ragsdale (zdr82) – HW6 – ditmire – (58335)
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Multiple-choice questions may continue on
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1
the direction of the current in the figure. Using the lower circuit in the figure, we get
E 2 + I2 R 2 = 0
(2)
so
001 (part 1 of 2) 10.0 points
The currents are flowing in the direction indicated by the arrows. A negative current
denotes flow opposite to the direction of the
arrow. Assume the batteries have zero internal resistance.
5.4 V
17.3 Ω
I2 =
−E2
−19.9 V
=
= −0.921296 A .
R2
21.6 Ω
Then, for the upper circuit
E 1 − I2 R 2 − I1 R 1 = 0 .
E 1 + E 2 − I1 R 1 = 0 .
E1 + E2
R1
5.4 V + 19.9 V
=
17.3 Ω
= 1.46243 A .
I1 =
I
21.6 Ω
I
19.9 V
I
Find the current through the 17.3 Ω resistor
and the 5.4 V battery at the top of the circuit.
Correct answer: 1.46243 A.
Explanation:
Let :
R1
R2
E1
E2
= 17.3 Ω ,
= 21.6 Ω ,
= 5.4 V , and
= 19.9 V .
E1
Alternate Solution: Using the outside
loop
−E1 − E2 + I1 R1 = 0
(4)
E1 + E2
I1 =
.
R1
002 (part 2 of 2) 10.0 points
Find the current through the 21.6 Ω resistor
in the center of the circuit.
Correct answer: −0.921296 A.
Explanation:
From Eq. (2)
E2
R2
19.9 V
=−
21.6 Ω
= −0.921296 A .
I2 = −
R1
I1
I2
I3
R2
003 (part 1 of 3) 10.0 points
C
9r
2r
E2
A
i1
9r
At nodes, we have
I1 − I2 − I3 = 0 .
(3)
(1)
Pay attention to the sign of the battery and
E
B
i3
2r
i5
i2
i4
D
I
ragsdale (zdr82) – HW6 – ditmire – (58335)
i1
, where I is the current
i2
entering and leaving the battery.
Hint: Apply the Kirchhoff’s law to the loop
ACDA.
Find the ratio
i1
2
=
i2
7
i1
5
2.
=
i2
9
4
i1
=
3.
i2
3
4
i1
=
4.
i2
9
i1
5
5.
=
i2
6
3
i1
=
6.
i2
2
i1
1
7.
=
i2
2
9
i1
= correct
8.
i2
2
i1
4
9.
=
i2
7
7
i1
=
10.
i2
6
Explanation:
2
⇒
1.
Let :
R1
R2
R3
R4
004 (part 2 of 3) 10.0 points
Find the magnitude of the current i5 which
flows from C to D.
i1
R2
E
1
I
4
1
2. i5 = I
5
7
3. i5 =
I correct
11
1
4. i5 =
I
11
1
5. i5 = I
8
3
6. i5 = I
7
3
I
7. i5 =
11
3
8. i5 =
I
13
1
9. i5 =
I
13
1
10. i5 = I
3
Explanation:
1. i5 =
= 2r,
= 9r,
= R2 = 9 r ,
= R1 = 2 r .
C
R1
A
i1
R2
=
i2
R1
9r
=
2r
9
=
.
2
and
R3
B
i3
i5
R4
i2
i4
D
I
Basic Concept: DC Circuit.
Solution: Based on Kirchhoff’s law, the
equation for the loop ACDA is given by
−i1 R1 + i2 R2 = 0
9
i2
2
I = i1 + i2
9
= i2 + i2
2
11
i2 , therefore
=
2
2
I , and
i2 =
11
9
I.
i1 =
11
Following a similar analysis, one finds that
9
2
9
i4
= , so that i3 =
I and i4 =
I.
i3
2
11
11
i1 =
ragsdale (zdr82) – HW6 – ditmire – (58335)
Note: The junction equation at D is
=
i2 + i5 = i4
⇒ i5 = i4 − i2 , or
= i3 − i1
2
9
=
I−
I
11
11
−7
I
=
11
7
|i5 | =
I .
11
006
28.4 V
005 (part 3 of 3) 10.0 points
Find the resistance RAB .
