midterm 03 – SAENZ, LORENZO – Due: Apr 3 2008, 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 – SAENZ, LORENZO – Due: Apr 3 2008, 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 – SAENZ, LORENZO – Due: Apr 3 2008, 11:00 pm
3
Question 1, chap 31, sect 3.
part 1 of 1
10 points
Question 2, chap 31, sect 4.
part 1 of 1
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
0.37 T.
If in 0.04 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: 1.55907 (choice number 9).
A long solenoid carries a current 30 A.
Another coil (of larger diameter than the
solenoid) is coaxial with the center of the
solenoid, as in the figure below.
The permeability of free space is 4π ×
10−7 N/A2 .
7m
3m
16.1 cm
Explanation:
6.1 cm
Let : r = 0.5 m ,
b = 0.37 T ,
∆t = 0.04 s .
and
Inside solenoid has 7600 turns
Outside solenoid has 380 turns
Find the mutual inductance of the system.
Correct answer: 0.00606064 (choice number 7).
The average induced emf E is given by
|∆Φ|
|∆Φ|
hEi = N
=
∆t
∆t
Explanation:
since N = 1, and
ℓ2
ℓ1
|∆Φ| = B (Acircle − Asquare )
= B (π r 2 − Asquare ) .
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
Asquare = L2 =
Thus
2
|∆Φ| = B π r −
"
π r 2
2
π r 2 = 0.37 T π (0.5 m)2 −
= 0.0623627 T · m2 .
A2
Inside solenoid has N2 turns
Outside solenoid has N1 turns
.
2
A1
π (0.5 m)
2
2 #
and the average induced emf is
|0.0623627 T · m2 |
= 1.55907 V .
hEi =
0.04 s
Let : r1
r2
N1
N2
ℓ1
ℓ2
= 16.1 cm = 0.161 m ,
= 6.1 cm = 0.061 m ,
= 380 ,
= 7600 ,
= 3 m,
= 7 m,
N1
= 126.667 turns/meter,
n1 =
ℓ1
N1
n2 =
= 1085.71 turns/meter,
ℓ1
I2 = 30 A , and
midterm 03 – SAENZ, LORENZO – Due: Apr 3 2008, 11:00 pm
µ0 = 4π × 10−7 N/A2 .
M is the mutual inductance
A1 = πr1 2 = π(0.161 m)2 = 0.0814332 m2
A2 = πr2 2 = π(0.061 m)2 = 0.0116899 m2
N2 I2
B2 = µ0
ℓ2
The flux Φ12 through coil 1 due to coil 2 is
Φ12 = B2 A2 . Thus, the mutual inductance is
N2 Φ12
M=
I1
N1 Φ21
=
I2
N1 N2 A2 I2
= µ0
ℓ 2 I2
N1 N2 A2
= µ0
ℓ2
(380)(7600)
= (4π × 10−7 N/A2 )
7m
× (0.0116899 m2 )
4
The inductive reactance is
XL = ω L
= (100 rad/s) (0.03 H)
= 3 Ω.
The maximum current is
Imax =
Vmax
Z
Vmax
=p
2
R + (XL − XC )2
127 V
=p
(13 Ω)2 + (3 Ω − 434.783 Ω)2
= 0.293996 A .
= 0.00606064 H .
Question 3, chap 32, sect 5.
part 1 of 1
10 points
Question 4, chap 31, sect 1.
part 1 of 1
10 points
In a series RLC ac circuit, the resistance is
13 Ω, the inductance is 30 mH, and the capacitance is 23 µF. The maximum potential is
127 V, and the angular frequency is 100 rad/s.
Calculate the maximum current in the circuit.
Correct answer: 0.293996 (choice number
2).
Given: Assume the bar and rails have negligible resistance and friction.
In the arrangement shown in the figure,
the resistor is 3 Ω and a 6 T magnetic field
is directed into the paper. The separation
between the rails is 9 m . Neglect the mass of
the bar.
An applied force moves the bar to the right
at a constant speed of 9 m/s .
m≪1 g
6T
3Ω
Let : R = 13 Ω ,
L = 30 mH = 0.03 H ,
C = 23 µF = 2.3 × 10−5 F ,
Vmax = 127 V , and
ω = 100 rad/s .
The capacitive reactance is
1
XC =
ωC
1
=
(100 rad/s) (2.3 × 10−5 F)
= 434.783 Ω .
9m
I
Explanation:
9 m/s
6T
At what rate is energy dissipated in the
resistor?
Correct answer: 78732 (choice number 10).
midterm 03 – SAENZ, LORENZO – Due: Apr 3 2008, 11:00 pm
Explanation:
Basic Concept: Motional E
Explanation:
E = Bℓv.
