Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Solutions toConceptual Practice Problems
PHYS 1112 In-Class Exam #2A+2B
Thu. Apr. 9, 2009, 11:00am-12:15pm and 2:00pm-3:15pm
CP 3.01: If a wire of some length L and a circular cross-section of diameter D has a
resistance of 36 kΩ, what will be the resistance of a wire made from the same material, at
the same temperature, of length 3L and the same diameter D ?
(A)
(B)
(C)
(D)
(E)
4 kΩ
12 kΩ
24 kΩ
72 kΩ
108 kΩ
Answer: (E)
Using R = ρL/A and A = π(D/2)2 , we get R ∝ L/D2 . So, R ∝ L for fixed D. Increasing
L → L0 = 3L thus increases R → R0 = 3R = 3 × 36Ω = 108Ω.
CP 3.02: If a wire of some length L and a circular cross-section of diameter D has a
resistance of 36 kΩ, what will be the resistance of a wire made from the same material, at
the same temperature, of the same length L and diameter 3D ?
(A)
(B)
(C)
(D)
(E)
4 kΩ
12 kΩ
24 kΩ
72 kΩ
108 kΩ
Answer: (A)
Using R = ρL/A and A = π(D/2)2 , we get R ∝ L/D2 . So, R ∝ 1/D2 for fixed L. Increasing
D → D0 = 3D thus changes R → R0 = R/32 = 36Ω/9 = 4Ω.
CP 3.03: If a wire of some length L and a circular cross-section of diameter D has a
resistance of 36 kΩ, what will be the resistance of a wire made from the same material, at
the same temperature, of length 18L and diameter 3D ?
(A) 4 kΩ
(B) 12 kΩ
(C) 24 kΩ
1
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
(D) 72 kΩ
(E) 108 kΩ
Answer: (D)
Using R = ρL/A and A = π(D/2)2 , we get R ∝ L/D2 . Increasing L → L0 = 18L and
D → D0 = 3D thus changes R → R0 = 18 × R/32 = (18/9) × 36Ω = 72Ω.
CP 3.04: Three different circuits, X, Y and Z, are built with the same three resistors,
R1 > 0, R2 > 0, and R3 > 0, and the same battery of battery voltage E, as shown in Fig.
3.04. Compare and rank the magnitude of the currents I1 through R1 , observed in the three
different circuits.
Fig. 3.04
Io
I2
I3
E
I1
(X)
(A)
(B)
(C)
(D)
(E)
Io
R2
R3
R1
I2
Io
R2
I1
E
I3
R1
I3
R3
I1
E
I2
R3
(Y)
R1
R2
(Z)
Ranking cannot be determined from information given.
I1 (X) > I1 (Z) > I1 (Y )
I1 (Y ) > I1 (X) > I1 (Z)
I1 (Y ) > I1 (Z) > I1 (X)
I1 (Z) > I1 (Y ) > I1 (X)
Answer: (D)
In circuit Y : R1 is directly connected to battery, hence I1 (Y ) = E/R1 .
In circuit Z: R1 and R2 in series are connected to battery. Hence I1 (Z) = E/(R1 + R2 ) <
I1 (Y ), since R1 + R2 > R1 .
2
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
In circuit X: R1 , R2 and R3 in series are connected to battery. Hence I1 (X) = E/(R1 +
R2 + R3 ) < I1 (Z) < I1 (Y ), since R1 + R2 + R3 > R1 + R2 .
CP 3.05: Three different circuits, X, Y and Z, are built with the same three resistors,
R1 > 0, R2 > 0, and R3 > 0, and the same battery of battery voltage E, as shown in Fig.
3.04. Compare and rank the magnitude of the voltage drops V3 across resistor R3 , observed
in the three different circuits.
(A)
(B)
(C)
(D)
(E)
Ranking cannot be determined from information given.
V3 (X) > V3 (Z) > V3 (Y )
V3 (Y ) > V3 (X) > V3 (Z)
V3 (Y ) > V3 (Z) > V3 (X)
V3 (Z) > V3 (Y ) > V3 (X)
Answer: (E)
Since in V3 = R3 I3 and R3 is the same in all three circuits, ranking V3 is equivalent to ranking
I3 . That is, for example, V3 (Z) > V3 (Y ) > V3 (X) is equivalent to I3 (Z) > I3 (Y ) > I3 (X).
So, let’s rank I3 !
In circuit Z: R3 is directly connected to battery, hence I3 (Z) = E/R3 .
In circuit Y : R3 and R2 in series are connected to battery. Hence I3 (Y ) = E/(R3 + R2 ) <
I3 (Z), since R3 + R2 > R3 .
In circuit X: R3 , R2 and R1 in series are connected to battery. Hence I3 (X) = E/(R3 +
R2 + R1 ) < I3 (Y ) < I3 (Z), since R3 + R2 + R1 > R3 + R2 .
CP 3.06: Three different circuits, X, Y and Z, are built with the same three resistors,
R1 > 0, R2 > 0, and R3 > 0, and the same battery of battery voltage E, as shown in Fig.
3.04. Compare and rank the equivalent resistance Req (from R1 , R2 and R3 combined) in
these three circuits.
(A)
(B)
(C)
(D)
(E)
Ranking cannot be determined from information given.
Req (X) < Req (Z) < Req (Y )
Req (Y ) < Req (X) < Req (Z)
Req (Y ) < Req (Z) < Req (X)
Req (Z) < Req (Y ) < Req (X)
Answer: (A)
Obviously, circuit X has the highest Req of all three circuits, regardless of the choices of R1 ,
R2 and R3 . However, the ranking of Req in circuits Y and Z depends on the specific choice
of values for R1 , R2 and R3 , which are not given.
3
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Example: Make R2 R1 and R2 R3 . Then Req (Y ) ∼
= R1 and Req (Z) ∼
= R3 , since, in
either Y - or Z- circuit, almost no current will flow through that branch which contains R2 :
R2 almost completely blocks the current flow through its own branch.
Now consider two possible combinations of R1 ad R3 , both with, say, R2 = 100000Ω R1 , R3 :
(1) If you choose, say, R1 = 1Ω and R3 = 3Ω, Req (Y ) ∼
= R1 will be lowest.
(2) If you choose, say, R1 = 3Ω and R3 = 1Ω, Req (Z) ∼
= R3 will be lowest.
So you can make either circuit Y or circuit Z have the lowest Req , depending on the values
of R1 and R3 you pick. So, the ranking of Req in circuits Y and Z depends on the specific
choice of values for R1 and R3 , which are not given.
CP 3.07: Three different circuits, X, Y and Z, are built with the same three resistors,
R1 > 0, R2 > 0, and R3 > 0, and the same battery of battery voltage E, as shown in Fig.
