Exam 2 Solutions - Department of Physics
[Pages:11]PHY2049 Fall 2015 ? Acosta, Woodard
Exam 2 Solutions
Note that there are several variations of some problems, indicated by choices in parentheses.
Problem 1: Two light bulbs have resistances of 100 and 300 . They are connected in parallel across a 120V line. What is the total power dissipated by the two bulbs (or the 100 or 300 bulbs)?
This problem was similar to Exercise 26-C done in class.
P = IE = E2 R-1 =
E2 ( R1-1 + R2-1)
=
(120)2
(
1 100
+
1 300
)W
=
192W
;
or 144 W and 48 W individually.
(1.1)
Problem 2: In the circuit shown in the figure both batteries have insignificant internal resistance and the idealized ammeter reads I1 = 4.0A in the direction shown. Find the E.M.F. of the battery (a negative answer
indicates that the E.M.F. polarity is opposite to what is shown).
This problem was similar to Exercise 26-D done in class.
75.0V = E1 = R1I1 + R2 (I1 - I2 ); - E = R3I2 + R2 (I2 - I1); I1 = 4.0A; R1 = 12.0; R2 = 48.0; R3 = 15.0; (1.2)
We note that the
first
equation in (1.2) implies
I2
=
1 R2
((R1 + R2 )I1 - E1) , which
can
be immediately
put
into the
2nd equation to yield E, as,
E = R2I1 - (R2 + R3 )I2
=
R2 I1
-
R2 +R3 R2
((R1
+
R2
)I1
-
E1 )
=
-24.5625V
;
(1.3)
Problem 3: A capacitor with an initial potential difference of V (0) = 150V is discharged through a resistor when a switch between them is closed at t = 0 . At t = 10.0 , the potential difference across the capacitor is V1 = 1.5V . What is the potential difference across the capacitor at t = 20s ?
This is based on homework problem 27.64
PHY2049 Fall 2015 ? Acosta, Woodard Recalling that q(t) = CV (t) = q0 + q1e-t/ in which the discharging-conditions q(0) = CE = CV (0) and q() = 0
give q0 = 0 and q1 = CV (0), we have V (t) = V (0)e-t/ . The third condition V (10.0s) = 1.5V = (150V )e-(10.0s)/
determines
to
be
= 10.0s
/
(- ln
) 1.5
150
=
2.1715s ,
which
determines
the
voltage
at
all
times,
and
so
we
calculate directly V (20.0s) = (150V )e-(20.0s)/(2.1715s) = 0.015V .
Problem 4: In the figure E = 14V , R1 = R3 = 1.00, and R2 = 2. What is the potential difference VA -VB ?
(1.4)
This is based on a Ch.27 homework problem (27.35), and Exercise 26-F done in class
Let the current through the EMF be i, that through R1 be i1, and the current through R2 be i2.
Junction rules: i = i1 + i2 i1 = i2 + i3 i3 = i1 - i2
Left loop rule, substituting R1 = 1 and R2 = 2 : 14 - i1 - 2i2 = 0 i1 = 14 - 2i2
Zig-zag loop rule: 14 - i1 - i3 - i1 = 0
14 - 2i1 - (i1 - i2 ) = 0
14 - 3i1 + i2 = 0
14 - 3(14 - 2i2 ) + i2 = 0
-28 + 7i2 = 0 i2 = 4A i1 = 14 - 2i2 = 6A i = i1 + i2 = 10A
Now VA - VB = i1R1 = (6A)(1) = 6V
PHY2049 Fall 2015 ? Acosta, Woodard Problem 5: A proton travels through uniform
magnetic
and
electric
fields.
The
magnetic
field
is
r B
=
-2.50i^
mT.
At one instant the velocity of the proton is vr = 2000 ^j m/s . At that instant what is the net force acting on the
proton if the electric field is 4.00k^ V/m?
This is problem 28.10, which was assigned in homework
! F
=
q(
! E
+
v!
?
! B)
=
q(
Ek^
+
v^j
?
