final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
E & M - Basic Physical Concepts
Current and resistance
Current: I = ddtQ = n q vd A
Ohm’s law: V = I R, E = ρJ
I , R = ρℓ
E = Vℓ , J = A
A
Electric force and electric field
Electric force between 2 point charges:
|q | |q |
|F | = k 1r2 2
k = 8.987551787 × 109 N m2 /C2
ǫ0 = 4 π1 k = 8.854187817 × 10−12 C2 /N m2
qp = −qe = 1.60217733 (49) × 10−19 C
mp = 1.672623 (10) × 10−27 kg
me = 9.1093897 (54) × 10−31 kg
~
~ =F
Electric field: E
2
Power: P = I V = VR = I 2 R
Thermal coefficient of ρ: α = ρ ∆ρ
0 ∆T
Motion of free electrons in an ideal conductor:
a τ = vd → qmE τ = nJq → ρ = n qm2 τ
|Q|
~2 + · · ·
~ =E
~1 + E
Point charge: |E| = k r2 , E
Field patterns: point charge, dipole, k plates, rod,
spheres, cylinders,. . .
Charge distributions:
Linear charge density: λ = ∆Q
∆x
Surface charge density: σsurf =
∆Q
Volume charge density: ρ = ∆V
∆Qsurf
∆A
Electric flux and Gauss’ law
~ · n̂∆A
Flux: ∆Φ = E ∆A⊥ = E
Gauss law: Outgoing Flux from S, ΦS = Qenclosed
ǫ0
Steps: to obtain electric field
~ pattern and construct S
–Inspect E
H
~ · dA
~ = Qencl , solve for E
~
–Find Φs = surf ace E
ǫ
0
Spherical: Φs = 4 π r2 E
Cylindrical: Φs = 2 π r ℓ E
Pill box: Φs = E ∆A, 1 side; = 2 E ∆A, 2 sides
σ
k
~ = 0, Esurf
Conductor: E
= 0, E ⊥ = surf
in
surf
Potential
ǫ0
Potential energy: ∆U = q ∆V 1 eV ≈ 1.6 × 10−19 J
Positive charge moves from high V to low V
Point charge: V = krQ V = V1 + V2 = . . .
1 q2
Energy of a charge-pair: U = k rq12
Potential difference: |∆V | = |E ∆sk |,
R
~ · ∆~s, V − V = − B E
~ · d~s
∆V = −E
B
A
A
¯
¯
d
V
∆V
E = − dr , Ex = − ∆x ¯
= − ∂V
∂x , etc.
f ix y,z
Capacitances
Q=CV
Series: V = CQ = CQ + CQ + CQ + · · ·, Q = Qi
eq
1
2
3
Parallel: Q = Ceq V = C1 V + C2 V + · · ·, V = Vi
ǫ A
Q
Parallel plate-capacitor: C = V
= EQd = 0d
2
RQ
Q
Energy: U = 0 V dq = 12 C , u = 12 ǫ0 E 2
2
1
2
Uκ = 21κ Q
C0 , uκ = 2 ǫ0 κ Eκ
Q
Q
Spherical capacitor: V = 4 π ǫ r1 − 4 π ǫ r2
0
0
~
Potential energy: U = −~
p·E
Dielectrics: C = κC0 ,
V =IR
Direct current circuits
q
Area charge density: σA = ∆Q
∆A
1
Series: V = I Req = I R1 + I R2 + I R3 + · · ·, I = Ii
V + V + V + · · ·, V = V
Parallel: I = RV = R
i
R2
R3
eq
1
Steps: in application of Kirchhoff’s Rules
–Label currents: i1 , i2 , i3 , . . .
P
P
i
–Node equations:
i =
P in
Pout
–Loop equations: “ (±E) + (∓iR)=0”
–Natural: “+” for loop-arrow entering − terminal
“−” for loop-arrow-parallel to current flow
RC circuit: if ddty + R1C y = 0, y = y0 exp(− RtC )
Charging: E − Vc − R i = 0, 1c ddtq + R ddti = ci + R ddti = 0
Discharge: 0 = Vc − R i = qc + R ddtq , ci + R ddti = 0
Magnetic field and magnetic force
µ0 = 4 π × 10−7 T m/A
µ a2 i
µ i
0
Wire: B = 2 π0 r
Axis of loop: B =
2 (a2 +x2 )3/2
~ → q ~v × B
~
Magnetic force: F~M = i ~ℓ × B
~ × B,
~
Loop-magnet ID: ~τ = i A
µ
~ = i A n̂
2
r
Circular motion: F = mrv = q v B, T = f1 = 2 π
v
~ + q ~v × B
~
Lorentz force: F~ = q E
~
Hall effect: V = FM d , U = −~
µ·B
H
q
~ and magnetism of matter
Sources of B
µ
~
µ
v ×r̂
0 q~
~ = 0 i ∆ℓ×r̂
Biot-Savart Law: ∆B
4 π r2 , B = 4 π r2
2
µ0 i ∆y
∆B = 4 π
sin θ, sin θ = ar , ∆y = r a∆θ
r2
H
~ · d~s = µ I
B
Ampere’s law: M =
L
0
encircled
Steps: to obtain magnetic field
~ pattern and construct loop L
–Inspect B
~
–Find M and Iencl , and solve for B.