2. RAB =
3. RAB =
4. RAB =
5. RAB =
24
r
7
21
r
5
80
r
13
16
r
3
84
r
13
4.3 Ω
Explanation:
B
36
r correct
11
63
8. RAB =
r
8
48
9. RAB =
r
11
35
10. RAB =
r
6
Explanation:
By inspection, the following circuit is equivalent to the original circuit.
R1
R3
RAB
5.3 Ω
Find the current through the 19.2 Ω lowerright resistor.
Correct answer: 0.95302 A.
7. RAB =
R2
10.0 points
19.2 Ω
6. RAB = 4 r
A
36
r .
11
keywords:
14.2 V
1. RAB =
3
R4
R1 R2
=2
R1 + R2
(2 r) (9 r)
=2
2r +9r
B
C
E1
r1
E2
i1
r2
A
D
i2
F
R
E
I
Let : E1 = 28.4 V ,
E2 = 14.2 V ,
r1 = 5.3 Ω ,
r2 = 4.3 Ω , and
R = 19.2 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)
(2)
ragsdale (zdr82) – HW6 – ditmire – (58335)
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 )
(28.4 V) (4.3 Ω) + (14.2 V) (5.3 Ω)
=
(4.3 Ω) (19.2 Ω) + (5.3 Ω) (19.2 Ω + 4.3 Ω)
I=
= 0.95302 A .
007
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
E
4
PA
1
= correct
PB
4
PA
4.
=2
PB
P
1
5. A =
PB
16
P
1
6. A =
PB
2
P
7. A = 16
PB
P
8. A = 8
PB
P
9. A = 4
PB
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
2
V
V2
2
PA = I R =
R=
.
4R
16 R
3.
Thus the total power consumed by all four
bulbs in circuit A is
PA,T otal = 4 PA =
V2
.
4R
In circuit B, the equivalent resistance is
1
1
1
1
+
=
=
RB
2R 2R
R
RB = R ,
Circuit B
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 = 1
PB
P
1
2. A = √
PB
8
so the electric current through each bulb is
V
2R
and the power of each bulb is
2
V2
V
2
.
R=
PB = I R =
2R
4R
iB =
Thus the total power consumed by all four
bulbs in circuit B is
PB,T otal
V2
= 4 PB =
R
ragsdale (zdr82) – HW6 – ditmire – (58335)
and
• Ohm’s Law.
PA,T otal
1
PA
= .
=
PB,T otal
PB
4
008
5
There are two rules for adding up resistances. If the resistances are in series, then
Rseries = R1 + R2 + R3 + · · · + Rn .
10.0 points
If the resistances are parallel, then
Consider the combination of resistors
shown in the figure.
2.2 Ω
3.7 Ω
b
What is the resistance between point a and
point b?
Correct answer: 8.16186 Ω.
R6
i
i
R1
R2
i
R7
i
i
R3
i
1
1
1
1
+
+
+···+
.
R1 R2 R3
Rn
i
i
i
i
R1
R2
i
b
R5
R4
Step 1: The three resistors on the right are
all in series, so
b
R367 = R3 + R6 + R7
= (4.3 Ω) + (2.2 Ω) + (3.7 Ω)
= 10.2 Ω .
R5
i
=
Solution: The key to a complex arrangements of resistors like this is to split the problem up into smaller parts where either all the
resistors are in series, or all of them are in
parallel. It is easier to visualize the problem
if you redraw the circuit each time you add
them.
R367
a
Explanation:
Let’s redraw the figure
a
Rparallel
4.3 Ω
7.8 Ω
1.5 Ω
9.6 Ω
a
4.5 Ω
1
R4
a
Let : R1
R2
R3
R4
R5
R6
R7
= 4.5
= 1.5
= 4.3
= 9.6
= 7.8
= 2.2
= 3.7
Basic Concepts:
• Equivalent resistance.
