de rt of Probiein le calcalateNpo radiatod by thes antenna.ar her radiates 3 W ofM (a) Fnd the rate of emis-the transatier fb) ln this case isdlne berween the correctKy in a circular(11.1) is modified by an extra factor of y2kg'utycase the motion would be relativistic(11.50)3cFind the rate of energy loss of the electron, and com-pare with that for a proton. (Your answer for the elec-tron should be enormously larger than for the proton.This explains why most electron accelerators are lin-car, not circular, since the acceleration in a linear ac-celeratorcentripetal acceleration considered here.)and t chassial picture in which thesmied contii lne at Esample 1l using SIphnstNr you ea caloulator and ex-n pe ange (uly 20) vound eponei separatelyonce ec - is far smaller than thetsaLlyar the power Pradated by andg be darved by the methodLad Since Pwold be experted toLmwe mght veasanably guns that11.8 Answer the same questions as in Problem 11.7, butassume that both the proton and electron have kinet-C energy 10 GeV and move in a circle of radius 20 m.(In this case both particles are relativistic and youmust use the relativistic formula (11.50)]11.9 The ring called PEP-II at Stanford in Californiastores clectrons orbiting around a circle of radius170 m Because of the centripetal acceleration,r. the particles lose energy in accordance withEq (1150) (which is the appropnate relativistic formof Eg (11.1)) (a) Find the rate of encrgy loss of a sin-e 9-GeV electron. (b) If a total of 2 x 10 clectronsare radiating at this rate, whal is the total power, inwattk needed to keep them at 9 GeV? (For compari-son, the power used by a typical household applianceis on the order of 100 W)(11 48)mumber of arder !ebdwhene Lm and pd &s the Coulumb forseBy aning theiral qantities in (11 48)at dtermanes the un-Show that yuu obtain theg thas the dmensanless(Sec 1adascal ahom Wouldgta y ough mlimate ofllapt coplenclyInd the cate at which*Note that this is for the case of circular motion. For linearmotion the facttor is yThe actual device contains hoth curved and straight sec-tons bus is teasonahly described as a single circle for theurposes of this problemS1.49)

Question
Asked Dec 28, 2019
1 views

How would I be able to solve Problem 11.9? The chapter that this problem is in is called "Atomic transitions and radiation." The section it resides in is named Radiation by Classical Charges.

de rt of Probiein le calcalate
Npo radiatod by thes antenna.
ar her radiates 3 W of
M (a) Fnd the rate of emis-
the transatier fb) ln this case is
dlne berween the correct
Ky in a circular
(11.1) is modified by an extra factor of y
2kg'uty
case the motion would be relativistic
(11.50)
3c
Find the rate of energy loss of the electron, and com-
pare with that for a proton. (Your answer for the elec-
tron should be enormously larger than for the proton.
This explains why most electron accelerators are lin-
car, not circular, since the acceleration in a linear ac-
celerator
centripetal acceleration considered here.)
and t chassial picture in which the
smied conti
i lne at Esample 1l using SI
phnstNr you ea caloulator and ex-
n pe ange (uly 20) vou
nd eponei separately
once ec - is far smaller than the
tsaLlyar the power Pradated by an
dg be darved by the method
Lad Since Pwold be experted to
Lmwe mght veasanably guns that
11.8 Answer the same questions as in Problem 11.7, but
assume that both the proton and electron have kinet-
C energy 10 GeV and move in a circle of radius 20 m.
(In this case both particles are relativistic and you
must use the relativistic formula (11.50)]
11.9 The ring called PEP-II at Stanford in California
stores clectrons orbiting around a circle of radius
170 m Because of the centripetal acceleration,
r. the particles lose energy in accordance with
Eq (1150) (which is the appropnate relativistic form
of Eg (11.1)) (a) Find the rate of encrgy loss of a sin-
e 9-GeV electron. (b) If a total of 2 x 10 clectrons
are radiating at this rate, whal is the total power, in
wattk needed to keep them at 9 GeV? (For compari-
son, the power used by a typical household appliance
is on the order of 100 W)
(11 48)
mumber of arder !
ebdwhene Lm and p
d &s the Coulumb forse
By aning their
al qantities in (11 48)
at dtermanes the un-
Show that yuu obtain the
g thas the dmensanless
(Sec 1
adascal ahom Would
gta y ough mlimate of
llapt coplencly
Ind the cate at which
*Note that this is for the case of circular motion. For linear
motion the facttor is y
The actual device contains hoth curved and straight sec-
tons bus is teasonahly described as a single circle for the
urposes of this problem
S1.49)
help_outline