d (E A)
ΦE = ǫ
Displ. current: Id = ǫ0 d dt
0
dt
Magnetism in atom:
Orbital motion: µ = i A = 2 em L
L = m v r = n h̄,
QA
= d dt
h̄ = 2hπ = 1.06 × 10−34 J s
h̄ = 9.27 × 10−24 J/T
µB = 2em
µspin = µB
Magnetism in matter:
0
B = B0 + BM = (1 + χ) B0 = (1 + χ) µ0 B
µ0 = κm H
Ferromagnetic: χ ≫ 1
Diamagnetic: −1 ≪ χ < 0
Paramagnetic: 0 < χ ≪ 1, M = C
TB
µorbit = n µB ,
Spin: S = h̄2 ,
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
Complete reflection: P = 2cU ,
Faraday’s law
φB
E = −N ddt
,
φB =
R
R
~M
~ = F
E
q
~ · dA
~,
B
d (B A⊥ )
d φB
A⊥
= ddtB A⊥ + B d dt
dt =
dt
´
³
d 1 R ·Rθ
Moving rods: ddtA = ℓ v, ddtA = dt
2
A⊥ = d (A cos ωt)
Rotating loop: d dt
dt
Cutting B lines → change φB → Eind → Eind
H
H
~ · dA
~ = 0,
B
~ · d~s = µ [I + ǫ d φE ]
B
0
0 dt
Inductance
M21 = M12 = N2i φ21
1
L = Ni φ , VL = L ddti
Mutual: E2 = −M21 ddti1 ,
Self: E = −L ddti ,
A
Long solenoid: L = N B
i , B = µ0 n i
Energies: UL = 21 L i2 , uB = 2 1µ B 2
0
UC = 21C q 2 , uE = 21 ǫ0 E 2
q
q = q0 cos(ω t + δ),
L C: VL + VC = 0 ⇒ L ddti = − C
q
ω = L1C , UC + UL = UC max = UL max = U0
Decay Equations: ddty = −a y, y = y0 exp(−a t)
VL + R VL = 0,
L R: E = VL + R i, ddt
Lh
´
³
´i
³
E 1 − exp −R t
,
i
=
VL = E exp − Rt
L
R
L
L R C:
r
³ ´2
R
Q ≈ Q0 e− 2 L t cos ωd t, ωd = L1C − 2RL
Underdamped, critically damped & overdamped
A C Circuits
q
Impedance: [Ohm ≡ Ω]
Z≡
R2 + (XL − XC )2
Inductive XL = ω L, Capactive XC = ω1C
R
Mean value: f¯(t) = T1 0T f (t) dt
1
1
2
2
[sin ω t]rms = [sin2 ω t] = [ 12 (1 − cos 2 ω t)] = √1
2
Electromagnetic waves
Properties of em waves:
E = Em cos(k z − ω t), B = E
c
λ
c
v = ddtz = ω
k = λf = T , n = v
speed of light: c = √ǫ1 µ = 2.99792458 × 108 m/s
0
P = 2cS
Reflection and Refraction
~ · d~s,
E= E
~ opposes change of Φ
Lenz law: Induced B
B
Maxwell equations:
H
~ · dA
~ = Q,
E
ǫ0
H
φB
~
,
E · d~s = − ddt
2
0
~ ⊥ E,
~ propagating along: E
~ ×B
~
B
u = uE + uB , uE = uB
~ B
~
~ = E×
Poynting vector: S
, S̄ = I¯ = Erms Brms
µ0
µ0
∆U
d
z
P
Intensity: I = A = A ∆z dt = u c
R
~ · dA
~ = dU + P
Energy conservation:
S
R
dt
Complete absorption: Momentum p = Uc
∆p 1
∆U 1
S
Pressure: P = F
A = ∆t A = c ∆t A = u = c
n1 = v2 = λ2
Index of refraction: n
v1
λ1
2
Snell’s law: n1 sin θ1 = n2 sin θ2
Critical angle: n2 > n1 , n2 sin θc = n1 sin 90◦
Total reflection: θ > θc
Mirrors and lenses
1
1
1
p+q = f
Ray tracing rules:
Mirror: At symm pt S, reflected symmetrically through
center of sphere, undeflected. Parallel to axis, converges
toward F (or diverges away from F ), f = R
2 .
Lens: Through center of lens, undeflected. Parallel to
axis, converges toward F (or diverges away from F )
Image: q > 0 (real), q < 0 (virtual)
Focal point F : at p = ∞, q = f
f = ±|f |, “+” convergent, “−” divergent
′
Magnification: M = hh = − pq
1
Refraction at spherical surface: np1 + nq2 = n2 −n
R
R is coordinate of center with origin at S, with
S the symmetry³point of´ ³
surface on ´the axis
n
1
2
Lens maker: f = n1 − 1 R1 − R1
1
′
2
1
Two media: M = hh = − pq n
n2
Huygen’s principles:
Points in wave front are sources of next wavelets
Forward tangent surface is next wave front
Interference
Maxima φ = 0, 2 π, 4 π, · · ·; Minima
³ φ´ = π, 3 π, 5 π, · · ·
Double slits: Iaverage = I0 cos2 φ
2 , φ = k∆.
y
for small θ, θ ≈ sin θ ≈ tan θ
sin θ = ∆
d , tan θ = L ,
~=A
~1 + A
~2 + A
~3 + · · ·
Phasor diagram: A
Ax = A1x +A2x +A3x +· · ·, Ay = A1y +A2y +· · ·
a
b
c
sin α = sin β = sin γ
First minimum for N slits: φ = 2Nπ
Thin film: φ = k ∆ + |φ1ref lected − φ2ref lected |, ∆ = 2 t
φref lected = π (denser medium); =0 (lighter medium)
·Diffraction
¸2
sin β2
Single slit: I = I0
, β = k∆,
β
∆ = a sin θ
2
λ
Resolution criterion: θcriterion = 1.22 D
Grating: Principle maxima ∆ = m λ
Polarization
Brewster (n1 < n2 ): n1 sin θbr = n2 sin( π2 − θbr )
Polarizer: Etransmit = E0 cos θ, I = I0 cos2 θ
I0
Unpolarized light: ∆I
∆θ = 2 π
Transmitted Intensity: ∆I ′ = ∆I cos2 θ
R
I ′ = 2I0π 02 π cos2 θ dθ = I20
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
3
.