3.04. Compare and rank the magnitude of the total battery current Io observed in the three
circuits.
(A)
(B)
(C)
(D)
(E)
Ranking cannot be determined from information given.
Io (X) > Io (Z) > Io (Y )
Io (Y ) > Io (X) > Io (Z)
Io (Y ) > Io (Z) > Io (X)
Io (Z) > Io (Y ) > Io (X)
Answer: (A)
Since Io = E/Req and E is the same in all three circuits, this problem is equivalent to ranking
the equivalent resistance Req . However, as shown in the solution to CP 3.06the ranking of
Req cannot be determined from the information given; hence the same applies for the ranking
of Io .
CP 3.08: Three different circuits, X, Y and Z, are built with the three resistors, R1 > 0,
R2 > 0, and R3 > 0, and a battery of battery voltage E, as shown in Fig. 3.04.
Let Io > 0, I1 > 0, I2 > 0, and I3 > 0 denote the currents through the battery, R1 , R2 , and
R3 , respectively. Let V1 > 0, V2 > 0, and V3 > 0 denote the voltage drops across R1 , R2 ,
and R3 , respectively.
Which of the following statements is false ?
(A)
(B)
(C)
(D)
In
In
In
In
circuit
circuit
circuit
circuit
X: Io = I1 and V1 + V3 = E − V2 .
Y : Io = I1 + I2 and V1 = V2 + V3 .
Z: Io = I1 + I3 and V2 = V1 + V3 .
Y : I1 + I2 = I3 + I1 and V2 = E − V3 .
4
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
(E) In circuit Z: I1 = I2 and V1 /V2 = R1 /R2 .
Answer: (C)
For circuit Z, the statement (F): V2 = V1 + V3 is false. To see this, use the Kirchhoff loop
rules to get: V1 + V2 = E = V3 , hence (L): V2 = V3 − V1 < V3 . By contrast (F) would imply
that V2 > V3 which contradicts the correct loop rule result (L).
All other statements in (A)-(E) are true and can be derived from Kirchhoff rules and Ohm’s
law: this is left mostly as an exercise for the reader!
For example, in (A) for circuit X: V1 + V2 + V3 = E by loop rule, hence V1 + V3 = E − V2 ,
as stated in (A).
For another example, in (E) for circuit Z: V1 /R1 = I1 = I2 = V2 /R2 by Ohm’s law (V/R=I)
and junction rule (I1 = I2 for resistors R1 and R2 in series). Hence, by cross multiplying:
V1 /V2 = R1 /R2 .
CP 3.09: Three different circuits, X, Y and Z, are built with the three resistors, R1 > 0,
R2 > 0, and R3 > 0, and a battery of battery voltage E, as shown in Fig. 3.04.
Let Io > 0, I1 > 0, I2 > 0, and I3 > 0 denote the currents through the battery, R1 , R2 , and
R3 , respectively. Let V1 > 0, V2 > 0, and V3 > 0 denote the voltage drops across R1 , R2 ,
and R3 , respectively.
Which of the following statements is false ?
(A)
(B)
(C)
(D)
(E)
In
In
In
In
In
circuit
circuit
circuit
circuit
circuit
X: I2 + I3 = I3 + I1 and V3 /E = R3 /(R1 + R2 + R3 ).
Y : Io = I1 + I3 and V1 /R1 = E[1/(R3 + R2 ) + 1/R1 ].
Z: I1 (R1 + R2 ) = I3 R3 and V3 = V1 + V2 .
Y : I1 /(R2 + R3 ) = I2 /R1 and V1 = E.
Z: 1/R3 + 1/(R1 + R2 ) = Io /E and V1 /R1 = E/(R1 + R2 ).
Answer: (B)
For circuit Y , the statement (F): V1 /R1 = E[1/(R3 + R2 ) + 1/R1 ] is false. To see this, use
Ohm’s law to write the LHS of (F) as: V1 /R1 = I1 .
Also, use the reciprocal of the circuits equivalent resistance 1/Req = [1/(R3 + R2 ) + 1/R1 ]
to write the RHS of (F) as: E[1/(R3 + R2 ) + 1/R1 ] = E/Req = Io .
Thus, statement (F), if true, would imply: I1 = Io . However this clearly contradicts the
Kirchhoff junction rule for this circuit, whereby (J): I1 = Io − I2 < Io . So, using the (true!)
junction rule result (J), we see that (F) must be false.
All other statements in (A)-(E) are true and can be derived from Kirchhoff rules and Ohm’s
law: this is left mostly as an exercise for the reader!
For example, in (A) for circuit X: V3 /R3 = I3 = Io = E/Req = E/(R1 + R2 + R3 ). Hence,
by cross-multiplying, V3 /E = R3 /(R1 + R2 + R3 ).
5
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
For another example, in (E) for circuit Z: 1/R3 + 1/(R1 + R2 ) = 1/Req = Io /E
For yet another example, in (D) for circuit Y : R1 I1 = V1 = E = V2 + V3 = (R2 + R3 )I2 .
Hence, by cross-multiplying, I1 /(R2 + R3 ) = I2 /R1 .
CP 3.10: In a Kirchhoff rule analysis, current arrows have been assigned to the I1 -, I2 - and
I3 -branches of a circuit for which a fragment (i.e., not the whole circuit) is shown in Fig.
3.10. Also assume that the voltrage drop across R is defined as
VR ≡ Va − Vb
where Va and Vb are the electric potential values at points a and b shown in Fig. 3.10
I1
I2
Fig. 3.10
a
I3
R
b
Suppose R = 10Ω and I1 = +5A and I2 = +2A for the arrow directions shown in Fig. 3.10.
What is VR and in which direction is the current actually flowing through R ?
(A)
(B)
(C)
(D)
(E)
VR
VR
VR
VR
VR
= +70V
= −70V
= +30V
= −30V
= +30V
and
and
and
and
and
|I3 |
|I3 |
|I3 |
|I3 |
|I3 |
flows
flows
flows
flows
flows
from
from
from
from
from
a to b.
b to a.
a to b.
b to a.
b to a.
Answer: (C)
For arrows as drawn: I1 = I2 + I3 . So I3 = I1 − I2 = [(+5) − (+2)]A = +3A. Since I3 -arrow
points from a to b and VR ≡ Va − Vb , we have VR = +RI3 = (10Ω) × (+3A) = +30V. I3 > 0
6
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
means |I3 | flows in direction of I3 -arrow, i.e. from a to b.
CP 3.11: In a Kirchhoff rule analysis, current arrows have been assigned to the I1 -, I2 - and
I3 -branches of a circuit for which a fragment (i.e., not the whole circuit) is shown in Fig.
