Version 086 – EX2 – ditmire – (58335)
This print-out should have 23 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
The current in a resistor decreases by 3.5 A
when the voltage applied across the resistor
decreases from 15.9 V to 5.29 V.
Find the resistance of the resistor.
1. 4.15917
2. 3.03143
3. 2.97519
4. 1.84091
5. 4.48551
6. 2.30201
7. 8.60588
8. 2.77591
9. 4.6943
10. 2.44527
Correct answer: 3.03143 Ω.
Explanation:
The figure below shows a cylindrical coaxial
cable of radii a, b, and c in which equal, uniformly distributed, but antiparallel currents i
exist in the two conductors.
a
b
iout ⊙
c
iin ⊗
O
and
V =IR
Call the initial voltage and current Vi and I,
respectively, and the final voltage and current
Vf and (I − ∆I), respectively.
Vf
V
Vi
R=
=
=
I
I
I − ∆I
Vi (I − ∆I) = Vf I
(Vi − Vf ) I = Vi ∆I
Vi − Vf
Vi
=
,
I
∆I
D
C
Vi − Vf
Vi
=
I
∆I
15.9 V − 5.29 V
=
3.5 A
= 3.03143 Ω .
002
10.0 points
r1
r2
r3
r4
Which expression gives the magnitude
B(r3 ) at D of the magnetic field in the region b < r3 < a?
2. B(r3 ) =
3. B(r3 ) =
4. B(r3 ) =
5. B(r3 ) =
6. B(r3 ) =
7. B(r3 ) =
8. B(r3 ) =
so
R=
F
E
1. B(r3 ) =
Let : ∆I = 3.5 A ,
Vi = 15.9 V ,
Vf = 5.29 V .
1
µ0 i (a2 − b2 )
2 π r3 (r32 − b2 )
µ0 i (r32 − b2 )
2 π r3 (a2 − b2 )
µ0 i r3
2 π a2
µ0 i (a2 − r32 )
correct
2 π r3 (a2 − b2 )
µ0 i (a2 + r32 − 2 b2 )
2 π r3 (a2 − b2 )
µ0 i r3
2 π b2
µ0 i
2 π r3
µ0 i
π r3
9. B(r3 ) = 0
10. B(r3 ) =
µ0 i r3
2 π c2
Explanation:
Ampere’s
Law states that the line inteI
~ · d~ℓ around any closed path equals
gral
B
Version 086 – EX2 – ditmire – (58335)
µ0 I, where I is the total steady current passing through any surface bounded by the closed
path.
Considering the symmetry of this problem,
we choose a circular path, so Ampere’s Law
simplifies to
B (2 π r3 ) = µ0 Ien ,
where r3 is the radius of the circle and Ien is
the current enclosed.
Aen
π (r32 − b2 )
Since
, when b <
=
Acylinder
π (a2 − b2 )
r3 < a for the cylinder,
µ0 Ien
2 π r3
π (r32 − b2 )
µ0 i − i
π (a2 − b2 )
=
2 π r3
2
a − r32
µ0 i
a2 − b2
=
2 π r3
B=
µ0 i (a2 − r32 )
.
=
2 π r3 (a2 − b2 )
003 10.0 points
A conductor consists of an infinite number
of adjacent wires, each infinitely long and
carrying a current I (whose direction is out-ofthe-page), thus forming a conducting plane.
A
C
If there are n wires per unit length, what is
~
the magnitude of B?
1. B = µ0 I
2. B = 2 µ0 n I
3. B =
µ0 n I
correct
2
4. B =
2
µ0 n I
4
5. B = 4 µ0 n I
6. B = µ0 n I
7. B =
µ0 I
4
8. B = 4 µ0 I
9. B =
µ0 I
2
10. B = 2 µ0 I
Explanation:
B
A
l
W
C
B
By symmetry the magnetic fields are equal
and opposite through point A and C and horizontally oriented. Following the dashed curve
in a counter-clockwise direction, we calculate
I
~ · d~s, which by Ampere’s law is proporB
tional to the current through the dashed loop
coming out of the plane of the paper. In
this problem this is a positive current. Hence
~ along the horizontal legs points in the diB
rection in which we follow the dashed curve.
