Download Exam 2 Solutions

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 100Ω and 300Ω . They are connected in parallel across a
120V line. What is the total power dissipated by the two bulbs (or the 100Ω or 300Ω bulbs)?
This problem was similar to Exercise 26-C done in class.
1
1
P = IE = E2 R −1 = E2 ( R1−1 + R2 −1 ) = (120)2 ( 100
+ 300
)W = 192W ;
(1.1)
or 144 W and 48 W individually.
Problem 2: In the circuit shown in the figure both batteries have insignificant internal resistance and the
idealized ammeter reads I1 = 4.0 A 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 − I 2 ); − E = R3 I 2 + R2 ( I 2 − I1 ); I1 = 4.0 A; R1 = 12.0Ω; R2 = 48.0Ω; R3 = 15.0Ω; (1.2)
We note that the first equation in (1.2) implies I 2 =
1
R2
(( R1 + R2 ) I1 − E1 ) , which can be immediately put into the
2nd equation to yield E , as,
E = R2 I1 − ( R2 + R3 ) I 2 = 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 /τ 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 /τ . The third condition V (10.0s) = 1.5V = (150V )e− (10.0 s )/τ
1.5
determines τ to be τ = 10.0s / (− ln 150
) = 2.1715s , which determines the voltage at all times, and so we
calculate directly V (20.0s) = (150V )e − (20.0 s )/(2.1715 s ) = 0.015V .
Problem 4: In the figure E = 14V , R1 = R3 = 1.00Ω , and R2 = 2Ω . 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 = 1Ω and R2 = 2 Ω:
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 = i1 R1 = ( 6A ) (1Ω ) = 6V
PHY2049 Fall 2015 – Acosta, Woodard
r
Problem 5: A proton travels through uniform magnetic and electric fields. The magnetic field is B = −2.50iˆ mT .
r
At one instant the velocity of the proton is v = 2000 ˆj m/s . At that instant what is the net force acting on the
proton if the electric field is 4.00kˆ V/m ?
This is problem 28.10, which was assigned in homework
!
! ! !
F = q( E + v × B) = q(Ek̂ + vĵ × B(− iˆ)) = q(E + vB) k̂
= (1.602 × 10
−19
(1.5)
−3
)(4.00 + (2000)(2.50 × 10 )) k̂N = 1.442 × 10
−18
Nk̂
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)
r
We immediately notice that the particle velocity v = v(− ˆj ) is deflected in the −iˆ direction by the magnetic field
r
B = Bkˆ pointing out of the page. Looking at the vector-value of the forces,
⎧⎪q = +e = 1.602 × 10−19 C;
!
! !
ˆ
ˆ
(1.7)
FB = FB (− i ) = qv × B = q(v(− ĵ)) × (Bk̂) = qvB(− i ) → qvB > 0 → q > 0 → ⎨
−27
⎪⎩ m = mp = 1.672 × 10 kg;
r
While in the circular region, the charged particle has constant speed v = v , and maintains this velocity. This is
r
r
r
r
r
because the magnetic force FB (due to magnetic field B = Bkˆ and velocity v = v ⋅ dθ = Rω ⋅ dθ ) 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) • Rdθ = qvBR( r̂ • dθ ) = qvBR(0) = 0 → dK = dW = 0 → K i = K f = K a ; (1.8)
r
Hence, no kinetic energy is added or subtracted to the particle of mass m . the velocity v is of constant
magnitude. In the circular trajectory, Newton’s 2nd Law then is,
=sin90°=1
"
!
mp v
−v 2
−v 2
−v 2
v2
! !
! !
! ↔ B=
!
F
=
m
a
=
m
=
−
F
→
m
=
−q
v
×
B
→
m
=
−q
v
×
B
→
m
=
qvBsin
θ
; (1.9)
∑
p
p
B
p
p
p
vB
R
R
R
R
Rq
In (1.9), the velocity is given by v =
2π R
T
=
πR
(1/2)T
, so,
PHY2049 Fall 2015 – Acosta, Woodard
πR
m p ( (1/2)
m pπ
2π R π R
(1.672 ×10−27 kg )π
kg
N
T)
v=
= 1 →B=
= 1 =
= 0.252
= 0.252 m
= 0.252T ; (1.10)
−19
−9
T
R ( + e)
e 2 T (1.602 ×10 C )(130 ×10 s)
C⋅s
2T
s ⋅C
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 = 12 mv 2 = 12 m(ωC R)2 =
1
2
−19
×10 C )(1.2T )
2
−27
1
m( eB
kg )( (1.6021.67
0.5m)2 1.602×101 −13
m R) = 2 (1.67 × 10
×10−27 kg
J
MeV
= 17.267 MeV ; (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 θ 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.
