midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 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 ,
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
Complete reflection: P = 2cU ,
Faraday’s law
φB
E = −N ddt
,
φB =
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
Maxwell equations:
H
~ · dA
~ = Q,
E
ǫ0
H
φB
~
,
E · d~s = − ddt
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
R
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
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
3
.
Flow Diagram for iCliker test Mode
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from your liker, and memorize your box number.
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ode using your liker. DO NOT PRESS THE \E" (enter) BUTTON YET.
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ther
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enter: E
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
The rms voltage across the transformer’s
secondary is
Question 1, chap 33, sect 3.
part 1 of 1
10 points
An electromagnetic wave in vacuum has an
electric field amplitude of 270 V/m.
The speed of light is 2.99792 × 108 m/s.
Calculate the amplitude of the corresponding magnetic field.
Correct answer: 9.00623 × 10−7 .
Explanation:
N2
V1
N1
24 turns
(125 Vrms )
=
30 turns
= 100 Vrms ,
V2 =
which is the same as the electric potential
across the load resistor.
Question 3, chap 31, sect 2.
part 1 of 1
10 points
Let : E = 270 V/m and
c = 2.99792 × 108 m/s .
For a plane electromagnetic wave, Maxwell’s
~ = kEk/c,
~
equations require that kBk
so
B=
4
E
270 V/m
=
c
2.99792 × 108 m/s
A circular coil enclosing an area of 102 cm2
is made of 236 turns of copper wire as shown
schematically in the figure. Initially, a 0.997 T
uniform magnetic field points perpendicularly
left-to-right through the plane of the coil. The
direction of the field then reverses to right-toleft. The field reversal takes 1.0 ms.
= 9.00623 × 10−7 T .
Question 2, chap 32, sect 6.
part 1 of 1
10 points
53 Ω
24 turns
30 turns
125 Vrms
A ideal transformer shown in the figure
below having a primary with 30 turns and
secondary with 24 turns.
The load resistor is 53 Ω.
The source voltage is 125 Vrms .
What is the rms electric potential across
the 53 Ω load resistor?
Correct answer: 100.
Magnetic
Field B(t)
R
During the time the field is changing its
direction, how much charge flows through the
coil if the resistance is 5.04 Ω?
Correct answer: 0.952372.
Explanation:
From Faraday’s Law for Solenoids
E = −N
and Ohm’s Law
I=
Explanation:
V
,
R
the current through R is
Let : N1 = 30 turns ,
N2 = 24 turns , and
V1 = 125 Vrms .
dΦB
dt
V
R
E
=
R
I=
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
N d ΦB
R dt
dBA 1
=N
dt R
dB A
.
=N
dt R
5
=
?
Spole
y
Integrating both sides of the equation above
yields
Z
t
Z
I dt =
t
N
t0
B
t0
Z
=
−B
x
v
A
dB
R
A
∆B
R
A
= N 2B.
R
=N
The left hand side of the above equation is
just the charge flowing through the R during
this period of time! So,
S
dipole
magnet
N
Npole
?
Ibelow
z
A dB
dt
R dt
N
Iabove
Determine the directions of the induced
currents Iabove and Ibelow in an imaginary loop
shown in the figure, as viewed from above,
when the loop is above the falling magnet and
when the loop is below the falling magnet.
1. Iabove = clockwise and
Ibelow = clockwise
2. Iabove = counter-clockwise and
Ibelow = counter-clockwise
Q=
Z
t
I dt
t0
A
2B
R
= 0.952372 C .
=N
3. Iabove = clockwise and
Ibelow = counter-clockwise correct
4. Iabove = counter-clockwise
Ibelow = clockwise
and
5. no current flow
Question 4, chap 31, sect 3.
part 1 of 1
10 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.
Explanation:
When the falling magnet is below the upper
→
loop, −
µ ind must be down to attract the falling
magnet and slow it down; i.e., clockwise as
viewed from above.
→
Before reaching the lower loop, −
µ ind must
be up to oppose the falling magnet; i.e.,
counter-clockwise as viewed from above.
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
B
6
B
+++++
Iabove
3.
−−−−−
Spole
y
x
S
v
z
N
v
B
B
dipole
magnet
Npole
B
B
−−−−−
4.
v
+++++
Ibelow
B
B
B
Question 5, chap 31, sect 3.
part 1 of 1
10 points
B
+
+
+
+
+
5.
A square piece of copper is pulled through
a magnetic field B (into the page ⊗, out of
the page ⊙). Shown below are different charge
configurations associated with this procedure.
Select the figure with an acceptable charge
distribution.
−
−
−
−
−
v
B
B
B
B
−−−−−
6.
