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

Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
Solution to PHYS 1112 In-Class Exam #3B
Thu. April 9, 2009, 2:00pm-3:15pm
Conceptual Problems
Problem 1: If a wire of some length L and with a circular cross-section of diameter D has
a resistance of 60 kΩ, what will be the resistance of a wire made from the same material, at
the same temperature, of length 24L and diameter 6D ?
(A)
(B)
(C)
(D)
(E)
10 kΩ
40 kΩ
90 kΩ
120 kΩ
240 kΩ
Answer: (B)
Using R = ρL/A and A = π(D/2)2 , we get R ∝ L/D2 . Increasing L → L0 = 24L and
D → D0 = 6D thus changes R → R0 = 24 × R/62 = (24/36) × R = (2/3) × 60Ω = 40Ω.
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 currents I3 through R3 , observed in the three
different circuits. (Hint: I3 = V3 /R3 .)
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
I3 (Z) > I3 (Y ) > I3 (X)
I3 (X) > I3 (Z) > I3 (Y )
I3 (Y ) > I3 (X) > I3 (Z)
I3 (Y ) > I3 (Z) > I3 (X)
I3 (Z) > I3 (X) > I3 (Y )
Answer: (A)
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 .
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 below, with B 6= 0 above
~ perpendicular to, and pointing out of, the plane of the
the lower horizontal line, and B
drawing. 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
~
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 = 1 : 2 : 3 : 4 ,
as indicated in Fig. 3.15. Assume all four ion types carry the same amount of charge per
2
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
~
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
=1 : 2 : 3 : 4 .
√
√
√
√
= 1 : 2 : 3 : 4 .
= 11 : 12 : 13 : 14 .
= √11 : √12 : √13 : √14 .
(E) mP : mQ : mR : mS = 1 : 4 : 9 : 16 .
Answer: (A)
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.
Problem 4: 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.
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
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:
3
Physics 1112
Spring 2009
(A)
(B)
(C)
(D)
(E)
University of Georgia
Instructor: HBSchüttler
(Bx , By , Bz ) = (0 , B1 + B2 ,
(Bx , By , Bz ) = (B1 − B2 , 0 ,
(Bx , By , Bz ) = (−B2 , −B1 ,
(Bx , By , Bz ) = (+B1 , −B2 ,
(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
~ 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 into 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 = (0, −B2 , 0).
drawing, i.e., B2 points in −y-direction. Therefore, B2y = −B2 < 0 and 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)
Numerical Problems
Problem 5: A wire 350m in length, made of dawgium alloy (a very poorly conducting
metal!), draws a 1.25mA current when connected to a 16V battery. The resistivity of
dawgium is 4 × 10−6 Ω · m. The area of a circle with a diameter D is A = (π/4)D2 .
What is the diameter of the wire ?
(A)
(B)
(C)
(D)
0.085mm
0.16mm
0.37mm
1.8mm
4
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
(E) 2.5mm
Answer: (C)
The resistance of the wire is R = V /I = 16V/0.00125A = 12800Ω. Also, R = ρL/A and A =
πD2 /4. So D = [(4/π)A]1/2 = [(4/π)ρL/R]1/2 = [(4/π)(4.0×10−6 Ω·m)(350m)/(12800Ω)]1/2 =
0.3732 × 10−3 m
Problem 6: How long must the current flow through the dawgium wire described in Problem
5 in order to produce 36J of heat ?
(A)
(B)
(C)
(D)
(E)
5min
15min
30min
2h
5h
Answer: (C)
P ≡ ∆U/∆t = IV , so ∆t = ∆U/P = ∆U/(IV ) = (36J)/[(0.00125A)(16.0V)] = 1800s ×
(1min/60s) = 30min
Problem 7: In circuit Y , shown in the center panel of Fig. 3.04 (see Problem 2), the
resistances are R1 = 8 Ω, R2 = 14 Ω, R3 = 10 Ω and the battery voltage is E = 72V. What
is the current Io flowing through the battery ?
(A)
(B)
(C)
(D)
(E)
3A
7A
10A
12A
22A
Answer: (D)
R2 and R3 are in series: replace by R23 = R2 + R3 . Then R23 and R1 are in parallel: replace by Req = (1/R1 + 1/R23 )−1 . Then Io = E/Req = E[1/R1 + 1/(R2 + R3 )] =
(72V)[1/(8Ω) + 1/(14 + 10)Ω] = 12A.
Problem 8: If I1 = +9A and I2 = +4A and the resistance is R = 8Ω in the circuit fragment
shown in Fig. 3.10, what is the voltage drop VR ≡ Va − Vb across the resistor ?
5
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
I1
I2
Fig. 3.10
a
I3
R
b
(A)
(B)
(C)
(D)
(E)
120V
80V
60V
40V
20V
Answer: (D)
By Kirchhoff’s junction rule: I1 = I2 + I3 ; so I3 = I1 − I2 = (9 − 4)A = 5A. Then
VR = RI3 = (8Ω)(5A) = 40V.