à
(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 τ B is,
4
4
4
i =1
i =1
i =1
ma = m ⋅ 0 = 0 = ∑ F = f − mg sin θ ; Iα = 0 = ∑τ = ∑ Ri Fi sin θi − fR = ∑ Ri Fi sin 90° − fR = ∑ Ri Fi − Rf ; (1.13)
1
2
3
A tesla is a unit of force per unit velocity per unit charge; essentially the units of electric field divided by velocity.
The radius of the wooden cylinder is R , but the wooden-material may be inhomogeneous, so assume I ≠ 12 mR 2 .
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 µ S , is max f = µS N , where N is a reaction force which has a known
contribution
mg cos θ 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
r
Figure. The magnitude of the force upon a wire of length l making angle θ with a magnetic field B carrying
r
current I = dq / dt is given by F = I l B sin θ . 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,
R1 F1 = Ri1l 1 B sin θ1 N = R(+i)(2 R) B sin(90° − θ ) N = +2iR 2 B cos θ N ;
R3 F3 = Ri3l 3 B sin θ3 N = R(+i )(2 R) B sin(270° − θ ) N = −2iR 2 B cos θ N = − R1F1;
R2 F2 = Ri2 l 2 B sin θ 2 N = (+ R)(+iLB sin θ ) N = RiLB sin θ N ;
(1.14)
R4 F4 = Ri4 l 4 B sin θ 4 N = (− R)(−iLB sin θ ) N = RiLB sin θ N ;
Combining (1.13) and the explicit forces (1.14), and noting the simplification R1F1 = − R3 F3 , we have,
f =mg sin θ
!##
"##$ 4
Rf = Rmg sin θ = ∑ Ri Fi = R1 F1 + R2 F2 + R3 F3 + R4 F4 = RiLBsin θ N + RiLBsin θ N + 0 = 2RiLBsin θ N; (1.15)
i=1
Solving (1.15) for i , we have,
solve for i
Rmg sin θ = 2 RiLB sin θ N ←⎯⎯⎯
→i =
(0.250kg )(9.81 sm2 )
mg
C
=
=
2.453
;
2 LBN 2(0.100m)(0.500 ( m /Ns )C )(10.0)
s
(1.16)
Afterword: We note that the net torque τ B due to the four branches of the loop has the property,
!#"#$
% %
%
τ B = ∑ Ri Fi = 2RiLBN sin θ = B 2RL ⋅ i ⋅ N sin θ = B a ⋅ i ⋅ N sin θ = B A⋅ i sin θ ≡ B µ sin θ = B × µ ; (1.17)
4
2 RL=a=area= A/ N
i=1
!
!
!
We introduced the magnetic dipole moment vector, µ = iA = iNa = iNan̂ , where nˆ = kˆ cos θ + ˆj sin θ is the
r
r
plane-normal defining the vector-area A = Na = Nanˆ = N 2LRnˆ . 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 θ 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)
r
r
The two magnetic fields decompose as B1 = B1 ˆj and B2 = B2 (sin θ2iˆ − cosθ2 ˆj ) , so the resultant of this, using
Ampère’s law to say B1 = 2µπ0iR1 and B2 =
µ0i2
2π R
B = ( B2 sin θ 2 )2 + ( B1 − B2 cos θ 2 )2 =
(in which we clearly have i1 = 32 i2 ), is,
µ0i2
2π R
sin 2 θ 2 + ( 23 − cos θ 2 ) 2 =
µi
µ0i2
2π R
sin 2 θ 2 + ( 23 ) 2 + cos 2 θ 2 − 2 23 cos θ 2
−2
0.2(80×10 T ) 2
2
−1 1 13
= 2π0 R2 1 + 94 − 3cos θ 2 ↔ θ 2 = cos −1 13 ( 134 − ( 2µπ0RB
i2 ) ) = cos
3 ( 4 − ( 2(40×10−3 A) ) ) = 104.4775° ;
(1.19)
Problem 10: In the figure a long straight wire carries a current i1 = 30.0 A and a rectangular loop carries
current i2 = 20.0 A . 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
r
r
Bb = 2µπ0ida (− ˆj ) , so we calculate the force per unit length fba ≡ dFdxba as,
!
