Download Solution to PHYS 1112 In-Class Exam #3A

Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Solution to PHYS 1112 In-Class Exam #3A
Thu. April 9, 2009, 11:00am-12:15pm
Conceptual Problems
Problem 1: If a wire of some length L and with a circular cross-section of diameter D has
a resistance of 50 kΩ, what will be the resistance of a wire made from the same material, at
the same temperature, of length 15L and diameter 5D ?
(A)
(B)
(C)
(D)
(E)
10 kΩ
30 kΩ
60 kΩ
83.3 kΩ
150 kΩ
Answer: (B):
Using R = ρL/A and A = π(D/2)2 , we get R ∝ L/D2 . Increasing L → L0 = 15L and
D → D0 = 25D thus changes R → R0 = 15 × R/52 = (3/5) × 50Ω = 30Ω.
Problem 2: 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 V1 across R1 , observed in the
three different circuits. (Hint: V1 = R1 I1 .)
Fig. 3.04
Io
I2
I3
E
I1
(X)
Io
R2
R3
R1
I2
Io
R2
I1
E
I3
R1
I3
R3
I1
E
I2
R3
(Y)
(Z)
1
R1
R2
Physics 1112
Spring 2009
(A)
(B)
(C)
(D)
(E)
University of Georgia
Instructor: HBSchüttler
V1 (X) > V1 (Y ) > V1 (Z)
V1 (X) > V1 (Z) > V1 (Y )
V1 (Y ) > V1 (X) > V1 (Z)
V1 (Z) > V1 (Y ) > V1 (X)
V1 (Y ) > V1 (Z) > V1 (X)
Answer: (E)
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 .
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 .
Since V1 = R1 I1 in all 3 circuits (with the same R1 in all 3 circuits!), I1 (X) < I1 (Z) < I1 (Y )
implies V1 (X) < V1 (Z) < V1 (Y ). That is the same as saying V1 (Y ) > V1 (Z) > V1 (X).
Problem 3: 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
~ perpendicular to, and pointing out of, the plane of the drawing.
lower horizontal line, and B
The incident beam, below the lower horizontal line, is in the plane of the drawing.
Fig. 3.15
B (out)
R
Q
P
S
B=0
~
From the semicircular ion trajectories in the 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.
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Physics 1112
Spring 2009
(A)
(B)
(C)
(D)
(E)
P :+,
P :+,
P :−,
P :+,
P :−,
University of Georgia
Instructor: HBSchüttler
Q:+,
Q:−,
Q:+,
Q:−,
Q:−,
R:−,
R:−,
R:+,
R:+,
R:+,
S:−,
S:+,
S:−,
S:−,
S:+,
Answer: (B)
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 R and Q are beginning to be deflected to the left, ions P 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 R and Q that center of the circle in
Fig. 3.15 is to the left of the point of entry; whereas for P 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
R and Q, F~ points to the left; for P 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 R 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 : R 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 : P and S ions have positive charges.
Problem 4: 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.
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Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
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)
Answer: (E)
~ 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 both I1
B
and I2 in clockwise direction, as seen in the x − y-plane view in Fig. 3.29. To see this, point
~ i -FL direction.
RH thumb in Ii direction; then RH 4 fingers point in B
~ 1 -FL and that circular B
~ 2 -FL (or a piece of each) which each passes
Now draw that circular B
~ 1 -vector and the B
~ 2 -vector at P , each tangential to its
through point P . Then draw the B
~ 1 pointing in its clockwise FL looping direction; and B
~ 2 pointing in its
respective FL, with B
clockwise FL looping direction.
~ i -vectors at P !!
And I really mean it: Draw these FLs into Fig. 3.29, with their respective B
~ 1 will then point vertically downward, i.e., in the −y-direction; and B
~ 2 will point horizonB
tally rightward, i.e., in the +x-direction
~
~ =B
~ 1 +B
~ 2 , therefore must have a horizontally rightward
The resultant total B-vector,
B
~ 2 ) component and a vertically downward (B
~ 1 ) component. Only arrow (E) in Fig. 3.29
(B
satisfies both of these conditions.
4
Physics 1112
Spring 2009
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Instructor: HBSchüttler
Numerical Problems
Problem 5: A wire made of ugalium alloy (a very poorly conducting metal!), with a circular
cross-section of 0.07mm diameter, draws a 2.8mA current when connected to a 15V battery.
