Physics 232
Exam 2
Name _________________
Multiple Choice (2 pts each)
Use the following diagram for questions 1-2.
6
Ω
9#48
Ω
A
B
69ΩΩ#48
9#13
Ω
6
C
#13
D
96ΩΩ
1. The brightest bulb is
a) A
b) B
c) C
d) D
e) A and B
f ) C and D
g) B and C
2. The dimmest bulb is
a) A
b) B c) C
d) D
e) A and B
f ) C and D
g) B and C
x
x
x
#1
x
x
x
x
B into paper
x
x
x
x
x
x
x
x
#3 x
x
x
x
x
x
x
x
x
x
#2
3. Three particles have identical charges and masses (diagram above). They enter a
constant magnetic field and follow the paths shown. Which particle is moving the
fastest?
a) particle #1 b) particle # 2
c) particle #3
d) cannot be determined
4. A wire carries current vertically upward. A 2.0 µC charge is located to the north of
the wire but moving south towards the wire. The direction of the force on the charge is
a) north
b) south
c) east
d) west
e) upward
f) downward
Problems
5. What is the maximum voltage that can be supplied to 5.4 KΩ resistor rated at ¼ Watt?
(5 pts)
From P = V 2/R, we see that the maximum voltage will produce the maximum power, so we have
W = Vmax2/(5.4 × 103 Ω), which gives Vmax =
37 V.
6. A15 kΩ resistor and a capacitor are connected in series with a 24.0 V battery. If the
time constant is measured to be 55 ms,
(a) Determine the total capacitance of the circuit and (5 pts)
(b) Determine the time for the voltage across the resistor to reach 16.0 V. (5 pts)
(a) We find the capacitance from
τ = RC;
55 × 10–6 s = (15 × 103 Ω)C,
which gives C = 3.7 × 10–9 F =
3.7 nF.
(b) The voltage across the capacitance will increase to the final
steady state value. The voltage across the resistor will start
at the battery voltage and decrease exponentially:
VR = e – t/τ ;
16.0 V = (24.0 V)e – t/(55 µs), or t/(55 µs) = ln(24.0 V/16.0 V) = 0.405, which gives t =
22 µs.
7. Two long parallel wires 7.00 cm apart carry 16.5 Amp currents in the same direction.
Determine the magnetic field at a point P 12.0 cm from one wire and 13.0 cm from the
other. (15 pts)
81.8
o
66.0
o
We find the direction of the field for each wire from
the tangent to the circle around the wire, as shown.
For their magnitudes, we have
B1 = (µ0/4π)2I1/L1
= (10–7 T · m/A)2(16.5 A)/(0.120 m) = 2.75 × 10–5 T.
B2 = (µ0/4π)2I2/L2
= (10–7 T · m/A)2(16.5 A)/(0.130 m) = 2.54 × 10–5 T.
From the vector diagram, we have
B = B1(– sin θ1 i + cos θ1 j) + B2(– sin θ2 i – cos θ2 j)
= (2.75 × 10–5 T)(– sin 81.8° i + cos 81.8° j) + (2.54 × 10–5
T)(– sin 66.0° i – cos 66.0° j)
=
(– 5.04 × 10–5 T) i + (– 6.41 × 10–6 T) j.
For the direction of the field, we have
tan θ = By/Bx = (– 6.41 × 10–6 T)/(– 5.04 × 10–5 T) = 0.127, θ =
187.2°.
We find the magnitude from
Bx = B cos θ;
– 5.04 × 10–6 T = B cos 187.2°,
which gives B =
5.08 × 10–5 T 187.2° from the x-axis.
θ1
B1
B
y
P
θ2
B2
α
r2
r1
θ1
I1
θ2
d
I2
x
8. A 20 loop rectangular coil with sides of 10 cm and 15 cm lies in the xz plane. The
current in each loop of the coil is 6.2 A counter clockwise when viewed from the positive
y direction. An external magnetic field B = 5.0Ti – 2.6Tj – 8.0Tk passes through the
coil. Determine the torque due to the external magnetic field. (10 pts)
µ = NI A = ( 20 )( 6 . 2 A )(. 10 m )(. 15 m ) ˆj = 1 . 86 Am 2 ˆj
i
τ = µxB =
j
0
1.87 Am
5.0T
− 2.6T
k
2
0 = −15mNiˆ − 9.35mNkˆ
− 8.0T
9. The current through the 4.0 kΩ resistor shown in the figure is 3.50 mA. What is the
terminal voltage Vba of he unknown battery? (10 pts)
If we assume the current in R4 is to the right, we have
Vcd = I4R4 = (3.50 mA)(4.0 kΩ) = 14.0 V.
We can now find the current in R8 :
Vba
I8 = Vcd/R8 = (14.0 V)/(8.0 kΩ) = 1.75 mA.
R5
From conservation of current at the junction c, we have
I = I4 + I8 = 3.50 mA + 1.75 mA = 5.25 mA.
c
a
b
If we go clockwise around the outer loop, starting at a,
we have
r
I
Vba – IR5 – I4R4 – – Ir = 0, or
Vba = (5.25 mA)(5.0 kΩ) + (3.50 mA)(4.0 kΩ) + 12.0 V + (5.25
mA)(1.0 × 10–3 kΩ) = 52 V.
If we assume the current in R4 is to the left, all currents are reversed, so we have
Vdc = 14.0 V; I8 = 1.75 mA, and I = 5.25 mA.
If we go counterclockwise around the outer loop, starting at a, we have
– Ir + – I4R4 – IR5 + Vab = 0, or Vba = – Vab = – Ir + – I4R4 – IR5 ;
Vba = – (5.25 mA)(1.0 × 10–3 kΩ) + 12.0 V – (3.50 mA)(4.0 kΩ) – (5.25 mA)(5.0 kΩ) =
The negative value means the battery is facing the other direction.
R4
I4
R8
d
I8
– 28 V.
10. A very long copper wire with a radius of 5.0 mm carries a current of 4.0 Amps
uniformly distributed across its cross section.
a) What is the magnetic field 20 cm from the center of the wire? (5 pts)
b) What is the magnetic field 2 mm from the center of the wire? (5 pts)
a)
B • dl = µ o I enc
B=
µ o I 4πx10 −7 (4.0 A)
=
= 4.0 x10 −6 T
2πr
2π (.20m)
B • dl = µ o I
I
πr 2
2
πR
µ o Ir (4πx10 −7 )(4.0 A)(.002m)
B=
=
= 6.4 x10 −5 T
2
2
2πR
2π (.005m)
b) B 2πr = µ o JA = µ o