Ampere’s Law is
I
~ · d~s = µ0 I .
B
To evaluate this line integral, we use the rectangular path shown in the figure. The rectangle has dimensions l and w. The net current
through the loop is n I l. Note that since there
~ in the direction of w, we
is no component of B
are only interested in the contributions along
sides l
I
~ · d~s = 2 B l = µ0 n l I
B
Version 086 – EX2 – ditmire – (58335)
B=
µ0 n I
.
2
3
x
y, ̂
α
ı̂
B
004 10.0 points
If a metal wire carries a current of 20.0 mA,
how long does it take for 4.80×1020 electrons
to pass a given cross-sectional area anywhere
along the wire? The magnitude of the charge
on an electron is 1.6 × 10−19 C.
1. 354.286
2. 12631.6
3. 2434.78
4. 430.769
5. 3840.0
6. 10857.1
7. 560.825
8. 1866.67
9. 253.333
10. 1177.01
Correct answer: 3840 s.
Explanation:
Let : qe = 1.6 × 10−19 C ,
I = 20.0 × 10−2 A and
N = 4.80 × 1020 .
I=
∆t =
=
=
N qe
∆Q
=
∆t
∆t
N qe
I
(4.8 × 1020 )(1.6 × 10−19 C)
0.02 A
3840 s .
k̂
z
What is the direction of the torque vector
~τ ?
1. b
τ = −ı̂
2. b
τ = + k̂ correct
3. b
τ = ı̂ + ̂ sin α
4. b
τ = +ı̂
5. b
τ = + ̂
k̂ − ı̂
6. b
τ= √
2
ı̂ + k̂
7. b
τ= √
2
8. b
τ = − k̂
9. b
τ = k̂ − ̂ sin α
10. b
τ = − ̂
Explanation:
Basic Concepts: Torque on a current
loop due to a magnetic field.
x
y, ̂
ı̂
I
µ
α
B
005 (part 1 of 2) 10.0 points
A circular current loop of radius R is placed
in a horizontal plane and maintains a current
~ in the
I. There is a constant magnetic field B
◦
xy-plane, with the angle α (0 < α < 90◦ )
defined with respect to y-axis. The current in
the loop flows clockwise as seen from above.
In this problem we determine the torque
vector τ which the field exerts on the current
loop.
I
z
k̂
Solution: We know that torque is
~
~τ = ~µ × B
= µ (−̂) × [Bx (+ı̂) + By (−̂)]
Version 086 – EX2 – ditmire – (58335)
and that ̂ × ̂ = 0 and −̂ × ı̂ = k̂, so
~τ = µ Bx k̂ ,
and the direction of the torque is +k̂ . This
agrees with the answer from the right-hand
rule.
006 (part 2 of 2) 10.0 points
Let I = 0.137 A, R = 4.36 cm, B = 4.92 µT,
and α = 38.2◦ .
What is the magnitude of the torque exerted on the current loop?
1. 1.74997e-09
2. 1.82876e-10
3. 4.38521e-11
4. 1.41458e-10
5. 2.23746e-09
6. 2.05647e-09
7. 2.48934e-09
8. 1.63982e-09
9. 3.55707e-10
10. 3.12427e-10
Correct answer: 2.48934 × 10−9 N m.
007 (part 1 of 2) 10.0 points
Consider two cylindrical conductors made out
of the same material (i.e. they have the same
density of charge carriers and the same resistivity).
~E 1
~ = kA
~ × Bk
~ = A B sin θ ,
kCk
~ and B
~.
where θ is the angle between A
~
The angle between ~µ and B is α , so the
magnitude of the torque (k~τ k ≡ τ ) is
τ = µ B sin θ = I π R2 B sin α
= µ B sin α
= (0.00081817 A m2 )(4.92 × 10−6 T) sin 38.2◦
= 2.48934 × 10
−9
N ·m .