The resistivity of ugalium is 3 × 10−6 Ω · m. The area of a circle with a diameter D is
A = (π/4)D2 .
How long is the wire ?
(A)
(B)
(C)
(D)
(E)
6.87m
76.3m
594m
4.17km
325km
Answer: (A)
The resistance of the wire is R = V /I = 15V/0.0028A = 5357.1Ω. Also, R = ρL/A and
A = πD2 /4. So L = AR/ρ = (π/4)D2 R/ρ = (π/4)(0.07 × 10−3 m)2 (5357.1Ω)/(3.0 × 10−6 Ω ·
m) = 6.872m.
Problem 6: How much heat does the current flow through the ugalium wire described in
Problem 5 produce during 10 hours ?
(A)
(B)
(C)
(D)
(E)
14.3J
266J
1.51kJ
33.5kJ
473kJ
Answer: (C)
P ≡ ∆U/∆t = IV , so ∆U = P ∆t = IV ∆t = (0.0028A)(15.0V)(10h)(3600s/h) = 1.512 ×
103 J
Problem 7: In circuit Z shown in the rightmost panel of Fig. 3.04 (see Problem 2), the
resistances are R1 = 5 Ω, R2 = 7 Ω, R3 = 4 Ω and the battery voltage is E = 36V.
What is the current Io flowing through the battery ?
(A) 3A
(B) 7A
(C) 10A
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Physics 1112
Spring 2009
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Instructor: HBSchüttler
(D) 12A
(E) 22A
Answer: (D)
R1 and R2 are in series: replace by R12 = R1 + R2 . Then R12 and R3 are in parallel: replace by Req = (1/R3 + 1/R12 )−1 . Then Io = E/Req = E[1/R3 + 1/(R1 + R2 )] =
(36V)[(1/4Ω + 1/(5 + 7)Ω] = 12A.
Problem 8: If I1 = +8A and I2 = +3A and the voltage drop VR ≡ Va − Vb across the
resistor is is VR = +20V, in the circuit fragment shown in Fig. 3.10, what is the resistance
R?
I1
I2
Fig. 3.10
a
I3
R
b
(A)
(B)
(C)
(D)
(E)
12Ω
8Ω
6Ω
4Ω
2Ω
Answer: (D)
By Kirchhoff’s junction rule: I1 = I2 + I3 ; so I3 = I1 − I2 = (8 − 3)A = 5A. Then
R = VR /I3 = 20V/5A = 4Ω.
6
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Problem 9: Last time you checked you weighed in at 170kg. Also, some space alien
frat brothers, as part of their annual pledging ritual, have implanted a point charge of +5
Coulomb in your head. Now, to make matters worse, they’ve ordered you to jog on a straight
east-west track to become suspended against gravity by the magnetic force exerted on you
by planet earth’s magnetic field. Just to be nice to you, they’ve cranked up earth’s field to
80 Tesla with the field vector pointing horizontally due north. But they left earth’s value of
g at 9.81m/s2 ...
In case you can still handle electromagnetic theory with all that electricity under your hairdo
– and you still want to join that fraternity:
How fast do you have to run, and in which direction, to achieve levitation ?
(A)
(B)
(C)
(D)
(E)
4.17m/s,
4.17m/s,
3.02m/s,
3.02m/s,
1.84m/s,
running
running
running
running
running
eastward
westward
eastward
westward
westward
Answer: (A)
To lift off, the upward magnetic force F = |q|vB sin θ must cancel the downward weight
~
force mg. Since you’re running on an east-west track, ~v points east or west, while B
o
~
points north; so: θ ≡ 6 (~v , B) = 90 and sin θ = 1. So, mg = |q|vB or v = mg/(|q|B) =
(170kg)(9.81m/s2 )/[(5C)(80T)] = 4.17m/s.
~ pointing
Also, if you run east, ~v points east and then q~v also points east, since q > 0. With B
~ by right-hand rule, then points upward, as required for lift-off. (Note: if
north, F~ = q~v × B,
you were running west, q~v would point west and F~ would point downward, i.e., in the wrong
direction.)
Problem 10: If a 20A current flows in a thin straight metal rod of length 3m through a
uniform magnetic field of 60T and the direction of the current flow is at an angle of 35o from
the direction of the magnetic field vector, what is the strength of the magnetic force exerted
on the rod ?
(A)
(B)
(C)
(D)
(E)
581N
1272N
2065N
2949N
4513N
Answer: (C)
7
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
F = ILB sin θ = (20A)(3m)(60T) sin(35o ) = 2065N.