B(-i^))
=
q(E + vB)k^
(1.5)
= (1.602 ? 10-19 )(4.00 + (2000)(2.50 ? 10-3))k^N = 1.442 ? 10-18 Nk^
r Problem 6: In the figure a charged particle moves into a region of uniform magnetic field B , goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It
r spends 130 ns in the region. What is the magnitude of B ?
This is problem 28.26, which was assigned in homework
(1.6)
We immediately notice that the particle velocity vr = v(- ^j) is deflected in the -i^ direction by the magnetic field
r B
=
Bk^
pointing
out
of
the
page.
Looking
at
the
vector-value
of
the
forces,
! FB
=
FB (-i^)
=
qv!
?
! B
=
q(v(-
^j))
?
( Bk^)
=
qvB(-i^)
qvB
>
0
q
>
0
qm==+mep==11.6.60722??1100-1-927Ck;g;
(1.7)
While in the circular region, the charged particle has constant speed v = vr , and maintains this velocity. This is
because
the magnetic force
r FB
(due to magnetic field
r B
=
Bk^
and velocity
vr
= v dr
= R dr) is
perpendicular to the displacement, and thus does no work. Letting the half-circle have radius R , and letting the
particle have charge q , we have,
dK = dW
=
! FB
?
ds!
=
(qv!
?
! B)
?
Rd!
=
qvBR(r^
?
d!)
=
qvBR(0)
=
0
dK
=
dW
=0
Ki = K f
= Ka ; (1.8)
Hence,
no
kinetic
energy
is
added
or
subtracted
to
the
particle
of
mass
m.
the
velocity
r v
is
of
constant
magnitude. In the circular trajectory, Newton's 2nd Law then is,
F
=
mpa
=
mp
-v2 R
=
-
! FB
mp
-v2 R
=
-qv!
?
! B
mp
-v2 R
=
-qv!
?
! B
mp
v2 R
" =sin 90?=1 = qvB sinv!B! B =
mpv ; (1.9) Rq
In
(1.9),
the
velocity
is
given
by
v
=
2 R T
=
R (1/ 2 )T
,
so,
PHY2049 Fall 2015 ? Acosta, Woodard
v
=
2 R T
=
R
1 2
T
B
=
m ( R p (1/2)T R(+e)
)
=
mp
e
1 2
T
=
(1.672?10-27 kg) (1.602?10-19C)(130?10-9 s)
=
0.252
kg Cs
= 0.252
N
m s
C
= 0.252T
; (1.10)
In the last steps of (1.10), we illustrate the units1 of magnetic field.
Problem 7: In a certain cyclotron a proton moves in a circle of radius 0.5 m. The magnitude of the magnetic field is 1.2 T. What is the kinetic energy of the proton in million electron-volts (MeV)?
This is problem 28-38, which was assigned in homework.
K
=
1 2
mv2
=
1 2
m(C R)2
=
1 2
m(
eB m
R)2
= (1.67?10 kg)( 0.5m) 1
-27
(1.602?10-19 C )(1.2T )
2
1
2
1.67?10-27 kg
1.602?10-13
J MeV
= 17.267MeV ; (1.11)
Problem 8: The figure shows a wood cylinder of mass m = 0.250 kg and length L = 0.100 m, with N = 10 turns of wire wrapped around it longitudinally, so that the plane of the wire contains the long central axis of the
cylinder. The cylinder is released on a plane inclined at an angle to the horizontal, with the plane of the coil
parallel to the incline plane. If there is a vertical uniform magnetic field of magnitude 0.500 T, what is the least current i through the coil that keeps the cylinder from rolling down the plane?
This is problem 28-51, which was assigned in homework.
?
(1.12)
Let the wood-cylinder be of mass m , and have moment of inertia2 I . Newton's 2nd Law for translational
equilibrium between the force of static3 friction f and magnetic force FB and rotational equilibrium between the torque of static friction Rf and magnetic torque B is,
4
4
4
ma = m 0 = 0 = F = f - mg sin; I = 0 = = RiFi sini - fR = RiFi sin 90? - fR = RiFi - Rf ; (1.13)
i =1
i =1
i =1
1 A tesla is a unit of force per unit velocity per unit charge; essentially the units of electric field divided by velocity.