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final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
c
R1
Question 1, chap 34, sect 3.
part 1 of 1
10 points
a
A 659 nm light beam hits the flat surface of
a certain liquid, and the beam is split into a
reflected ray and a refracted ray.
If the reflected ray is completely polarized
at grazing angle 26◦ , what is the wavelength
of the refracted ray?
Correct answer: 321.415.
EB
Let :
a
λ
n
659 nm
=
2.05031
= 321.415 nm .
λrefracted =
58 V
d
c
R3
R4
b
d
9Ω
S1
Find the current in the path from a to c.
Correct answer: 1.34884.
EB
58 V
=
= 1.34884 A .
R13
43 Ω
Question 3, chap 25, sect 2.
part 1 of 1
10 points
Points A (3 m, 2 m) and B (3 m, 10 m) are in
a region where the electric field is uniform and
~ = Ex ı̂ + Ey ̂, where Ex = 2 N/C
given by E
and Ey = 3 N/C.
What is the potential difference VA − VB ?
Correct answer: 24.
Explanation:
Let :
Explanation:
R1
and
Therefore,
b
4
= 12 Ω ,
= 28 Ω ,
= 31 Ω ,
= 49 Ω ,
= 58 V .
R13 = R1 + R3 = 12 Ω + 31 Ω = 43 Ω .
In the figure below consider the case where
switch S1 is closed and switch S2 is open.
c
31
Ω
Ω
12
S2
S1
R1 and R3 are in series, so
Question 2, chap 28, sect 4.
part 1 of 1
10 points
Ω
R1
R2
R3
R4
EB
R2
EB
I1 = I13 =
28
b
R4
Redrawing the figure, we have
Therefore the wavelength of the refracted ray
(i.e. in the liquid) will be
a
R3
d
Explanation:
From the formula for the polarizing angle
θp we obtain that the refraction index of the
liquid is
n = tan θp
= tan(90◦ − θg )
= tan(64◦ )
= 2.05031 .
S2
R2
4
Ex = 2 N/C ,
Ey = 3 N/C ,
(xA , yA ) = (3 m, 2 m) , and
(xB , yB ) = (3 m, 10 m) .
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
We know
Z
A
~ · d~s
V (A) − V (B) = −
E
B
Z B
~ · d~s
=
E
5
From Faraday’s law, the self-induced emf is
proportional to the time rate of change of the
current, with the proportionality constant L
is the inductance of the inductor.
Thus the self inductance is
A
For a uniform electric field
|ε| = L
~ = Ex ı̂ + Ey ̂ .
E
Now consider the term Ex ı̂ · d~s in the integrand. Ex is just a constant and ı̂ · d~s may be
interpreted as the projection of d~s onto x so
that
Likewise
∆I
∆t
= (2.26 H)
0.68 A
0.0136 s
= 113 V .
Ex ı̂ · d~s = Ex dx .
Question 5, chap 23, sect 2.
part 1 of 1
10 points
Ey ̂ · d~s = Ey dy .
A circular arc has a uniform linear charge
density of −8 nC/m.
The value of the Coulomb constant is
8.98755 × 109 N · m2 /C2 .
y
Or more simply, d~s = dxı̂ + dy ̂ so dotting
it with Ex ı̂ + Ey ̂ gives the same result as
above.
Therefore
Z yB
Z xB
dy
dx + Ey
VA − VB = Ex
xA
98 ◦
yA
= (2 N/C) (3 m − 3 m)
+ (3 N/C) (10 m − 2 m)
2. 3
m
x
= 24 V .
Note that the potential difference is independent of the path taken from A to B.
Question 4, chap 31, sect 4.
part 1 of 1
10 points
A 2.26 H inductor carries a steady current of 0.68 A. When the switch in the circuit is thrown open, the current disappears in
13.6 ms.
What is the average induced emf in the
inductor during this time?
Correct answer: 113.
Explanation:
Let :
L = 2.26 H ,
I = 0.68 A , and
t = 13.6 ms .
What is the magnitude of the electric field
at the center of the circle along which the arc
lies?
Correct answer: 47.186.
Explanation:
Let : λ = −8 nC/m = −8 × 10−9 C/m ,
∆θ = 98◦ , and
r = 2.3 m .
θ is defined as the angle in the counterclockwise direction from the positive x axis as
shown in the figure below.
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
◦
49 ◦
r
49
θ
~
E
First, position the arc symmetrically
around the y axis, centered at the origin. By
symmetry (in this rotated configuration) the
field in the x direction cancels due to charge
from opposites sides of the y-axis, so
Ex = 0 .
For a continuous linear charge distribution,
Z
dq
~
E = ke
r̂ .
r2
In polar coordinates
dq = λ (r dθ) ,
E = −2 (−31.261 N/C)
× [cos(41◦ ) − cos 90◦ ] ̂
= (47.186 N/C) ̂
~ = 47.186 N/C .
kEk
~ in a
Alternate Solution: Solve for kEk
straightforward manner, positioning the beginning of the arc on the positive x axis (as
shown in the original figure in the question).
θ is still defined as the angle in the counterclockwise direction from the positive x axis.
Ex = −
ke λ
r
Note: By symmetry, each half of the arc
about the y axis contributes equally to the
electric field at the origin. Hence, we may just
consider the right-half of the arc (beginning
on the positive y axis and extending towards
the positive x axis) and double that result.
Note: The upper angular limit θ = 90◦ .
The lower angular limit θ = 90◦ − 49◦ = 41◦ ,
is the angle from the positive x axis to the
right-hand end of the arc.
!