I2
V2
ℓ2
r2
If ℓ2 = 3 ℓ1 , r2 = 2 r1 , V2 = 4 V1 , and ρ2 =
vd,2
ρ1 , what is the ratio
of the magnitudes
vd,1
of the drift velocities?
1.
2.
3.
6.
7.
The magnetic dipole moment is
The magnitude of a cross product is
b
r1
5.
µ = I A = I π R2 .
V1
ℓ1
4.
and
~E 2
I1
b
Explanation:
Let : I = 0.137 A ,
R = 4.36 cm ,
B = 4.92 µT ,
α = 38.2◦ .
4
8.
9.
10.
vd,2
vd,1
vd,2
vd,1
vd,2
vd,1
vd,2
vd,1
vd,2
vd,1
vd,2
vd,1
vd,2
vd,1
vd,2
vd,1
vd,2
vd,1
vd,2
vd,1
=
=
=
=
=
64
3
16
3
3
4
1
2
3
16
=3
4
correct
3
1
=
3
3
=
64
=
=2
Explanation:
~J = n q ~vd
Since the conductors are made out of the same
material, the density of charge carriers must
Version 086 – EX2 – ditmire – (58335)
be the same (n1 = n2 ), so
vd,2
=
vd,1
=
=
=
=
J2
n2 q
J1
n1 q
J2
J1
ρ V2 ℓ1
ρ V1 ℓ2
4 V1 ℓ1
3 V1 ℓ1
4
,
3
R=ρ
5
ℓ
, so
A
ℓ2
R2
A2
=
ℓ1
R1
ρ
A1
ℓ2
ρ
π r22
=
ℓ1
ρ
π r12
2
ℓ2
r1
=
r2
ℓ1
2
r1
3 ℓ1
=
2 r1
ℓ1
ρ
=
V
E
=
and the two conductors
ρ
lρ
have the same resistivity.
3
.
4
since J =
008 (part 2 of 2) 10.0 points
R2
of the resistances?
What is the ratio
R1
R2
64
=
1.
R1
3
R2
2.
=2
R1
R2
4
3.
=
R1
3
R2
3
4.
=
R1
64
R2
1
5.
=
R1
3
R2
1
6.
=
R1
2
R2
3
7.
= correct
R1
4
R2
3
8.
=
R1
16
R2
9.
=3
R1
16
R2
=
10.
R1
3
Explanation:
009
10.0 points
Four identical light bulbs are connected either in series (circuit A), or in a parallel-series
combination (circuit B), to a constant voltage
battery with negligible internal resistance, as
shown.
Circuit A
Circuit B
E
E
Assuming the battery has no internal resistance and the resistance of the bulbs is
temperature independent, what is the ratio of
the total power consumed bycircuit A to that
PA,T otal
?
consumed by circuit B; i.e.,
PB,T otal
P
1. A = 8
PB
Version 086 – EX2 – ditmire – (58335)
2.
3.
4.
5.
6.
7.
PA
PB
PA
PB
PA
PB
PA
PB
PA
PB
PA
PB
PA
PB
PA
PB
Thus the total power consumed by all four
bulbs in circuit B is
V2
PB,T otal = 4 PB =
R
and
PA,T otal
1
PA
= .
=
PB,T otal
PB
4
= 16
=1
=
1
correct
4
=2
=
1
2
010
20 V
=4
1
16
1
9.
=√
8
Explanation:
In circuit A, the equivalent resistance is
RA = 4 R, so the electric current through
each bulb is
V
iA =
4R
and the power of each bulb is
8.
6
10.0 points
5Ω
=
2
PA = I R =
V
4R
2
R=
V2
.
16 R
Thus the total power consumed by all four
bulbs in circuit A is
PA,T otal = 4 PA =
V2
.
