic field to the
velocity selector. The charged particle (ion) nowvelocity selector. The charged particle (ion) now
will experience only a deflecting force, so:will experience only a deflecting force, so:
Fc = FM
mv2
/r= qv B ⊥ , r is measured in the second fieldis measured in the second field
m = qrB⊥ /v The mass can be found since allThe mass can be found since all
other quantities are known; the value ofother quantities are known; the value of q is foundis found
from the velocity selector.from the velocity selector.
Δ d
D O M+ q
+ + + + +
- - - - -
UP
S e c o n d
M a g n e t ic F ie ld
V e lo c ity S e le c to r r
M a s s S p e c t r o m e t e r

10. Practice Questions
Nelson Textbook:
Page 401 # 1, 2, 4Page 401 # 1, 2, 4
Page 404 # 2 – 6Page 404 # 2 – 6
{For #2 on page 404, you must draw a{For #2 on page 404, you must draw a
picture and then try to relate the equationspicture and then try to relate the equations
you have learned to the information given.you have learned to the information given.
For question #4, ignore the initial speedFor question #4, ignore the initial speed
and direction.}and direction.}
Page 419 # 42, 44, 47Page 419 # 42, 44, 47

11. From the McGraw-Hill TB:
An isotope of hydrogen having a proton and aAn isotope of hydrogen having a proton and a
neutron in its nucleus is ionized and theneutron in its nucleus is ionized and the
resulting positive ion travels in a circular pathresulting positive ion travels in a circular path
of radiusof radius 36.0 cm in a perpendicular magneticin a perpendicular magnetic
field of strengthfield of strength 0.80 T ..
1.1. Calculate the speed of the isotope.Calculate the speed of the isotope.
2.2. Calculate the accelerating potential differenceCalculate the accelerating potential difference
that gave the isotope its speed.that gave the isotope its speed.
[ Ans 1.4 x 107
m/s, 2.0 x 106
V ]