sadani (~~) - Computer hw#14 - matney -- (523602)
This print-out should have 19 questions .
Multiple-choice questioIls may continue on
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before answering.
001
10.0 points
An alpha particle has a mass of 6.6 x 10- 27 kg
and is accelerated by a voltage of 1.35 k V.
The charge on a proton is 1.60218 x 10- 19 C.
If a uniform magnetic field of 0.073 T is
maintained on the alpha particle and perpen­
dicular to its velocity, what will be particle's
radius of curvature?
1
Correct answer: 62.963 ftT.
Explanation:
E = 17000 N/ C and
v = 2.7 x 10 8 m I s.
Let:
If the electrons move undeftected through the
crossed fields
E
V= ­
B
then
Correct answer: 0.102155 m.
H= E
Explanation:
v
17000 N/C
Let:
B = 0.073 T ,
V = 1.35 k\l = 1350 \I ,
m = 6.6 x 10- 27 kg) and
q = 2 e = 3.20435 x 10- 19 C.
From Newton's second law,
mv 2
F=qvB=-­
T
V =
gBT
--.
m
The kinetic energy is
I
2
K=-mv =
2
q2 B2 r2
2m
=gV,
so the particle's radius of curvature is
,~ ~J2~rn
.--------------------­
1
2 (1350 V) (6.6 x 10- 27 kg)
= 0.073 T \
3.20435 x 10- 19 C
= 10. 102155 mi.
002
10.0 points
The electrons in a beam are moving at 2.7 x
K
10 mls in an electric field of 17000 N IC.
\Vhat valnc must the magnetic fiein have if
the electrons pass through the crossed fi elds
undeftected?
2.7 x 10 8
= 162.963
mls
TI·
003 (part 1 of 4) 10.0 points A bar of negligible resistance and mass of 34 kg in the figure is pulled horizontally across frictionless parallel rails, also of negligi ble re­
sistance, by a massless string that passes over an ideal pulley and is attached to a suspended mass of 420 g. Thc uniform magnetic field has a magnitude of 350 mT, and the distance between the rails is 91 cm. The rails are con­
nected at one end by a load resistor of 1:3 mO. 34 kg
/­
91 cm
~~-------7'
/­
t
t
350 mT
l
350 mT
I
a
t
l
t
l
350 mT
What is the magnitude of the terminal ve­
locity; i.e., the eventual steady-state speed
1)00) reached by th e bar? The acceleration
due to gravity is 9.8 rnls 2 .
Correct answer: 0.527473 m/s.
sadani (sen
2
5) - Computer hw#14 - matney - (523602)
Explanation:
004 (part 2 of 4) 10.0 points
What is the acceleration when the velocity is
0.94 m/s?
Let : m = 34 kg ,
/vI = 420 g,
e = 91 cm,
B = 350 rnT)
R = 13
Correct answer: - 0.0935228 m/s2.
and
mn.
From Lem's law the magneti c force opposes
the motion of the bar. When the wire acquires
steady-state speed, the gravitational force Fg
is counter-balanced by the magnetic force Fm.
m
/­
L£
t I T I Bt Bt Bt Bt
a@F.
t
1 ~1 t 1 1 1 1 B
Fg =Fm Mg=£IB 1= Mg
PB .
d<t>
The induced emf is £ = - dt . Applying
Ohm's law,
I = ~ = ~ d <I> _ ~ . d A
R
Since
R dt - R B dt .
BA)
<I> =
d <I>
dA
Bdt
= Bev
so for a terminal velocity of Voo
eVoo
R
=
'
,
/vI 9 ma = T-Fm
M a = Fg - T
.
B2 £2 V
(m+M)a= Fg -Fm = Mg - - ­
R
JW,9
B2 £2 v
a=----.:.::..­
m M R (m M)
(420 g)(9.8 m/ s2)
34 kg 420 g
(:550 mT)2(91 cm) 2
13 mn
0.226175 m/s
x----~
34 kg + 420 g
= '1--0-.0-9-3-52- 2-8-m
- /- s-2 1·
005 (part 3 of 4) 10.0 points
\;\That is the tim e constant T?
dv
a(1J) == - .
dt
From the previous part,
eB = ._._-".­
00
Frn =
second law t.o the bar and the suspeIlded mass
sepamtely,
Explanation:
MgR
v
R
R
B2 e2 v
R ,so applyi ng applying Newton's
Correct answer: 4.41098 s.
1£1 = -dt =
1= B
Explanation:
Let T be t.he tension in the string.
