PHY1160C
Exam #2
June 25, 1997
Possibly useful information:
Fmag = |q| v B sin ϕ
F = I B l sin ϕ
µ I
B= o
2πr
τ = N I A B sin θ
E=–N∆ Φ
∆t
f = ω/2π
E = – L ∆i
∆t
F =l IxB
µ I
Bcenter = o
2a
Φ = B A cos θ
E=vBl
E = Eo sin ωt
i = If [ 1 – e–(R/L)t ]
XC = 1
2π f C
v = Vo sin ωt
Fmag = q v x B
B = µo n I
E=–∆ Φ
∆t
Φ = BA cos θ = BA cos ωt
Eo = N ω B A
V1 = V2
N1 N2
Z = R2 + (XL - XC )2
V = IZ
XL = 2π f L
1
fo = 1
2π
LC
F = ma
F = m v2/r
F=qE
p = mv
KE = (1/2) m v2
PE = mgy
If you think of other equations that might be useful, come ask me for them and I will
put them on the board for everyone to see and use.
Please read Chapter 23, Reflection and Refraction of Light. Tomorrow we will go
over this exam as a final review for our study of Electromagnetism and then we will
begin our study of Light or Optics.
For all questions, also consider the option of
E) none of the above
A)
B)
C)
D)
Key #2, p - 1 -
For all questions, also consider the option of
E) none of the above
1. (20.12)
An electron moves in a circular path, perpendicular to a magnetic
field of B = 1.25 T, with a a speed of 1.2 x 10 4 m/s. What is the radius of the electron’s
circular path?
Fmag = q v B sin ϕ
Since ϕ = 90, sin ϕ= 1
Fmag = q v B
This is also the centripetal force,
Fc = m v2 / r
q v B = m v2 / r
r = m v / [q B]
r=(9.11x10–31kg)(1.2x104m/s)/[(1.6x10–19C)(1.25T)]
r = 5.47 x 10 –8 m
A)
B)
C)
D)
r = 3.45 x 10–4 m
r = 3.45 x 10–6 m
r = 5.47 x 10–6 m
r = 5.47 x 10–8 m <<<<<<<<<<
In a region where Earth’s magnetic field is 5.0 x 10–5 T and makes an
angle of 60° with Earth’s surface, determine the force on a straight 2.0 m wire that
carries a current of 8.0 A straight up.
2. (20.24)
B
I
30°
Current-carrying wire
60°
Earth’s surface
F=I
l B sin ϕ
F = (8.0 A)(2.0 m)(5.0 x 10–5 T)(sin 30°)
F = (8.0 A)(2.0 m)(5.0 x 10–5 T)(0.50)
F = 4 x 10 –4 N
Key #2, p - 2 -
The force is perpendicular to both the magnetic field and the
current. The force is directed out of the paper (or screen).
A)
B)
C)
D)
F = 2 x 10–2 N
F = 3 x 10–3 N
F = 4 x 10–4 N <<<<<<<<<<
F = 4 x 10–5 N
3. (20.30)
An electric transmission line carries 2,500 A of current from east to
west. What is the magnetic field 10 m below the line?
From Equation 20.10 and our magnetism lab from last week,
B = µo I
2πr
–7
(4
π
x
10
T-m/A) (2 500 A)
B=
2 π (10 m)
B = 5 x 10 –5 T
A)
B)
C)
D)
B = 4 x 10–4 T
B = 5 x 10–5 T <<<<<<<<<<
B = 6 x 10–6 T
B = 7.5 x 10–6 T
Key #2, p - 3 -
4.
(21.3)
In a region where the magnetic field is 0.02 T, what can be the
maximum magnetic flux through a 20 cm x 30 cm rectangular coil? What can be the
minimum flux?
Φ = A B cos θ
Φ = Φmax for θ = 0 and cos θ = 1
Φmax = A B = (0.20 m)(0.30 m)(0.02 T)
Φmax = 0.0012 webers
Φ = Φmin = 0 for θ = 90° and cos θ = 0
The orientation of the coil and the magnetic field make the
difference:
A) Φmax = 0.0012 webers, Φmin = 0 <<<<<<<<<<
B) Φmax = 0.0024 webers, Φmin = 0.0012
C) Φmax = 0.0036 webers, Φmin = 0
D) Φmax = 0.0048 webers, Φmin = 0.0012
Key #2, p - 4 -
5. (21.7)
A rectangular loop has dimensions of 2.0 cm by 3.0 cm and is initially
between the poles of a large magnet, where the magnetic field strength is B = 0.050 T.