Z ◦
λ 90
sin θ dθ ̂
E = −2 ke
r 41◦
λ
= −2 ke [cos (41◦ ) − cos 90◦ ] ̂ .
r
Since
λ
ke = (8.98755 × 109 N · m2 /C2 )
r
−8 × 10−9 C/m
×
2.3 m
= −31.261 N/C ,
Z
98◦
Z
98 ◦
cos θ dθ
0◦
!
ı̂
ke λ
(sin 98◦ − sin 0◦ )ı̂
r
= −(−31.261 N/C) (sin 98◦ − sin 0◦ )ı̂
= (30.9568 N/C) ı̂ ,
=−
where λ is the linear charge density. The
positive y axis is θ = 90◦ , so the y component
of the electric field is given by
dEy = dE sin θ .
6
Ey = −
ke λ
r
sin θ dθ
0◦
!
̂
ke λ
(cos 0◦ − cos 98◦ ) ̂
r
= −(−31.261 N/C) (cos 0◦ − cos 98◦ ) ̂
= (35.6118 N/C) ̂ ,
=−
q
~ = E2 + E2
kEk
x
y
h
= (30.9568 N/C)2
+ (35.6118 N/C)
= 47.186 N/C .
2
i1/2
Question 6, chap 31, sect 4.
part 1 of 1
10 points
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
R
The distance from the slits to the screen is
L.
r
θ
y
S1
I
x
θ
d
S2
Inside coil has N turns
Outside solenoid has n turns per meter
What is the mutual inductance M between
the solenoid and the coil?
1. M = µ0 n N π r 2 sin θ
L
At the fifth dark fringe on the screen (position y on the screen), δ (the corresponding
path length difference) and φ (the phase angle
difference) is given by
13
λ
2
2. M = µ0 n N π r 2 cos θ
1. δ =
3. M = µ0 n N π R2 cos θ correct
2. δ = 6 λ and φ = 12 π .
4. M = µ0 n N π R2 sin θ
3. δ =
π r2
5. M = µ0 n N
2
4. δ = 4 λ and φ = 8 π .
6. M = µ0 n N π R2
5. δ =
π R2
2
π r2
8. M = µ0 n N
cos θ
2
π r2
9. M = µ0 n N
sin θ
2
6. δ = 7 λ and φ = 14 π .
7. M = µ0 n N
10. M = µ0 n N π r 2
Explanation:
The mutual inductance of the coil satisfies
N ΦB
, which
M I = N ΦB . Thus, M =
I
yields,
M = µ0 n N π R2 cos θ .
Question 7, chap 35, sect 3.
part 1 of 1
10 points
A screen is illuminated by monochromatic
light whose wave lenght is λ, as shown in the
figure below.
viewing
screen
A long solenoid has a tilted coil inside it
made of fine wire, as shown in the figure
below.
7
3
λ
2
5
λ
2
and φ = 13 π .
and φ = 3 π .
and φ = 5 π .
9
λ and φ = 9 π . correct
2
11
8. δ =
λ and φ = 11 π .
2
7. δ =
9. δ = 5 λ and φ = 10 π .
10. δ = 3 λ and φ = 6 π .
Explanation:
Basic Concepts: For bright fringes, we
have
δ = d sin θ = m λ ,
and for dark fringes, we have
1
δ = d sin θ = m +
λ,
2
where m = 0 , ±1 , ±2 , ±3 , · · · .
Solution:
1
λ
δ = m+
2
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
=
Since
9
λ .
2
8. 2 Ee−1
φ
δ
=
, we have
λ
2π
2πδ
φ=
λ
1
2π m +
λ
2
=
λ
= 2 π (4.5)
9. 0.5 Ee−2
10. 1.5 Ee−1.5
Explanation:
When S is at a, the loop equation is given
by
q
− i (R1 + R2 ) = 0
E−
C
This gives
= 9π .
i=
Question 8, chap 28, sect 7.
part 1 of 1
10 points
−t
−t
E
E
e (R1 +R2 )C =
e2RC
R1 + R2
2R
At t = RC,
i=
R3
C
R2
R1
S b
E
a
Let
R1 = R2 = R3 = R .
Initially the switch is in position “b” and the
capacitor is uncharged. At time t = 0 the
switch is moved to position “a”.
Determine the potential difference across
the resistor R1 at t = R1 C.
1. 0.5 Ee−0.5 correct
2. 1 Ee−1
−1
3. 0.5 Ee
4. 1 Ee−0.5
5. 2 Ee−2
6. 1 Ee−2
7. 2 Ee−0.5
8
E −0.5
e
2R
Thus the corresponding potential across R1 is
E −0.5
iR =
e
.
2
Question 9, chap 32, sect 4.
part 1 of 1
10 points
A 65 µF capacitor is connected in series
with a 30 mH inductance and a switch. The
capacitor is first charged to a voltage of 330 V.
The charging battery is then removed. As
soon as the switch is closed, the current begins to oscillate back and forth between one
direction and the reversed direction.
What is the maximum current in the circuit?
Correct answer: 15.3607.
Explanation:
Let :C = 65 µF = 6.5 × 10−5 F ,
V = 330 V , and
L = 30 mH = 0.03 H .
The initial charge is
Qmax = C V
= (6.5 × 10−5 F) (330 V)
= 0.02145 C .
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
p
n1 = 1.63
Qmax
Imax = ω Qmax = √
LC
0.02145 C
=p
(0.03 H)(6.5 × 10−5 F)
p
q
R
d
= 15.3607 A .
Derivation of Imax = ω Qmax :
1. Initially, there is an electric field energy
stored in the capacitor,
UC =
1 Q2max
.
2 C
When the current achieves its maximum, all
the energy in the system is stored as magnetic
energy in the inductor,
1
2
UL = L Imax
.