Ω,
Ω,
Ω,
Ω,
Ω,
Ω,
Ω.
i
i
i
R1
R2
i
b
R3675
R4
Step 2: R5 and R367 are connected parallel, so
and
R3675
−1
1
1
=
+
R5 R367
R5 R367
=
R5 + R367
(7.8 Ω) (10.2 Ω)
=
18 Ω
= 4.42 Ω .
ragsdale (zdr82) – HW6 – ditmire – (58335)
i
i
R1
R36752
i
b
E1
R4
Step 3: R2 and R3675 are in series, so
R3
h
b
R1
R367524
Step 5: Finally, R1 and R236754 are in series, so the equivalent resistance of the circuit
is
Req = R1 + R236754
= 4.5 Ω + 3.66186 Ω
f
g
= 22.6 Ω ,
= 8.4 Ω ,
= 42.5 Ω ,
= 91 Ω ,
= 21.4 V ,
= 20.7 V .
and
Basic Concepts: Kirchhoff’s Laws
X
V = 0 around a loop.
X
I = 0 at a circuit junction.
Solution: A key simplification is to realize
that R3 and R4 are connected in parallel and
can be combined immediately, before applying
Kirchoff’s rules.
c d
e
b
I1
I2
R34
−1
1
1
=
+
R4 R23675
R4 R23675
=
R4 + R23675
(9.6 Ω) (5.92 Ω)
=
15.52 Ω
= 3.66186 Ω .
i
E2
R2
Let : R1
R2
R3
R4
E1
E2
i
e
R4
a
Step 4: R23675 and R4 are parallel, so
a
d
R1
R23675 = R2 + R3675
= (1.5 Ω) + (4.42 Ω)
= 5.92 Ω .
R236754
c
b
E1
R1
= 8.16186 Ω .
I34
a
6
E2
R2
a
f
h g
The combined resistance is given by
22.6 Ω
20.7 V
91 Ω
42.5 Ω
21.4 V
009 (part 1 of 2) 10.0 points
8.4 Ω
What is the current through 8.4 Ω bottomright resistor?
Correct answer: 1.5511 A.
Explanation:
1
1
1
R3 + R4
=
+
=
R34
R3 R4
R3 R4
R3 R4
R34 =
R3 + R4
(42.5 Ω) (91 Ω)
=
42.5 Ω + 91 Ω
= 28.97 Ω .
This now gives 2 loop equations and 1 junction
equation. The loop equations are
V1 − I34 R34 − I1 R1 = 0
V2 + I34 R34 − I2 R2 = 0
ragsdale (zdr82) – HW6 – ditmire – (58335)
and the junction equation is
The voltage across R4 is just
I34 + I2 = I1 .
Substituting I34 = I1 − I2 in to the loop equations, we have
V4 = V34 = R34 I34
so the power is
V42
R4
(−7.67076 V)2
=
91 Ω
= 0.6466 W .
P =
(R1 + R34 ) I1 − R34 I2 = V1
−R34 I1 + (R2 + R34 ) I2 = V2 .
Rewriting and multiplying by factors, we have
2
(R1 + R34 ) R34 I1 − R34
I2 = R34 V1
−(R1 + R34 ) R34 I1 + (R1 + R34 ) (R2 + R34 ) I2
= (R1 + R34 ) V2 .
Subtracting, we have
2
[(R1 + R34 ) (R2 + R34 ) − R34
] I2
= R34 V1 + (R1 + R34 ) V2 ,
for
011 (part 1 of 2) 10.0 points
The switch S has been in position b for a
long period of time.
R3
2
D = (R1 + R34 ) (R2 + R34 ) − R34
= (22.6 Ω + 28.97 Ω) (8.4 Ω + 28.97 Ω)
− (28.97 Ω)2
= 1087.91 Ω2 ,
we have
R34 V1 (R1 + R34 ) V2
+
D
D
(28.97 Ω)(21.4 V)
=
1087.91 Ω2
(22.6 Ω + 28.97 Ω)(20.7 V)
+
1087.91 Ω2
= 1.5511 A .
R1
I34
R2 V1 − R1 V2
=
R1 R2 + R34 (R1 + R2 )
(8.4 Ω)(21.4 V) − (22.6 Ω)(20.7 V)
=
(22.6 Ω)(8.4 Ω) + (28.97 Ω)(22.6 Ω + 8.4 Ω)
= −0.264783 A .
S b
E
a
When the switch is moved to position “a”,
find the characteristic time constant.