3.10. Also assume that the voltrage drop across R is defined as
VR ≡ Va − Vb
where Va and Vb are the electric potential values at points a and b shown in Fig. 3.10
Suppose R = 10Ω and I1 = +2A and I2 = +5A for the arrow directions shown in Fig.
3.10.
What is VR and in which direction is the current actually flowing through R ?
(A)
(B)
(C)
(D)
(E)
VR
VR
VR
VR
VR
= +70V
= −70V
= +30V
= −30V
= +30V
and
and
and
and
and
|I3 |
|I3 |
|I3 |
|I3 |
|I3 |
flows
flows
flows
flows
flows
from
from
from
from
from
a to b.
b to a.
a to b.
b to a.
b to a.
Answer: (D)
For arrows as drawn: I1 = I2 + I3 . So I3 = I1 − I2 = [(+2) − (+5)]A = −3A. Since I3 -arrow
points from a to b and VR ≡ Va − Vb , we have VR = +RI3 = (10Ω) × (−3A) = −30V. I3 < 0
means |I3 | flows against direction of I3 -arrow, i.e. from b to a.
CP 3.12: In a Kirchhoff rule analysis, current arrows have been assigned to the I1 -, I2 - and
I3 -branches of a circuit for which a fragment (i.e., not the whole circuit) is shown in Fig.
3.10. Also assume that the voltrage drop across R is defined as
VR ≡ Va − Vb
where Va and Vb are the electric potential values at points a and b shown in Fig. 3.10
Suppose R = 10Ω and I1 = −5A and I2 = +2A for the arrow directions shown in Fig.
3.10.
What is VR and in which direction is the current actually flowing through R ?
(A)
(B)
(C)
(D)
VR
VR
VR
VR
= +70V
= −70V
= +30V
= −30V
and
and
and
and
|I3 |
|I3 |
|I3 |
|I3 |
flows
flows
flows
flows
from
from
from
from
a to b.
b to a.
a to b.
b to a.
7
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
(E) VR = +30V and |I3 | flows from b to a.
Answer: (B)
For arrows as drawn: I1 = I2 + I3 . So I3 = I1 − I2 = [(−5) − (+2)]A = −7A. Since I3 -arrow
points from a to b and VR ≡ Va − Vb , we have VR = +RI3 = (10Ω) × (−7A) = −70V. I3 < 0
means |I3 | flows against direction of I3 -arrow, i.e. from b to a.
CP 3.13: In a Kirchhoff rule analysis, current arrows have been assigned to the I1 -, I2 - and
I3 -branches of a circuit for which a fragment (i.e., not the whole circuit) is shown in Fig.
3.10. Also assume that the voltrage drop across R is defined as
VR ≡ Va − Vb
where Va and Vb are the electric potential values at points a and b shown in Fig. 3.10
Suppose R = 10Ω and VR = −20V and I2 = −5A for the arrow directions shown in Fig.
3.10.
What is I1 ? Is the current |I1 | actually flowing towards or away from point a ?
(A)
(B)
(C)
(D)
(E)
I1
I1
I1
I1
I1
= +7A
= −7A
= +3A
= −3A
= −3A
with
with
with
with
with
|I1 |
|I1 |
|I1 |
|I1 |
|I1 |
flowing
flowing
flowing
flowing
flowing
towards a.
away from a.
towards a.
away from a.
towards a.
Answer: (B)
Since I3 -arrow points from a to b and VR ≡ Va −Vb , we have VR = +RI3 , thus I3 = +VR /R =
(−20V)/(10Ω) = −2A. For arrows as drawn: I1 = I2 + I3 = [(−5) + (−2)]A = −7A. I1 < 0
means |I1 | flows against direction of I1 -arrow, i.e. away from a.
CP 3.14: In a Kirchhoff rule analysis, current arrows have been assigned to the I1 -, I2 - and
I3 -branches of a circuit for which a fragment (i.e., not the whole circuit) is shown in Fig.
3.10. Also assume that the voltrage drop across R is defined as
VR ≡ Va − Vb
where Va and Vb are the electric potential values at points a and b shown in Fig. 3.10
Suppose |VR | = 120V and I1 = +5A and I2 = −7A for the arrow directions shown in
Fig. 3.10.
What is R and what is the sign of VR ?
8
Physics 1112
Spring 2009
(A)
(B)
(C)
(D)
(E)
University of Georgia
Instructor: HBSchüttler
R = 10 Ω and VR > 0.
R = 10 Ω and VR < 0.
R = 60 Ω and VR > 0.
R = 60 Ω and VR < 0.
R = −60 Ω and VR < 0.
Answer: (A)
For arrows as drawn: I1 = I2 + I3 . So I3 = I1 − I2 = [(+5) − (−7)]A = +12A. Since I3 -arrow
points from a to b and VR ≡ Va − Vb , we have VR = +RI3 . Thus, since R can never be
negative, I3 > 0 implies VR > 0. Also then R = VR /I3 = (+120V)/(+12A) = 10Ω
CP 3.15: A molecular ion beam containing four different types of ions, called P , Q, R and
S here, enters a uniform magnetic field, as shown in Fig. 3.15, with B 6= 0 above the lower
~ perpendicular to, and pointing out of, the plane of the drawing. The
horizontal line, and B
incident beam, below the lower horizontal line, is in the plane of the drawing.
Fig. 3.15
B (out)
P
Q
R
S
B=0
~
From the semicircular ion trajectories in B-field,
in Fig. 3.15, find the signs (+ or −) of the
charges qP , qQ , qR and qS of each of the four different ion types in the beam.
(A)
(B)
(C)
(D)
P :+,
P :+,
P :−,
P :+,
Q:+,
Q:−,
Q:+,
Q:−,
R:−,
R:−,
R:+,
R:+,
S:−,
S:+,
S:−,
S:−,
9
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
(E) P :−, Q:−, R:+, S:+,
Answer: (E)
For simplicity consider the magnetic force F~ acting on the ions only at, or immediately
~
after, their point of entry into the B-field.
At/immediately after their point of entry into the
~
B-field, ions P and Q are beginning to be deflected to the left, ions R and S are beginning
to be deflected to the right. Also, from the mechanics of circular motion, we know that
for (semi-)circular trajectories the respective forces F~ acting on the ions must always point
towards the center of that (semi-)circle. And for P and Q that center of the circle in
Fig. 3.15 is to the left of the point of entry; whereas for R and S that center of the circle is
to the right of the point of entry. Thus, at/immediately after the point of entry we find: for
P and Q, F~ points to the left; for R and S, F~ points to the right.