Problem 11: If you are standing below of a very long horizontal straight wire which is
carrying a current of 70A in northward direction and the tip of your nose is exactly 14m
below the wire, what is the strength and direction of the magnetic field produced by that
wire at the tip of your nose ?
~ pointing northward
(A) 0.4µT, B
~ pointing westward
(B) 1.0µT, B
~ pointing eastward
(C) 7.0µT, B
~ pointing downward
(D) 12.0µT, B
~ pointing upward
(E) 18.0µT, B
Answer: (B)
For an infinitely (≡very) long wire, B = (µo /2π)(I/r) = (2 × 10−7 Tm/A)(70A)/(14m) =
1.0 × 10−6 T.
~ must point westward at your observation position below the wire, using the concenAlso B
~
tric circular B-field
lines (FLs) centered at the wire: By right-hand (RH) rule, the FLs loop
around current I in clockwise direction, when viewed facing northward, i.e. looking in the
~
direction of current flow. Hence, the B-vector
points: westward below the wire, eastward
above the wire, downward to the right of (≡ east of) the wire, and upward to the left of (≡
west of) the wire. Draw it !!!
Problem 12: Two very thin circular rings, labeled 1 and 2 in the following, with radii
R1 = 9m and R2 = 6m, 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 = 18A 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 = 24A
flowing around the ring in the direction indicated in panel (C) of Fig. 3.23.
8
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
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
x
z
y
y
y-z-Plane View
x
x-z-Plane View
What is the angle between the +x-direction and the total magnetic field vector produced by
the two rings at the origin O ? (Hint: You only need to calculate the ratio B2 /B1 !)
(A)
(B)
(C)
(D)
(E)
153.4o
116.6o
90.0o
63.4o
26.6o
Answer: (D)
~ 1 and B
~ 2 generated
By right-hand rule applied to the magnetic field vector contributions B
by ring 1 and 2, respectively, (see also, e.g., CP 3.19 through CP 3.23) have the following
~ 1 = (+B1 , 0, 0) points in +x-direction and B
~ 2 = (0, −B2 , 0) points in −ydirections: B
~ =B
~1 + B
~ 2 = (+B1 , −B2 , 0) points at an
direction. So, the total magnetic field vector B
o
angle φ < 90 below the +x direction, given by tan φ = B2 /B1 . Draw it!!
Also, Bi = (µo /2)(Ii /Ri ) for ring i = 1 and i = 2. So, in the ratio B2 /B1 , the pre-factors
µo /2 cancel out and thus: tan φ = B2 /B1 = (I2 /R2 )/(I1 /R1 ) = (24A/6m)/(18A/9m) = 2.0,
hence φ = arctan(2.0) = 63.43o .
9
Physics 1112
Spring 2009
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Instructor: HBSchüttler
Formula Sheet
Capacitors and Capacitance
(1) Definition of Capacitance: For two oppositely charged metallic objects a and b,
with −Q stored on a and Q stored on b, their electric potential difference V ≡ Vb − Va is
proportional to the charge Q. The capacitance of the two metallic objects is then defined
as:
Q
Q
C≡
,
hence
Q = CV
or
V =
V
C
(2) Voltage and Capacitance of a Planar Capacitor: For two oppositely charged,
parallel planar metallic plates, each of opposing surface area A, closely spaced with distance
~ inside the capacitor, and capacitance C
d, the voltage V , the electric field strength E ≡ |E|
are related by
|Q|d
A
|Q|
|V | = Ed =
;
C≡
= κo ≡ κ Co
κo A
|V |
d
where κ is the dielectric constant of the dielectric (insulating) material between the plates
and κ = 1 for vacuum or air, and Co ≡ o A/d is the capacitance without dielectric.
(3) Electric Field Energy Storage in a Capacitor: The energy UE required to build up
a charge Q and a voltage V = Q/C in a capacitor is stored as electric field energy between
the capacitor plates and it is given by
UE =
1 2 1
Q = CV 2
2C
2
Current, Resistance, Ohm’s Law
(1) Definition of Current: For a charge ∆Q flowing through a wire or, more generally,
some cross-sectional area of a conducting object, over a time interval ∆t, the current I is
I≡
∆Q
∆t
(2) Ohm’s Law: The current I flowing through an ”ohmic” conductor (e.g., metallic wire)
and the applied voltage drop V ≡ Va − Vb across the conductor are proportional:
V = RI ∝ I
with R ≡
V
= constant ,
I
i.e., the conductor’s resistance R is independent of I or V . Here Va and Vb denote the
electric potential at current’s point of entry a and current’s point of exit b into/from the
conductor, respectively.