1. τ =
1
R2 C
p
2. τ = R1 R2 C
3. τ =
4. τ =
Explanation:
In a similar way we can solve for the current
through R34 giving
C
R2
I2 =
010 (part 2 of 2) 10.0 points
What is the power dissipated in 91 Ω rightcentered resistor?
Correct answer: 0.6466 W.
7
5. τ =
6. τ =
7. τ =
1
(R1 + R2 ) C
R1 + R2
C
2
2
(R1 + R2 ) C
1
√
R1 R2 C
1
R1 C
8. τ = R1 C
9. τ = R2 C
10. τ = (R1 + R2 ) C correct
Explanation:
ragsdale (zdr82) – HW6 – ditmire – (58335)
In charging an R C circuit, the characteristic time constant is given by
τ = RC ,
where in this problem R is the equivalent
resistance, or
R = R1 + R2 .
012 (part 2 of 2) 10.0 points
3.7 MΩ
1 µF
1 MΩ
S b
1.5 V
1.5 MΩ
a
S has been left at position “a” for a long
time. It is then switched from “a” to “b” at
t = 0.
Determine the energy dissipated through
the resistor R2 alone from t = 0 to t = ∞.
Correct answer: 0.239362 µJ.
Explanation:
Let : E
R1
R2
R3
C
= 1.5 V ,
= 1.5 MΩ ,
= 1 MΩ ,
= 3.7 MΩ ,
= 1 µF .
The total energy dissipated
R2 +R3
=
Udissip
series they share a common current, I. The
corresponding power consumptions by R2 and
R3 are respectively P2 = I 2 R2 and P3 = I 2 R3 .
This shows the correctness of the fraction, i.e.
P2 /(P2 + P3 ) = R2 /(R2 + R3 ). Alternate
solution:
More formally, noting that the initial current
E
is I0 =
, the total energy dissipated
R2 + R3
by R2 is
Z ∞
R2
U =
I(t)2 R2 dt
Z0 ∞
=
I02 R2 e−2t/[C(R2 +R3 )] dt
0
2
E
C (R2 + R3 )
=
R2 −
R2 + R3
2
∞
× e−2t/[C(R2 +R3 )] 0
R2
1
=
C E2
R2 + R3 2
1
(1 MΩ)
(1 µF) (1.5 V)2
=
(1 MΩ) + (3.7 MΩ) 2
= 0.239362 µJ .
013
and
We observe that power consumption consideration provides an independent check on the
fraction used. Since the two resistors are in
10.0 points
In the figure below the battery has an emf
of 13 V and an internal resistance of 1 Ω .
Assume there is a steady current flowing in
the circuit.
13 V
1Ω
1
C E2
2
. Since R2 and R3 are in series, the energy
R2
of
dissipated by R2 is only a fraction
R2 + R3
the total energy:
1
R2
R2
2
CE
Udissip =
R2 + R3
2
8
9Ω
6Ω
4 µF
Find the charge on the 4 µF capacitor.
Correct answer: 19.5 µC.
Explanation:
Let :
R1 = 9 Ω ,
ragsdale (zdr82) – HW6 – ditmire – (58335)
R2
rin
V
C
= 6 Ω,
= 1 Ω,
= 13 V ,
= 4 µF .
V
.
R2
V
3. I0 =
. correct
R1
V
4. I0 =
.
R1 + R2
2. I0 =
and
The equivalent resistance of the three resistors
in series is
Req = R1 + R2 + rin
= (9 Ω) + (6 Ω) + (1 Ω)
= 16 Ω ,
so the current in the circuit is I =
the voltage across R2 is
V
, and
Req
V2 = I R2
R2
V
=
Req
(6 Ω)
=
(13 V)
(16 Ω)
= 4.875 V .
Since R2 and C are parallel, the potential
difference across each is the same. Hence the
charge on the capacitor is
Q = C V2
= (4 µF) (4.875 V)
= 19.5 µC .
5. I0 = 0 .
Explanation:
Before the switch is closed, there is no
charge on the capacitor, so the voltage is zero
across the capacitor at this time. Because it
is not possible to change the charge on the
capacitor like a step function (or the current
should be infinitely large), immediately after
the switch is closed, the voltage across the capacitor (and R2 ) is still zero. Therefore, the
voltage across R1 is V ; i.e., think of the capacitor as being a short-circuit for this instant
of time.