~ the right-hand rule, and the fact that B
~ points out of paper, it’s then easy
Using F~ = q~v × B,
~
to see that F pointing leftward (for P and Q ions) requires q~v pointing downward. Thus q~v
points opposite to the direction of the velocity ~v which is upward at the point of entry, in
the travel direction of the incident beam. So, q must be negative in order for q~v to point in
opposite direction as ~v : P and Q ions have negative charges.
Likewise, F~ pointing rightward (for R and S ions) requires q~v pointing upward, i.e., in same
direction as the velocity vector ~v . So, q must be positive in order for q~v to point in same
direction as ~v : R and S ions have positive charges.
CP 3.16: A molecular ion beam containing four different types of ions, called P , Q, R and
S here, enters a uniform magnetic field, as shown in Fig. 3.15, with B 6= 0 above the lower
~ perpendicular to, and pointing out of, the plane of the drawing. The
horizontal line, and B
incident beam, below the lower horizontal line, is in the plane of the drawing.
~
The diameters of the semicircular ion trajectories in the B-field,
denoted by dP , dQ , dR and
dS , respectively are observed to be in a ratio of
dP : dQ : dR : dS = 3 : 2 : 1 : 4 ,
as indicated in Fig. 3.15. Assume all four ion types carry the same amount of charge per
~
ion, |q|, and they all enter the B-field
with the same speed v. What is the ratio of the four
ion masses, denoted by mP , mQ , mR and mS , respectively ?
(A)
(B)
(C)
(D)
mP
mP
mP
mP
:
:
:
:
mQ
mQ
mQ
mQ
:
:
:
:
mR
mR
mR
mR
:
:
:
:
mS
mS
mS
mS
=3 : 2 : 1 : 4 .
√
√
√
√
= 3 : 2 : 1 : 4 .
= 13 : 12 : 11 : 14 .
= √13 : √12 : √11 : √14 .
(E) mP : mQ : mR : mS = 9 : 4 : 1 : 16 .
Answer: (A)
10
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
From d = 2r = 2mv/(|q|B), we get d ∝ m for fixed v, B and |q|. So, the masses m must be
in the same ratios as diameters d if all ions have the same v and |q| and travel in the same B.
CP 3.17: A molecular ion beam containing four different types of ions, called P , Q, R and
S here, enters a uniform magnetic field, as shown in Fig. 3.15, with B 6= 0 above the lower
~ perpendicular to, and pointing out of, the plane of the drawing. The
horizontal line, and B
incident beam, below the lower horizontal line, is in the plane of the drawing.
~
The diameters of the semicircular ion trajectories in the B-field,
denoted by dP , dQ , dR and
dS , respectively are observed to be in a ratio of
dP : dQ : dR : dS = 3 : 2 : 1 : 4 ,
as indicated in Fig. 3.15. Assume all four ion types carry the same amount of charge per
~
ion, |q|, and they all enter the B-field
with the same kinetic energy K. What is the ratio of
the four ion masses, denoted by mP , mQ , mR and mS , respectively ?
(A)
(B)
(C)
(D)
(E)
mP
mP
mP
mP
mP
:
:
:
:
:
mQ
mQ
mQ
mQ
mQ
:
:
:
:
:
mR
mR
mR
mR
mR
:
:
:
:
:
mS
mS
mS
mS
mS
=3 : 2 : 1 : 4 .
√
√
√
√
= 3 : 2 : 1 : 4 .
= 13 : 12 : 11 : 14 .
= √13 : √12 : √11 : √14 .
= 9 : 4 : 1 : 16 .
Answer: (E)
√
√
From d = 2r = 2 2mK/(|q|B),
we
get
d
∝
m for fixed K, B and |q|. So, the square
√
roots of the masses, m, must be in the same ratios as diameters d if all ions have
the same K and |q| and travel in the same B. So, for the d-ratios given, this implies:
√
√
√
√
mP : mQ : mR : mS = 3 : 2 : 1 : 4. Squaring both sides of these ratio equations, we
thus get: mP : mQ : mR : mS = 32 : 22 : 12 : 42 = 9 : 4 : 1 : 16.
CP 3.18: In a series of four magnetic deflection experiments, referred to as experiment P ,
Q, R and S, respectively, a beam of electrons, all of the same kinetic energy K, enters a
uniform magnetic field, as shown in Fig. 3.15, with B 6= 0 above the lower horizontal line
~ perpendicular to the plane of the drawing. The incident electron beam, below the
and B
lower horizontal line, is in the plane of the drawing.
In each of these four experiments, a different magnetic field strength and field direction is
~ is pointing out of the plane of the
used. In two of the experiments (which ones ?) B
drawing, i.e., in the direction indicated in Fig. 3.15; in the other two experiments (which
~ is pointing into the plane of the drawing, i.e., opposite to the direction
ones ?) B
indicated in Fig. 3.15.
The diameters of the four semicircular electron trajectories denoted by dP , dQ , dR and dS ,
respectively are observed to be in a ratio of
dP : dQ : dR : dS = 3 : 2 : 1 : 4 ,
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as indicated in Fig. 3.15. Find the ratio of the four magnetic field strengths, denoted by
BP , BQ , BR and BS , respectively, that were used in these four deflection experiments ?
(A)
(B)
(C)
(D)
BP
BP
BP
BP
:
:
:
:
BQ
BQ
BQ
BQ
:
:
:
:
BR
BR
BR
BR
:
:
:
:
BS
BS
BS
BS
=3 : 2 : 1 : 4 .
√
√
√
√
= 3 : 2 : 1 : 4 .
= 13 : 21 : 11 : 41 .
= √13 : √12 : √11 : √14 .
(E) BP : BQ : BR : BS = 9 : 4 : 1 : 16 .
Answer: (C)
√
From d = 2r = 2 2mK/(|q|B), we get d ∝ 1/B for fixed K, m and |q|. So, reciprocals of
the field strengths, 1/B, must be in the same ratios as diameters d since all electrons have
the same m and |q|; and the same K in all four experiments.
So, for the d-ratios given, this implies: (1/BP ) : (1/BQ ) : (1/BR ) : (1/BS ) = 3 : 2 : 1 : 4.
Taking the reciprocals on both sides of these ratio equations, we thus get: BP : BQ : BR :
BS = 13 : 12 : 11 : 14 .
