Download Homework 7 Solutions Ch. 28: #28 à 28)

1
Physics 132 Homework #7
Homework 7 Solutions
Ch. 28: #28
Ch. 29: #1, 9, 18, 21, 24, 37
à 28)
E = 6.0 V
R1 = 100 Ω
R2 = R3 = 50 Ω
R4 = 75 Ω
ü a)
resistors R2 , R3 and R4 are in parallel, so their equivalent resistance, R234 , is :
1
R2 R3 R4
R234 =
=
1
1
1
R2 R3 + R2 R4 + R3 R4
+
+
R
R
R
2
3
4
R234 and R1 are in series, so their equivalent resitance, Req , is :
R2 R3 R4
Req = R1 + R234 = R1 +
R2 R3 + R2 R4 + R3 R4
plugging in the given values for resistance yields :
H50 ΩL H50 ΩL H75 ΩL
=
Req = 100 Ω +
H50 ΩL H50 ΩL + H50 ΩL H75 ΩL + H50 ΩL H75 ΩL
100 Ω +
1.875 × 105 Ω3
=
2.5 × 103 Ω2 + 3.75 × 103 Ω2 + 3.75 × 103 Ω2
1.875 × 105 Ω3
= 100 Ω + 1.875 × 101 Ω = 118.75 Ω
100 Ω +
104 Ω2
= 120 Ω
H2 significant figuresL
2
Physics 132 Homework #7
ü b)
using Kirchoff' s voltage law,
E − i1 R1 − i2 R2 = 0
the current out of the battery is equal to the current into R1
E
6.0 V
i1 =
=
= 5.0 × 10−2 A
Req
120 Ω
solving the loop equation from above for i2 yields :
E − i1 R1
i2 =
R2
=
6.0 V − H5.0 × 10−2 AL H100 ΩL
= 2.0 × 10 −2 A
50 Ω
=
6.0 V − 5.0 V
50 Ω
=
1.0 V
50 Ω
using Kirchoff' s voltage law on a loop with
just R2 and R4 Hand choosing i4 to point away from R1 L gives :
i2 R2 − i4 R4 =
0
solving for i4 yields :
i4 =
i2 R2
R4
=
H2.0 × 10 −2 AL H50 ΩL
H75 ΩL
= 1.3 × 10−2 A
similarly for R3 and R4 ,
i3 R3 − i4 R4 = 0 and
i3 =
i4 R4
R3
=
H1.3 × 10−2 AL H 75 ΩL
H50 ΩL
= 2.0 × 10−2 A
you could also use Kirchoff' s current law
i1 = i2 + i3 + i4
Physics 132 Homework #7
à 1)
Ӵ = 550 m
∞v
s
”
∞B¥ = .045 T
qα = 3.2 × 10−9 C
mα = 6.6 × 10−27 kg
φ = 52°
ü a)
the magnitude of the force on a moving charged particle due a magnetic field is given by
÷÷÷”
”
”¥ ∞B
∞FB ¥ = … q … ∞v
¥ sin HφL
m
= H3.2 × 10−19 C L J5.50 × 10−2
N H4.5 × 10−2 TL sin H52° L
s
= 6.2 × 10−18 N
ü b)
Newton' s second law
”
”
F = ma
thus, the accelaration of the particle due to the magnetic field is
÷÷÷”
Ӵ
∞FB ¥ = m ∞a
÷÷÷”
−18
N
m
”¥ = ∞FB ¥ = 6.2 × 10
∞a
= 9.5 × 108
mα
s2
6.6 × 10−27 kg
ü c)
”
÷÷÷”
÷”
” ”
”
since FB ¶ ”
v , it does no work on the particle IW = F ⋅ d = 0 if F and d are perpendicularM
by the work − energy theorem, if the work done is zero, the change in kinetic energy is zero;
if the change in kinetic energy is zero,
the change in the magnitude of the velocity is zero sinc e the mass stays the same
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Physics 132 Homework #7
à 9)
q = − e = −1.6 × 10−19 C
me = 9.1 × 10−31 kg
∆Vexternal = 1.0 × 103 V
d = 2.0 ∗ 10−2 m
∆Vplates = 1.0 ∗ 102 V
à a)
by Newton' s first law, for the velocity to remain constant,
the net force on the electron must be zero
Fnet = 0
Fnet = FB + FE = » q » v B sin φ + q E
the magnetic field is perpendicular to the direction the
π
π
electron is traveling and hence its velocity, so φ =
and sin
= 1
2
2
the electric field between two parallel plates is given by
∆Vplates
E =
HEq. 25 − 42 of textL
d
by energy conservation,
Ui + Ti = Uf + Tf
H where U is the potenetial energy of an object and T the kinetic energyL
before moving through the potential the electron has an initial kinetic energy, Ti = 0
Uf − Ui = − Tf
putting in the definition of Tf
− ∆U =
1
2
m vf 2
∆U = q ∆V
− 2 q ∆V
⇒ vf = $%%%%%%%%%%%%%%%%%%%%%
m
putting in our values for q, m, ∆V and calling vf , v
2 e ∆Vexternal
v = $%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
me
putting this into the equation for Fnet and setting Fnet = 0, we get
∆Vplates
2 e ∆Vexternal
0 = e $%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% B − e
me
d
4
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Physics 132 Homework #7
solving for B
∆Vplates
B =
d
me
$%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%
2 e ∆Vexternal
checking the units of this answer to make sure they are units of magnetic field HTL
V $%%%%%%%%
kg %%
kg
N m $%%%%%%%%
kg %%%
C
=
&''''''''