2 The radius of the wooden cylinder is
R , but the wooden-material may be inhomogeneous, so assume
I
1 2
mR2 .
3 CAUTION: The force of static friction is a reaction force, and its magnitude is unknown. The maximum value the force of friction
could take on, if we knew the static-friction-coefficient ?S , is max f = ?S N , where N is a reaction force which has a known contribution mg cos from gravity, but an unknown contribution from the net magnetic force.
PHY2049 Fall 2015 ? Acosta, Woodard
The problem is to find F1, F2, F3, F4 : the magnetic forces upon the four sections of the square-loop shown in the
Figure. The magnitude of the force upon a wire of length l
making angle
with a magnetic field
r B
carrying
current
I
= dq / dt
is given by
F = Il
r B
sin
.
There
are
N
such wires producing identical and superimposing
forces. Thus, let the wood-cylinder be of radius R . Sections 1 and 3 are of length 2R , while Sections 2 and 4
are of length L . Then,
R1F1 = Ri1l 1B sin1N = R(+i)(2R)B sin(90? - )N = +2iR2B cos N;
R3F3 = Ri3l 3B sin3N = R(+i)(2R)B sin(270? - )N = -2iR2B cos N = -R1F1; R2F2 = Ri2l 2B sin2N = (+R)(+iLB sin )N = RiLB sin N; R4F4 = Ri4l 4B sin4N = (-R)(-iLB sin )N = RiLB sin N;
(1.14)
Combining (1.13) and the explicit forces (1.14), and noting the simplification R1F1 = -R3F3, we have,
!##f =m"gs#in #$ 4
Rf = Rmg sin = Ri Fi = R1F1 + R2F2 + R3F3 + R4F4 = RiLBsin N + RiLBsin N + 0 = 2RiLBsin N ; (1.15) i=1
Solving (1.15) for i , we have,
Rmg sin = 2RiLB sin N solveforii = mg =
(0.250kg
)(9.81
m s2
)
=
C 2.453
;
2LBN
2(0.100m)(0.500
(
N m/s
)C
)(10.0)
s
(1.16)
Afterword: We note that the net torque B due to the four branches of the loop has the property,
B =
4
2!RL#=a="are#a=$A/ N Ri Fi = 2RiLBN sin = B 2RL i N sin =
B a i N sin = B
A i sin
B ?% sin =
% B
?
?%
;
(1.17)
i=1
We introduced the magnetic dipole moment vector,
?!
=
! iA
=
iNa!
=
iNan^
,
where
n^ = k^ cos +
^j sin
is the
plane-normal
defining
the
vector-area
r A
=
Nar
=
Nan^
=
N
2LRn^
.
Recall,
also,
that
we
encountered
vector
area
in
our study of the flux that naturally occurred in Gauss's law.
Problem 9: The figure shows, in cross section, two long straight wires held against a plastic cylinder of radius R = 20cm . Wire 1 carries current i1 = 60mA out of the page and is fixed in place at the left side of the cylinder. Wire 2 carries current i2 = 40mA out of the page and can be moved around the cylinder. At what (positive) angle 2 should wire 2 be positioned such that, at the origin, the net magnetic field due to the two currents has magnitude B = 80nT ?
