Transcript Magnetic field II

Magnetic field II
Plan
• Motion of charge particle in
electric and magnetic fields
• Some applications of magnetic
fields
Recap………
Force due to magnetic
field

 
FB  q v  B


 

FB   Id l  B
Recap………
Work done by FB
W 0
Magnetic forces can only alter the
direction
Recap………
Lorentz force




F  qE  q(v  B)
Combined E and B
fields


 
F  qE  q(v  B)
Acts on any
particle, whether
moving or in rest
Acts on moving
particle only
Motion of charge
particle in the
crossed B and E fields
  
E Bv
If
E
v
B
Particle for which v = E/B, passes un deflected.
Particles with other fields are deflected.
Calculation of e/m for
electron
Motion of charged particle in
constant uniform Electric field
xvt
1 2
y  at
2
m a  eE  m g
Neglecting
gravity
1 eE  x 
Y
 
2 m v
 yx
2
2
A Parabolic
path
1 eE  x 
Y
 
2 m v
 yx
• Substituting
2
2
E
v  and x  L
B
e 2 yE
 2 2
m B L
e
 1.7  1011 C / kg
m
Constant uniform
Magnetic field
• B does not change the speed
• It changes direction of motion
• Thus particle should move in a circle,
if moves in a plane perpendicular to B.
2
v
| q | Bv  m
r
mv
r
|q|B
mv
r
|q|B
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FB
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Angular
frequency
v Bq
 
r m

qB
f 

2 2m
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FB
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Applications of magnetic
fields
Application I- Mass
spectrometer
To separate the ionized atoms
Rm
Application II - Cyclotron
1 2 q2B2R2
K  mv 
2
2m
E. O.
Lawerence
(1930)
World’s largest cyclotron
• This is at TRIUMF (CANARA)
• Accelerates protons to energies up
to 520 MeV
• Diameter of machine = 18 m
• During the course of acceleration
proton travels 45 km
Magnetic Mirror
• Non uniform magnetic field
Strong B
Weak B
Strong B
The Hall effect
• Provides a way to determine sign and
density of charge carriers.
conductor
w
If electrons are majority charge
carriers
E
++++++++++++
conductor
---------------B
EH
If the electrons are majority
charge carriers
++++++++++++
conductor
----------------
VH
+

 
qEH  q (vd  B)  0 E  VH  IB
H
w
nqwt
EH  vd B  0
j
E H  vd B 
B
nq
j
IB
EH 
B
nq
nqwt
VH
I
EH 

B
w
nqwt
Density of charge carriers
I
n
B
qtVH
Convention
• If the voltage of lower terminal is
negative, electrons are the majority
charge carriers.
• If the voltage of lower terminal is
positive, holes are the majority charge
carriers.
Show that , in terms of hall electric field
EH, and the current density j, the number
of charge carriers per unit volume is given
by
jB
n
eEH
Show that the ratio of hall electric
field to the electric field is given by
EH
B

EC ne
Ec  j
j
EH 
B
ne
A metal strip 10 cm long, 1 cm wide and
1mm thick moves with constant speed v
through a magnetic field of 1mT
perpendicular to the strip. A potential
difference of 4 V is measured between
the points x and y across the strip.
Calculate the speed v
VH
vd B 
w
Vd is the velocity of strip now
Vd = 0.4 m/s
A strip of copper 150 m thick is
placed in a magnetic field B = 0.65 T
perpendicular to the plane of the
strip. A current 23 A is setup in the
strip. What hall potential difference
will appear across the width of the
strip
if there were 8.5 x 1028
electrons/m3 ?
I
VH 
B
net
I
n
B
etVH
VH  7.3 V