PHYS 2421 - Fields and Waves - Academics Portal Index 2421/class material... · 6 31 19 2 2 2450 10...
Transcript of PHYS 2421 - Fields and Waves - Academics Portal Index 2421/class material... · 6 31 19 2 2 2450 10...
![Page 1: PHYS 2421 - Fields and Waves - Academics Portal Index 2421/class material... · 6 31 19 2 2 2450 10 9.11 10 0.0877 1.6 10 fm B q Hz kg T C. Summary of Section 27.4 Hmwk Sect. 27.4:](https://reader031.fdocuments.in/reader031/viewer/2022022509/5ad454297f8b9a177c8b8768/html5/thumbnails/1.jpg)
PHYS 2421 - Fields and Waves
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Consider a charge moving in a magnetic field
Since F is
perpendicular
to v, the force
leads to a
circular motion
B field
into plane
F=ma acceleration change of direction of velocity
0
2
cos90mv
RqB
F qvB qvB
vF m
RCentripetal
Radius of
circular motion
Take F as
centripetal force:
Angular velocity:q Bv
R m
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Homework :
Problem 27.15 and 27.21 (both 11th and 12th Eds.)
Radiation is produced by oscillating electrons
Angular frequency of electrons = radiation frequency
22
q B fmf B
m q
6 31
19
2 2450 10 9.11 1020.0877
1.6 10
fmB
q
Hz kg T
C
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Summary of Section 27.4
Hmwk Sect. 27.4: Probls 27.15 and 27.21 (11th and 12th Eds.)
mvR
qB
q Bv
R m
Charge in a magnetic field move in circular path
Radius:
Angular frequency:
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Charges pass by an area
with a B and an E crossed
The E field
deflects +
charges down
The B field
deflects +
charges up
Particles will go through undeflected if Fe and Fm balance each other
m eF F qvB qE
Ev
B
Only those particles with v = E/B will go through undeflected
Crossed B and E fields act as a velocity selector
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2
2
2
2
1 1
2 2
2
EKE eV mv m
B
e E
m VB
Hmwk : Probl. 27.27 (11th Ed.) or 27.25 (12th Ed.)
Velocity selectors are used in
analog TV screens
He received the Nobel prize for discovering that the electron was a particle
J.J. Thomson
used a velocity
selector to study
the electron:
• Accelerated electrons through a potential V
• Sent them through a v-selector with E and B
adjusted to let them to go undeflected
• Calculated the ratio of the charge to the mass
of the electron
His son received it for discovering that it was a wave . . .
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Homework : Problem 27.31 (both 11th or 12th Eds.)
2
21 1;
2 2
E EKE eV mv m v
B B
231 62
19
9.1 10 6 100.6825 0.826
2 2 1.6 10 150
mEB
eV
kg N/C T
C V
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Summary of Section 27.5
Hmwk Sect. 27.5: Problems 27.27 and 27.31 (11th Ed.) or 27.25 and 27.31 (12th Ed.)
Ev
BVelocity selector
2
22
e E
m VBRatio of charge to mass of electron
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Consider charges moving inside a conductor
placed in a magnetic field B
Each charge feels a forcedF qv B
Magnitude:
Direction: to the left
Number of charges
in length of wire nlA= n = density of charges
A = cross sectional area
Now obtain the total force on the whole length of wire
Total force in
length of wire d dF nAl qv B nqv A lB IlB
dF qv B
Vd = drift velocity
Force is to the left
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For cases in which B is not to perpendicular to wire
F Il B sinF IlB
Some examples
For a segment of a wire
dF Idl B
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Hmwk: Probl. 27.34 (11th Ed.) or 27.36 (12th Ed.)
Solution: a) b) and c): 7.06x 10-3 N
Force on wire?0sin 50 1 1.2 sin45
42.4
F IlB A m T
N
Again using vectors:0 0
0
0 0
ˆ ˆ ˆcos45 sin 45
ˆˆ ˆ
ˆ ˆ0 0 sin 45 42.4
cos45 sin 45 0
F Il B I li B i B j
i j k
I l IlB k k
B B
N
Force pointing out of the plane
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Find the force on the
three segments of wire
Segment 1 Segment 2
Segment 3: length
L into the plane
Segment 1:
Segment 2 is more interesting . . .
0 ˆ ˆsin90F ILB j ILBj
Segment 3: 0sin180 0F IlB
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Homework : Problem 27.35 and 27.39 or
27.37 and 27.39 (12th Ed.)
Segment 3:
• x components of force cancel
• net force will be in the y direction
0 0
0 0
0
0
180 180
0 0
180
0
sin
sin 2
y yF dF IB dl
IBR d IRB0 0
0 0
0
0
180 180
0 0
180
0
cos
cos 0
x xF dF IB dl
IBR d
Total force on all three segments: ˆ ˆ ˆ2 2F ILBj IRBj IB L R j
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Summary of Section 27.6
Hmwk Sect. 27.6: Probls 27.34, 27.35 and 27.39 (11th Ed.) or
27.36, 27.37 and 27.39 (12th Ed.)
Forces on current carrying conductors
F Il B dF Idl B
Add forces
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Consider a current
loop in a B field
perpendicular to
current
• x components of force cancel
• y components of force cancel
• net force on loop will be zero
Now consider a loop tilted respect to the B field
Again
• x and y components of force cancel
• net force on loop will be zero
• but there will be a torque on loop
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Forces in x and y cancel
Forces in x form a couple
F Il B F IaB
Magnetic dipole moment is:
Loop tilted respect to the B field
Segments in y direction feel forces:
in ± x
sinbSeparation between these forces is
and torque is then:
sin sin sin sinFb IaB b IAB B
IA
Or in vector form: B
r F
Valid for all geometries
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Some examples
What are the directions of and ?IA B
A current loop in a B field has potential energy:
cosU B B
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Torque, method 1:0
1.18 1.2 sin 90 1.41sin 2
Am T NmB
2
30 5 0.05 1.182
Total A m AmNIA
Torque, method 2:2 0
30 5 1.2 0.05 sin 90 1.41sin A T m NmNIBA
Torque tends to rotate
Magnetic moment:
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Homework:
Probls 27.42 and 27.44 (both 11th and 12th Eds.)
Soln. 27.42: a), b) and c): 7.06x 10-3 N
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Homework : Problem 27.47 (both 11th and 12th eds.)
Initial potential energy: 0cos90 0InitialU BFinal potential energy:
0cos0 1.18 1.2 1.412
Final Am T JU B
Change in potential energy:
1.41Final Initial JU U
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If for any loop in a uniform B field
• The forces in x and y cancel
• Forces form a couple that produces only a torque
Then, how do magnets attract or repel?
• Magnets affect atomic loops in materials
• B field of magnets is not uniform
• Forces produced are not zero
How do magnets work?
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Summary of Section 27.7
Hmwk Sect. 27.7: Probls 27.42, 27.44 and 27.47
(both 11th Ed and 12th Eds.)
In loops :
•Forces in x and y cancel
•Forces in x form a couple
•Loops feel a torque
Magnetic dipole moment is: IA
B
cosU B BPotential energy:
Magnets work by exerting forces on atomic loops
in materials thanks to their not uniform B field
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PHYS 2421 - Fields and Waves
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PHYS 2421 - Fields and Waves