This is problem 29-34, which was worked in class on Oct.14
PHY2049 Fall 2015 ? Acosta, Woodard
(1.18)
The two magnetic
fields
decompose as
r B1
=
B1 ^j
and
r B2
= B2(sin2i^ - cos2 ^j) , so the resultant of
this, using
Amp?re's law to say
B1 =
?0i1 2 R
and
B2
=
?0i2 2 R
(in which we clearly have
i1 =
3 2
i2
),
is,
B
= =
(B2 sin2 )2
+ (B1 - B2 cos2 )2
=
?0i2 2 R
?0i2 2 R
1+
9 4
- 3cos2
2
= cos-1
(1 13
34
sin2 2
+
(
3 2
-
cos2
)2
=
?0i2 2 R
sin2 2
- ( ) ) = cos ( - ( ) ) 2 RB 2 ?0i2
-1 1 13 34
0.2(80?10-2 T ) 2 2(40?10-3 A)
+ =
(
3 2
)2
+
cos2
2
104.4775? ;
-
2
3 2
cos2
(1.19)
Problem 10: In the figure a long straight wire carries a current i1 = 30.0A and a rectangular loop carries current i2 = 20.0A. Take the dimensions to be a = 1.00cm , b = 8.00cm , and L = 30.0cm . In unit vector notation,
what is the force on the loop due to i1 ?
This is problem 29-41, which was assigned in homework.
(1.20) The horizontal wires: Consider two typical wires, a and b , a distance d apart, and carrying respective currents
r ia and ib . A differential element of force dFba acts upon wire- b and is due to wire- a ,
(1.21)
PHY2049 Fall 2015 ? Acosta, Woodard
r
By the Lorentz force law, the differential element of force dFba per unit length dx is due to magnetic field4
r Bb
=
?0ia 2 d
(-
^j) ,
so
we
calculate
the
force
per
unit
length
r
r
f dFba ba dx
as,
( ) ( ) !
fba
=
! dFba dz
=
dqb v!b ? dz
! Ba
=
dqb
dz! dt
?
?0ia 2 d
(-
^j)
dz
=
dqb dt
(k^ dz) ?
dz
?0ia 2 d
(-
^j)
=
ib
?0ia 2 d
k^
?
(-
^j)
=
?0iaib 2 d
i^;
(1.22)
Evidently, the force between wires a and b is in the +i^ direction, and thus is attractive. Using this result
(1.22) upon the two horizontal wires in the Figure (numbered 1 and 3 (note the different coordinates!)),
!
F1
=
! dF1
=
L 0
! dF1 dx
dx
=
L 0
?0i1i2 2 a
^j
dx
=
?0i1i2 L 2 a
^j;
! F3
=
L 0
! dF3 dx
dx
=
L 0
?0i1i2 2 (a + b)
(-
^j)
dx
=
-
?0i1i2 L 2 (a + b)
^j;
(1.23)
The vertical wires: The total forces upon wires 2 and 4 due to wire 0 (of infinite length) are,
!
F2
=
! dF2
=
a+b a
! dF2 dy
dy
=
a+b a
dq2
v!2 ? dy
! B0
dy;
! F4
=
a+b a
! dF2 dy
dy
=
a a+b
dq4
v!4 ? dy
! B0
dy;
(1.24)
The integrands in (1.24) (i.e., the forces per unit y -length) are,
! dF2 dy
=
dq2
v!2 ? dy
! B0
=
i2
(-
^j dy) ? dy
?0i1 2 y
(-
k^)
=
?0i1i2 2 y
i^;
! dF4 dy
=
dq4
v!4 ? dy
! B0
=
i2
(+
^j dy) ? dy
?0i1 2 y
(-
k^)
=
- ?0i1i2 2 y
i^; (1.25)
Combining (1.24) and (1.25), we have,
r
F2
=
a+b a
?0i1i2 2 y
i^ dy
=
?0i1i2 2
a+b
i^
a
dy y
=
?0i1i2 2
i^ ln
a
+b ;
a
r F4
=
a+b a
-?0i1i2 2 y
i^ dy
=
-?0i1i2 2
a+b
i^
a
dy y
=
-?0i1i2 2
i^
ln
a
+ a
b
;(1.26)
Looking
at
(1.26),
we
see
r F2
= -Fr4 ,
so
the
superposition
of
these
two
forces
make
no
contribution.