£
Bev
Fg = M 9 , Fm = I e B , I = - = - - and
a=
£2 B2
_ (0.42 kg)(9.8 m/s2)( O.013 D)
(0. 91 m)2(350 mT)2
= 10.527473 m / s I·
dv
dt
Mg
m+M
MgR
B 2 e2
(m + ~1) R
e
B2 2
v
(m
+ ~1) R
B2 £2
3
sadani (si!U!fiJ - Computer hw#14 - matney - (523602)
-
-
dt
dv
--
MgR
B 2f2
-
-~:-
2.6Y­
dt
dv
vx -
R(m+M)
8 2 £2
V
V
so
T
R(m+M)
B2f2
(0.013 52)( 34 kg + 0.42 kg )
(350 mT)2(0.!H m)2
T=
1. both will move toward the right.
\4.41098 s \.
=
I+-+i
0.27 m When the rods are let go, they 2. will heat up, and remain motionless.
006 (part 4 of 4) 10.0 points
What is the bori'L;onLal speed of the bar at
2.47015 s, assuming tbat the bar was at rest
at t = 0 s?
3. will change in a way not determined by
the informa.tion given.
4. will move away from each other.
Correct answer: 0.226175 m / s.
5. will move toward each other. correct
Explanation:
6. both will move toward the left.
dv
Vx -
dt
V
- - -=e
voo
v =
Voo
(1 - e- t /
-tiT
T
)
= (0.527473 m / s)
x
[1 _
e-(247015 8)/ (441098
S)]
I
= 0.226175 m l s \.
007
10.0 points
Two identical parallel sections of metal rods
are connected parallel to a battery as showIl.
The two sections of metal rods are initially
held in place.
Explanation:
Both currents in the rods move downwa.rd ,
so they are parallel currents that attract,
causing them to accelerate toward each other.
008
10.0 points A magnetic dipole is falling in a conducting metallic tube. Consider the induced current in an imaginary current loop when the magnet is moving away from the upper loop and when the magnet is moving toward the lower loop. sadani (8.
[\) - Computer hw#14 - matney - (523602)
..... ,---~ "
"
?
I
,,
/~"'---""',
). JabO'l)c
~
\-1-[
i
X
4
X
1ab01le
,
\-1-[
i
Spole
.. I
~
S'[!ol~
J- z­
J-z-­
r
s
v
~
y
r
dipole
magnet
I
N po1e
S
I
v
dipole
magnet
~
y
I
N p(J/r-:
~~"'---"'"
",
?
)
helolll
~
~
.
,J
-"
Determine the directions of the induced
currenLs 1abo'Ut: and h elolll in an imaginary loop
shown in the figure , as viewed from above,
wh en the loop is above the falling magnet and
when the loop is iJelow the falling magnet.
1. l abo1'('
helow
= counter-clockwise
=
009
10.0 points A negatively charged particle moving parallel to the x-axis enters a magnetic field (pointing into of the page), as shown in the figur e below. ®
and counter-clockwise ®~
®
®
®
B ®
®
®
2. 1abmw = clockwise and Jbelolll = clockwise ®
o-- v ....... ®
- q
3. 1 above
helow
= counter-clockwise
=
®
®
clockwise ®
Explanation:
Wh en the falling magnet is below the upper
loop , Tl ind mllst be down to attract the falling
magnet and slow it down; i.e., clockwise as
viewed from above.
Before reaching th e lower loop, Tl ina must
be up to oppose the falling magnet; z.e.,
counter-clockwise as viewed from above.
x
®
B
®
®
®
®
Figure: i is in the x-dir~ctioI1, J is
in Lhe y-direction, and k is in the
z-direction.
4. 1 ab01If: = clockwise and hdu'U) = counter-clockwise correct 5. no current flow
z
®
®
®
and yL
What is the initial direction of deflection?
1. F
= +J
2. F
=
~
-i
A
3. F = +k
4. F =-k
5.
F=
0; no deflec tion
sadani
~
6. F
= -) correct
7. F
= +i
(~£i_Rfe.t)
The magnetic force is the centripetal force
which keeps t he proton in circular motion.
From the centripetal force equation, we have
Tn v 2
Explanation:
Basic Concepts:
Cha rged P art icle:
F=-P­
r
(1.67 x 10- 27 kg)(2.63
(0.813 m)
MagneLic Force on a
= 11.42081 x 10-
Right-h_and rule for cross-products.
F
F == -_-; i.e., CL unit vector in the F direc~
I FII
tion.
Solution: The forc e is F = q iJ x B.
-
-
B=B
(-k) ,
-Iql
= -Iql
=
B
B =
vB [(+i) x
(-k)]
This IS the third of eight versions of the
problem.
010 (part 1 of 2) 10.0 points
A proton in a cyclotron is moving with a speed
of 2.63 x 10 7 ml s in a circle of radius 0.813 m .