Let : r = 6 cm = 0.06 m ,
ℓ = 27 cm = 0.27 m ,
N = 190 turns ,
I = 0.4 A , and
µ0 = 1.25664 × 10−6 N/A2 .
Ohm’s Law
I=
V
.
R
Solution: The motional E induced in the
circuit is
The inductance is
A
π r2
= µ0 N 2
ℓ
ℓ
−6
= (1.25664 × 10 N/A2 )
π (0.06 m)2
× (190 turns)2
0.27 m
= 0.00190023 H .
E = Bℓv
= (6 T) (9 m) (9 m/s)
= 486 V .
L = µ0 N 2
From Ohm’s law, the current flowing through
the resistor is
E
R
Bℓv
=
R
(6 T) (9 m) (9 m/s)
=
R
= 162 A .
I=
5
The energy stored in an inductor is
1
L I2
2
1
= (0.00190023 H)(0.4 A)2
2
= 0.000152018 J .
U=
The power dissipated in the resistor is
P = I2 R
B 2 ℓ2 v 2
R
=
R2
B 2 ℓ2 v 2
=
R
(6 T)2 (9 m)2 (9 m/s)2
=
(3 Ω)
= 78732 W .
Note: Second of four versions.
Question 5, chap 31, sect 5.
part 1 of 1
10 points
A device (“source”) emits a bunch of
charged ions (particles) with a range of velocities (see figure). Some of these ions pass
through the left slit and enter “Region I” in
which there is a vertical uniform electric field
(in the −̂ direction) and a B uniform magnetic field (aligned with the ±k̂-direction) as
shown in the figure by the shaded area.
+V
Region of
Magnetic
Field
B
q
m
y
r
An inductor of 190 turns has a radius of
6 cm and a length of 27 cm.
The permeability of free space is
1.25664 × 10−6 N/A2 .
Find the energy stored in it when the current is 0.4 A.
Correct answer: 0.000152018 (choice number 10).
Question 6, chap 30, sect 1.
part 1 of 1
10 points
d
x
z
Region I
Region II
midterm 03 – SAENZ, LORENZO – Due: Apr 3 2008, 11:00 pm
Br
E
B
q B2 r
=
E
Figure: ı̂ is in the direction +x
(to the right), ̂ is in the direction
+y (up the page), and k̂ is in the
direction +z (out of the page).
=q
The ions that make it into “Region II” are
observed to be deflected downward and then
follow a circular path with a radius of r.
The charge on each ion is q.
What is the mass of the ions?
qBr
1. m =
E2
qEr
2. m =
B2
q E2
3. m =
rB
qBr
4. m =
E
qE
5. m =
rB
qEr
6. m =
B
q E2 r
7. m =
B
q B2
8. m =
rE
qB
9. m =
rE
q B2 r
correct
10. m =
E
Explanation:
Since the electric and magnetic forces on
the ion are equal,
qE = qvB
E
v=
.
B
The radius of a circular path taken by a
charged particle in a magnetic field is given
by
mv
.
qB
Br
m=q
v
r=
6
Question 7, chap 30, sect 1.
part 1 of 1
10 points
A cosmic-ray proton in interstellar space
has an energy of 77 MeV and executes a circular orbit having a radius equal to that of Mercury’s orbit around the sun (5.8 × 1010 m).
The proton has a charge of 1.60218 ×
10−19 C and a mass of 1.67262 × 10−27 kg.
What is the magnetic field in that region of
space?
Correct answer: 2.18601 × 10−11 (choice
number 9).
Explanation:
Let : q = 1.60218 × 10−19 C ,
m = 1.67262 × 10−27 kg ,
r = 5.8 × 1010 m , and
K = 77 MeV = 7.7 × 107 eV .
From the kinetic energy K, we get the speed
of the proton is
r
2K
v=
.
m
The centripetal force is
m v2
qvB =
,
r
so the field is
r
m 2K
mv
=
B=
qr
qr
m
1 √
2mK .
=
qr
=
10−19
1
C) (5.8 × 1010 m)
(1.60218 ×
q
× 2 (1.67262 × 10−27 kg) (7.7 × 107 eV)
= 2.18601 × 10−11 T .
midterm 03 – SAENZ, LORENZO – Due: Apr 3 2008, 11:00 pm
Question 8, chap 31, sect 6.
part 1 of 1
10 points
time, move the switch from “a” to “b”. There
will be current oscillations.
L
A large electromagnet has an inductance of
60 H and a resistance of 8 Ω. It is connected
to a DC power source of 400 V.