2
From the conservation of energy, UL = UC ,
we get
Q2
2
L Imax
= max .
C
Therefore
Qmax
Imax = √
.
LC
2.
Q = Qmax cos ω t .
I=
dI
= −ω Qmax sin ω t .
dt
Therefore
Imax = ω Qmax .
Question 10, chap -1, sect -1.
part 1 of 1
10 points
There is a tube of transparent material with
a refraction index n1 = 1.63 .
The tube is emersed in air with a refraction
n0 = 1 .
One end of the tube is convex and semispherical (with a radius of curvature of R =
5 m).
An object is on axis at a distance p = 9 m
from the rounded end of the tube.
n0 = 1
Therefore,
9
Figure: Drawn to scale.
How deep d does it appear to be when
viewed from the rounded end?
Note: The “apparent depth” in the present
context is a positive quantity.
Correct answer: 18.1452.
Explanation:
Let : n0 = 1 ,
n1 = 1.63 ,
p = 9 m , and
R = −5 m .
At the rounded end, we have
n1 n0
n0 − n1
+
=
p
q
R
n0
n0 n1 n1
=
−
−
, so
q
R
R
p
−1
n1
n1
1
−
−
q=
R R n0 p n0
1
(1.63)
=
−
(−5 m) (−5 m) (1)
(1.63)
−
(9 m) (1)
= (−0.2 m−1 ) − (0.326 m−1 )
= −18.1452 m ,
−1
−1
−(−0.181111 m−1 )
the apparent depth is
d = 18.1452 m .
Question 11, chap 28, sect 4.
part 1 of 1
10 points
Consider the combination of resistors
shown in the figure.
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
4.1 Ω
2.9 Ω
1.1 Ω
7.9 Ω
3.3 Ω
9.2 Ω
a
3.2 Ω
b
What is the resistance between point a and
point b?
Correct answer: 6.76245.
Explanation:
Let’s redraw the figure
i
R1
R2
i
R5
i
= 3.3
= 1.1
= 4.1
= 9.2
= 7.9
= 2.9
= 3.2
a
Ω,
Ω,
Ω,
Ω,
Ω,
Ω,
Ω.
and
Basic Concepts:
• Equivalent resistance.
• Ohm’s Law.
There are two rules for adding up resistances. If the resistances are in series, then
Rseries = R1 + R2 + R3 + · · · + Rn .
If the resistances are parallel, then
1
Rparallel
=
i
i
R1
R2
i
i
i
b
R5
R367 = R3 + R6 + R7
= (4.1 Ω) + (2.9 Ω) + (3.2 Ω)
= 10.2 Ω .
b
R4
Let : R1
R2
R3
R4
R5
R6
R7
a
R7
i
i
R3
i
i
Solution: The key to a complex arrangements of resistors like this is to split the problem up into smaller parts where either all the
resistors are in series, or all of them are in
parallel. It is easier to visualize the problem
if you redraw the circuit each time you add
them.
R367
R4
Step 1: The three resistors on the right are
all in series, so
R6
a
10
1
1
1
1
+
+
+···+
.
R1 R2 R3
Rn
i
i
R1
R2
i
i
b
R3675
R4
Step 2: R5 and R367 are connected parallel, so
−1
1
1
R3675 =
+
R5 R367
R5 R367
=
R5 + R367
(7.9 Ω) (10.2 Ω)
=
18.1 Ω
= 4.45193 Ω .
a
i
i
R1
R36752
i
R4
Step 3: R2 and R3675 are in series, so
R23675 = R2 + R3675
= (1.1 Ω) + (4.45193 Ω)
= 5.55193 Ω .
b
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
Step 4: R23675 and R4 are parallel, so
−1
1
1
R236754 =
+
R4 R23675
R4 R23675
=
R4 + R23675
(9.2 Ω) (5.55193 Ω)
=
14.7519 Ω
= 3.46245 Ω .
a
i
i
Question 13, chap 24, sect 1.
part 1 of 1
10 points
A (7.43 m by 7.43 m) square base pyramid
with height of 5.27 m is placed in a vertical
electric field of 80.4 N/C.
5.27 m
b
R1
R367524
Step 5: Finally, R1 and R236754 are in series, so the equivalent resistance of the circuit
is
Req = R1 + R236754
= 3.3 Ω + 3.46245 Ω
= 6.76245 Ω .
Question 12, chap -1, sect -1.
part 1 of 1
10 points
A glass sphere (index of refraction of 1.24)
of radius 18.7 cm has a tiny air bubble located
13.7 cm from the center. The sphere is viewed
in air along a direction parallel to the radius
containing the bubble, on the side where the
bubble is closest to the viewer.
What is the apparent depth of the bubble
below the surface of the sphere?
Note: The “apparent depth” in the present
context is a positive quantity.
Correct answer: 4.25232.
Explanation:
We can use refracitve index equation for
this system,
n1 n2
n2 − n1
+
=
p
q
R
Solving this for q, we find,
n2 R p
q=
p (n2 − n1 ) − n1 R
In this case we have, n1 = 1.24, n2 =
1.00, R = −18.7 cm , and p = (18.7 cm −
13.7 cm) cm. Then the apparent depth, |q|, is
|q| = 4.25232 cm .
11
b
7.43 m
80.4 N/C
Calculate the total electric flux which goes
out through the pyramid’s four slanted surfaces.
Correct answer: 4438.47.
Explanation:
Let : s = 7.43 m ,
h = 5.27 m , and
E = 80.4 N/C .
~ · A.
~ Since there is no
By Gauss’ law Φ = E
charge contained in the pyramid, the net flux
through the pyramid must be 0 N/C. Since
the field is vertical, the flux through the base
of the pyramid is equal and opposite to the
flux through the four sides. Thus we calculate
the flux through the base of the pyramid,
which is
Φ = E A = E s2
= (80.4 N/C) (7.43 m)2
= 4438.47 N m2/C .