4R
10 V
4Ω
18 Ω
Find the current through the 18 Ω lowerright resistor.
1. 0.714286
2. 0.931298
3. 0.959016
4. 1.33663
5. 1.72289
6. 0.535714
7. 0.903614
8. 0.857143
9. 0.836735
10. 1.4375
In circuit B, the equivalent resistance is
Correct answer: 0.714286 A.
1
1
1
1
=
+
=
RB
2R 2R
R
Explanation:
RB = R ,
B
E1
r1
E2
i1
r2
so the electric current through each bulb is
iB =
V
2R
C
PB = I 2 R =
V
2R
2
R=
D
i2
and the power of each bulb is
A
V2
4R
F
.
R
I
E
Version 086 – EX2 – ditmire – (58335)
Let : E1 = 20 V ,
E2 = 10 V ,
r1 = 5 Ω ,
r2 = 4 Ω , and
R = 18 Ω .
From the junction rule, I = i1 + i2 .
Applying Kirchhoff’s loop rule, we obtain
two equations:
E1 = i1 r1 + I R
E2 = i2 r2 + I R
= (I − i1 ) r2 + I R
= −i1 r2 + I (R + r2 ) ,
(1)
of magnitude 0.9 T directed perpendicular to
the rod and the rails?
1. 1.12119
2. 5.58018
3. 2.19144
4. 3.83494
5. 2.39472
6. 4.51777
7. 1.3741
8. 0.687023
9. 4.25162
10. 1.56374
Correct answer: 4.51777 m/s.
Explanation:
(2)
Let :
Multiplying Eq. (1) by r2 , Eq. (2) by r1 ,
E1 r2 = i1 r1 r2 + r2 I R
E2 r1 = −i1 r1 r2 + I r1 (R + r2 )
Adding,
E1 r2 + E2 r1 = I [r2 R + r1 (R + r2 )]
E 1 r2 + E 2 r1
r2 R + r1 (R + r2 )
(20 V) (4 Ω) + (10 V) (5 Ω)
=
(4 Ω) (18 Ω) + (5 Ω) (18 Ω + 4 Ω)
I=
= 0.714286 A .
011 (part 1 of 2) 10.0 points
A rod of mass 0.4 kg and radius 0.055 m rests
on two parallel rails that are 0.17 m apart and
0.58 m long. The rod carries a current of 69 A
(in the direction shown) and rolls along the
rails without slipping.
i
B
7
d
L
If it starts from rest, what is the speed of
the rod after moving a distance 0.58 m to
the right if there is a uniform magnetic field
m = 0.4 kg ,
r = 0.055 m ,
d = 0.17 m ,
L = 0.58 m ,
I = 69 A , and
B = 0.9 T .
The rod feels magnetic force F = i B d. The
work-energy theorem is
(Ttrans + Trot )i + W = (Ttrans + Trot )f
1
1
0 + 0 + F L = m v2 + I ω2 ,
2
2
where I is the moment of inertia of the rod of
radius R. The moment of inertia is
1
I = m R2 .
2
Since
W = idBL,
and
v 2
1 1
1
2
2
mR
W = mv +
2
2 2
R
3
= m v2 ,
4
the speed is
r
4idBL
v=
s 3m
=
4(69 A)(0.17 m)(0.9 T)(0.58 m)
3(0.4 kg)
= 4.51777 m/s .
Version 086 – EX2 – ditmire – (58335)
8
I = 6.6 A .
012 (part 2 of 2) 10.0 points
The magnetic field needs to be directed upwards for the rod to be accelerated to the
right.
1. False correct
From Ohm’s law, the total resistance of the
circuit is
V
E1 − E2
=
I
I
9 V − 3.11 V
=
6.6 A
= 0.892424 Ω .
Rtotal =
2. True
Explanation:
If the magnetic field is directed upward,
the rod is directed to the left. Hence the
statement is False.
013 (part 1 of 2) 10.0 points
See the circuit below.