Find the time for the current to reach 20 A.
Correct answer: 3.83119 (choice number 2).
C
E
Explanation:
Let : L = 60 H ,
R = 8 Ω,
I = 20 A , and
Eo = 400 V ,
The current constant of this inductance is
If =
400 V
ε0
=
= 50 A ,
R
8Ω
and the time constant
τ=
L
60 H
=
= 7.5 s .
R
8Ω
If the current is initially zero in this LR circuit, the current at time t is
I = If 1 − e−t/τ
I
= 1 − e−t/τ
If
I
e−t/τ = 1 −
If
t
1−I
− = ln
τ
If
I
t = −τ ln 1 −
If
20 A
= −(7.5 s) ln 1 −
50 A
= 3.83119 s .
Question 9, chap 32, sect 4.
part 1 of 1
10 points
Consider the following circuit. After leaving the switch at the position “a” for a long
7
S b
a
R
The maximum current will be given by
r
E
L
1. Imax =
R C
√
2. Imax = E L C
r
1
3. Imax = E
LC
r
L
4. Imax = E
C
r
C
E
5. Imax =
R L
r
C
6. Imax = E
correct
L
E
7. Imax =
R
E √
LC
8. Imax =
R
Explanation:
1
1 2
2
L Imax
=
q
2
2 C max
1
= C E2
2
r
C
E.
Imax =
L
Question 10, chap 31, sect 3.
part 1 of 1
10 points
A metal bar with length OA = L is rotating in a counter-clockwise manner about the
point O with a constant angular velocity ω.
midterm 03 – SAENZ, LORENZO – Due: Apr 3 2008, 11:00 pm
There is a constant magnetic field B directed
into the paper.
v
A
ω
Let : OC = R = 4 m ,
L = m,
v = 79 m/s , and
B = 7 T.
C
+q
A
L
B
8
O
B
v
The length of the bar is 10 m and the
magnitude of the magnetic field is 7 T .
The speed of the bar at point C is 79 m/s ,
and the length of OC = 4 m .
Determine the magnitude of the potential
difference |VO − VA | .
395
V
2
2765
V
2. |VO − VA | =
4
1106
3. |VO − VA | =
V
5
13825
4. |VO − VA | =
V correct
2
1. |VO − VA | =
5. |VO − VA | = 27650 V
2765
V
2
3318
7. |VO − VA | =
V
5
6. |VO − VA | =
8. |VO − VA | = 2765 V
9. |VO − VA | = 13825 V
10. |VO − VA | =
Explanation:
+q
B
553
V
5
FB
B
O
From the figure above we can see that the
force on +q is directed radially inward . The
sign of the charge can reverse the direction of
the force FB . This may be obtained from the
equation
~ = +q ~v × B
~.
F
Because of this magnetic force, the positive
charges begin to accumulate at O (or normally, the negative charges begin to accumulate at A), producing an electric field that
points from O to A. Hence, VO > VA .
Recall that the induced electric field at a
point r is given by
~ = vB = rωB,
Eind = |~v × B|
where in this problem, the angular velocity is
ω=
(79 m/s)
79
v
=
=
rad/s .
R
(4 m)
4
Hence the magnitude of the induced E is
Z R
Eind =
r ω B dr
0
1
= ω B R2
2 1 79
rad/s (7 T) (10 m)2
=
2
4
=
13825
V .
2
midterm 03 – SAENZ, LORENZO – Due: Apr 3 2008, 11:00 pm
y
Question 11, chap 30, sect 5.
part 1 of 1
10 points
a
P1
~
B
L
V
P2
q vd B = q E ,
x
cm
B = 1.2 T
or
E = vd B .
P1
Also we know
I = n q vd A
I
vd =
,
nqA
6.3 A
4.2 cm
~
B
I
For the Hall effect the magnetic force balances the electric force which means
y
6
~
B
b
Assume: The mobile charge carriers are
either electrons or holes. The holes have
the same magnitude of charge as the electrons. The number of mobile charge carriers for this particular material is n =
8.49 × 1028 electrons/m3 .
Note: In the figure, the point at the upper
edge P1 and at the lower edge P2 have the
same x coordinate.
A constant magnetic field of magnitude
points out of the paper. There is a steady
flow of a horizontal current flowing from left
to right in the x direction.
x
9
or
so that the magnitude of the electric field is
V
P2
2.2 m
The charge on the electron is 1.6021 ×
10−19 C.
What is the magnitude of the electric field
between the upper and lower surfaces?
Correct answer: 2.20559 × 10−7 (choice
number 3).