B
v
B
+++++
+++++
1.
B
v
−−−−−
B
B
correct
B
2.
B
Explanation:
Using the right-hand-rule, the only acceptable charge distribution is
B
B
−
−
−
−
−
v
B
+
+
+
+
+
B
+++++
v
−−−−−
B
B
B
Question 6, chap 33, sect 3.
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
part 1 of 1
7
10 points
A microwave that has a wavelength of
4.1 cm.
The speed of light is 2.99792 × 108 m/s.
What is its frequency?
Correct answer: 7.31201 × 109 .
Explanation:
Question 8, chap 33, sect 3.
part 1 of 1
10 points
An unpolarized light beam with intensity
of I0 passes through 2 polarizers shown in the
picture.
41◦
71◦
Let : λ = 4.1 cm = 0.041 m and
c = 2.99792 × 108 m/s .
I0
I
f=
108
c
2.99792 ×
m/s
=
λ
0.041 m
= 7.31201 × 109 Hz .
Question 7, chap 32, sect 2.
part 1 of 1
10 points
A certain capacitor in a circuit has a capacitive reactance of 33 Ω when the frequency is
122 Hz.
What capacitive reactance does the capacitor have at a frequency of 5040 Hz?
Correct answer: 0.79881.
Explanation:
Let :
XC = 33 Ω ,
flow = 122 Hz , and
fhigh = 5040 Hz .
The capacitive reactance is XC =
C
, so
2πf
XC(high)
2 π flow C
flow
=
=
.
XC(low)
2 π fhigh C
fhigh
Thus
flow
XC(high) = XC(low)
fhigh
122 Hz
= (33 Ω)
5040 Hz
= 0.79881 Ω .
If the first polarizer is rotated at an angle
of 41 ◦ with respect to vertical and the second at an angle 71◦ with respect to vertical,
what is the beam intensity I after the second
polarizer?
5
I0
8
1
2. I =
I0
16
3
3. I = I0 correct
8
1
4. I = I0
8
7
5. I =
I0
16
9
I0
6. I =
16
1
7. I = I0
4
3
8. I =
I0
16
1
9. I = I0
2
5
I0
10. I =
16
Explanation:
1. I =
Let : θ1 = 41◦ and
θ2 = 71◦ .
The beam intensity after the first polarizer
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
is
I0
I1 =
,
2
since the initial beam is unpolarized.
We use the formula for the intensity of the
transmitted (polarized) light. Thus the beam
intensity after the second polarizer is
I = I1 cos2 (∆θ)
= I1 cos2 (θ2 − θ1 )
= I1 cos2 (71◦ − 41◦ )
I0
=
cos2 (30◦ )
2
√ !2
3
I0
=
2
2
3 I0
=
.
8
Question 9, chap 31, sect 3.
part 1 of 2
10 points
A horizontal circular wire loop of radius
0.5 m lies in a plane perpendicular to a uniform magnetic field pointing from above into
the plane of the loop, has a magnitude of
0.4 T.
If in 0.05 s the wire is reshaped from a circle
into a square, but remains in the same plane,
what is the magnitude of the average induced
emf in the wire during this time?
Correct answer: 1.34838.
Explanation:
Let : r = 0.5 m ,
b = 0.4 T , and
∆t = 0.05 s .
The average induced emf E is given by
∆Φ
∆Φ
hEi = N
=
∆t
∆t
since N = 1, and
∆Φ = B (Acircle − Asquare )
= B (π r 2 − Asquare ) .
8
Also, the circumference of the circle is 2 π r,
so each side of the square has a length
L=
2πr
πr
=
,
4
2
so
2
Asquare = L =
Thus
2
∆Φ = B π r −
"
π r 2
π r 2 2
.
2
= 0.4 T π (0.5 m)2 −
π (0.5 m) 2
2
!#
= −0.0674191 T · m2 .
and the average induced emf is
hEi = −
−0.0674191 T · m2
= 1.34838 V .
0.05 s
Question 10, chap 31, sect 3.
part 2 of 2
10 points
The current in the loop during the deformation
1. does not arise.
2. flows clockwise when viewed from above.
correct
3. flows in a direction that cannot be determined from given information.
4. flows counter-clockwise when viewed from
above.
Explanation:
The deformation causes the flux through
the loop to decrease since the area of the loop
is reduced. By Lenz’s law, the induced emf
will cause the current to flow in the loop so
as to induce a magnetic field that attempts
to resist the change of magnetic flux through
the loop. A clockwise flow of current, when
viewed from above tends to increase the existing downward magnetic field through the
loop, thereby resisting the decrease of magnetic flux through the loop.