Question 14, chap 29, sect 1.
part 1 of 1
10 points
A negatively charged particle moving at 45◦
angles to both the z-axis and x-axis enters a
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
magnetic field (pointing towards the bottom
of the page), as shown in the figure below.
x
~
B
y
z
v
~
B
−q
Figure: ı̂ is in the x-direction, ̂ is
in the y-direction, and k̂ is in the
z-direction.
12
~ = q ~v × B.
~
Solution: The force is F
~ = B (−ı̂) ,
B
1
~v = √ v −k̂ + ı̂ , and
2
q < 0 , therefore,
~ = −|q| ~v × B
~
F
h
i
1
= −|q| √ v B −k̂ + ı̂ × (−ı̂)
2
1
= −|q| √ v B (−̂)
2
b = −̂ .
F
This is the seventh of eight versions of the
problem.
What is the initial direction of deflection?
b = −̂ correct
1. F
b = √1
2. F
+̂ − k̂
2
1
b=√
3. F
−k̂ + ı̂
2
1
b = √ (+̂ − ı̂)
4. F
2
1
b
√
5. F =
+k̂ − ı̂
2
b = √1
6. F
−̂ + k̂
2
b
7. F = +̂
b = √1 (−̂ + ı̂)
8. F
2
~ = 0 ; no deflection
9. F
1 b
√
10. F =
−̂ − k̂
2
Explanation:
Basic Concepts: Magnetic Force on a
Charged Particle:
~ = q ~v × B
~
F
Right-hand rule for cross-products.
~
b ≡ F ; i.e., a unit vector in the F direcF
~k
kF
tion.
Question 15, chap 29, sect 4.
part 1 of 1
10 points
A solenoid 81.4 cm long has 580 turns and
a radius of 1.94 cm.
If it carries a current of 2.38 A, find the
magnetic field along the axis at its center.
Correct answer: 0.00213103.
Explanation:
Let : l
N
I
µ0
= 81.4 cm = 0.814 m ,
= 580 turns ,
= 2.38 A , and
= 1.25664 × 10−6 N/A2 .
Since l >> R, the magnetic field at the center
is approximately
B ≈ µ0 n I
N
= µ0 I
l
= (1.25664 × 10−6 N/A2 )
580 turns
(2.38 A)
×
0.814 m
= 0.00213103 T .
Question 16, chap 29, sect 5.
part 1 of 1
10 points
Consider two radial legs extending to infinity and a circular arc carrying a current I as
shown below.
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
13
Z
µ0 I
=
r dθ
y
4 π r2
Z 6π
6
7
µ0 I
π
dθ
=
7
4πr 0
6 π
µ0 I 7
=
θ
I
r
4 π r 0
I
I
6
µ0 I
π−0
=
4πr 7
x
O
I
3 µ0 I
=
.
What is the magnitude of the magnetic field
14 r
BO at the origin O due to the current through
this path?
Question 17, chap 29, sect 2.
part 1 of 1
10 points
5 µ0 I
1. BO =
28 r
A beam of protons moves along the x axis
1 µ0 I
in the positive x direction with a speed of
2. BO =
5 r
13.2 km/s through a region of crossed fields
3. BO = 0
4. BO =
5. BO =
6. BO =
7. BO =
8. BO =
9. BO =
10. BO =
1 µ0 I
7 r
3 µ0 I
correct
14 r
5 µ0 I
32 r
5 µ0 I
24 r
3 µ0 I
16 r
3 µ0 I
20 r
5 µ0 I
36 r
Explanation:
Using the Biot-Savart law, the magnetic
field due to the two radial legs is zero since
d~s × r̂ = 0 .
However, around the arc d~s×r̂ = d~s = r d~θ .
The magnetic field at at the center of an
arc with a current I is
Z
d~s × r̂
µ0 I
B=
4π
r2
Z
µ0 I
=
ds
4 π r2
balanced for zero deflecton.
If there is a magnetic field of magnitude
0.75 T in the positive y direction, find the
magnitude of the electric field.
Correct answer: 9.9.
Explanation:
Let : ~v = 13.2 km/s ı̂ = 13200 m/s ı̂
~ = 0.75 T ̂ .
B
and
The field strengths have equal magnitudes:
~ = q ~v B
~
qE
~ = ~v B
~
E
= (13200 m/s) ı̂ × (0.75 T) ̂
1 kV
×
1000 V
= −(9.9 kV/m) k̂ .
Question 18, chap 25, sect 1.
part 1 of 1
10 points
It takes 134 J of work to move 2.6 C of
charge from a positive plate to a negative
plate.
What voltage difference exists between the
plates?
Correct answer: 51.5385.
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
14
Explanation:
Question 20, chap 35, sect 3.
part 1 of 1
10 points
Let : W = 134 J and
q = 2.6 C .
Consider a double slit experiment as shown
in the sketch.
W
134 J
=
= 51.5385 V .
q
2.6 C
S1
An object is placed 10 cm to the left of
a diverging lens of focal length −7.4 cm. A
converging lens of focal length 11 cm is placed
a distance d to the right of the diverging lens.
Find the distance d that places the final
image at infinity.
Correct answer: 6.74713.
Explanation:
Given : p1 = 10 cm ,
f1 = −7.4 cm ,
f2 = 11 cm .
and
Applying the thin lens equation to the first
lens gives
p1 f1
p1 − f1
(10 cm) (−7.4 cm)
=
10 cm − (−7.4 cm)
= −4.25287 cm
q1 =
The virtual image formed by the first lens
serves as the object for the second lens. If
the image formed by that converging lens is
to be located at infinity, the image formed by
the diverging lens must be located at the focal
point in front of the converging lens. Thus
p2 = d + |q1 | = f2
d = f2 − |q1 |
= 11 cm − 4.25287 cm
= 6.74713 cm .
