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CHAPTER 27
Magnetic Field and Magnetic Forces
PRODUCING A MAGNETIC FIELD
Permanent Magnet
Current
MAGNETIC FIELD LINES,
BAR MAGNET EXAMPLE
The compass can be used to trace the field lines
 The lines outside the magnet point from the North pole
to the South pole

MAGNETIC FIELD LINES, BAR MAGNET
MAGNETIC FIELD LINES OF CURRENT
GEOGRAPHIC NORTH VS MAGNETIC NORTH
MAGNETIC FIELD B
Unit: T (Tesla)
1T=1Ns/(Cm)
Magnetic field of earth: 10-4T
(Reverses every 250 000 years)
Poles moves 30 miles per year
Non SI unit: Gauss (G)
1 G = 10-4T
FORCE ON A CHARGE
FB = qv ´ B (Vector cross product)
Þ FB = qvBsinf (Magnitude)
f is the angle between v and B.
Special cases :
v = 0 Þ FB = 0
f = 0 or 180 Þ FB = 0
f = 90 Þ FB = qvB
DIRECTION OF MAGNETIC FORCE
FB = qv ´ B
q>0
q<0
RIGHT HAND RULE
EXAMPLES OF DIRECTION OF MAGNETIC FORCE
EXAMPLES OF DIRECTION OF MAGNETIC FORCE
MAGNETIC FORCE ON A PROTON
Find the magnetic force on a proton (q = 1.6 ´ 10 -19 C) moving at 3 ´ 10 5 m /s
through a uniform magnetic field of 2T.
Magnitude :
F = qvBsin f = (1.6 ´ 10 -19 C)(3 ´ 10 5 m /s)(2T)sin 30 = 4.8 ´ 10 -14 N
Right hand rule tells us the force is point in the negative y - axis.
VECTOR CROSS PRODUCT
ìïu = ux iˆ + uy ˆj + uz kˆ
í
ïîv = v x iˆ + v y ˆj + v z kˆ
u ´ v º (uyv z - uzv y )iˆ + (uzv x - ux v z ) ˆj + (uxv y - uyv x ) kˆ
iˆ
= ux
vx
ˆj
uy
vy
kˆ
uz (determinant)
vz
Useful facts :
(1) u ´ v = - v ´ u
(2) u ´ v = uv sin q
where u = u , v = v
q is the angle between u and v .
CROSS PRODUCT EXAMPLE
ìu = - iˆ + 2 ˆj + 3kˆ
í
ˆ + 5 ˆj - 6kˆ
v
=
4
i
î
u ´ v = (uyv z - uzv y )iˆ + (uzv x - ux v z ) ˆj + (uxv y - uyv x ) kˆ
iˆ
ˆj
kˆ
= -1 2 3 = -27iˆ + 6 ˆj - 13kˆ
4 5 -6
VECTOR CROSS PRODUCT
iˆ ´ ĵ = k̂
iˆ ´ iˆ = 0
k̂
If C = A ´ B, then we have:
C^A
iˆ
ĵ
and
C^B
EXAMPLE
q = -1C,v = (2 ĵ)m / s, B = (3iˆ - 2 ĵ - 9 k̂)T
What is F ?
k̂
F = qv ´ B = (-1)(2 ĵ) ´ (3iˆ - 2 ĵ - 9 k̂)N
= (-6 ĵ ´ iˆ + 4 ĵ ´ ĵ + 18 ĵ ´ k̂)N
= (6 k̂ + 0 + 18iˆ )N
= (18iˆ + 6 k̂)N
iˆ
ĵ
ANOTHER METHOD
F = qv ´ B = (-1)(2 ĵ) ´ (3iˆ - 2 ĵ - 9 k̂)N
iˆ
ĵ
k̂
= (-1) 0
2
0 N
3 -2 -9
= (18iˆ + 6 k̂)N
IN AND OUT OF THE PLANE
Symbols for B field going in/out of paper
Going into the paper
Going out of the paper
EXAMPLE
B = 1.2mT pointing out.
v
Proton : v = 3.2 ´ 10 7 m /s ,
m = 1.67 ´ 10 -27 kg, q = 1.6 ´ 10 -19 C
What is the force and acceleration.
What is the angle f?
f = 90 Þ sin f = 1
F = qvBsin f = 6.1 ´ 10 -15 N
Points to the RIGHT
a = F / m = 3.7 ´ 1012 m / s 2
B
EXAMPLE
v
What if ϕ is not 90o ?
f = 30 Þ sin f = 1 / 2
F = qvBsin f = 3 ´ 10 -15 N
Points into the screen
a = F / m = 1.8 ´ 10 m / s
12
2
30o
B
CIRCULAR MOTION IN A MAGNETIC FIELD
What happens when a positively charged particle enters
with velocity v into a region with B field pointing down
(into the screen).
FB = qvB
mv 2
Fcentripetal =
r
mv 2
Þ qvB =
r
mv
Þr=
qB
BENDING OF AN ELECTRON BEAM
MEASURING CHARGE TO MASS RATIO Q/M
Measuring r gives you q/m.
mv
r=
qB
q
v
Þ =
m Br
What if the charge is negative?
A BETTER WAY TO GET Q/M
The velocity v is often difficult to measure, but we can
figure out v using the potential difference V.
From the first equation :
qBr
v=
m
Put into the second :
1 qBr 2
m(
) = qV
2
m
q B 2r 2
Þ
=V
m 2
q
2V
Þ = 2 2
m Br
ì mv 2
ï r = qvB
í1
ï mv 2 = qV
î2
FIND THE PARTICLE VELOCITY
Mass : 12m p
Charge : q = 3e
Field : 0.25T
(m p = 1.67 ´ 10 -27 kg, e = 1.6 ´ 10 -19 C)
Charge is negative.
r = 95cm /2 = 0.475m
qBr 3eBr
v=
=
= 2.84 ´ 10 6 m /s
m 12m p
CROSSED FIELDS
q vB = q E when there is not deflection
Þv= E/B
FORCE ON CURRENT
F = IL ´ B
(V ector)
Þ F = ILBsinf (Magnitude)
Does it matter whether the charge carriers are positive or
negative?
EXAMPLES OF MAGNETIC FORCE ON A WIRE
F = IL ´ B
DERIVATION: MAGNETIC FORCE ON A WIRE
If there are N number of charges in a wire of length L,
each with charge q moving at v d . The force on the wire is :
F = Nqv d ´ B
But J = nqv d , n : number of charge per unit volume
N
Þ J = ( )qv d
AL
Þ Nqv d = ALJ = IL, L points along the current.
Put together, we have :
F = IL ´ B
MAGNETIC DIPOLE MOMENT
A loop of current carries magnetic dipole moment (or
magnetic moment):
m = IA
SI unit: Am2
The direction of the area vector is
determined by the right hand rule.
DIPOLE IS A SOURCE OF B FIELD
The B field generated by a magnetic dipole is roughly
the same as that of a bar magnet.
μ
EXAMPLE
Find the magnetic dipole of a loop of wire of radius 10cm
carrying a current of 3A.
m = IA = 3p r 2 = 0.094 Am2
What if the wire has N=30 loops?
m = NIA = 30(3p r 2 ) = 2.8Am2
TORQUE ON A DIPOLE
t =m´B
Magnitude : t = m ´ B = mBsinf
μ and B tends to align
m
EXAMPLE: TORQUE ON A DIPOLE
POTENTIAL ENERGY OF A DIPOLE
U = -m × B = -mBcosf
μ and B tends to align in order to attain the lowest energy.
m