Explanation:
Let :
a = 6 cm = 0.06 m ,
b = 4.2 cm = 0.042 m ,
B = 1.2 T ,
n = 8.49 × 1028 electrons/m3 ,
q = 1.6021 × 10−19 C ,
I = 6.3 A , and
L = 2.2 m .
E=
=
IB
nqA
(6.3 A) (1.2 T)
n (1.6021 × 10−19 C) (0.00252 m2 )
= 2.20559 × 10−7 N/C ,
where the area
A= a·b
= (0.06 m) (0.042 m)
= 0.00252 m2 .
Question 12, chap 31, sect 3.
part 1 of 1
10 points
Given: Two coils are suspended around
a central axis as shown in the figure below.
One coil is connected to a resistor with ends
labeled “a” and “b”. The other coil is connected to a battery E. The coils are moving
midterm 03 – SAENZ, LORENZO – Due: Apr 3 2008, 11:00 pm
relative to each other as indicated by the velocity vectors v.
Use Lenz’s law to answer the following question concerning the direction of induced currents and magnetic fields.
− +
v
Using the right-hand rule, the magnetic flux
through the coil with the battery attached has
a magnetic field direction right to left.
Question B1 & B2:
Note: The induced magnetic field depends
on whether the flux is increasing or decreasing.
− +
E
10
Binduced
E
a
R
b
v
The direction of the magnetic field in the
coil with the battery attached is
A1: from right to left (⇐= Bprimary ).
A2: from left to right (Bprimary =⇒).
The direction of the induced magnetic field
in the coil with the resistor attached is
B1: from right to left (⇐= Binduced ).
B2: from left to right (Binduced =⇒).
The direction of the induced current in resistor R is
C1: from “b” through R to “a” (← I).
C2: from “a” through R to “b” (I →).
Choose the appropriate answer.
1.
A2, B2, C2
2.
A1, B2, C1
3.
A2, B1, C1
4.
A1, B1, C1 correct
5.
A1, B2, C2
6.
A1, B1, C2
7.
A2, B2, C1
8.
A2, B1, C2
Explanation:
Question A1 & A2: The helical coil with
the battery attached (when viewed from either end) is wound clockwise (as you go into
the coil).
a
R
b
Bprimary
The magnetic flux through the coil is from
right to left. When the coils are moving
towards each other, the magnetic flux through
the coil with the attached resistor increases.
The induced current in the coil must produce an induced magnetic field from right
to left (⇐= Binduced ) to resist any change of
magnetic flux in the coil (Lenz’s Law).
Question C1 & C2: The helical coil with
the resistor attached (when viewed from either end) is wound counter-clockwise (as you
go into the coil).
Since the induced field is right to left
(⇐= Binduced ) the induced current in the coil
flows clockwise when viewing the coil from the
right-hand side.
Therefore the current flows from “b”
through R to “a” (←− I).
Note: There are eight different presentations of this problem and this is the seventh.
Question 13, chap 31, sect 2.
part 1 of 1
10 points
A solenoid with circular cross section produces a steadily increasing magnetic flux
through its cross section. There is an octagonally shaped circuit surrounding the solenoid
as shown below.
The increasing magnetic flux gives rise to a
counterclockwise induced emf E.
Initial Case: The circuit consists of two
midterm 03 – SAENZ, LORENZO – Due: Apr 3 2008, 11:00 pm
identical light bulbs of equal resistance, R,
connected in series, leading to a loop equation
E − 2 i R = 0.
i
Figure 1:
B
B
X
Y
B
B
i
Primed′ Case: Now connect the points
C and D with a wire CAD, (see the figure
below).
i3
Figure 2:
D
i1
B
A
Y
B
i3
6. Bulb X goes out and bulb Y remains at
the same brightness.
Explanation:
Basic Concepts: Induced emf.
Solution: Let E and R be the induced emf
and resistance of the light bulbs, respectively.
For the first case, since the two bulbs are
in series, the equivalent resistance is simply
Req = R + R = 2 R and the current through
the bulbs is
E
i=
.
2R
Hence, for the first case, the power consumed
by bulb X is
2
E
PX =
R
2R
E2
=
4R
For the second (primed) case, since bulb Y is
shorted, the current through bulb X is now
B
X
i′ =
i2
B
C
What happens after the points C and D are
connected by a wire as in the second (primed)
case?
1. Bulb Y goes out and bulb X gets dimmer.
2. Bulb X goes out and bulb Y gets
brighter.
3. Bulb Y goes out and bulb X remains at
the same brightness.