CP 3.19: A very thin circular metallic ring lies in the y − z-plane and it is centered at the
coordinate origin O ≡ (0, 0, 0), as shown in Fig. 3.19. A current I flows around the ring in
~ ≡ (Bx , By , Bz )
the direction indicated in panel (B) of Fig. 3.19. The magnetic field vector B
produced by this current at the center of the ring has cartesian components Bx , By and Bz
~ by:
given in terms of the field strength B ≡ |B|
(A)
(B)
y
Fig. 3.19
z
I
x
y
y
z
x-y-Plane View
z
x
x
y
y-z-Plane View
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(A) (Bx , By , Bz ) = (+B, 0, 0)
(B) (Bx , By , Bz ) = (−B, 0, 0)
(C) (Bx , By , Bz ) = (0, √B2 , √B2 )
(D) (Bx , By , Bz ) = (0, −B, 0)
(E) (Bx , By , Bz ) = (0, 0, +B)
Answer: (A)
~ at center of loop (or more generally at any point inside
For circular loop current I, vector B
~ must
the loop!) must be perpendiclar to the plane of the loop. So, in Fig. 3.19 panel (B), B
be perpendicular to plane of drawing, i.e., parallel to x-axis. Therefore, Bx = ±B 6= 0 and
By = Bz = 0.
~
Also, the loop current direction and B-direction
(at loop center or more generally at any
point inside the loop) are related by a right-hand (RH) rule: point RH 4 fingers around
~ perpendicular to plane of
loop in I-direction; then the RH thumb points in direction of B,
~ must be pointing out of the plane of
loop. So, in Fig. 3.19 panel (B), by this RH rule, B
~ points in +x-direction. Therefore, Bx = +B > 0 and B
~ = (+B, 0, 0). Also,
drawing, i.e., B
~ points rightward
when viewed in the x − y-plane, as in panel (A) of Fig. 3.19, note that B
(≡ +x-direction!).
~ inside a closed current loop are more
Note: The foregoing rules concerning direction of B
generally applicable than just for circular loops: they apply for any ”planar” loop current,
regardless of the shape of the loop, as long as the entire loop is ”flat”, i.e., lies within a single plane. Thus, these rules can be applied, for example, to loops of rectangular, triangular,
hexagonal or any other polygonal shape lying in a single plane.
CP 3.20: A very thin circular metallic ring lies in the x − z-plane and it is centered at the
coordinate origin O ≡ (0, 0, 0), as shown in Fig. 3.20. A current I flows around the ring in
~ ≡ (Bx , By , Bz )
the direction indicated in panel (B) of Fig. 3.20. The magnetic field vector B
produced by this current at the center of the ring has cartesian components Bx , By and Bz
~ by:
given in terms of the field strength B ≡ |B|
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(A)
(B)
y
Fig. 3.20
z
I
x
x
y
z
x-y-Plane View
(A)
(B)
(C)
(D)
(E)
z
x
y
x
x-z-Plane View
(Bx , By , Bz ) = (+B, 0, 0)
(Bx , By , Bz ) = (−B, 0, 0)
(Bx , By , Bz ) = (0, √B2 , √B2 )
(Bx , By , Bz ) = (0, −B, 0)
(Bx , By , Bz ) = (0, 0, +B)
Answer: (D)
~ at center of loop (or more generally at any point inside
For circular loop current I, vector B
~ must
the loop!) must be perpendiclar to the plane of the loop. So, in Fig. 3.20 panel (B), B
be perpendicular to plane of drawing, i.e., parallel to y-axis. Therefore, By = ±B 6= 0 and
Bx = Bz = 0
~
Also, the loop current direction and B-direction
(at loop center or more generally at any
point inside the loop) are related by a right-hand (RH) rule: point RH 4 fingers around
~ perpendicular to plane of
loop in I-direction; then the RH thumb points in direction of B,
~ must be pointing out of the plane of
loop. So, in Fig. 3.20 panel (B), by this RH rule, B
~
~ = (0, −B, 0). Also,
drawing, i.e., B points in −y-direction. Therefore, By = −B < 0 and B
~ points downward
when viewed in the x − y-plane, as in panel (A) of Fig. 3.20, note that B
(≡ −y-direction!).
~ inside a closed current loop are more
Note: The foregoing rules concerning direction of B
generally applicable than just for circular loops: they apply for any ”planar” loop current,
regardless of the shape of the loop, as long as the entire loop is ”flat”, i.e., lies within a single plane. Thus, these rules can be applied, for example, to loops of rectangular, triangular,
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hexagonal or any other polygonal shape lying in a single plane.
CP 3.21: A very thin circular ring lies initially in the x − z-plane and it is centered at the
coordinate origin O ≡ (0, 0, 0), as shown in Fig. 3.21. A current I flows around the ring in
the direction indicated in panel (B) of Fig. 3.21.
(A)
(B)
z
Fig. 3.21
z
I
y
x
z
x
y-z-Plane View
z
y
y
x
x-z-Plane View
Suppose this ring is now rotated by 45o around the x-axis, with the rotation direction being
counter-clockwise in the y−z-plane view shown in panel (A) of Fig. 3.21. The magnetic
~ ≡ (Bx , By , Bz ) produced by the current I at the center of the ring, after the
field vector B
o
~ by:
45 -rotation, is then given in terms of the field strength B ≡ |B|
B
√ ,√
(A) (Bx , By , Bz ) = (0, −B
)
2
2
√ , −B
√ )
(B) (Bx , By , Bz ) = (0, −B
2
2
(C) (Bx , By , Bz ) = (0, √B2 , √B2 )
(D) (Bx , By , Bz ) = (0, −B, 0)
(E) (Bx , By , Bz ) = (0, 0, +B)
Answer: (B)
~
Consider first the B-vector
before the loop is rotated by 45o around x-axis.
~ at center of loop (or more generally at any point inside
For circular loop current I, vector B
~ must
the loop!) must be perpendiclar to the plane of the loop. So, in Fig. 3.21 panel (B), B
be perpendicular to plane of drawing, i.e., parallel to y-axis. Therefore, By = ±B 6= 0 and
Bx = Bz = 0 before 45o rotation of loop around x-axis.
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~
Also, the loop current direction and B-direction
(at loop center or more generally at any
point inside the loop) are related by a right-hand (RH) rule: point RH 4 fingers around loop
~ perpendicular to plane of loop.
in I-direction; then the RH thumb points in direction of B,
~ must be pointing out of the plane of drawing,
So, in Fig. 3.21 panel (B), by this RH rule, B
~ points in −y-direction, before 45o rotation of loop around x-axis. Therefore, when
i.e., B
~ points leftward, (≡ −y-direction!),
viewed in the y − z-plane, as in panel (A) of Fig. 3.21, B
o
before 45 rotation of loop around x-axis.
Now consider, in Fig. 3.21 panel (A), what happens if you rotate the loop by 45o . The
crucial point to note is this: rotating the loop by 45o counter-clockwise [in Fig. 3.21,
~ at its center,
panel (A)] around x-axis will also rotate its magnetic field, including B
o
~ points into the 3rd
by the same 45 counter-clockwise around x-axis. So, after rotation, B
o
quadrant of the y − z-plane, with 45 angle to both −z- and −y-direction. Draw it!!
