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Magnetic Fields
Certain iron containing materials have been known to attract or repel each other.Compasses align to the magnetic field of earth.Analogous to positive and negative charges, every magnet has a north and a south pole (but always as a pair).Magnetic fields can be created by electrical currents.
Magnetic Field and Magnetic Force
vF rr,qB ∝
0// =⇒ BFBvrrr
vBF rrr,0 ⊥⇒≠ Bθ
θsin∝BFr
θ
B
v
( )vF rr,0<qB ( )vF rr
,0>qB
oppositedirection
FB
+q
Magnetic ForceBvFrrr
×= qBθsinvBqFB =
Magnetic Field, B: Tesla (T)=N/(C.m/s)=N/(A.m)
Right Hand Rule Magnetic Field Representation
Electric and Magnetic Forces
Electric forces act in the direction of the electric field, magnetic forces are perpendicular to the magnetic field.A magnetic force exists only for charges in motion.The magnetic force of a steady magnetic field does no work when displacing a charged particle.The magnetic field can alter the direction of a moving charged particle but not its speed or its kinetic energy.
BvFrrr
×= qB EFrr
qE =
Motion of Charged Particles in a Uniform Magnetic Field
Consider a positive charge moving perpendicular to a magnetic field with an initial velocity, v.The force FB
is always at right angles to v and its magnitude is,
qvBFB =
So, as q moves, it will rotate about a circle and FB
and v
will always be perpendicular. The magnitude of v
will always be
the same, only its direction will change.
Cyclotron FrequencyTo find the radius and frequency of the rotation:
∑ = rmaF
rvmqvBFB
2
==
qBmvr =
mqB
rv==ω
qBm
vrT π
ωππ 222===
The radial force
The angular speed
The period
Cyclotron frequency
Example 29.3Electron beam in a magnetic field
ΔV = 350 Vr = 7.5 cmq = e = 1.6 x 10-19 Cm = me = 9.11 x 10-31 kg
B = ?ω = ?
v
r Vevm
UK
e Δ=
=Δ
2
21
smm
Veve
/1011.12 7×=Δ
=
qBmvr = T
ervmB e 4104.8 −×==
sradmeB
mqB
e
/105.1 8×===ω
FBΔV
B
Guiding Charged Particles
Cosmic Rays in the Van Allen Belt
Helical Motion – vx , vy and vz
Complex Fields –Magnetic Bottles
Lorenz Force and ApplicationsEssentially it is the total electric and magnetic force acting upon a charge.
BvEFrrrr
×+= qqVelocity Selector
BEv
FF Be
=
=
Mass Spectrometer
EBrB
qm 0=
Magnetic Force on a Current Carrying Conductor
Magnetic force acts upon charges moving in a conductor.The total force on the current is the integral sum of the force on each charge in the current.In turn, the charges transfer the force on to the wire when they collide with the atoms of the wire.
Magnetic Force on a Current Carrying Conductor
Magnetic Force on a Current Carrying Conductor
( )BvFrrr
×= dqB q,
( )nALq dB BvFrrr
×=
qnAvI d=
BLFrrr
×= IB
For a straight wire in a uniform field
BsFrrr
×= Idd B
∫ ×=b
aB dI BsF
rrr For an arbitrary wire in an arbitrary field
A Special Case – arbitrary wire in a uniform field
∫ ×=b
aB dI BsF
rrr
BsFrrr
×⎟⎟⎠
⎞⎜⎜⎝
⎛= ∫
b
aB dI
BLFrrr
×′= IB
The net magnetic force acting on a curved wire in a uniform magnetic field is the same as that of a straight wire between the same end points.
Closed Loop in a Magnetic Field
( ) BsFrrr
×= ∫ dIB
0=∫ srd
0=BFr
Net magnetic force acting on any closed current loop in a uniform magnetic field is zero.
Forces on a Semicircular Conductor
IRBILBF 21 ==
On the straight wire
dsIBdIdF θsin2 =×= Bsrr
On the curved wire
θθRdds
Rs==
θθdIRBdF sin2 =
∫∫ ==ππ
θθθθ00
2 sinsin dIRBdIRBF
IRBF 22 =
Directed out of the board
Directed in to the board
021 =+= FFFrrr
Torque on a Current Loop in a Uniform Magnetic Field
0
0
=
=×
BF
BLr
rr
For 1 and 3,
IaBFF == 42
( ) ( )
( )bIaB
bIaBbIaBbFbF
=
+=+=
max
42max 2222τ
τ
If the field is parallel to the plane of the loop
IAB=maxτ
For 2 and 4,
Torque on a Current Loop in a Uniform Magnetic Field
If the field makes an angle with a line perpendicular to the plane of the loop:
θτ
θθθτ
θθτ
sin
sinsin2
sin2
sin2
sin2 42
IAB
IabBbIaBbIaB
aFaF
=
=+=
+=
BAτrrr
×= I
Concept QuestionA rectangular loop is placed in a uniform magnetic field with the plane of the loop perpendicular to the direction of the field.If a current is made to flow through the loop in the sense shown by the arrows, the field exerts on the loop:
1. a net force.2. a net torque.3. a net force and a net torque.4. neither a net force nor a net torque.
Magnetic Dipole Moment
Aμrr I=
Bμτrrr
×=
(Amperes.m2)
If the wire makes N loops around A,
BμBμτrrrrr
×=×= coilloopN
Bμrr⋅−=U
The potential energy of the loop is:
Hall Effect
BvEqEBqv
dH
Hd
==
BdvV dH =Δ
nqAIvd =
tIBR
nqtIBV H
H ==Δ
SummaryThe right hand ruleMagnetic forces on charged particlesCharged particle motionMagnetic forces on current carrying wiresTorque on loops and magnetic dipole moments
For Next ClassReading Assignment
Chapter 30 – Sources of the Magnetic Field
WebAssign: Assignment 7