10
Physics 1112
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(3) Resistance and Resistivity: For current flow through an ohmic conductor, over a
constant cross-sectional area A and a length L, the conductor’s resistance R is proportional
to L and inversely proportional to A:
R=ρ
L
L
∝
A
A
where the resistivity ρ is independent of L or A or, more generally, independent of the size
and shape of the conductor: ρ depends on the material, the chemical composition and the
temperature of the conductor.
Electric Power Dissipation
(1) General Electric Power Dissipation Law: In any electric device with current I
flowing from entry point a to exit point b, subject to a potential drop V ≡ Va − Vb , the
electric power dissipated, i.e., amount of electric energy per time consumed and heat per
time generated is
P = IV .
(2) Power Dissipation in Ohmic Resistors: For the special case of I flowing through
an ohmic resistor with voltage drop V = RI,
P = RI 2 =
1 2
V .
R
Electric Circuits
(1) Equivalent Capacitance: For any combination of capacitors C1 , C2 , ... having only one
point of entry a and one point of exit b of stored charge, the total voltage drop V ≡ Va − Vb
across the capacitor combination is proportional to the total stored charge Q, injected into
the capacitor combination at point a, with equal charge Q extracted at point b,
V =
1
Q∝Q
Ceq
with Ceq ≡
Q
= constant ,
V
i.e., the capacitor combination’s equivalent capacitance Ceq is independent of Q or V .
(2) Capacitors C1 , C2 , ... in Series:
1
V
1
1
≡
=
+
+ ...
Ceq
Q
C1 C2
or
Ceq ≡
−1
Q 1
1
=
+
+ ...
V
C1 C2
and
Q = Q1 = Q2 = ... ;
V = V1 + V2 + ...
where Q is total charge injected at a and V ≡ Va − Vb is total voltage drop; Q1 , Q2 , ... are
the charges stored on a-side plates of C1 , C2 , ... respectively; and V1 = Q1 /C1 , V2 = Q2 /C2 ,
... are the voltage drops across C1 , C2 , ... respectively.
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Physics 1112
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(3) Capacitors C1 , C2 , ... in Parallel:
Ceq ≡
Q
= C1 + C2 + ...
V
and
Q = Q1 + Q2 + ... ;
V = V1 = V2 = ...
where Q is total charge injected at a and V ≡ Va − Vb is total voltage drop; Q1 , Q2 , ... are
the charges stored on a-side plates of C1 , C2 , ... respectively; and V1 = Q1 /C1 , V2 = Q2 /C2 ,
... are the voltage drops across C1 , C2 , ... respectively.
(4) Equivalent Resistance: For any combination of ohmic resistors R1 , R2 , ... having
only one point of entry a and one point of exit b of current flow, the total voltage drop
V ≡ Va − Vb across the resistor combination is proportional to the total current I flowing
through the resistor combination
V = Req I ∝ I
with Req ≡
V
= constant ,
I
i.e., the resistor combination’s equivalent resistance Req is independent of I or V .
(5) Resistors R1 , R2 , ... in Series:
Req ≡
V
= R1 + R2 + ...
I
and
I = I1 = I2 = ... ;
V = V1 + V2 + ...
where I is total current; V ≡ Va − Vb is total voltage drop; I1 , I2 , ... are currents through
R1 , R2 , ... respectively; and V1 , V2 , ... are the voltage drops across V1 = R1 I1 , V2 = R2 I2 , ...
are the voltage drops across R1 , R2 , ... respectively.
(6) Resistors R1 , R2 , ... in Parallel:
1
I
1
1
≡
=
+
+ ...
Req
V
R1 R2
or
Req ≡
1
−1
V
1
=
+
+ ...
I
R1 R2
and
I = I1 + I2 + ... ;
V = V1 = V2 = ...
where I is total current; V ≡ Va − Vb is total voltage drop; I1 , I2 , ... are currents through
R1 , R2 , ... respectively; and V1 = R1 I1 , V2 = R2 I2 , ... are the voltage drops across R1 , R2 ,
... respectively.