Hence,
! F
=
! F1
+
! F2
+
! F3
+
! F4
=
! F1
+
! F2
+
! F3
-
! F2
=
! F1
+
! F3
=
?0i1i2 L 2 a
^j -
?0i1i2 L 2 (a + b)
^j
=
?0i1i2 L 2
^j
1 a
-
a
1 +
b
=
(4
?
10-7
T m A
)(30.0 A)(20.0 2
A)(30.0cm)
^j
1 1.00cm
-
1.00cm
1 +
8.00cm
=
0.0032N ^j ;
(1.27)
Problem
11:
The
current
density
r J
inside
a
long,
solid,
cylindrical
wire
of
radius
a
= 3.1?10-3 m
is
in
the
direction of the central axis, and its magnitude varies linearly with radial distance r from the axis according to
r J
=
r J (r)
=
J0 (r
/
a)k^
,
where
J0
=
310
A m2
.
What
is
the
magnitude
of
the
magnetic
field
at
r
/
a
=1/
2
?
You
may
need the Jacobian term r dr d for integration in polar coordinates.
This is problem 29-47, which was assigned in homework.
4 Proof: from Amp?re's law:
r Bwire
?
r dl
=
?0iencl
r Bwire
? r^
ds
=
?0iencl
B wire
2
d
=
?0iencl
Brwire
=
?0iencl 2 d
r^ .
In
this
problem, iencl = ia and (by the right hand rule) r^ = - ^j .
PHY2049 Fall 2015 ? Acosta, Woodard
The current enclosed by an Amp?rian-loop of radius 0 b a is,
iencl = iencl (b) =
diencl
=
di dA = dA wire x.s.
b 2 00
J 0 (r
/ a) dr r d
=
2 J0 a
b
r 2dr
0
=
2 J0 a
1 (b3 3
- 03) =
2 J0b3 3a
;(1.28)
The magnetic field at a distance b from the center of the wire, then, is given from the enclosed current (1.28)
via the law of Amp?re,
" " ! B
?
ds!
=
?0iencl
(b)
! B
?
ds!
=
? 0
2
J
b3
0
3a
B^ ? 2b^ =
? 0
2
J b3 0
solve
for B
3a
B=
? 0
J b2 0 3a
;
(1.29)
From (1.29) follows,
Bb
=
B(
1 2
a)
=
?0
J0
(
1 2
3a
a)2
=
1 12
?0aJ0
=
1 12
(4
?10-7
T m A
)(3.1?10-3
m)(310
A m2
)
=
1.0064 ?10-7 T
;
(1.30)
Problem 12: A solenoid that is L = 95cm long has a radius R = 2.0cm and a winding of N = 1200 turns; it carries a current of i = 3.6A . What is the magnitude of the magnetic field inside the solenoid?
This is problem 29-50, which was assigned in homework.
r The magnetic field B at the center of a solenoid made of a wire carrying current i with n = N / L turns per
meter is of magnitude B = ?0ni .
Therefore,
B=
?0(N / L)i
= 4 ?10-7
T m A
1200 0.95cm
(3.6
A)
=
5.7144mT
.
Problem 13: A wire
loop of lengths
L = 40cm and
W
= 25cm
lies in a magnetic
field
r B
=
(0.08
T ms
)
yt
k^
.
What
are the magnitude and direction of the induced E.M.F.?
This is problem 30-12, which was worked in class on Oct.19
E
=
-
d dt
W
dy
0
L 0
b0
yt
dx
=
-b0
(
1 2
W
2
-
1 2
02
)(L
-
0)
=
-
1 2
b0W
2
L
=
-1.00mV
;
clockwise;
Problem 14: The figure shows a rod of length L = 10cm that is forced to move at constant speed v = 5m / s
along the horizontal rails. The rod, rails and connecting strip at the right form a conducting loop. The rod has a resistance R = 0.4 ; the rest of the loop has negligible resistance. A current i = 100A through the long straight wire at a distance a = 10mm from the loop sets up a nonuniform magnetic field through the loop. At
what rate (in ?W) is thermal energy generated in the rod?
This is problem 30-33, which was worked in class on Oct.21
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