So the current supplied by the battery,
which is the same as the current going through
V
R1 , is I0 =
.
R1
015 (part 2 of 2) 10.0 points
A long time after the switch has been closed,
the current I∞ supplied by the battery is
014 (part 1 of 2) 10.0 points
The switch has been open for a long period
of time.
R2
C
V
S
Immediately after the switch is closed, the
current supplied by the battery is
1. I0 =
V (R1 + R2 )
.
R1 R2
V
.
R2
V (R1 + R2 )
=
.
R1 R2
1. I∞ =
2. I∞
R1
9
3. I∞ = 0 .
V
. correct
R1 + R2
V
5. I∞ =
.
R1
Explanation:
After a long time, the capacitor has been
charged and remained stable. That means
the current going through R1 is the same as
the current going through R2 ; i.e., think of
the capacitor as being a open-circuit for this
time.
So we can write down the equation
4. I∞ =
V = I∞ R 1 + I∞ R 2 ,
ragsdale (zdr82) – HW6 – ditmire – (58335)
which gives the current I∞ as
V
I∞ =
.
R1 + R2
016 (part 1 of 2) 10.0 points
The circuit has been connected as shown in
the figure for a “long” time.
R2
b
Ib
“After a long time” implies that the capacitor C is fully charged, so it acts as an open
circuit with no current flowing to it. The
equivalent circuit is
Ib
R2
R1
R3
R2
r
R4
Ir
b
R4
Ib
Ieq
R1
R3
ℓ
C
a
Ir
Iℓ
Iℓ
It
It
017 (part 2 of 2) 10.0 points
If the battery is disconnected, how long does
it take for the voltage across the capacitor to
E0
drop to a value of V (t) = , where E0 is the
e
initial voltage across the capacitor?
Correct answer: 110 µs.
Explanation:
With the battery removed, the circuit is
S
b
Since E1 and E3 are “measured” from the same
point “a”, the potential across C must be
Req
C
R4
Ib
E3 = Ib R3 = (1 A) (1 Ω) = 1 V .
|EC | = 21 V .
t
It
E
48 V
= 1 A.
=
Rb
48 Ω
EC = E3 − E1
= 1 V − 22 V = −21 V
= 11 Ω ,
= 13 Ω ,
= 1 Ω,
= 47 Ω , and
= 11 µF = 1.1 × 10−5 F .
R1
48 V
E
=
= 2 A and
Rt
24 Ω
and across R3
Explanation:
E
It =
E1 = It R1 = (2 A) (11 Ω) = 22 V
S
It
R3
so
Across R1 ,
What is the magnitude of the electric potential across the capacitor?
Correct answer: 21 V.
a
Rb = R3 + R4 = 1 Ω + 47 Ω = 48 Ω ,
Ib =
47 Ω
48 V
Let : R1
R2
R3
R4
C
and
C
11 µF
1Ω
Rt = R1 + R2 = 11 Ω + 13 Ω = 24 Ω
13 Ω
11 Ω
10
ragsdale (zdr82) – HW6 – ditmire – (58335)
What is the initial direction of deflection?
b = √1
1. F
+̂ + k̂
2
1 b
2. F = √
+k̂ + ı̂
2
b = √1 (−̂ + ı̂)
3. F
2
~ = 0 ; no deflection
4. F
where
Rℓ = R1 + R3 = 11 Ω + 1 Ω = 12 Ω,
Rr = R2 + R4 = 13 Ω + 47 Ω = 60 Ω
and
Req =
1
1
+
Rℓ Rr
−1
1
1
+
=
12 Ω 60 Ω
= 10 Ω ,
−1
b = √1
5. F
2
b = √1
6. F
2
1
b=√
7. F
2
1
b=√
8. F
2
b = √1
9. F
2
1
b=√
10. F
2
so the time constant is
τ ≡ Req C = (10 Ω) (11 µF)
= 110 µs .