(7) Kirchhoff Rule Circuit Analysis: Any circuit can be analyzed, i.e., unknown voltages, currents and/or resistances, etc., can be calculated from known ones, by the following
steps:
(K1) Find all ”junctions” (≡where more than two wires meet); label them (e.g., a, b, ...).
12
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(K2) Break up the circuit into ”branches” by cutting off all wires at each junction; assign
”current arrows” and current labels (e.g. I1 , I2 , ...) to each branch.
(K3) Mark ”check points” on all wires in circuit so that each circuit element (resistor,
battery, ...) is separated from every other element by at least one intervening check point;
add more check points for convenience, as needed; label all check points (e.g. f, g, ...); all
junctions also serve as check points.
(K4) Write down Kirchhoff’s ”Junction Rule” for enough junctions j:
X
X
I=
j,in
I
j,out
P
where j,in I is the sum of all currents with branches connected to j and arrows pointing
P
towards j and j,out I is the sum of all currents with branches connected to j and arrows
pointing away from j.
(K5) Write down Kirchhoff’s ”Loop Rule” for enough closed paths (”loops”) L in the circuit:
X
∆V = 0
L
P
where L ∆V is the sum of all voltage drops ∆V between adjacent check points, accumulated
as you walk around the closed loop from one check point to the next.
(K6) Express, as needed, each voltage drop ∆V ≡ Vx − Vy between two check points x and
y [used in Loop Rule (K5)], in terms of the voltage (EMF) E of intervening battery; in terms
of current I through intervening resistor R; or in terms of charge Q stored on intervening
capacitor C; etc. as follows:
For battery E between x and y,
Vx − Vy = +E if x conn. to + of battery;
Vx − Vy = −E if x conn. to − of battery .
For resistor R with current I between x and y,
Vx − Vy = +RI if I−arrow from x to y;
Vx − Vy = −RI if I−arrow from y to x .
For capacitor C between x and y, with charge Q stored on x-connected plate and −Q stored
on y-connected plate,
1
Vx − Vy = Q .
C
For wire only, without any other circuit element, between x and y,
Vx − Vy = 0 .
(K7) Solve system of coupled Junction Rule equations (K4) and Loop Rule equations (K5)
for all unknown quantities, with voltage drops ∆V in (K5) expressed in terms of unknown
currents, battery voltages, resistancesetc., using (K6) as needed.
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(K8) To find the electric potential difference Vp − Vq between any two check points p and
q in the circuit (incl. non-adjacent check points), walk through the circuit from q to p,
along any open path L connecting q and p, and add up the potential drops ∆V across each
element, as you step from one check point to the next:
X
Vp − Vq =
∆V .
L, q→p
Use (K6) to express voltage drops ∆V in terms of currents, resistances, battery voltages,
etc., as needed.
Forces from Magnetic Fields
(1) Lorentz Force Law: the magnetic force F~ on particle of charge q moving with velocity
~ is
~v at angle θ to magnetic field B
~ ;
F~ = q~v × B
~ ;
F~ ⊥ ~v , B
F = |q|vB sin θ (with 0 ≤ θ ≤ 180o );
~
with F~ -dir. given by right-hand (RH) turn of q~v into B.
(2) Motion of a Charged Particle in Uniform Magnetic Field: a particle of charge
~ in a
q and mass m moving with speed v ≡ |~v |, subject only to magnetic force F~ = q~v × B
~ has a circular trajectory of radius r, period T and frequency f of
uniform magnetic field B,
revolution, given by:
mv
1
m
r=
; T = = 2π
|q|B
f
|q|B
(3) Magnetic Force on Straight Wire Segment: the magnetic force F~ on I-wire seg~ pointing along wire, at angle θ to uniform magnetic field B,
~
ment, with length vector L
~
~
current I > 0 flowing in L-dir., L ≡ |L| =length of wire, is given by
~ ×B
~ ;
F~ = I L
~ B
~ ;
F~ ⊥ L,
F = ILB sin θ (with 0 ≤ θ ≤ 180o );
~ into B.
~
with F~ -dir. given by right-hand (RH) turn of I L
(4) Magnetic Torque on Current Loop: the magnetic torque ~τ on a planar I-wire loop
~ at angle θ to uniform magnetic field B,
~ A
~ perpendicular to plane of
with loop area vector A
~ (i.e.,
loop, with current I > 0 flowing around loop in right-handed (RH) dir. relative to A
~ must be chosen so that, if RH thumb points in A-dir.,
~
A
then RH 4 fingers point in I-dir.
