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)
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)
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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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81ed2a7d-ef58-4ac0-a29f-af1e2992bb8e%2F4fba6c54-0c10-4508-b1ba-77aca3abea6a%2Fn4jkzp.jpeg&w=3840&q=75)
Transcribed Image Text: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)
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