!
µ0ia
dqb
µi
! !
dz
⋅( k̂ ⋅ dz) × 2π0 da (− ĵ)
!
dFba dqb ⋅ vb × Ba dqb ⋅ dt × 2π d (− ĵ)
µi
µii ˆ
dt
f ba =
=
=
=
= ib 0 a k̂ × (− ĵ) = 0 a b i;
(1.22)
dz
dz
dz
dz
2π d
2π d
(
)
(
)
Evidently, the force between wires a and b is in the +iˆ 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!)),
!
!
L
L
!
! L dF1
! L dF3
⎛ µ0 i1i2 ⎞
⎛ µ0 i1i2
⎞
µ0 i1i2 L
µ ii L
F1 = ∫ dF1 = ∫
dx = ∫ ⎜
ĵ ⎟ dx =
ĵ; F3 = ∫
dx = ∫ ⎜
(− ĵ)⎟ dx = − 0 1 2
ĵ; (1.23)
dx
2π a
dx
2π (a + b)
⎝ 2π a ⎠
⎝ 2π (a + b)
⎠
0
0
0
0
The vertical wires: The total forces upon wires 2 and 4 due to wire 0 (of infinite length) are,
!
!
! !
! !
a+b
a
!
! a+b dF2
! a+b dF2
dq2 ⋅ v2 × B0
dq4 ⋅ v4 × B0
F2 = ∫ dF2 = ∫
dy = ∫
dy; F4 = ∫
dy = ∫
dy;
dy
dy
dy
dy
a
a
a
a+b
(1.24)
The integrands in (1.24) (i.e., the forces per unit y -length) are,
!
!
! !
! !
µi
µi
dF2 dq2 ⋅ v2 × B0 i2 ⋅(− ĵ ⋅ dy) × 2π0 1y (− k̂) µ0 i1i2 ˆ dF4 dq4 ⋅ v4 × B0 i2 ⋅(+ ĵ ⋅ dy) × 2π0 1y (− k̂) − µ0 i1i2 ˆ
=
=
=
i;
=
=
=
i; (1.25)
dy
dy
dy
2π y
dy
dy
dy
2π y
Combining (1.24) and (1.25), we have,
a +b
r a +b µ0i1i2
µ0i1i2 ˆ a +b dy µ0i1i2 ˆ a + b r a +b − µ0i1i2 ˆ
− µ0i1i2 ˆ dy − µ0i1i2 ˆ a + b
ˆ
F2 = ∫
i ⋅ dy =
i
=
i ln
; F4 = ∫
i ⋅ dy =
i ∫
=
i ln
; (1.26)
2π y
2π ∫a y
2π
a
2π y
2π
y
2π
a
a
a
a
r
r
Looking at (1.26), we see F2 = − F4 , so the superposition of these two forces make no contribution. Hence,
! ! ! ! !
! ! ! !
! ! µ ii L
µ ii L
µ ii L
F = F1 + F2 + F3 + F4 = F1 + F2 + F3 − F2 = F1 + F3 = 0 1 2 ĵ − 0 1 2
ĵ = 0 1 2
2π a
2π (a + b)
2π
=
(4π × 10
−7 T ⋅m
A
)(30.0 A)(20.0 A)(30.0cm)
2π
⎛1
1 ⎞
ĵ ⎜ −
⎝ a a + b ⎟⎠
⎛ 1
⎞
1
ĵ ⎜
−
= 0.0032N ⋅ ĵ ;
⎝ 1.00cm 1.00cm + 8.00cm ⎟⎠
(1.27)
r
Problem 11: The current density 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 r
J = J (r ) = J 0 (r / a)kˆ , where J 0 = 310 mA2 . What is the magnitude of the magnetic field at r / a = 1/ 2 ? You may
need the Jacobian term rdrdθ for integration in polar coordinates.
This is problem 29-47, which was assigned in homework.
4
Proof: from Ampère’s law:
problem,
r
r
r
r⊥
µ0iencl
⊥
ˆ
ˆ . In this
B
•
d
l
=
µ
i
→
B
•
r
ds
=
µ
i
→
B
2
π
d
=
µ
i
→
B
0 encl
wire
⊥∫
0 encl
wire
0 encl
wire = 2π d r⊥
∫ wire
iencl = ia and (by the right hand rule) r̂⊥ = − ˆj .