1.67 x 10- 27 kg is the mass of the pro­
ton, and 1.60218 x 10- 19 C it; itt; fundamental
charge.
What is the magnitude of the force exerted
on the proton by the magnetic field of the
cyclotron?
Correct a nswer: 1.42081 x 10- 12 N.
Explanation:
v = 2.63 x 10 7 mi s,
T
qe
= 1.67 x 10 - '27 kg ,
= 0.813 m)
Bpprp qf' V,
~
= 10.337187 T I.
qe v
012
vB (+))
F=GJ.
mp
N I.
Correct answer: 0.337187 T.
F8 =
x
10 7 m/t;)2
011 (part 2 of 2) 10.0 points
What is the magnitude of the magnetic fiel d
required to keep it moving in this circle?
and therefore , < 0,
F = - Iql iJ
q
12
X
Explanation:
The force it; due to the magnet ic field.
·v = 'V (+ i) ,
Let:
5
- Computer hw#14 - mat ney - (5 23602 )
10.0 points
In the arrangement shown in the fi gure, the
resistor is 1 n and a 8 T magnetic field is
directed out of the paper. The separation
between the rails is 7 rn. An CLpplied force
moves the bar to the left at a constant speed
of 9 m / s.
8T
"
·I1
6,-<
8T
P
9 ml s
C
¥
.--!
~
'~
Calculate the applied force required to
move the bar to the left at. a constant sp(;(;d of
9 m / s . Assume the bar and rails have negligi­
ble resistance and fri ction . Neglect the mass
of t he bar.
and
= 1.60218 x 10- 19 C.
b.O
.--!
Correct answer:
2~224
N.
sadani (§tiilJiiU) - Computer hw#14 - matney ~ (523602)
The area of the circular coil is
Explanation:
A
R
Let:
= 1
n,
B.e v = (8 T) (7 m) (9 m js)
= 504. V.
From Ohm's law , the current flowing through
the resistor is
Ilill = 11;1 x BII = 11. B
T
504 V
= - - = 504 A .
R
1n
I.
013
10.0 points
A circular coil consisting of a single loop of
wire has a radius of 21.1 cm and carries a
current of 12 A. It is placed in an external
magnetic field of 0.23 T.
Find the magnitude of the torque on the
wire when the plane of the coil makes an
angle of 21.'t with the direct.ion of t.he field.
Correct answer: 0.358675 N . m.
Explanation:
Let :
21.1 ern = 0.211 m,
B = 0.23 T,
1= 12 A, and
'f"
=
e=21.'t.
= I A B sine
= (12 A) (0.139867 m 2) (0.23 T)
x sin(90° - 21.'t)
Thus, the magnitude of the force exerted on
the bar due to the magnet.ic field is
F = Fa = 128224 1
sin (),
where () is the angle that the normal to the
loop makes with t.he magm~t.ic field, and I.L =
I A is the magnetic moment of the loop.
Thus the magnitude of the torque is
[
To maintain the motion of the bar, a force
must be applied on the bar to balance the
magnetic force
Jr
The torque is
=-
Fa = I.e B = (504 A)(7 m)(8 T)
= 28224 N.
Jr '(2
(0.211 m)2
2
= 0.139867 m .
The motional emf inuuceu in the circuit is
I
=
=
B=8T,
.e = 7 m, and
v = 9 mjs.
[ =
6
I
= 0.358675 N . ill
I.
014
10.0 points
Suppose a new particle is discovered, and it is
found that a beam of these particles passes
undeflected through "crossed" electric and
magnetic fields , where E = 452 V j m and
B = 0.00125 T. If the electric field is t.urned
ofT, the particles move in the magnetic field in
circular paths of radius '( = 1.6 cm .
Determine q for the particles from these
data.
Tn
Correct answer: 1.808 x 1010 C j kg.
Explanation:
=
=
me =
TrLp =
B =
T =
Let: E
qe = qp
452 V jm,
1.60218 X 10-
19
C,
31
9.10939 x 10- kg,
1.67262 x 10 - 27 kg,
0.00125 T,
0.016 m.
and
Since the particle passes undeflected, the
electric force on it is equal to the magnetic
force. When t.he electric field is turned off,
sadani (ss:t!sIlO_.') -- Computer hw#14 - matney - (523602)
only the magnetic force is exerted on it as the
centripetal force. So we get two equations
'U
=
)2:
2(1.5 MeV) 1 x 10 6 MeV j eV
1.67 x 10- 27 kg
1 MeV
qE = Bqv
rnv 2
Bqv = - - .
T
So, the charge to mass ratio for this new
particle is
q
E
Tn
TB2
X
1010 C j kg
I.