~ must satisfy: By = Bz < 0 and Bx = 0 (in
So, after 45o rotation of loop, the components of B
order q
to point into 3rd y −
quadrant and be at 45o angle to −y- and −z-axes). Hence
q z-plane√
√
√
B = Bx2 + By2 + Bz2 = 2By2 = 2|By |; and hence, By = −B/ 2 and Bz = −B/ 2, so
that Bz = By < 0.
CP 3.22: A very thin circular ring lies initially in the x − y-plane and it is centered at the
coordinate origin O ≡ (0, 0, 0), as shown in Fig. 3.22. A current I flows around the ring in
the direction indicated in panel (B) of Fig. 3.22.
(A)
(B)
z
Fig. 3.22
y
I
y
x
z
x
y-z-Plane View
y
y
z
x
x-y-Plane View
Suppose this ring is now rotated by 45o around the x-axis, with the rotation direction being
clockwise in the y−z-plane view shown in panel (A) of Fig. 3.22. The magnetic field
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~ ≡ (Bx , By , Bz ) produced by the current I at the center of the ring, after the
vector B
~ by:
45o -rotation, is then given in terms of the field strength B ≡ |B|
B
√ ,√
(A) (Bx , By , Bz ) = (0, −B
)
2
2
√ , −B
√ )
(B) (Bx , By , Bz ) = (0, −B
2
2
(C) (Bx , By , Bz ) = (0, √B2 , √B2 )
(D) (Bx , By , Bz ) = (0, −B, 0)
(E) (Bx , By , Bz ) = (0, 0, +B)
Answer: (C)
~
Consider first the B-vector
before the loop is rotated by 45o around x-axis.
~ at center of loop (or more generally at any point inside
For circular loop current I, vector B
~ must
the loop!) must be perpendiclar to the plane of the loop. So, in Fig. 3.22 panel (B), B
be perpendicular to plane of drawing, i.e., parallel to z-axis. Therefore, Bz = ±B 6= 0 and
Bx = By = 0 before 45o rotation of loop around x-axis.
~
Also, the loop current direction and B-direction
(at loop center or more generally at any
point inside the loop) are related by a right-hand (RH) rule: point RH 4 fingers around loop
~ perpendicular to plane of loop.
in I-direction; then the RH thumb points in direction of B,
~ must be pointing out of the plane of drawing,
So, in Fig. 3.22 panel (B), by this RH rule, B
~ points in +z-direction, before 45o rotation of loop around x-axis. Therefore, when
i.e., B
~ points upward, (≡ +z-direction!),
viewed in the y − z-plane, as in panel (A) of Fig. 3.22, B
o
before 45 rotation of loop around x-axis.
Now consider, in Fig. 3.22 panel (A), what happens if you rotate the loop by 45o . The
crucial point to note is this: rotating the loop by 45o clockwise [in Fig. 3.22, panel
~ at its center, by the
(A)] around x-axis will also rotate its magnetic field, including B
o
~ points into the 1st quadrant of the
same 45 clockwise around x-axis. So, after rotation, B
o
y − z-plane, with 45 angle to both +z- and +y-direction. Draw it!!
~ must satisfy: By = Bz > 0 and Bx = 0 (in
So, after 45o rotation of loop, the components of B
order q
to point into 1st y −
quadrant and be at 45o angle to +y- and +z-axes). Hence
qz-plane √
√
√
B = Bx2 + By2 + Bz2 = 2By2 = 2|By |; and hence, By = +B/ 2 and Bz = +B/ 2, so
that Bz = By > 0.
CP 3.23: Two very thin circular rings, labeled 1 and 2 in the following, with radii R1 and
R2 , respectively, and R1 > R2 , are both centered at the coordinate origin O ≡ (0, 0, 0), as
shown in Fig. 3.23.
Ring 1 lies in the y − z-plane with current I1 flowing around the ring in the direction indicated in panel (B) of Fig. 3.23. Ring 2 lies in the x − z-plane with current I2 flowing around
the ring in the direction indicated in panel (C) of Fig. 3.23.
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Fig. 3.23
(A)
(B)
(C)
y
z
z
I1
1
2
x
I2
2
2
y
x
1
1
y
z
x-y-Plane View
z
x
z
y
x
y
y-z-Plane View
x
x-z-Plane View
Let B1 > 0 and B2 > 0 denote the magnetic field strengths produced at the origin O by only
~ ≡ (Bx , By , Bz )
I1 alone and only I2 alone, respectively. The total magnetic field vector B
produced by both currents at the origin O is then given by:
(A)
(B)
(C)
(D)
(E)
(Bx , By , Bz ) = (0 , B1 + B2 ,
(Bx , By , Bz ) = (B1 − B2 , 0 ,
(Bx , By , Bz ) = (−B2 , −B1 ,
(Bx , By , Bz ) = (+B2 , −B1 ,
(Bx , By , Bz ) = (+B1 , −B2 ,
0)
0)
0)
0)
0)
Answer: (E)
~ i at center of loop i (or more
For single circular loop current Ii , its magnetic field vector B
generally at any point inside the loop!) must be perpendiclar to the plane of the loop.
~ 1 ≡ (B1x , B1y , B1z ) generated by I1 must be
So, in Fig. 3.23 panel (B), the magnetic field B
perpendicular to plane of drawing, i.e., parallel to x-axis. Therefore, B1x = ±B1 6= 0 and
B1y = B1z = 0.
~ 2 ≡ (B2x , B2y , B2z ) generated by I2
Likewise, in Fig. 3.23 panel (C), the magnetic field B
must be perpendicular to plane of drawing, i.e., parallel to y-axis. Therefore, B2y = ±B2 6= 0
and B2x = B2z = 0.
~ i -direction (at loop center or more generally at
Also, the loop current (Ii ) direction and B
any point inside the loop) for a single loop are related by a right-hand (RH) rule: point
18
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~ i,
RH 4 fingers around loop in Ii -direction; then the RH thumb points in direction of B
perpendicular to plane of loop.
~ 1 must be pointing out of the plane of drawing,
So, in Fig. 3.23 panel (B), by this RH rule, B
~ 1 points in +x-direction. Therefore, B1x = +B1 > 0 and B
~ 1 = (+B1 , 0, 0).
i.e., B
~ 2 must be pointing out of the plane of
Likewise, in Fig. 3.23 panel (C), by this RH rule, B
~ 2 points in −y-direction. Therefore, B2y = −B2 < 0 and B
~ 2 = (0, −B2 , 0).
drawing, i.e., B
~ is the vector sum of B
~ 1 and B
~ 2 , i.e., B
~ =B
~1 + B
~ 2 . So,
Lastly, the total magnetic field B
~ ≡ (Bx , By , Bz ) = (B1x + B2x , B1y + B2y , B1z + B2z ) =
adding component-to-component: B
(B1 +0, 0−B2 , 0+0) = (+B1 , −B2 , 0)
CP 3.24: Two very thin circular rings, labeled 1 and 2 in the following, with radii R1 and
R2 , respectively, and R1 > R2 , are both centered at the coordinate origin O ≡ (0, 0, 0), as
shown in Fig. 3.24.