9 V 3.11 V
0.4 Ω
0.21 Ω
X
1.86 Ω
R
6.6 A
Y
Find the resistance R.
1. -1.356
2. -1.54098
3. -1.27373
4. 1.67667
5. 1.30611
6. -0.294444
7. 6.56385
8. 0.239474
9. 1.10387
10. -1.57758
Explanation:
E1
R2
X
E2
I
R3
R
Y
Let : E1
E2
R1
R2
R3
= 9 V,
= 3.11 V ,
= 0.4 Ω ,
= 1.86 Ω ,
= 0.21 Ω ,
R = Rtotal − R1 − R2 − R3
= 0.892424 Ω − 0.4 Ω − 1.86 Ω − 0.21 Ω
= −1.57758 Ω .
014 (part 2 of 2) 10.0 points
Find the potential difference VXY = VX − VY
between points X and Y .
1. 12.064
2. -0.191
3. 8.424
4. -7.844
5. 0.825
6. -5.916
7. -3.872
8. 11.348
9. -9.5
10. 0.45
Correct answer: −5.916 V.
Correct answer: −1.57758 Ω.
R1
Therefore the resistance R is
Explanation:
The current in the circuit goes counterclockwise, so the potential difference between
X and Y is
VXY = E2 + R3 I + R I
= 3.11 V + (0.21 Ω + −1.57758 Ω) (6.6 A)
= −5.916 V .
015
and
10.0 points
In the figure below the battery has an emf of
8 V and an internal resistance of 1 Ω . Assume
there is a steady current flowing in the circuit.
Version 086 – EX2 – ditmire – (58335)
1Ω
11 Ω
9
8V
Since R2 and C are parallel, the potential
difference across each is the same. Hence the
charge on the capacitor is
4Ω
Q = C V2
= (2 µF) (2 V)
2 µF
= 4 µC .
Find the charge on the 2 µF capacitor.
1. 4.0
2. 8.14286
3. 8.57143
4. 6.66667
5. 23.5714
6. 24.0
7. 47.1579
8. 43.75
9. 30.0
10. 18.0
016 (part 1 of 2) 10.0 points
An electron is projected into a uniform mag~ = Bz k̂ + Bx ı̂, where
netic field given by B
Bz = 3.7 T and Bx = 1.5 T.
The magnitude of the charge on an electron
is 1.60218 × 10−19 C .
x
1.5 T
3.7 T
Correct answer: 4 µC.
Explanation:
Let : R1
R2
rin
V
C
B
z
v = 3.9 × 105 m/s
y
= 11 Ω ,
= 4 Ω,
= 1 Ω,
= 8 V , and
= 2 µF .
The equivalent resistance of the three resistors
in series is
Req = R1 + R2 + rin
= (11 Ω) + (4 Ω) + (1 Ω)
= 16 Ω ,
so the current in the circuit is I =
the voltage across R2 is
V2 = I R2
R2
V
=
Req
(4 Ω)
=
(8 V)
(16 Ω)
= 2 V.
V
, and
Req
electron
Find the direction of the magnetic force
when the velocity of the electron is v ı̂, where
v = 3.9 × 105 m/s.
1 b
ı̂ − k̂
1. F = √
2
b = −̂
2. F
b = ı̂
3. F
b = −k̂
4. F
b = k̂
5. F
b = −ı̂
6. F
b = √1
7. F
k̂ + ı̂
2
b
8. F = ̂ correct
b = √1
k̂ − ı̂
9. F
2
Version 086 – EX2 – ditmire – (58335)
Explanation:
Let : q = 1.60218 × 10−19 C ,
Bz = 3.7 T , and
Bx = 1.5 T .
Basic Concepts: Magnetic force on a moving charge is given by
~ = q ~v × B
~.
F
Solution:
~ = (3.7 T) k̂ + (1.5 T) ı̂
B
v = (3.9 × 105 m/s) ı̂ for the electron.
Find: The vector expression for the force on
the electron. This solves both part 1 and part
2.
