-magnetic force on a charged particle -magnetic force on a current-carrying wire -torque on a...
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-Magnetic Force on a Charged Particle-Magnetic Force on a Current-Carrying Wire-Torque on a Current-Carrying Loop
AP Physics CMrs. Coyle
Fermi Lab, Chicago Illinois Circumference 6.3km
Mass Spectrometer
DC Motor
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• A magnetic field can exert a force on a charged particle that moves in it.
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Three ways we talk about “magnetic field”.
• Magnetic Field: Regions surrounding a magnet where another magnet or a moving electric charge will feel a force of attraction or repulsion.
• Magnetic Field Lines: exit the north pole and enter the south.
• Magnetic Field Strength, B
• Vector, Unit: Tesla, T• Named after Nikola Tesla
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Poles• Law of Poles: Like poles repel, unlike poles attract.• The force between two poles varies as the inverse
square of the distance between them.• A single pole (monopole) has not been isolated.
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Force on a Charged Particle Moving in a Magnetic Field
-Magnetic fields only exert forces on moving charged particles or other magnets.
• F = |q|(v x B) = |q|vB sin
• vector cross-product
is the angle between v and B• F=0 for =0 or 1800
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Remember: Cross Product Using Determinants
or
ˆ ˆ ˆ
ˆ ˆ ˆy z x yx zx y z
y z x yx zx y z
A A A AA AA A A
B B B BB BB B B
i j k
A B i j k
ˆ ˆ ˆy z z y x z z x x y y xA B A B A B A B A B A B A B i j k
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Remember: Properties of Cross Remember: Properties of Cross Products of Unit VectorsProducts of Unit Vectors
ixi=0
jxj=0
kxk=0
ixj=k
jxk=i
kxi=j
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To find the direction of the force use the
Right Hand Rule
For a positive test charge:
Thumb v
Fingers B
Out of palm F
The force is always perpendicular to the vB plane
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For a negative particle the F is opposite to what it would be for a positive particle
(use left hand)
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Alternate Rule: Right Hand Curl Rule
•Curl fingers from v to B F = q(v x B)
•F is in the direction of the thumb
•Similarly used for the direction of torque (=r x F)
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Graphical Representation of the Magnetic Field Vector (Strength), B
x field lines pointing into the page
● field lines pointing out of the page
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Question• Find the direction of the magnetic force
acting on the +charged particle entering the magnetic field with a velocity v perpendicular to B.
x x x x x
x x x x x
V x x x x x
x x x x x
Answer: Upwards
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What motion will the particle in the previous example undergo (particle entered the B-
field in a direction perpendicular to B?• Circular Motion
• Magnetic force will represent the centripetal force
• http://online.cctt.org/physicslab/content/applets/JavaPhysMath/java/partmagn/index.html
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Examples:
http://physicslearning.colorado.edu/PiraHome/PhysicsDrawings.htm
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Does the magnetic force do work?
• F is always perpendicular to the displacement
• F can change the direction of v not the magnitude
• F cannot do work, cannot change KE
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Mass Spectrometer U.S. Department of Energy
http://doegenomestolife.org
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Motion of Alpha- Beta-Gamma Particles in a Magnetic Field
1) Alpha particles, positive helium nuclei, charge +2e
2) Gamma rays, (no charge) electromagnetic radiation
3) Beta particles, electrons charge -1e
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Problem 1A proton is accelerated through a constant electric field
(parallel plates) and acquires kinetic energy of 4eV.
It enters perpendicularly to the 2T field of a detector as shown.
Charge of a proton=1.6x10-19 C, mass of proton=1.67x10-27 kg, ignore gravity.
a) Draw the path of the positive charge as it enters the magnetic field.
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Problem 1 cont’db) Calculate the force acting on the charge due to the
magnetic field.
Charge of a proton=1.6x10-19 C , mass of proton=1.67x10-27 kg
KE= 4eV B= 2T
Ans: v= 2.77x104 m/s, F= 8.86x10-15 N
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Problem 1 cont’dc) Calculate the distance on the detector where the particle will
land (radius of the circular path). Ignore gravity.
Charge of a proton=1.6x10-19 C, mass of proton=1.67x10-27 kg KE= 4eV B= 2T
Ans: 1.45x10-4 m
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Large Hadron ColliderCERN (Conseil Européen pour la Recherche
Nucléaire) Switzerland- France 2008
Circumference: 27km
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• Particle Accelerators use electric and magnetic fields to accelerate charged particles.
• Cyclotron Applet
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Magnetic Force on a Current Carrying WireF = I (L x B) = I L B sin
What is the direction of the magnetic force acting on this wire?
F
• I is the current
• L is a vector of magnitude of the length of the wire and direction that of the current
•is the angle formed between I and B
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The force acting on a wire of arbitrary shape is the same as if it were a
straight wire with the same ends• The total force is:
I sin
bd ILB θ
aF s B
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What is the net force acting on this current-carrying loop?
I bd
aF s B
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Which way will a loop turn?
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Forces on a Current-Carrying Loop
For B as shown:F 1 = F3 = 0
F 2 = F4 = IaB
Electric DC Motor
(Fendt Applet)
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Top View
(I ) (I )
I
I
2 42 2 2 2
, where A is the area of the loop
b b b bτ F F aB aB
τ abB
τ AB
F 2 = F4 = IaB
Torque acting on the loop:
Torque Acting on the Current-Carrying Loop
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Torque Acting on the Current-Carrying Loop
sin
sin
In general:
, where A is the area of the loop
=IA x B
Magnetic Dipole Moment
τ IabB θ
τ IAB θ
τ
μ IA
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Magnetic Force on a Current Carrying Wire
• Lorentz Force- Magnetic Force on a Current Carrying Wire (Fendt Applet)
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DC Motor
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DC Motor
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Problem #9
A proton moves with a velocity of v=(2i-4j+k)m/s
in a region in which the magnetic field is
B=(i+2j-3k)T. What is the magnitude of the magnetic force this charge experiences?
Ans: 2.34x10-18 N
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Problem #13
A wire 2.80 m in length carries a current of 5.00 A in a region where a uniform magnetic field has a magnitude of 0.390 T. Calculate the magnitude of the magnetic force on the wire assuming the angle between the magnetic field and the current is (a) 60.0°, (b) 90.0°, (c) 120°.
Ans: a)4.73N, b)5.46N, c)4.73N
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Problem # 21
A small bar magnet is suspended in a uniform 0.250-T magnetic field. The maximum torque experienced by the bar magnet is 4.60 × 10–3 N · m. Calculate the magnetic moment of the bar magnet.
Ans: 18.4mA m2
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Problem #54
A 0.200-kg metal rod carrying a current of 10.0 A glides on two horizontal rails 0.500 m apart. What vertical magnetic field is required to keep the rod moving at a constant speed if the coefficient of kinetic friction between the rod and rails is 0.100?
Ans: F=0.196N, B=0.039T