PHY2049 Fall 2015 – Acosta, Woodard
The current enclosed by an Ampèrian-loop of radius 0 ≤ b ≤ a is,
iencl = iencl (b) = ∫ diencl =
b 2π
2π J 0 2
2π J 0 1 3 3 2π J 0b3
di
dA
=
J
(
r
/
a
)
⋅
dr
⋅
r
⋅
d
φ
=
r
dr
=
(b − 0 ) =
; (1.28)
0
∫
∫
∫
∫
dA
a
a
3
3
a
wire x . s .
0 0
0
b
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 Ampère,
! !
!
2π J 0 b3
2π J 0 b3 solve
J 0 b2
!
for B
B
•
d
s
=
µ
i
(b)
→
B
•
d
s
=
µ
→
B
φ̂
•
2
π
b⋅
φ̂
=
µ
←
⎯⎯
→
B
=
µ
;
0 encl
0
0
0
"∫
"∫
3a
3a
3a
(1.29)
From (1.29) follows,
Bb = B( 12 a) =
µ0 J 0 ⋅ ( 12 a)2
3a
=
1
12
µ0 aJ 0 = 121 (4π ×10−7 TA⋅m )(3.1×10−3 m)(310 mA ) = 1.0064 ×10−7 T ;
2
(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 = 3.6 A . 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
1200
meter is of magnitude B = µ0 ni . Therefore, B = µ0 ( N / L)i = 4π ×10−7 TA⋅m 0.95
.
cm (3.6 A) = 5.7144mT
r
Problem 13: A wire loop of lengths L = 40cm and W = 25cm lies in a magnetic field B = (0.08 mT⋅s ) yt ⋅ kˆ . 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
W
L
0
0
E = − dtd ∫ dy ∫ b0 yt ⋅ dx = −b0 ( 12 W 2 − 12 02 )( L − 0) = − 12 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 = 0.4Ω ; the rest of the loop has negligible resistance. A current i = 100 A 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 µW) is thermal energy generated in the rod?
This is problem 30-33, which was worked in class on Oct.21
PHY2049 Fall 2015 – Acosta, Woodard
(1.31)
r
r
The magnetic field B due to the long straight wire at a distance y ∈ [a, L + a] is B =
kˆ , and the differential
r
r r
vector area through which it fluxes is dA( y) = dy(vt + x0 )kˆ , and so the E.M.F. generated is E = − dtd ∫ B • dA,
µ0 i
2π y
yielding the power P = I E = E2 / R . Putting all this together,
1⎛ d
P = ⎜−
R ⎝ dt
=
L+a
∫
a
2
2
2
L+a
µ0 i ⎞
1 ⎛ µ0 iv ⎞ ⎛ dy ⎞
1 ⎛ µ0 iv ⎞
(vt + x0 )
dy ⎟ = ⎜
⎜ ∫
⎟ = ⎜
⎟
2π y ⎠
R ⎝ 2π ⎠ ⎝ a y ⎠
R ⎝ 2π ⎟⎠
1
( 42ππ × 10−7
(
0.4Ω
2
⎛ L+a⎞
⎜ ln
⎟
a ⎠
⎝
2
(1.32)
2
T ⋅m
A
2 ⎛
0.11 ⎞
)(100 A)(5 ms ) ) ⎜ ln
⎟ = 0.14375µW ;
⎝ 0.01 ⎠
Problem 15: The current i = i (t ) through a L = 4.6 H inductor varies with time t as shown in the graph, where
the vertical axis scale is set by is = 8.0 A and the horizontal axis scale is set by ts = 6.0 ×10−3 s . The inductor has
a resistance of R = 12Ω . What is the magnitude of the induced E.M.F. during the time interval
2 ×10−3 s < t ≤ 5 ×10−3 s ?
This is problem 30-46, which was assigned in homework.