We will go through two methods of doing
the problem.
The first is more mathematically oriented
and the second uses more of a reasoning argument.
Method 1: The force acting on a charge q
with velocity ~v in the presence of an external
~ is given by
magnetic field B
~ = q ~v × B
~
F
~ we
Taking the cross product of ~v with B
obtain
~ = q ~v × B
~
F
ı̂
̂ k̂
=q v 0 0
Bx 0 Bz
n
= q [(Bx )(0) − (Bz )(v)] ̂
− [(0)(0) − (Bz )(0)] ı̂
+ [(v)(0) − (Bz )(0)] k̂
o
= −q Bz v ̂
= −(−1.60218 × 10−19 C)(3.7 T)×
(3.9 × 105 m/s) ̂
= (2.31194 × 10−13 N) ̂ ,
and the direction is +k̂ .
10
Method 2: The other method is to realize that the only component of the magnetic
field which affects the electron is the component perpendicular to its velocity. Therefore,
~ = q v B⊥ with the direction
F = q |~v × B|
given by the right hand rule to be in the negative k̂ direction; but recalling to reverse the
direction because the electron has a negative
instead of positive charge.
F = q v B⊥
= (1.60218 × 10−19 C)×
(3.9 × 105 m/s) (3.7 T)
= 2.31194 × 10−13 N in the ̂ direction.
017 (part 2 of 2) 10.0 points
What is the magnitude of this force?
1. 2.75575e-13
2. 2.37443e-13
3. 2.31194e-13
4. 2.43531e-13
5. 2.61796e-13
6. 2.4994e-13
7. 2.55708e-13
8. 2.25266e-13
9. 2.94801e-13
10. 2.81182e-13
Correct answer: 2.31194 × 10−13 N.
Explanation:
See above.
018 (part 1 of 4) 10.0 points
In the circuit shown, the capacitor is initially
uncharged. At t1 = 0, the switch S is moved
to position “a”.
R2
C
R1
V0
S b
a
Find VR1 , the voltage drop across R1 , as a
function of time t1 .
Version 086 – EX2 – ditmire – (58335)
−(R1 +R2 ) t1 /(R1 R2 C)
1. VR1 = V0 e
h
i
−t1 /[(R1 +R2 ) C]
2. VR1 = V0 1 − e
By Kirchhoff’s law, the sum of voltage
around a closed circuit must be zero. Consequently, moving clockwise around the circuit
as drawn
3. VR1 = V0 e−t1 /[(R1 +R2 ) C]
4. VR1 = V0 e−t1 /(R2 C)
i
h
−(R1 +R2 ) t1 /(R1 R2 C)
5. VR1 = V0 1 − e
i
h
−t1 /(R1 C)
6. VR1 = V0 1 − e
−t1 /(R1 C)
correct
7. VR1 = V0 e
i
h
−t1 /(R2 C)
8. VR1 = V0 1 − e
Explanation:
For an “RC” circuit,
I = I0 e−t1 /(RC)
V0 −t1 /(R1 C)
=
e
.
R1
Since I R1 = VR1 ,
VR1 = V0 e−t1 /(R1 C) .
019 (part 2 of 4) 10.0 points
Find VC , the voltage across C, as a function
of time t1 .
i
h
1. VC = V0 1 − e−t1 /(R1 C) correct
i
h
2. VC = V0 1 − e−(R1 +R2 ) t1 /(R1 R2 C)
−t1 /(R1 C)
3. VC = V0 e
i
h
4. VC = V0 1 − e−t1 /(R2 C)
5. VC = V0 e−t1 /(R2 C)
6. VC = V0 e−t1 /[(R1 +R2 ) C]
h
i
7. VC = V0 1 − e−t1 /[(R1 +R2 ) C]
8. VC = V0 e−(R1 +R2 ) t1 /(R1 R2 C)
Explanation:
11
V0 − VR1 − VC = 0 .