~ =area enclosed by I-wire loop, is given by
around loop), and A ≡ |A|
~×B
~ ;
~τ = I A
~ B
~ ;
~τ ⊥ A,
τ = IAB sin θ (with 0 ≤ θ ≤ 180o );
~ into B.
~ This results holds for any shape of I-loop (not
with ~τ -dir. given by RH turn of I A
just rectangular!), as long as the loop is planar. Note that ~τ acts to rotate the loop, with
~τ -dir. as rotation axis, in RH dir. (i.e., if RH thumb points in ~τ -dir., then RH 4 fingers
point in τ -induced rotation dir.).
14
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Magnetic Fields from Electric Currents
(1) Magnetic Field from a Very Short Straight Wire (Biot-Savart Law): magnetic
~ generated at observation point P by a short I-wire segment, with
field contribution ∆B
vector ∆~s pointing along wire and ∆s ≡ |∆~s| =length of wire segment, with distance vector
~r pointing from wire segment to P at angle θ to ∆~s, with current I > 0 flowing in ∆~s-dir.,
~ =
∆B
µo
I ∆~s × ~r ;
4πr3
~ ⊥ ∆~s, ~r ;
∆B
~ =
∆B ≡ |∆B|
µo
I ∆s sin θ
4πr2
~ at oberva(2) Magnetic Field from Infinitely Long Straight Wire: magnetic field B,
tion point P , at distance r measured perpendicular to the line of the wire carrying current
I>0
~ = µo I ,
B ≡ |B|
2π r
~
B-dir.
is given by concentric circular field line (FL) passing through P and looping around
I in right-handed (RH) dir. (i.e., if RH thumb points in I-dir., then RH 4 fingers point in
FL dir.).
(3) Magnetic Force Between Two Long Straight Parallel Wires: Two long straight
parallel wires, both of length L, spaced a distance d apart with d L, and carrying currents
I1 and I2 , will attract or repel each other with force F on each wire, given by
F ≡ |F~ | =
µo I1 I2
L ,
2π d
Force between wires is attractive if I1 and I2 flow in the same direction; else it is repulsive.
~ at observation point
(4) Magnetic Field from Finite Straight Wire: magnetic field B
P , at distance r measured perpendicular to the line of the wire carrying current I > 0
~ = µo I cos α + cos β ,
B ≡ |B|
2π r
2
where α ≡ 6 (P, a, b) and β ≡ 6 (P, b, a) are the angles enclosed between the wire and lines
~
drawn from observation point P to the endpoints a and b of the wire, respectively. B-dir.
is given by concentric circular field line (FL) passing through P and looping around I in
right-handed (RH) dir. (i.e., if RH thumb points in I-dir., then RH 4 fingers point in FL
dir.).
~ at of center of
(5) Magnetic Field at Center of Circular Wire Loop: magnetic field B
wire loop of radius R, with N turns (i.e., wire is wrapped around loop N times), each turn
carrying same current I > 0
~ = µo N I .
B ≡ |B|
2 R
~
B-dir. is perpendicular to the plane of the I-loop and in in right-handed (RH) dir. rel. to I
~
(i.e., if RH 4 fingers point in I-dir., then RH thumb points in B-dir.).
15
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
~ inside of a cylindrical
(6) Magnetic Field Inside Long Thin Solenoid: magnetic field B
(or, more generally, prismatic) solenoid of height L with N turns of wire carrying current I
wrapped around the mantle:
~ = µo N I.
B ≡ |B|
L
~
B-dir.
inside is along the cylinder (prism) axis (of length L) in right-handed (RH) dir. rel.
to I [i.e., if RH 4 fingers point in I-dir., looping around the cylincer (prism) axis, then RH
~
thumb points in B-dir.].
~ is being generated
(7) Superposition Principle of Magnetic Field: If a magnetic field B
~ at any observation point P is the
by multiple current-carrying objects (I1 , I2 , ...), then B
~ 1, B
~ 2 , ... that would be
vector sum (resultant vector) of the magnetic field contributions B
generated by each of the current-carrying objects in isolation at that point P :
~ =B
~1 + B
~ 2 + ...
B
~ generated by electric currents flowing in closed
(8) Ampere’s Law: For a magnetic field B,
~
loops or circuits, without time-dependent accumulation of charges anywhere, the B-field
circulation C(L), around a closed, directed path L, is
C(L) = µo I(L) .