For comparison, the electron charge-to-mass
ratio is
1.60218 X 10­ 19 C
9.10939 x 10- 31 kg
qc
me
1.609 x 10- 19 J / eV
1 eV = 11.70012 X 10 7 mls ]. x
If the proton is shot into the magnetic field
452 V 1m
(0.016 m) (0.00125 T)2
= 11.808
= 1.75882
X
with a velocity at right angJes to the directioIl
of the field, we will get the largest radius
mv
r = -
Bq
(1.67 x 10 - 27 kg) (1.70012 X 10 7 m/s)
(3 T) (1.609 x 10- 19 C)
= 0-.0-5-88-1-9-3-m---'l.
r
I
­
10 11 C j kg,
016
and the proton charge-to-mass ratio is
1.60218 x 10 ' 10 C
mp
1.67262 x 10- 27 kg
= 9.57883 X 10 7 C/kg.
qp
015
10.0 points
What is the radius of the largest possible cir­
cuJ(tr orbit that a proton with energy 1.5 MeV
can have in a 3 T magnetic field? The mass
of a proton is 1.67 x 10- 27 kg and its charge
is 1.609 x 10- 19 C.
Correct answer: 0.0588193 m.
Explanation:
Let: m = 1.67 x 10- 27 kg,
Q = 1.609 X 10- 19 C,
R = 3 T, and
E = 1.5 MeV.
The energy of a protoll is
1
2
E = -
'((L 'u
7
10.0 points
Two long, parallel wires a re separated by a
distance 2 d, as shown below . Wire #1 carries
a steady current I into the plane of the page
while wire #2 carries a steady current 1 into
the page.
5
I
I
0--- d -~--- d - 0
wire #1
p
wire #2
, 5'
At what points in the plane of the page
(besides points at infinity), is t.he magnetic
field due to the currents zero?
1. At all points on the line connecting the
two wires.
2. At only point P. correct
3. At. all points on a circle of radius d cen­
tered at either wire.
2
4. At no points.
sadani (1111: E) - Computer hw#14 - matney - (523602)
5. At all points on the line 55', a perpen­
dicular bisector of a line connecting the two
wires.
8
Find the current in the solenoid if it has
11.7 turns/em.
Correct answer : 1.30768 rnA .
Explanation:
The only way that the tot.al magnetic field
would be zero is if the magnetic fields due t.o
the two wires have the same magnitude but
opposite directions at the same point.
Only at points on the line 55' do the mag­
neti c fields have the same magnitude. Only
at point Pare t.he magnet.ic fields parallel
(aligned with the vertical axis). Using the
right hand rule, they are in opposite direc­
Lion.
Thus, at only point P (besides points at in­
finity) is the m ag netic field due to the currents
zero.
Explanation:
The current in the solenoid is from B =
we get
017 (part 1 of 2) 10.0 points
An electron circles at a speed of 7710 m / s in a
radius of 2.28 em in a solenoid. The magnet ic
field of the solenoid is perpendicular to the
plane of the electron's path .
The charge on an electron is ] .60218 x
10- 19 C and its mass is 9.10939 x 10 - 31 kg.
Find t.he st.rength of the magnetic field in­
side the solenoid.
019
10.0 points
A wire carries a current of 11.6 A in a direction
that makes an angle of 13.SO with the direction
of a magnetic fi eld of st.rength 0.218 T.
Find the magnetic force on a 6.5 m length
of the wire.
Correct answer: 1.92264 x 10- 6 T.
Explanation:
Let:
2
= 1.25664 x 10- 6 N / A,
and
n = 11.7 turns / em = 1170 turns / m.
fl,o
fLO
n J,
1.92264 X 10- 6 T
_. (1.25664 X 10- 6 N / A~) (1170 turns/m)
=
0.00130768 A = 11.30768 rnA
I.
Correct. answer: 3.92082 N.
Explanation:
Let :
Let :
T
= 2.28 cm =
=
and
The sLrengLh of the magnetic field is
Tn
1)
B = ­
qr
-
= 11.6 A ,
B -=-- 0.218 T,
(j = 13.8° ,
and
0.0228 m,
7710 m/s,
q = 1.60218 x 10- 19 C ,
31
Tn = 9.10939 x 10kg.
'U 1
(9.10939 x 10- 31 kg) (7710 m/ s)
(1.60218 x 10- 19 C) (0.0228 m)
= 11.92264 x 10- 6 T I.
018 (part 2 of 2) 10.0 points
L
= 6.5 rn.
F = B J L Sill (j
= (0.218 T)(11.6 A)( 6.5 m) sin 13.8°
= 13.92082 N I.