Ring 1 lies in the x − z-plane with current I1 flowing around the ring in the direction indicated in panel (B) of Fig. 3.24. Ring 2 lies in the x − y-plane with current I2 flowing around
the ring in the direction indicated in panel (C) of Fig. 3.24.
Fig. 3.24
(A)
(B)
(C)
z
z
y
I1
1
2
y
I2
2
2
x
x
1
1
z
x
z
y
y-z-Plane View
y
y
x
z
x-z-Plane View
x
x-y-Plane View
Let B1 > 0 and B2 > 0 denote the magnetic field strengths produced at the origin O by only
~ of the total magnetic field
I1 alone and only I2 alone, respectively. The strength B ≡ |B|
~ ≡ (Bx , By , Bz ) produced by both currents at the origin O is then given by:
vector B
(A) B = B1 + B2
(B) B = (B1 B2 )1/2
19
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(C) B = (B12 + B22 )1/2
(D) B = |B1 − B2 |
(E) B = (B12 B22 )1/4
Answer: (C)
~ i at center of loop i (or more
For single circular loop current Ii , its magnetic field vector B
generally at any point inside the loop!) must be perpendiclar to the plane of the loop. Also,
~ i -direction (at loop center or more generally at any point
the loop current (Ii ) direction and B
inside the loop) for a single loop are related by a right-hand (RH) rule: point RH 4 fingers
~ i , perpendicular to
around loop in Ii -direction; then the RH thumb points in direction of B
plane of loop.
~ 1 ≡ (B1x , B1y , B1z ) generated by I1 must be
So, in Fig. 3.24 panel (B), the magnetic field B
perpendicular to plane of drawing, i.e., parallel to y-axis, and out of the plane of drawing,
~ 1 = (0, −B1 , 0).
i.e., in −y-direction Therefore, B1y = −B1 < 0 and B1x = B1z = 0 and B
~ 2 ≡ (B2x , B2y , B2z ) generated by I2 must
Likewise, in Fig. 3.24 panel (C), the magnetic field B
be perpendicular to plane of drawing, i.e., parallel to z-axis. and out of the plane of drawing,
~ 2 = (0, 0, +B2 ).
i.e., in +z-direction Therefore, B2z = +B2 > 0 and B2x = B2y = 0 and B
~ is the vector sum of B
~ 1 and B
~ 2 , i.e., B
~ =B
~1 + B
~ 2 . So, adding
The total magnetic field B
~
component-to-component: B ≡ (Bx , By , Bz ) = (B1x+B2x , B1y+B2y , B1z+B2z ) = (0+0, −B1+
0, 0+B2 ) = (0, −B1 , +B2 ).
~ = (Bx2 + By2 + Bz2 )1/2 = (B12 + B22 )1/2 .
Then B ≡ |B|
Note: There’s a faster way to get this last result. Unlike CP 3.23, you don’t actually have to
~ 1 and B
~ 2 and the signs of the components. It’s sufficient
figure out all the components of B
~1 ⊥ B
~2 !
to figure out only that B
~ 2 is easy to infer from the fact that each, B
~ 1 and B
~ 2 , is generated
~ 1 being perpendicular to B
B
by, and at the center of, a circular loop; and from the fact that the respective two loop planes
are perpendicular to each other. If the two planes of loops 1 and 2 are at an angle of 90o
~ 1 and B
~ 2.
from each other, then so must be B
~ 2 , then by geometric vector addition, you see that B
~ 1, B
~ 2 and
~1 ⊥ B
If you know only that B
~
~
~
~
B = B1 + B2 form a rectangular triangle with B as the hypotenuse. Hence by Pythagoras:
B 2 = B12 + B22 .
Draw it all!! ...in the y − z-plane, Fig. 3.24, panel (A).
CP 3.25: A very long thin straight wire running along the z-axis through the coordinate
origin, carries a current I1 in the +z-direction, as shown in Fig. 3.25.
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Fig. 3.25
P
y
(A)
(B)
(E)
(C)
(D)
x
I1
y
x-y-Plane View
z
x
Which arrow drawn at point P in the x − y-plane could correctly represent the magnetic
~ produced by I1 at P ?
field vector B
(A)
(B)
(C)
(D)
(E)
Answer: (A)
~
The long straight I1 wire surrounds itself by concentric circular B-field
lines (FLs). By
right-hand (RH) rule, these FLs loop around I1 in counter-clockwise direction, as seen in the
x − y-plane view in Fig. 3.25: point RH thumb in I1 direction; then RH 4 fingers point in
~
B-FL
direction.
Now draw that circular FL (or a piece of it) which passes through point P and draw the
~
~ pointing in the counter-clockwise FL looping
B-vector
at P , tangential to that FL, with B
~
direction: B then points in the direction of arrow (A).
CP 3.26: A very long thin straight wire running along the z-axis through the coordinate
origin, carries a current I1 in the +z-direction, as shown in Fig. 3.25.
Suppose a second wire running parallel to the z-axis through point P carries a current
I2 (not shown in Fig. 3.25) in the −z-direction. Which arrow drawn at point P in the
x − y-plane could correctly represent the magnetic force vector F~ on the I2 -wire due to the
magnetic field produced by I1 ?
(A)
(B)
(C)
(D)
(E)
Answer: (B)
Since I1 and I2 run in parallel wires, but in opposite directions, they repel each other. So
21
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the force vector F~ on the I2 -wire must point away from the I1 wire, i.e., in the direction of
arrow (B).
~2 × B
~ 1 . Here, B
~ 1 is
You can also get the same result by taking the cross-product F~ = I2 L
~ 1 pointing in the direction of arrow (A) (see
the magnetic field produced at P by I1 , with B
~
~
CP 3.25); and L2 is the L-vector of I2 , pointing along the I2 -wire in I2 -direction, i.e., in
−z-direction.
CP 3.27: A very long thin straight wire running parallel to the z-axis and cutting through
the x-axis, carries a current I1 in the −z-direction, as shown in Fig. 3.27.