Recall the definition of inductance: L is the E.M.F.-magnitude E per unit rate-of-change of current
per unit i , and specialize it to the case of
di
dt
=
Δi
Δt
L≡
di
dt
, or Φ B
,
E Φ B di Δi
≡
;
= ;
i
i
dt Δt
d
dt
(1.33)
Within the time-interval 2 × 10−3 s < t ≤ 5 × 10−3 s , we have,
(5.0 − 7.0 ) A
(5.0 − 7.0 ) A
⎛ Δi ⎞
⎛ Δi ⎞
→ E = L ⎜ ⎟ = ( 4.6 H )
= 3.1× 103 V ;
⎜ ⎟=
−3
−3
Δ
t
Δ
t
(5.0 − 2.0 ) ×10 s
⎝ ⎠ ( 5.0 − 2.0 ) × 10 s
⎝ ⎠
(1.34)
PHY2049 Fall 2015 – Acosta, Woodard
Problem 16: The switch in the figure is closed on a at time t = 0. What fraction of the total voltage drop E
occurs across the inductor at time t = 2 L / R ?
(1.35)
This is problem 30-52, which was worked in class on Oct.23
Recall that an LR-circuit has a current i (t ) = i0 + i1e − t /τ , where τ = L / R . The conditions i (0) = 0 and
i(∞) = E / R respectively yield i0 = −i1 and i0 = E / R , yielding i(t ) = (E / R)(1 − e−t /τ ) , meaning that at a time
of t = 2 L / R = 2τ the inductor’s voltage in units of E is,
vL 1 di L E d
L
−1
L 1 −2 L 1 −2
= L =
(1 − e − t /τ ) = (0 − e −2τ /τ ) =
e =
e = e −2 = 0.1353 ;
E E dt E R dt
R
τ
Rτ
R L/R
(1.36)
Problem 17:
An LC circuit has a capacitance of 20µF and an inductance of 10 mH. At time t = 0 the charge on the capacitor
is 27µC and the current is 80 mA. What is the maximum possible charge in µC (or what is the maximum
possible current)?
q
q
i
Q
q2
1
⌠ dI
U C = ∫ V (Q) ⋅ dQ = ⌠
⋅
dQ
=
; U L = ∫ P ⋅ dt = ∫ EI ⋅ dt = ⎮ L
I ⋅ dt = ∫ ( LI ⋅ dI ) = Li 2 ;
⎮
⌡C
2C
2
⌡ dt
0
0
(1.37)
0
Thus, the total energy in the circuit, by energy-conservation, is U = U C + U L =
1
2C
q 2 + 12 Li 2 , yielding a
maximum charge on the capacitor of max UC = U = 21C qmax 2 ↔ qmax = 2CU , which, explicitly, is,
qmax = 2CU = 2C( 21C q 2 + 12 Li 2 ) =
q 2 + LCi 2 = (27µC )2 + (10mH )(20µ F )(80mA)2 = 44.822µC ; (1.38)
imax = 2U / L = 100mA
Problem 18: A sinusoidally varying source of E.M.F. with an amplitude of 10V and a cyclic frequency of
5GHz is applied across a 100µ H inductor. What is the current amplitude through the inductor?
This is based on a HITT clicker question given in class Oct.30
I max =
Vmax Vmax
V
10V
1
=
= max =
=
× 10−3 = 3.18 × 10−6 A ;
X L ωd L 2π f d L 2π (5GHz )(100µ H ) 2π (5)(10)
(1.39)
PHY2049 Fall 2015 – Acosta, Woodard
Problem 19: A 218Ω resistor, a L = 0.775H inductor, and a 6.50 µ F capacitor are connected in series across a
sinusoidally varying source of E.M.F. that has voltage amplitude 31.0V and a cyclic frequency of 37.5Hz .
What is the magnitude of the phase difference between the current in the resistor and the E.M.F.?
This is similar to Example 31-E done in class.
φ = tan
−1
−1
−1
rad
rad
X L − XC
−1 ωd L − (ωd C )
−1 (235 s )(0.775 H ) − ((235 s )(6.50 µ F ))
= tan
= tan
= −65.234° ;
R
R
218Ω
Problem 20: A transformer connected to a VpRMS = 120V AC line is to supply VsRMS = 12, 000V for a neon sign.
To reduce shock hazard, a fuse is to be inserted in the primary circuit; the fuse is to blow when the R.M.S.
current in the secondary circuit exceeds isRMS = 3.0mA . What current rating should the fuse in the primary
circuit have?
The power delivered into the primary is the same as that on the secondary (or less for realistic transformers)
i
RMS
p
N s RMS VsRMS RMS 1.2 ×104 V
=
is = RMS is
=
(3.0mA) = 300mA ;
Np
Vp
120V
(1.40)