Here, I(L) the total current passing through a surface area bounded by L. A current
contribution Ii enclosed by L counts as a positive contribution to I(L) if L loops around Ii
in a right-handed (RH) direction (i.e., if RH 4 fingers point in L-dir., then RH thumb points
~ around a directed,
in Ii -dir.); else, Ii is a negative contribution to I(L). The circulation of B
closed path L is defined as
X
C(L) ≡
Bk ∆s
L
where L is broken up into small length vectors ∆~s pointing around L, ∆s ≡ |∆~s| and Bk is
~
the B-field
vector component parallel to ∆~s at the location of ∆~s.
Mechanics Memories: Velocity, Acceleration, Force, Energy, Power
(1) Velocity
~v =
∆~r
∆t
if constant; else ~v = lim
~a =
∆~v
∆t
if constant; else ~a = lim
∆t→0
∆~r
∆t
(2) Acceleration
∆~v
∆t→0 ∆t
(3) Constant-Acceleration Linear Motion: for ∆~r ≡ ~rf − ~ri and ∆~v ≡ ~vf − ~vi
∆~r =
1
(~vi + ~vf ) t ;
2
∆~r = ~vi t +
16
1
~a t2 ;
2
∆~v = ~a t .
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
(4) Constant-Speed Circular Motion: for motion at constant speed v ≡ |~v | around a
circular trajectory of radius r.
The velocity vector ~v is always tangential to trajectory and perpendicular to acceleration
vector ~a: ~v ⊥ ~a.
The acceleration vector ~a always points towards the center of the circular trajectory.
Period T and frequency f of revolution, angular velocity ω, and orbital speed v:
T =
2πr
2π
1
=
=
f
v
ω
ω = 2πf =
2π
v
=
T
r
v = ωr = 2πf r =
2πr
T
Circular centripetal acceleration:
v2
= ω2r
r
Orbital angle ∆φ and arc of circumference ∆s covered during time interval ∆t:
a=
∆φ = ω ∆t =
v ∆t
∆s
=
r
r
∆s = v ∆t = ω r ∆t = ∆φ r
(5) Newton’s 2nd Law:
m~a = F~
(6) Kinetic Knergy:
1
K = m v2
2
(7) Energy Conservation Law for ∆K ≡ Kf − Ki and ∆U ≡ Uf − Ui :
Ki + Ui = Kf + Uf
or
∆K + ∆U = 0
(8) Mechanical Power: P =rate of work done by force F~ on an object moving at speed
~v , with ~v pointing at an angle θ from F~ and 0o ≤ θ ≤ 180o :
P = F v cos θ
Algebra and Trigonometry
2
az + bz + c = 0
sin θ =
opp
,
hyp
⇒
cos θ =
z=
adj
,
hyp
−b ±
√
b2 − 4ac
2a
tan θ =
opp
sin θ
=
adj
cos θ
sin2 θ + cos2 θ = 1
For very small angles θ (with |θ| 90o ):
sin θ ∼
= tan θ ∼
= θ (in radians)
17
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Numerical Data
Acceleration of gravity (on Earth):
Speed of light in vacuum:
Biot-Savart’s constant:
Permittivity of vacuum:
Permeability of Vacuum:
Electron mass:
Proton mass:
c = 3.00 × 108 m/s
k = 8.99 × 109 Nm2 /C2
Coulomb’s constant:
Elementary charge:
g = 9.81m/s2
km ≡
µo
4π
= 1 × 10−7 Tm/A (exact)
o ≡ 1/(4πk) = 8.85 × 10−12 C2 /Nm2
µo ≡ 4πkm = 4π × 10−7 Tm/A (exact)
e = 1.60 × 10−19 C
me = 9.11 × 10−31 kg
mp = 1.67 × 10−27 kg
Other numerical inputs will be provided with each problem statement.
SI numerical prefixes:
y = yocto =10−24 , z = zepto =10−21 , a = atto =10−18 , f = femto =10−15 , p = pico =10−12 ,
n = nano =10−9 , µ= micro =10−6 , m = milli =10−3 , c = centi =10−2 , d = deci =10−1 ,
da = deca =10+1 , h = hecto =10+2 , k = kilo =10+3 , M = Mega =10+6 , G = Giga =10+9 ,
T = Tera =10+12 , P = Peta =10+15 , E = Exa =10+18 , Z = Zetta =10+21 , Y = Yotta =10+24 .
18