θ
d
Question 19, chap 34, sect 5.
part 1 of 1
10 points
viewing
screen
V =
y
The voltage difference is
S2
δ
θ
L
Denote the intensity for zero path difference
to be I0 .
I
at the path difFind the intensity ratio
I0
λ
ference δ = .
6
I
1
1.
=√
I0
2
√
I
3
2.
=
I0
2
I
1
3.
=
I0
4
I
2
4.
=√
I0
3
√
3
I
=
5.
I0
4
I
1
6.
=
I0
2
I
3
7.
= correct
I0
4
I
=1
8.
I0
√
I
9.
= 3
I0
1
I
=√
10.
I0
3
Explanation:
λ
The phase angle difference for δ = is
6
φ = kδ =
2π λ
π
=
λ 6
3
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
⇒
φ
2 π/3
I = I0 cos
= I0 cos
2
2
!2
√
3
2 π
= I0
= I0 cos
6
2
2
=
3
I0 .
4
Question 22, chap 30, sect 3.
part 1 of 1
10 points
A rigid, circular loop of radius R and mass
m carries a current I and lies in the xy plane
on a rough flat table. There is a horizontal
magnetic field of magnitude B.
What is the minimum value of B so that
one edge of the loop will lift off the table?
Question 21, chap 30, sect 3.
part 1 of 1
10 points
1. B =
A circular coil consisting of a single loop
of wire has a radius of 23.4 cm and carries a
current of 22.1 A. It is placed in an external
magnetic field of 0.234 T.
Find the magnitude of the torque on the
wire when the plane of the coil makes an
angle of 27.9◦ with the direction of the field.
Correct answer: 0.786189.
2. B =
Explanation:
Let : r
B
I
θ
= 23.4 cm = 0.234 m ,
= 0.234 T ,
= 22.1 A , and
= 27.9◦ .
The area of the circular coil is
A = π r2
= π (0.234 m)2
= 0.172021 m2 .
15
3. B =
4. B =
5. B =
mgR
Iπ
mg
IR
mg
I π R2
2mg
IπR
mg
correct
IπR
Explanation:
The loop will start to lift off the table when
the magnetic torque equals the gravitational
torque:
τmag = τgrav
µB = mgR
I π R2 B = m g R ,
mg
B=
.
IπR
The torque is
~ = µ B sin θ ,
k~τ k = k~µ × Bk
where θ is the angle that the normal to the
loop makes with the magnetic field, and µ =
I A is the magnetic moment of the loop.
Thus the magnitude of the torque is
τ = I A B sin θ
= (22.1 A) (0.172021 m2 ) (0.234 T)
× sin(90◦ − 27.9◦ )
= 0.786189 N · m .
Question 23, chap 26, sect 3.
part 1 of 1
10 points
Consider the two cases shown below. In
Case One two identical capacitors are connected to a battery with emf V . In Case Two,
a dielectric slab with dielectric constant κ fills
the gap of capacitor C2 . Let C be the resultant capacitance for Case One and C ′ the
resultant capacitance for Case Two.
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
Case One
C1
C2
16
Explanation:
Let :
C1 = C2 = C and
C2′ = κ C2 = κ C ,
where κ is dielectric constant.
V = constant. C1 and C2 are in series, so
V
Case Two
C1
C′ 2
κ
1
1
1
C2 + C1
=
+
=
C12
C1 C2
C1 C2
C1 C2
C12 =
.
C1 + C2
For Case One,
C12 =
C2
C
C1 C2
=
= .
C1 + C2
2C
2
For Case Two,
V
C′
The ratio 12 of the resultant capacitances is
C12
′
C
1+κ
1. 12 =
.
C12
2κ
C′
2. 12 = 1 + κ .
C12
C′
3. 12 = 2 κ .
C12
κ
C′
4. 12 = .
C12
2
′
1+κ
C
.
5. 12 =
C12
2
C′
6. 12 = κ .
C12
7. None of these
′
C12
1
= .
C12
κ
′
2κ
C
. correct
9. 12 =
C12
1+κ
′
2
C12
=
.
10.
C12
1+κ
8.
′
C12
=
Therefore
C1 C2′
κ C2
κC
=
=
.
′
C1 + C2
(1 + κ) C
1+κ
′
2κ
C12
=
.
C12
1+κ
Question 24, chap 24, sect 3.
part 1 of 1
10 points
A uniformly charged conducting plate with
area A has a total charge Q which is positive.
The figure below shows a cross-sectional view
of the plane and the electric field lines due to
the charge on the plane. The figure is not
drawn to scale.
+Q
+
E
E
+
+
+
+
+
+ P
+
+
+
+
Find the magnitude of the field at point P ,
which is a distance a from the plate. Assume
that a is very small when compared to the
dimensions of the plate, such that edge effects
can be ignored.
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
17
vertices of the triangle are equal. A fourth
charge 7 µC is placed below the triangle on
its symmetry-axis, and experiences a zero net
force from the other three charges, as shown
in the figure below.
−4.4 µC
~ = 4 π ǫ0 a2 Q
1. kEk
~ = ǫ0 Q A
2. kEk
Q
4 π ǫ0 a
Q
~ =
4. kEk
4 π ǫ0 a2
~ = 4 π ǫ0 a Q
5. kEk
~ =
3. kEk
2. 7
m
~ = ǫ0 Q a2
6. kEk
Q
correct
2 ǫ0 A
Explanation:
Consider the Gaussian surface shown in the
figure.