Fig. 3.27
(C)
y
(D)
(B)
(E)
(A)
I1
P
x
y
x-y-Plane View
z
x
Which arrow drawn at point P in the x − y-plane could correctly represent the magnetic
~ produced by I1 at P ?
field vector B
(A)
(B)
(C)
(D)
(E)
Answer: (E)
~
The long straight I1 -wire surrounds itself by concentric circular B-field
lines (FLs). By righthand (RH) rule, these FLs loop around I1 in clockwise direction, as seen in the x − y-plane
~
view in Fig. 3.27: point RH thumb in I1 direction; then RH 4 fingers point in B-FL
direction.
Now draw that circular FL (or a piece of it) which passes through point P and draw the
~
~ pointing in the clockwise FL looping direction:
B-vector
at P , tangential to that FL, with B
~
B then points in the direction of arrow (E).
CP 3.28: A very long thin straight wire running parallel to the z-axis and cutting through
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the x-axis, carries a current I1 in the −z-direction, as shown in Fig. 3.27.
Suppose a second wire running parallel to the z-axis through point P carries a current
I2 (not shown in Fig. 3.27) in the −z-direction. Which arrow drawn at point P in the
x − y-plane could correctly represent the magnetic force vector F~ on the I2 -wire due to the
magnetic field produced by I1 ?
(A)
(B)
(C)
(D)
(E)
Answer: (A)
Since I1 and I2 run in parallel wires, and in the same direction, they attract each other. So
the force vector F~ on the I2 -wire must point towards the I1 wire, i.e., in the direction of
arrow (A).
~2 × B
~ 1 . Here, B
~ 1 is
You can also get the same result by taking the cross-product F~ = I2 L
~
the magnetic field produced at P by I1 , with B1 pointing in the direction of arrow (E) (see
~ 2 is the L-vector
~
CP 3.27); and L
of I2 , pointing along the I2 -wire in I2 -direction, i.e., in
−z-direction.
CP 3.29: Two very long thin straight wires running parallel to the z-axis and cutting
through the y-axis and through the x-axis, respectively, carry currents I1 in the +z-direction
and I2 in the −z-direction, respectively, as shown in Fig. 3.29.
Fig. 3.29
(C)
y
I1
(D)
(B)
(E)
P
(A)
I2
x
y
x-y-Plane View
z
x
Which arrow drawn at point P in the x − y-plane could correctly represent the total magnetic
~ produced jointly by I1 and I2 at P ?
field vector B
(A)
(B)
(C)
(D)
(E)
23
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Answer: (D)
~ i generated at P by each single
First figure out the direction of the magnetic field vector B
current Ii (with i = 1 or 2 for wire 1 or wire 2, respectively), using the concentric circular
~ i -field lines (FLs) centered at Ii . By right-hand (RH) rule, these FLs loop around I1 in
B
counter-clockwise direction, and around I2 in clockwise direction, as seen in the x − y-plane
view in Fig. 3.29. To see this, point RH thumb in Ii direction; then RH 4 fingers point in
~ i -FL direction.
B
~ 1 -FL and that circular B
~ 2 -FL (or a piece of each) which each
Now draw that circular B
~ 1 -vector and the B
~ 2 -vector at P , each tangential
passes through point P . Then draw the B
~
~2
to its respective FL, with B1 pointing in its counter-clockwise FL looping direction; and B
pointing in its clockwise FL looping direction.
~ i -vectors at P !!
And I really mean it: Draw these FLs into Fig. 3.29, with their respective B
~ 2 will point horizontally
~ 1 will then point vertically upward, i.e., in the +y-direction; and B
B
rightward, i.e., in the +x-direction
~
~ =B
~1 + B
~ 2 , therefore must have a horizontally rightThe resultant total B-vector,
B
~ 2 ) component and a vertically upward (B
~ 1 ) component. Only arrow (D) in Fig.
ward (B
3.29 satisfies both of these conditions.
CP 3.30: Two very long thin straight wires running parallel to the z-axis and cutting
through the x-axis and through the coordinate origin, respectively, carry currents I1 in the
−z-direction and I2 in the +z-direction, respectively, as shown in Fig. 3.30.
Fig. 3.30
y
(D)
(C)
(E)
(B)
(A)
I2
I1
P
x
y
x-y-Plane View
z
x
Which arrow drawn at point P in the x − y-plane could correctly represent the total magnetic
24
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
~ produced jointly by I1 and I2 at P ?
field vector B
(A)
(B)
(C)
(D)
(E)
Answer: (D)
~ i generated at P by each single
First figure out the direction of the magnetic field vector B
current Ii (with i = 1 or 2 for wire 1 or wire 2, respectively), using the concentric circular
~ i -field lines (FLs) centered at Ii . By right-hand (RH) rule, these FLs loop around I1 in
B
clockwise direction, and around I2 in counter-clockwise direction, as seen in the x − y-plane
view in Fig. 3.30. To see this, point RH thumb in Ii direction; then RH 4 fingers point in
~ i -FL direction.
B
~ 1 -FL and that circular B
~ 2 -FL (or a piece of each) which each
Now draw that circular B
~
~ 2 -vector at P , each tangential
passes through point P . Then draw the B1 -vector and the B
~ 1 pointing in its counter-clockwise FL looping direction; and B
~2
to its respective FL, with B
pointing in its clockwise FL looping direction.
~ i -vectors at P !!
And I really mean it: Draw these FLs into Fig. 3.30, with their respective B
~ 2 will point into
~ 1 will then point horizontally rightward, i.e., in the +x-direction; and B
B
o
the 2nd quadrant of the x − y-plane, at 45 angles from both the −x- and the +y-direction.
~ 1 has no (zero!) vertical component, ı.e., B1y = 0; but B
~ 2 does have a
Therefore, B
non-zero upward vertical component, i.e., B2y > 0.
~
~ =B
~1 + B
~ 2 , therefore must also have a non-zero upward
The resultant total B-vector,
B
~ 2 which can not be canceled by B
~ 1 . That is, By = B1y + B2y =
vertical component from B
~ can only point into the 1st or 2nd quadrant of the x − y-plane,
B2y > 0. In other words, B
above the horizontal; it can not point horizontally or into the 3rd or 4th quadrant
below the horizontal. Only arrow (D) in Fig. 3.30 satisfies this condition.
~ 1 and B
~ 2 have opposing horizontal components, with B1x > 0 and B2x < 0.
Note here that B
~ =B
~ 1 +B
~ 2 , given by Bx = B1x +B2x = |B1x |−|B2x |,
Therefore, the horizontal component of B
~ could have horizontally either a rightward
could have either positive or negative sign, i.e., B
~ 1 and B
~ 2 . The arrrow (D)
or a leftward component, depending on the relative strengths of B
~ pointing into the 1st
shown in Fig. 3.30 represents the case Bx > 0 or B1x > |B2x |, with B
~ pointing into
quadrant. On the other hand, the case Bx < 0 or B1x < |B2x | would have B
the 2nd quadrant of the x − y-plane.
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