The capacitor discharges according to
Qt
= e−t/τ
Q0
1
V (t)
= e−t/τ =
E0
e
t
1
− = ln
= − ln e
τ
e
t = τ (ln e) = −(110 µs) (−1)
(+̂ + ı̂) correct
(−̂ − ı̂)
−̂ + k̂
−̂ − k̂
−k̂ − ı̂
Magnetic Force on a
~ = q ~v × B
~
F
10.0 points
A negatively charged particle moving at 45◦
angles to both the x-axis and y-axis enters a
magnetic field (pointing out of of the page),
as shown in the figure below.
−q
~
B
v
z
~
B
−k̂ + ı̂
Explanation:
Basic Concepts:
Charged Particle:
= 110 µs .
018
11
×
y
Figure: ı̂ is in the x-direction, ̂ is
in the y-direction, and k̂ is in the
z-direction.
x
Right-hand rule for cross-products.
~
b ≡ F ; i.e., a unit vector in the F direcF
~k
kF
tion.
~ = q ~v × B.
~
Solution: The force is F
~ = B −k̂ ,
B
1
~v = √ v (−ı̂ + ̂) , and
2
q < 0 , therefore,
~
~
F = −|q| ~v × B
h
i
1
= −|q| √ v B (−ı̂ + ̂) × −k̂
2
1
= −|q| √ v B (+̂ + ı̂)
2
b = √1 (+̂ + ı̂) .
F
2
ragsdale (zdr82) – HW6 – ditmire – (58335)
This is the seventh of eight versions of the
problem.
019
10.0 points
A negatively charged particle moving at 45◦
angles to both the z-axis and x-axis enters a
magnetic field (pointing towards the bottom
of the page), as shown in the figure below.
x
~
B
y
z
v
~
B
−q
Figure: ı̂ is in the x-direction, ̂ is
in the y-direction, and k̂ is in the
z-direction.
What is the initial direction of deflection?
b = √1
1. F
−k̂ + ı̂
2
1
b=√
2. F
+̂ − k̂
2
1
b = √ (+̂ − ı̂)
3. F
2
~ = 0 ; no deflection
4. F
b = −̂
5. F
Right-hand rule for cross-products.
~
b ≡ F ; i.e., a unit vector in the F direcF
~k
kF
tion.
~ = q ~v × B.
~
Solution: The force is F
~ = B (−ı̂) ,
B
1
~v = √ v +k̂ + ı̂ , and
2
q < 0 , therefore,
~
~
F = −|q| ~v × B
h
i
1
= −|q| √ v B +k̂ + ı̂ × (−ı̂)
2
1
= −|q| √ v B (+̂)
2
b = +̂ .
F
This is the third of eight versions of the
problem.
020 (part 1 of 6) 10.0 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 0.3 T uniform magnetic field (aligned with the ±k̂-direction) as
shown in the figure by the shaded area.
+2300 V
Region of
Magnetic
Field
0.3 T
q
m
1.5 cm
y
cm
Explanation:
Basic Concepts: Magnetic Force on a
Charged Particle:
~ = q ~v × B
~
F
32
b = +̂ correct
6. F
b = √1
+̂ + k̂
7. F
2
1
b
8. F = √
−̂ + k̂
2
b = √1
+k̂ − ı̂
9. F
2
1
b = √ (−̂ + ı̂)
10. F
2
12
x
z
Region I
Region II
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).
ragsdale (zdr82) – HW6 – ditmire – (58335)
In order for an ion to pass through both
slits on a straight line, which of the following
conditions must be true for the forces on the
~ E is the electric force vector and F
~ B is
ion? (F
magnetic force vector.)
~E = 2 F
~B
1. F
~B = 0
F
~ E = 0,
4. F
~ B 6= 0
F
~E = 1 F
~B
5. F
2
~E = F
~B
6. F
~ E = −F
~ B correct
7. F
~E ⊥ F
~ B , thus it is impossible.
8. F
~B
~E = −1 F
9. F
2
Explanation:
To obtain a straight orbit, the upward and
downward forces need to cancel. The force on
a charged particle is
~ =F
~E + F
~ B = q (E
~ + ~v × B)
~ .
F
For the force to be zero, we need
~E + F
~B = 0 ,
F
~
B
= −k̂ correct
~
kBk
~
B
3.
= +k̂
~
kBk
2.