Therefore,
VC = V0 − VR1
i
h
−t1 /(R1 C)
.
= V0 1 − e
020 (part 3 of 4) 10.0 points
Much later (t1 ≈ ∞), at some time t2 = 0
(the clock is restarted at t2 = 0), the switch is
moved from position “a” to position “b”.
Find the voltage drop, VR1 , across R1 , as a
function of time t2 .
i
h
R1
1. VR1 = V0
1 − e−t2 /(R2 C)
R1 + R2
R1
2. VR1 = V0
e−t2 /(R2 C)
R1 + R2
3.
h
i
R1
1 − e−(R1 +R2 )t2 /(R1 R2 C)
R1 + R2
h
i
R1
4. VR1 = V0
1 − e−t2 /(R1 C)
R1 + R2
R1
5. VR1 = V0
e−t2 /(R1 C)
R1 + R2
n
o
R1
−t2 /[(R1 +R2 ) C]
6. VR1 = V0
1−e
R1 + R2
VR1 = V0
7. VR1 = V0
rect
R1
e−t2 /[(R1 +R2 ) C] corR1 + R2
R1
e−(R1 +R2 ) t2 /(R1 R2 C)
R1 + R2
Explanation:
Now the switch moves to position “b”,
thereby excluding the battery from the circuit. Note: The equivalent resistance of the
circuit is Req = R1 + R2 , because R1 and R2
are in series.
8. VR1 = V0
I = I0 e−t2 /(RC)
V0
e−t2 /(R1 +R2 )C ,
=
R1 + R2
Version 086 – EX2 – ditmire – (58335)
because the capacitor has an initial potential
across it of V0 .
Thus
VR1 = I R1
= V0
R1
R1 + R2
e−t2 /(R1 +R2 )C .
021 (part 4 of 4) 10.0 points
Find VC as a function of time t2 .
o
n
1. VC = V0 1 − e[−t2 /(R1 +R2 ) C]
h
i
2. VC = V0 1 − e−t2 /(R2 C)
3. VC = V0 e−(R1 +R2 ) t2 /(R1 R2 C)
4. VC = V0 e−t2 /[(R1 +R2 ) C] correct
i
h
5. VC = V0 1 − e−(R1 +R2 ) t2 /(R1 R2 C)
h
i
6. VC = V0 1 − e−t2 /(R1 C)
7. VC = V0 e−t2 /(R2 C)
8. VC = V0 e−t2 /(R1 C)
12
Correct answer: 1.05263 × 10−7 T.
Explanation:
Let :
ℓ = 5.7 m
I = 7.0 A in the + x direction
Fm = 4.2 × 10−6
N in the − y direction
The magnetic force is
Fm = I ℓ B
Fm
Iℓ
4.2 × 10−6 N
=
(7 A) (5.7 m)
B=
= 1.05263 × 10−7 T
023 (part 2 of 2) 10.0 points
b) What is its direction?
Explanation:
Just as before, apply Kirchhoff’s law of
voltage to find
1. None of these
VC = VR1 + VR2 = V0 e−t2 /[(R1 +R2 )C] .
3. +x direction
022 (part 1 of 2) 10.0 points
A 5.7 m wire carries a current of 7.0 A toward
the +x direction. A magnetic force of 4.2 ×
10−6 N acts on wire in the −y direction.
a) Find the magnitude of the magnetic field
producing the force.
1. 9.08163e-07
2. 2.05714e-07
3. 3.50649e-07
4. 9.27318e-08
5. 5.32468e-07
6. 9.67033e-08
7. 3.80952e-07
8. 4.59184e-08
9. 2.2449e-07
10. 1.05263e-07
2. −z direction
4. −y direction
5. +z direction correct
6. +y direction
7. −x direction
Explanation:
Apply right-hand rule; force directed out of
the palm of the hand, fingers in the direction
of the field, thumb in the direction of the
current.
Palm faces in the negative y direction,
thumb points in the positive x direction, so
the fingers point in the +z direction.