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
9
Question 11, chap 30, sect 3.
part 1 of 1
10 points
Question 12, chap 31, sect 2.
part 1 of 1
10 points
A small rectangular coil composed of
51 turns of wire has an area of 36 cm2 and
carries a current of 2.4 A. When the plane
of the coil makes an angle of 29◦ with a uniform magnetic field, the torque on the coil is
0.06 N m.
What is the magnitude of the magnetic
field?
Correct answer: 0.155685.
A coil with N1 = 16 turns and radius r1 =
15.1 cm surrounds a long solenoid of radius
N2
= n2 = 1600 m−1 (see
r2 = 1.98 cm and
ℓ2
the figure below). The current in the solenoid
changes as I = I0 sin(ω t), where I0 = 5 A
and ω = 120 rad/s.
ℓ2
ℓ1
Explanation:
Let : N = 51 turns ,
I = 2.4 A ,
θ = 29◦ ,
A = 36 cm2 = 0.0036 m2 ,
τ = 0.06 N m .
A1
and
The magnetic force on the current is
~ = I ~ℓ × B
~
F
and the torque is
~,
~τ = ~r × F
so the torque on the loop due to the magnetic
field is
τ = 2 F r cos θ
= (N I ℓ B) w cos θ
= N I B (ℓ w) cos θ
= N I B A cos θ ,
where A is the area of the loop and θ is the
angle between the plane of the loop and the
magnetic field.
The magnetic field from above is
B=
=
τ
N I A cos θ
0.06 N m
(51 turns) (2.4 A) (0.0036 m2 ) cos(29◦ )
= 0.155685 T .
a
R
b
A2
E
Inside solenoid has N2 turns
Outside solenoid has N1 turns
What is the magnitude of the induced emf,
EAB , across the 16 turn coil at t = 1000 s?
Correct answer: 0.0198143.
Explanation:
Faraday’s Law for solenoid
dΦ
dt
d (BA)
.
= −N
dt
E = −N
Magnetic field induced by solenoid
B = µ0 n I .
Faraday’s Law for solenoid
dΦ
dt
d (BA)
.
= −N
dt
E = −N
Magnetic field induced by inner solenoid
B = µ0 n I .
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
So the induced emf is
dΦ
dt
d (B A2 )
= −N1
dt
dB
= −N1 A2
dt
d (µ0 n2 I)
= −N1 A2
dt
dI
= −µ0 N1 n2 A2
dt
= 0.0198143 V .
Question 13, chap 30, sect 5.
part 1 of 1
10 points
A copper strip (8.47 × 1022 electrons per
cubic centimeter) 14 cm wide and 0.12 cm
thick is used to measure the magnitudes of
unknown magnetic fields that are perpendicular to the strip.
The charge on the electron is 1.6 × 10−19 C.
Find the magnitude of B when the current
is 20 A and the Hall voltage is 2.6 µV.
Correct answer: 2.11411.
Explanation:
= 2.11411 T .
Question 14, chap 32, sect 5.
part 1 of 1
10 points
In a series RLC ac circuit, the resistance is
22 Ω, the inductance is 38 mH, and the capacitance is 16 µF. The maximum potential is
197 V, and the angular frequency is 100 rad/s.
Calculate the maximum current in the circuit.
Correct answer: 0.316929.
Explanation:
Let : R = 22 Ω ,
L = 38 mH = 0.038 H ,
C = 16 µF = 1.6 × 10−5 F ,
Vmax = 197 V , and
ω = 100 rad/s .
The capacitive reactance is
22
−3
Let : n = 8.47 × 10 cm ,
= 8.47 × 1028 m−3 ,
q = 1.6 × 10−19 C ,
t = 0.12 cm = 0.0012 m ,
w = 14 cm = 0.14 m ,
I = 20 A , and
VH = 2.6 µV = 2.6 × 10−6 V .
The current in the metal strip is
I = n q vd A = n q vd (w t)
I
vd w =
nqt
VH = vd w B
VH
B=
vd w
n q t VH
I
(8.47 × 1028 m−3 ) (1.6 × 10−19 C)
=
20 A
× (0.0012 m) (2.6 × 10−6 V)
B=
E = −N1
The Hall voltage is
10
XC =
1
ωC
1
(100 rad/s) (1.6 × 10−5 F)
= 625 Ω .
=
The inductive reactance is
XL = ω L
= (100 rad/s) (0.038 H)
= 3.8 Ω .
The maximum current is
Vmax
Imax =
Z
Vmax
=p
2
R + (XL − XC )2
197 V
=p
(22 Ω)2 + (3.8 Ω − 625 Ω)2
= 0.316929 A .