Explanation:
The force due to the magnetic field provides
the centripetal force that causes the positive
ions to move in the semicircle.
As the negatively charged ion exits the re~ B = q ~v × B,
~ so by
gion of the electric field, F
the right-hand rule the
magnetic field must
point out of the page or in the −z-direction
~ is in the direction
−k̂ , since the force F
down the page; i.e., “−̂ ”
~ E = −2 F
~B
2. F
~ E 6= 0,
3. F
13
or
~ E = −F
~B .
F
Therefore, the forces are equal and opposite
and the magnitude of forces are equal; i.e.,
~ E k = kF
~Bk .
kF
021 (part 2 of 6) 10.0 points
In which direction (relative to the coordinate
system shown above) should the magnetic
field point in order for negatively charged ions
to move along the path shown by the dotted
line in the diagram above?
~ = 0 ; direction undetermined
1. kBk
q
=−
|q|
~v
= +ı̂
k~v k
~
B
=?
~
kBk
~B
F
= −̂ ,
~B k
kF
and
the vector product
ı̂ × k̂ = −̂ , and since
~ = q ~v × B
~ = kF
~ k (−̂)
F
"
#
~
~
~v
B
q
F
×
=
~k
~
|q| k~v k kBk
kF
h
i
= − (+ı̂) × −k̂
= −̂ ,
~
B
= −k̂
~
kBk
consequently
is correct.
022 (part 3 of 6) 10.0 points
If the ions are positively charged, the electric
field must be downward for a charge to move
through Region-I undeflected.
1. Cannot be determined since FB ⊥ FE .
2. False correct
ragsdale (zdr82) – HW6 – ditmire – (58335)
14
by
3. True
Explanation:
The above statement is F alse because, as
~
we see from previous discussion, that the E
field is downward, independent of the charge
of the beam and given in the statement of the
question.
023 (part 4 of 6) 10.0 points
In “Region I”, the electric potential between
the plates is 2300 V, the distance between the
plates is 1.5 cm, and the magnetic field in
both “Regions I and II” is 0.3 T .
What is the speed of a singly charged ion
that passes through both slits and makes it
into “Region II”?
Correct answer: 5.11111 × 105 m/s.
Explanation:
Let : B = 0.3 T , and
V
(2300 V)
E≡
=
d
(1.5 cm)
= 1.53333 × 105 N/C .
Since the electric and magnetic forces on the
ion are equal,
mv
.
qB
Br
m=q
v
= (2 × 10−18 C)
(0.3 T)(0.32 m)
×
5.11111 × 105 m/s
r=
= 3.75652 × 10−25 kg .
025 (part 6 of 6) 10.0 points
An ion with the same mass and with twice the
charge magnitude, only positively charged,
gets through Region-I.
1. The ion will be deflected upward in
Region-II with twice the circular radius.
2. The ion will be deflected upward in
Region-II with half the circular radius.
3. The ion will be deflected downward in
Region-II with half the circular radius. correct
4. The ion will pass through Region-II undeflected.
qE = qvB
5. The ion will be deflected upward in
E
1.53333 × 105 N/C
v=
=
= 5.11111 × 105 m/s .Region-II with the same circular radius.
B
0.3 T
024 (part 5 of 6) 10.0 points
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 =
0.32 m.
The charge on each ion is 2 × 10−18 C.
What is the mass of the ions?
Correct answer: 3.75652 × 10−25 kg.
Explanation:
Let : r = 0.32 m = 0.32 m
q = 2 × 10−18 C .
and
The radius of a circular path taken by a
charged particle in a magnetic field is given
6. The ion will be deflected downward in
Region-II with the same circular radius.
7. The ion will be deflected downward in
Region-II with twice the circular radius.
Explanation:
In Region-II the magnetic force FB is now
~ = m~a .
FB = +2 |q| (~v × B)
Thus, the force is twice the magnitude but opposite in sign. Hence, the centripetal accelerv2
has the opposite direction (upward)
ation
r
so that the deflection is in that direction and
m v2
= 2qvB
r
ragsdale (zdr82) – HW6 – ditmire – (58335)
r=
mv
.
2qB
Thus the radius of the orbit is half of what
it was before;
mv
.
i .e., when the radius was r =
qB
15