+Q
+
+
+
+
E
E
+
+
+
S
+
+
+
+
~ =
9. kEk
1 Q
S,
ǫ0 A
Q
.
E=
2 ǫ0 A
ΦTOTAL = 2 E S =
7 µC
Find q .
Correct answer: 0.86295.
Explanation:
Let : Q1 = −4.4 µC ,
Q2 = q ,
Q3 = q ,
Q4 = 7 µC ,
a = 2.7 m , and
d = −1.2 m .
Q1
a
Due to the symmetry of the problem, there
is an electric flux only through the right and
left surfaces and these two are equal. If the
cross section of the surface is S, then Gauss’
Law states that
q
1.2 m
q
~ = Q
7. kEk
ǫ0 A
~ = 2 ǫ0 Q A
8. kEk
so
Question 25, chap 22, sect 2.
part 1 of 1
10 points
Three point charges are located at the vertices of an equilateral triangle. The charge
at the top vertex of the triangle is given in
the figure. The two charges q at the bottom
Q2
θ
Q3
d
7 µC
The force on Q4 is due to the Coulomb
forces from Q1 , Q2 , and Q3 . Because Q2 and
Q3 have equal charge, the x-components of
their forces cancel out (by symmetry). Thus
we only need to consider the y-components of
the forces.
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
Qi Q4
is the
r2
ith force on Q4 from the ith charge. The forces
from Q2 and Q3 are equal to each other, and
opposite of the direction of the force from Q1 ,
since otherwise they could not cancel. The
total force on Q4 is
Coulomb’s law tells us Fi = k
FQ4 = k
Q2 Q4
Q1 Q4
+2k 2
sin θ
2
r14
r24
with r14 the distance between Q4 and Q1 ,
r24 = r34 the distance between Q4 and either
Q2 or Q3 , and θ indicated in the sketch above.
Remember that this force FQ4 will be set equal
to zero since the problem tells us the forces
are in equilibrium.
Because Q1 , Q2 , and Q3 form an equilateral
triangle with sides of length a, it can be seen
that
!2
!2
√
√
3
3 a + 2 |d|
2
r14
=
,
a + |d| =
2
2
a2
a 2 + 4 d2
+ d2 =
, and
4
4
2 |d|
|d|
.
=√
sin θ =
r24
a 2 + 4 d2
Thus our force equation becomes
"
4 Q1
0 = k Q4
√
2
3 a + 2 |d|
4 Q2
2 |d|
+2 · 2
·√
a + 4 d2
4 d2 + a 2
−4 Q1
Q2 = √
2
3 a + 2 |d|
√
4 d2 + a 2
a 2 + 4 d2
×
16 |d|
3/2
a 2 + 4 d2
−Q1
·
=
4 |d| √3 a + 2 |d|2
2
r24
=
=
−(−4.4 µC)
4 |−1.2 m|
3/2
(2.7 m)2 + 4 (−1.2 m)2
× √
2
3 (2.7 m) + 2 |−1.2 m|
= 0.86295 µC .
18
Note: Neither the sign nor the magnitude
of the charge Q4 (given in the problem as
7 µC) enters into this equation.
Actually, “the resultant force on Q4 is zero”
means that the resultant electric field is zero.
Because the electric filed is independent of the
test charge Q4 , the answer is also independent
of the sign or magnitude of the charge Q4 .
Question 26, chap 26, sect 1.
part 1 of 1
10 points
A coaxial cable with length l of 45 m has
an inner conductor that has a radius a of
1.789 mm and carries a charge of 7.52 µC. The
surrounding conductor has an inner radius b
of 9.557 mm and a charge of −7.52 µC.
Assume the region between the conductors is air. The permittivity of a vacuum
is 8.85419 × 10−12 C2 /N · m2 .
What
field at the halfway
is the electric
a+b
point r =
of the conductors?
2
Correct answer: 529498.
Explanation:
Let :
ǫ0 = 8.85419 × 10−12 C2 /N · m2 ,
Q = 7.52 µC = 7.52 × 10−6 C ,
a = 1.789 mm = 0.001789 m ,
b = 9.557 mm = 0.009557 m , and
l = 45 m .
The halfway point is
a+b
2
0.001789 m + 0.009557 m
=
2
= 0.005673 m .
r=
The electric field of a cylindrical capacitor is
given by
λ
2 π r ǫ0
Q
=
l 2 π r ǫ0
7.52 × 10−6 C
=
(45 m) 2 π (0.005673 m)
E=
final 01 – JYOTHINDRAN, VISHNU – Due: Dec 15 2007, 6:00 pm
×
19
1
8.85419 ×
10−12
= 529498 V/m .
C2 /N · m2
10. Qr3 = −q1 − q2 − q3
Question 27, chap 24, sect 5.
part 1 of 1
10 points
A point charge q1 is concentric with two
spherical conducting thick shells, as shown
in the figure below. The smaller spherical
conducting shell has a net charge of q2 and
the larger spherical conducting shell has a net
charge of q3 .
r4
q3
r3
r2
r1
q2
q1
R1
R2 R3 R4 R5
What is the charge Qr3 on the inner surface of the larger spherical conducting shell?
Under static conditions, the charge on a conductor resides on the surface of the conductor.
1. Qr3 = −q1 − q2 correct
2. Qr3 = −q1 + q2
3. Qr3 = −q1 − q2 + q3
4. Qr3 = +q1 + q2
5. Qr3 = +q1
6. Qr3 = 0
7. Qr3 = −q1
8. Qr3 = +q1 − q2
9. Qr3 = +q1 + q2 + q3
Explanation:
The net charge inside a Gaussian surface
located at r = R4 must be zero, since the field
in the conductor must be zero. Therefore,
the charge on the inner surface of the large
spherical conducting shell must be −(q1 + q2 ).