4. Bulb Y goes out and bulb X gets brighter.
correct
5. Bulb X goes out and bulb Y gets dimmer.
11
E
,
R
and the power consumed by bulb X is
2
E
′
R
PX =
R
E2
=
.
R
Hence the ratio is
PX′
PX
E2
= R2 = 4 .
E
4R
For the second (primed) case, bulb Y is
shorted; i.e., PY′ = 0 . Hence the ratio is 0.
Therefore bulb X gets brighter and bulb Y
goes out.
Question 14, chap 32, sect 6.
part 1 of 1
10 points
A ideal transformer shown in the figure
below having a primary with 40 turns and
secondary with 32 turns.
midterm 03 – SAENZ, LORENZO – Due: Apr 3 2008, 11:00 pm
The load resistor is 33 Ω.
The source voltage is 90 Vrms .
12
Iabove
33 Ω
32 turns
90 Vrms
40 turns
Npole
1.
v
dipole magnet
Spole
Ibelow
What is the rms electric potential across
the 33 Ω load resistor?
Correct answer: 72 (choice number 8).
Iabove = 0
Explanation:
Npole
2.
Let : N1 = 40 turns ,
N2 = 32 turns ,
V1 = 90 Vrms .
v
Ibelow
and
Iabove
The rms voltage across the transformer’s
secondary is
Npole
3.
N2
V1
N1
32 turns
=
(90 Vrms )
40 turns
= 72 Vrms ,
v
V2 =
Iabove
Npole
4.
Imagine that Galileo had dropped a bar
magnet (permanent magnetic dipole) and
an unmagnetized bar of the same mass and
shape down identical resistive metallic (copper) tubes from the Tower of Pisa.
Which of the following figures best represents the direction of the induced currents
“ I ” in the metallic tube (above and/or below
the dipole magnet) as the dipole falls through
the tube?
dipole magnet correct
Spole
Ibelow
which is the same as the electric potential
across the load resistor.
Question 15, chap 31, sect 3.
part 1 of 1
10 points
dipole magnet
Spole
v
dipole magnet
Spole
Ibelow
midterm 03 – SAENZ, LORENZO – Due: Apr 3 2008, 11:00 pm
Iabove = 0
Npole
5.
v
dipole magnet
Spole
Ibelow
Iabove
Npole
6.
v
dipole magnet
Spole
Ibelow
Explanation:
When the bar magnet is falling, there will
be a change of magnetic flux around any
closed loop (which we take at constant height
and viewed from above) that lies in the wall
of the metal shell.
Iabove
Npole
v
dipole magnet
Spole
Ibelow
If this loop is below the bar magnet, the
magnetic flux through it points upward and
increases in magnitude as the leading south
pole of the bar magnet heads to the loop
and therefore the current induced in the wall
at this loop must produce downward magnetic fields to counteract this change of flux.
Such magnetic fields require currents that flow
clockwise when viewing the falling bar magnet from above as shown by the bottom loop
in the figure.
For the loop above the bar magnet the magnetic field is from the falling north pole which
13
then points upward and decreases in magnitude during the bar magnet’s fall. Therefore
the current induced in the wall at this loop
must produce upward magnetic fields to counteract the decreasing magnetic flux through
this loop. Such an induced magnetic field requires a current that flows counter-clockwise
when viewing the bar magnet from above as
shown in the top curve in the figure.
An alternative explanation is that the loop
below the bar magnet must produce a magnetic moment to repel the falling bar magnet
to try to prevent the bar magnet from increasing the magnetic flux through the loop.
The required direction of the magnetic moment must then be downward (which repels
a falling south pole), as produced by a clockwise current viewed from above. Alternatively, when the falling bar magnet is below
the loop, the induced magnetic moment of the
loop must attract the north pole of the falling
bar magnet, which requires the direction of
the magnetic moment to be upward, as produced by a counter-clockwise current when
viewed from above.
Question 16, chap 30, sect 3.
part 1 of 1
10 points
A small rectangular coil composed of
69 turns of wire has an area of 36 cm2 and
carries a current of 1.5 A. When the plane
of the coil makes an angle of 46◦ with a uniform magnetic field, the torque on the coil is
0.15 N m.
What is the magnitude of the magnetic
field?
Correct answer: 0.579532 (choice number
2).
Explanation:
Let : N = 69 turns ,
I = 1.5 A ,
θ = 46◦ ,
A = 36 cm2 = 0.0036 m2 ,
τ = 0.15 N m .
and
midterm 03 – SAENZ, LORENZO – Due: Apr 3 2008, 11:00 pm
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
0.15 N m
(69 turns) (1.5 A) (0.0036 m2 ) cos(46◦ )
= 0.579532 T .
14