Problem 9: Last time you checked you weighed in at 150kg. Also, some space alien sorority
sisters, as part of their annual pledging ritual, have implanted a positive point charge in your
brain. Now, to make matters worse, they’ve ordered you to jog on a straight north-south
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 50 Tesla
and turned the field vector to make it point horizontally due west. But they left earth’s
value of g at 9.81m/s2 ...
In case you can still handle electromagnetic theory with all that electricity between your
ears – and you still want to join that sorority:
How much positive charge do you have in your head if you just barely achieve levitation at
4m/s running speed? And in which direction are you running?
6
Physics 1112
Spring 2009
(A)
(B)
(C)
(D)
(E)
1.68C,
3.27C,
3.27C,
7.36C,
7.36C,
University of Georgia
Instructor: HBSchüttler
running
running
running
running
running
northward
southward
northward
southward
northward
Answer: (E)
To lift off, the upward magnetic force F = |q|vB sin θ must cancel the downward weight
~
force mg. Since you’re running on a north-south track, ~v points north or south, while B
~ = 90o and sin θ = 1. So, mg = |q|vB or |q| = mg/(vB) =
points west; so: θ ≡ 6 (~v , B)
2
(150kg)(9.81m/s )/[(4.0m/s)(50T)] = 7.36C.
~
Also, if you run north, ~v points north and then q~v also points north, since q > 0. With B
~ by right-hand rule, then points upward, as required for lift-off.
pointing west, F~ = q~v × B,
(Note: if you were running south, q~v would point south and F~ would point downward, i.e.,
in the wrong direction.)
Problem 10: If a 30A current flows in a thin straight metal rod of length 7m through a
uniform magnetic field of 50T and the direction of the current flow is at an angle of 55o 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)
12260N
8601N
6023N
2489N
725N
Answer: (B)
F = ILB sin θ = (30A)(7m)(50T) sin(55o ) = 8601N.
Problem 11: If you are standing on top of a very long horizontal straight wire which is
carrying a current of 80A in eastward direction and the tip of your nose is exactly 1.6m
above the wire, what is the strength and direction of the magnetic field produced by that
wire at the tip of your nose ?
~ pointing eastward
(A) 4µT , B
~ pointing southward
(B) 10µT , B
~ pointing northward
(C) 70µT , B
~ pointing downward
(D) 120µT , B
7
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
~ pointing upward
(E) 180µT , B
Answer: (B)
For an infinitely (≡very) long wire, B = (µo /2π)(I/r) = (2 × 10−7 Tm/A)(80A)/(1.6m) =
10. × 10−6 T.
~ must point southward at your observation position above 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 eastward, i.e. looking in the
~
direction of current flow. Hence, the B-vector
points: northward below the wire, southward
above the wire, downward to the right of (≡ south of) the wire, and upward to the left of
(≡ north of) the wire. Draw it !!!
Problem 12: Two very (”infinitely”) long thin straight wires running parallel to the z-axis
and cutting through the y-axis at y1 = 3m and through the x-axis at x2 = 4m, respectively,
carry currents I1 = 5A in the −z-direction and I2 = 6A in the −z-direction, respectively, as
shown in Fig. 3.31.
y
Fig. 3.31
P
I1
I2
x
y
x-y-Plane View
z
x
What is the angle between the +x-direction and the total magnetic field vector produced
by the two wires at point P ≡ (4m, 3m, 0m) in the x − y-plane ? (Hint: You only need to
calculate the ratio B1 /B2 !)
(A) 148.0o
(B) 122.0o
(C) 90.0o
8
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
(D) 58.0o
(E) 32.0o
Answer: (E)
By right-hand rule applied to concentric circular field lines surrounding the respective cur~ 1 and B
~ 2 generated by I1 and I2 , respectively,
rents, the magnetic field vector contributions B
~ 1 = (0, −B1 , 0)
(see also, e.g., CP 3.25 through CP 3.30) have the following directions: B
~ 2 = (+B2 , 0, 0) points in +x-direction. So, the total magnetic
points in −y-direction and B
~ =B
~1 +B
~ 2 = (+B2 , −B1 , 0) points at an angle φ < 90o below the +x direction,
field vector B
given by tan φ = B1 /B2 . Draw it!!
Also, Bi = (µo /2π)(Ii /ri ) for current Ii , with i = 1 and i = 2. Here, the distances from
P to the respective wires are r1 = xP = x2 = 4m and r2 = yP = y1 = 3m (see Fig.
3.31). In the ratio B1 /B2 , the pre-factors (µo /2π) cancel out and thus: tan φ = B1 /B2 =
(I1 /r1 )/(I2 /r2 ) = (5A/4m)/(6A/3m) = 0.625, hence φ = arctan(0.625) = 32.01o .
9
Physics 1112
Spring 2009
University of Georgia
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
Spring 2009
University of Georgia
Instructor: HBSchüttler
(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.
11
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
(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
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
(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.
13
Physics 1112
Spring 2009
University of Georgia
Instructor: HBSchüttler
(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