Image Transcriptionclose

de rt of Probiein le calcalate Npo radiatod by thes antenna. ar her radiates 3 W of M (a) Fnd the rate of emis- the transatier fb) ln this case is dlne berween the correct Ky in a circular (11.1) is modified by an extra factor of y 2kg'uty case the motion would be relativistic (11.50) 3c Find the rate of energy loss of the electron, and com- pare with that for a proton. (Your answer for the elec- tron should be enormously larger than for the proton. This explains why most electron accelerators are lin- car, not circular, since the acceleration in a linear ac- celerator centripetal acceleration considered here.) and t chassial picture in which the smied conti i lne at Esample 1l using SI phnstNr you ea caloulator and ex- n pe ange (uly 20) vou nd eponei separately once ec - is far smaller than the tsaLlyar the power Pradated by an dg be darved by the method Lad Since Pwold be experted to Lmwe mght veasanably guns that 11.8 Answer the same questions as in Problem 11.7, but assume that both the proton and electron have kinet- C energy 10 GeV and move in a circle of radius 20 m. (In this case both particles are relativistic and you must use the relativistic formula (11.50)] 11.9 The ring called PEP-II at Stanford in California stores clectrons orbiting around a circle of radius 170 m Because of the centripetal acceleration, r. the particles lose energy in accordance with Eq (1150) (which is the appropnate relativistic form of Eg (11.1)) (a) Find the rate of encrgy loss of a sin- e 9-GeV electron. (b) If a total of 2 x 10 clectrons are radiating at this rate, whal is the total power, in wattk needed to keep them at 9 GeV? (For compari- son, the power used by a typical household appliance is on the order of 100 W) (11 48) mumber of arder ! ebdwhene Lm and p d &s the Coulumb forse By aning their al qantities in (11 48) at dtermanes the un- Show that yuu obtain the g thas the dmensanless (Sec 1 adascal ahom Would gta y ough mlimate of llapt coplencly Ind the cate at which *Note that this is for the case of circular motion. For linear motion the facttor is y The actual device contains hoth curved and straight sec- tons bus is teasonahly described as a single circle for the urposes of this problem S1.49)

fullscreen
check_circle

Expert Answer

Step 1

(a) Write the expression for Power loss

Advanced Physics homework question answer, step 1, image 1
fullscreen
Step 2

Write the expression for energy of the electron

Advanced Physics homework question answer, step 2, image 1
fullscreen
Step 3

Rewrite the above equation for...

Advanced Physics homework question answer, step 3, image 1
fullscreen

Want to see the full answer?

See Solution

Check out a sample Q&A here.

Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*

See Solution
*Response times may vary by subject and question.

Related Advanced Physics Q&A

Find answers to questions asked by student like you
Show more Q&A
add
question_answer

Q: An air capacitor is made from two flat parallel plates 1.50 mm apart. The magnitude of charge on eac...

A: A) Let the charge on the capacitor be denoted as Q, and the potential difference as VThe capacitance...

question_answer

Q: This is the 3rd time I have uploaded this question under ADVANCED PHYSICS, not physics stop rejectin...

A: There are different types of models given for waves. As you can see in the figure. (i) Sinusoidal wa...

question_answer

Q: A 70-kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1...

A: Let the system be isolated. Then according to the conservation of energy principle the net energy of...

question_answer

Q: Starting from rest, a 12-cm-diameter compact disk takes 2.8 s to reach its operating angular velocit...

A: Given,

question_answer

Q: The Solar Nebula model introduces the concept of ‘frost line’. Where are asteroids and comets found ...

A: The frost line can be defined as the distance from the protostar (Sun in the case of our galaxy) wit...

question_answer

Q: The position of a rabbit along a straight tunnel as a function of time is plotted in the figure. Wha...

A: Draw a tangent line at t=10 s and t=30 s on the given plot that represents the position of the rabbi...

question_answer

Q: Explain how the frequency of the source emf affects the impedance of a) pure resistance, b) a pure c...

A: Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and s...

question_answer

Q: Need help on this problem. See attatched picture.

A: a) Free body diagram is given by,

question_answer

Q: I am stuck on this problem.

A: Write the velocity as the rate of change of displacement and integrate both the sides.