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
Question 15, chap 31, sect 2.
part 1 of 1
10 points
1 µF
A coil is wrapped with 665 turns of wire on
the perimeter of a circular frame (of radius
31 cm). Each turn has the same area, equal
to that of the frame. A uniform magnetic
field is directed perpendicular to the plane of
the coil. This field changes at a constant rate
from 24 mT to 42 mT in 43 ms.
What is the magnitude of the induced E in
the coil at the instant the magnetic field has
a magnitude of 41 mT?
Correct answer: 84.0425.
Q
S
ΦB ≡
Z
Explanation:
C
Solution:
d ΦB
dt
∆B
= −N A
∆t
(B2 − B1 )
= −N π r 2
∆t
= −(665) π (31 cm)2
(42 mT) − (24 mT)
×
43 ms
= −84.0425 V
|E| = 84.0425 V .
E = −N
Let : C = 1 µF = 1 × 10−6 F ,
L = 3 H , and
Q0 = 31 µC = 3.1 × 10−5 C .
Basic concept: Oscillation in LC circuit:
q = qmax cos(ω t + δ)
dq
= −ω qmax sin(ω t + δ)
dt
1
2π
, where T is the period
=√
with ω =
T
LC
of oscillation.
Solution: When the switch closes, initially
there is zero current in the circuit. Thus the
solution has the correct phase at t = 0
I=
dQ
= −ω qmax sin(ω t) .
dt
T
Now at t = ,
4
2π T
×
I = −q0 ω sin
T
4
= −q0 ω
= −0.0178979 A .
I=
Question 16, chap 32, sect 4.
part 1 of 1
10 points
In the LC circuit below the capacitor is
charged to its maximum 31 µC while the
switch S is open. The switch is closed at
t = 0.
L
Q
S
d ΦB
dt
~ · dA
~ =B·A
B
3H
Find the energy stored in the inductor at
T
t = , where T is the period of circuit oscil4
lations.
Correct answer: 480.5.
Explanation:
Basic Concepts:
E = −N
11
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
12
and the maximum current is
Thus
1
L I2
2
1
= L (−q0 ω)2
2
= ULmax
1 2
= UCmax =
q
2C 0
1
(3.1 × 10−5 C)2
=
2 (1 × 10−6 F)
106 µJ
×
J
= 480.5 µJ .
UL =
T
Observe that at t = , the magnitude of
4
the current flow through the inductor is at a
maximum, and the charge q = qmax cos ω t is
T
π
zero at t =
=
. Thus, at this moment
4
2ω
there is no energy stored in the capacitor and
all the energy is stored in the inductor. At
any other time, one can show that for an
ideal LC circuit where there is no dissipation
from resistance, the total energy stored is the
sum of the energy in the capacitor and the
energy in the inductor, and this energy is
constant in time (although the energy in each
one oscillates).
Question 17, chap 32, sect 3.
part 1 of 1
10 points
√
2 Vrms
2 Vrms
=
XL
2πf L
The current at time t is
I = Imax sin(ω t)
√
2 Vrms
=
sin(2 π f t)
2πf L
√
2 (189 V)
=
2 π (49 Hz) (0.059 H)
× sin[(2 π (49 Hz) (0.0073 s)]
= 11.4722 A ,
so the potential energy stored in the inductor
is
1
L I2
2
1
= (0.059 H) (11.4722 A)2
2
= 3.88251 J .
U=
Question 18, chap 31, sect 1.
part 1 of 2
10 points
The figure below shows one of the blades
of a helicopter which rotates around a central
hub. The vertical component of the Earth’s
magnetic field is into the plane of the paper.
5
m
s
2. 3
v/
re
3 × 10−5 T
1.
A 59 mH inductor is connected to a outlet where the rms voltage is 189 V and the
frequency is 49 Hz.
Determine the energy stored in the inductor
at t = 7.3 ms, assuming that this energy is
zero at t = 0.
Correct answer: 3.88251.
Imax =
√
Explanation:
Let :
t = 7.3 ms = 0.0073 s ,
f = 49 Hz ,
L = 59 mH = 0.059 H ,
Vrms = 189 V .
The inductive reactance is
XL = ω L = 2 π f L
and
3 × 10−5 T
What is the magnitude of the emf E induced
between the blade tip and the central hub?
Correct answer: 0.000747856.
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
13
Explanation:
ω
ℓ
f
B
θ
ℓ
B
The area inside the triangle is A =
ℓ2 d θ
ω ℓ2
dA
=
=
.
and
dt
2 dt
2
Therefore,
Let : ℓ = 2.3 m ,
f = 1.5 rev/s , and
B = 3 × 10−5 T .