'''' =
m
CV
m
Cm
Nm
C JC
J
=
kg
N
N $%%%%%%%%
s2 %
=
&''''''''
'''''''''' =
m
kg
m
2
m
C
C
C
m
s
s2
N
= T
plugging in our numbers
B =
H1.0 × 102 VL
H2.0 × 10−2
H9.1 × 10−31 kgL
&''''''''''''''''''''''''''''''''
''''''''''''''''''''''''''''''''
'''''''''''''
2 H1.6 × 10−19 kgL H1.0 × 103 VL
mL
= 2.7 × 10−4 T
à 18)
B = 4.50 × 10−2 T
q = e Hsingly charged ion positive ionL
= 1.60 × 10−19 C
number of revolutions = 7.00
t7 revolutions = 1.9 × 10−3 s
frequency is defined as one revolution per unit time so we
divide the number of revolutions by the time it took to make those
f =
7.00
1.9 × 10−3 s
= 5.43 × 103 Hz
usinging Eq. 29 − 18 from the text for frequency of oscillation in a magnetic field
qB
f =
2πm
and solving for m yields :
qB
m =
2πf
=
H1.60 × 10−19 CL H 4.50 × 10−2 TL
2 π H5.43 × 103 HzL
= 2.11 × 10−25 kg
converting to atomic mass units H1 u = 1.661 × 10−27 kgL
m = 1.27 × 10−27 u
Physics 132 Homework #7
à 21)
we want the electrons to travel on a circular path of
r ≤ d
for charged particles traveling perpendicular to a magnetic field,
the radius of curvature for their path is given by
mv
HEq. 29 − 16 of textL
r =
»q» B
the kinetic energy of a particle is given by
1
K =
m v2
2
solving for v,
2K
v = $%%%%%%%%%%
m
putting the equation for v into the equation
for r and the equation for r into the condition on r gives
2 K%%%
$%%%%%%%%
≤ d
»q »B
m
m
solving for B and putting in » q » = » − e » = e
2 me K
2 m2 K
B ≤ $%%%%%%%%%%%%%%%%%% = $%%%%%%%%%%%%%%%%%%
e2 m d2
e2 d 2
this equation holds true only for B pointing out of the page because
then the magnetic force points toward the center of the circular path
ü 24)
Tp = Td = Tα ≡ T Hkinetic energyL
qp = e; qd = e; qα = 2 e HchargeL
mp = u; md = 2 u; mα = 4 u HmassL
π
φ =
2
as in the previous problem
r =
mv
qB
=
m $%%%%%%%%%%
2T
1 è!!!!!!!!!!!
=
2mT
m
qB
qB
T and B are the same for each particle
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Physics 132 Homework #7
7
1 è!!!!!!!!!!!
2uT;
eB
1 è!!!!!!!!!!!!!!!!!!!!
è!!!! 1 è!!!!!!!!!!!
è!!!!
=
2 H2 uL T =
2
2uT =
2 rp
eB
eB
è!!!!
1 è!!!!!!!!!!!!!!!!!!!!
4
1 è!!!!!!!!!!!
2
=
2 H4 uL T =
2uT =
rp = rp
2eB
2 eB
2
rp =
rd
rα
è!!!!
since 2 > 1,
rd > rp = rα
à 37)
m = 1.0 kg
L = 1.0 m
i = 50 A
µs = .60
÷÷÷”
the force of static friction, fs , against the force generated by the magnetic
”
field and the current will be in the same plane as the magnetic field , B,
” ”
”
”
”
and the force generated by the magnetic field FB = i L × B = i L B for B ¶ L
÷÷
÷
”
”
FB makes an angle θ with fs
÷÷÷”
÷÷÷”
”
”
”
”
since fs ¶ N , the normal vector, and B ¶ FB , then the angle between B and N is also θ
in order for the rod to move their must be just above zero
acceleration and therefore the net force must be just above zero
we ' ll call the direction of the static friction force x and the direction of the normal force z
Fnet x = i L B cos θ − µs N
Fnetz = i L B sin θ + N − m g
since there is no acceleration upwards or downwards
Fnetz = 0 ⇒ N = mg − i L B sin θ
plugging this value for N into Fnetx and setting
Fnetx = 0 yields
i L B cos θ − µs Hmg − i L B sin θL = 0
solving for B gives
µs m g
B =
i L Hcos θ + µs sin θL
the minimal value of B will occur when
cos θ + µs sin θ is maximized
to find the maximum of cos θ + µs sin θ, take
θ
Hcos θ + µs sin θL = 0
− sin θ + µs cos θ = 0
solving for θ gives
θ = arctan Hµs L @also denoted : tan−1 Hµs LD
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Physics 132 Homework #7
putting in our value for µs gives
θ = .55 radians or 31°
then,
B =
H.60L H1.0 kgL H 9.8
m
s2
L
H50 AL H1.0 mL Hcos 31 + H.60L sin 31° L
°
= .10 T
28-40
Without the voltmeter current : V = iR
With the voltmeter current we replace R with the equivalent
resistance R ′ then: V ′ = i ′R ′
Since the equivalent resistance is just Rv and R in parallel then
1 1 1
1 1
1
or rearranging
= +
= −
R′ R Rv
R R ′ Rv