The loop is oriented so the normal to the plane of the coil lies parallel to the direction
of the magnetic field. The coil is removed from that region to a region with no
magnetic field in 0.15 seconds. What is the average induced emf?
Φo = B A = (0.05 T)(0.02 m)(0.03 m)
Φo = 0.000 030 T m2 = 0.000 030 W = 3 x 10 –5 W
Φf = 0
∆Φ f = Φf – Φo = – 3 x 10–5 W
E=
–∆ Φ
∆t
E=
3 x 10–5 W
0.15 s
E = 0.000 2 V = 2 mV
A)
B)
C)
D)
E = 0.000 1 V = 1 mV
E = 0.000 2 V = 2 mV <<<<<<<<<<
E = 0.000 3 V = 3 mV
E = 0.000 5 V = 5 mV
Key #2, p - 5 -
6—8. (21.19)
The loop in the figure below has a radius r of 4.42 cm. The
magnetic field strength is uniform and has a value of 33.0 mT. The angle between the
field and a line normal to the plane of the loop is 53°.
6) Determine the magnetic flux through the loop.
7) When the magnetic field is initially turned on, the magnetic field
strength rises to the value given above in 10.5 ms. What average voltage is induced
in the coil?
8) If the resistance of the coil is 27.3 Ω, what average current is
induced in the coil?
B
53°
B
r
Figure 21.30
Problem 21.19
Φ = B A cos θ
A = π r2 = π (0.0442 m) 2 = 0.006 14 m2 =6.14 x 10 –3 m2
Φ = (33.0 x 10 –3 T) (6.14 x 10–3 m2) (0.60)
Φ = 0.000 122 W
Φ = 0.122 mW
E=
–∆ Φ
∆t
E=
0.122 mW
10.5 ms
E = 0.012 V
I = V/R
I = 0.012 V / 27.3 Ω
I = 0.000 43 A
I = 0.43 mA
Key #2, p - 6 -
6. ——
A) Φ = 0.275 mW
B) Φ = 0.2502 mW
C) Φ = 0.122 mW <<<<<<<<<<
D) Φ = 0.061 mW
7. ——
A) E = 0.036 V
B) E = 0.025 V
C) E = 0.012 V <<<<<<<<<<
D) E = 0.006 V
8. ——
A) I = 0.27 mA
B) I = 0.43 mA <<<<<<<<<<
C) I = 0.86 mA
D) I = 1.43 mA
Key #2, p - 7 -
9, 10. (22.7)
An electric light, connected to an ordinary 120 V AC wall outlet,
uses 60 W of power. What are
9) the rms current through the light, and
10) the resistance of the light?
P = IV
60 W = (I) (120 V)
I = 60 W = 0.5 A
120 V
I rms = 0.5 A
I= V
R = V = 120 V = 240 Ω
R , or
I
0.5 A
R = 240 Ω
9. ——
A) Irms = 0.25 A
B) Irms = 0.50 A
C) Irms = 0.75 A
D) Irms = 1.5 A
10.——
A) R = 120 Ω
B) R = 240 Ω
C) R = 360 Ω
D) R = 480 Ω
Key #2, p - 8 -
11. (22.21) If 0.8 ampere flows through a capacitor when it is connected across
a 120-volt, 60 Hz AC voltage source, what is the capacitance of the capacitor?
I= V
XC
XC = V
I
X c = 120 V / 0.8 A = 150 Ω
XC =
C=
C=
1
2π f C
1
2π f X C
1
2π (60
s –1 )(150
Ω)
C = 1.77 x 10 –5 F
C = 17.7 x 10 –6 F = 17.7 µF
A)
B)
C)
D)
C = 56.5 x 10+3 F
C = 12.3 x 10–3 F = 12.3 µF
C = 17.7 x 10–6 F = 17.7 µF
C = 37.8 x 10–6 F = 37.8 µF
Key #2, p - 9 -
12. (22.31)
If 0.20 ampere flows through a 200 mH inductor when connected
across a 30-volt AC voltage source, what is the frequency of the source?
I= V
XL
XL = V
I
X L = 120 V = 600 Ω
0.20 A
XL = 2 π f L
600 Ω = 2 π f (200 mH)
f = 600 Ω / [2 π (200 x 10 –3 H)]
f = 478 Hz
A)
B)
C)
D)
f = 256 Hz
f = 478 Hz
f = 1230 Hz
f = 1580 Hz
Key #2, p - 10 -