For a point on the blade, the velocity with
which the point moves changes linearly with
the distance from the point to the center of
the hub. Then the effective velocity for the
whole blade is the mean velocity,
ω·ℓ
2
2πf · ℓ
=
2
2π (1.5 rev/s) (2.3 m)
=
2
= 10.8385 m/s ,
veff =
and the induced emf in the blade is
E = B ℓ veff
1
= B ω ℓ2
2
1
= (3 × 10−5 T)
2
× (9.42478 rad/s) (2.3 m)2
= 0.000747856 V .
Alternative Solution: An alternative
dA
d ΦB
=B
, where
method is to calculate
dt
dt
A is the area enclosed in the diagram below
and ΦB the enclosed flux in the figure.
ℓ2
θ,
2
d ΦB
dt
dA
= −B
dt
ℓ2
= −B ω
2
1
|E| = B ω ℓ2 ,
2
E =−
the same result as obtained above.
Question 19, chap 31, sect 1.
part 2 of 2
10 points
The tip of the blade is
1. charged positive. correct
2. charged, but sign cannot be determined.
3. charged negative.
4. uncharged.
Explanation:
The magnetic force on a charge carrier of
~ where ~v is the local veloccharge q is q ~v × B
ity of the blade. If q is positive, a force (use
the cross product rule) is induced in the direction from the hub to the tip of the blade.
Hence, positive charges move to the tip (leaving negative charges at the hub).
Eventually an electric field induced by this
charge separation points from the positively
charged tip to the hub and the total force
balance between magnetic and electric forces
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
is established on the charge carriers in the
blade.
The net result is the tip is charged positively. If the charge q is negative, the magnetic force is towards the hub; negative charge
accumulates there, leaving positive charge at
the tip, with the same conclusion that the tip
is positively charged.
Question 20, chap 33, sect 5.
part 1 of 1
10 points
Consider an electromagnetic wave pattern
as shown in the figure below.
B
~ ×B
~ is along the negative z axis.
Then E
Therefore, the electromagnetic wave is
traveling right to left.
Question 21, chap 33, sect 5.
part 1 of 1
10 points
Radiation from a 316 W point isotropic
source falls normally on the surface of a perfect absorber which is 2 m away.
What is the radiation pressure?
Correct answer: 2.09699 × 10−8 .
Explanation:
Let :
E
The wave is
1. traveling left to right.
2. a standing wave and is stationary.
3. traveling right to left. correct
Explanation:
~ vector and B
~ vector are not at the
The E
same point on the velocity axis.
Pick an instant in time, where the B and
E fields are at the same point on the velocity
axis.
B
y
x
v
z
E
For instance, let us choose the point where
~ vector is along the y axis, as shown in
the B
~
the above figures. At this same instant, the E
vector is along the negative x axis (at a point
with a phase difference of 360◦ from the place
~ vector is
on the velocity (z) axis where the B
drawn).
14
P = 316 W
d = 2 m.
and
The intensity I at a distance d away from
P
, where
a point source is given by I =
4 π d2
P is the power of the source. The radiation
pressure at that distance is given by
Pressure =
=
=
I
c
P
4 π d2 c
316 W
4 π (2
m)2 (2.99792 × 108
m/s)
= 2.09699 × 10−8 Pa .
Question 22, chap 31, sect 1.
part 1 of 1
10 points
Given: Assume the bar and rails have negligible resistance and friction.
In the arrangement shown in the figure, the
resistor is 7 Ω and a 4 T magnetic field is
directed out of the paper. The separation
between the rails is 8 m . Neglect the mass of
the bar.
An applied force moves the bar to the left
at a constant speed of 5 m/s .
midterm 03 – JYOTHINDRAN, VISHNU – Due: Nov 15 2007, 11:00 pm
5 m/s
m≪1 g
7Ω
8m
I
4T
4T
At what rate is energy dissipated in the
resistor?
Correct answer: 3657.14.
Explanation:
Basic Concept: Motional E
E = Bℓv.
Ohm’s Law
V
.
R
Solution: The motional E induced in the
circuit is
I=
E = Bℓv
= (4 T) (8 m) (5 m/s)
= 160 V .
From Ohm’s law, the current flowing through
the resistor is
E
I=
R
Bℓv
=
R
(4 T) (8 m) (5 m/s)
=
R
= 22.8571 A .
The power dissipated in the resistor is
P = I2 R
B 2 ℓ2 v 2
=
R
R2
B 2 ℓ2 v 2
=
R
(4 T)2 (8 m)2 (5 m/s)2
=
(7 Ω)
= 3657.14 W .
Note: Third of four versions.
15