PHYSICS 10320
EXAM 3
11 April 2006
INSTRUCTIONS: Write your NAME IN CAPITAL LETTERS and your LECTURE
(9:30=Balsara) on the front of the blue exam booklet. The exam is closed book, and you
may have only a pen/pencil and a calculator (no stored equations or programs and no
graphing). Show all of your work in the blue book. For problems II-V, an answer alone
is worth very little credit, even if it is correct - so show how you get it.
Suggestions: Draw a diagram when possible, circle or box your final answers, and cross
out parts which you do not want us to consider.
________________________________________________________________________
Some useful equations:
G
G
Force on a charge in electric field: F = qE
G
G G
Force on a moving charge in magnetic field: F = qv × B
G G
G
Force on a current element in magnetic field: dF = Idl × B
Newton’s second law in magnetic field: qvB = mv 2 / r
Radius of charged particle’s orbit in magnetic field: r = mv /(qB )
G
Magnetic dipole moment of a current-carrying coil: µ = NIAn̂
G G G
G G
Torque in magnetic field: τ = µ × B ; Potential energy of a magnetic dipole: U = − µ ⋅ B
Hall voltage: V = v d Bw
G µ 0 qvG × nˆ
Magnetic field due to a moving charge: B =
;
µ 0 = 4π × 10 −7 T ⋅ m / A 2
2
4π r
G
G µ Idl × nˆ
Biot-Savart law: dB = 0
ε 0 = 8.85 × 10 −12 F / m
2
4π r
µ 2I
Magnetic field: In a long solenoid B = µ 0 NI / l ; due to a long wire B = 0
4π r
G G
Ampere’s law: ∫ B ⋅ dl = µ 0 I C where C is any closed curve and IC is the net current through the
C
area enclosed by C.
G
Magnetic flux: Φ m = ∫ B ⋅ n̂dA ; for uniform field and flat surface bounded by coil of N turns
S
Φ m = NBA cos θ ; units 1Wb = 1T ⋅ m 2 ;
due to a current in a coil: Φ m = LI
Faraday’s law (for both induction and motional emf!): E = −
dΦ m
dI
; back-emf: E = − L
dt
dt
Self-inductance of a solenoid: L = µ 0 n 2 Al = µ 0 N 2 A / l
Magnetic energy density: u m = B 2 /( 2 µ 0 ) ; Electric energy density: u e = ε 0 E 2 / 2
Magnetic energy (in a coil): U m = LI 2 / 2 ; electric energy (in a capacitor): U e = Q 2 /( 2C )
E
Energizing an inductor with a battery: I = (1 − e −t / τ ) with τ = L / R ; for short times the
R
inductor is like a very large resistor; after a long time – like a short circuit.
De-energizing an inductor through a resistor: I = I 0 e − t / τ with τ = L / R .
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I. Multiple Choice Questions: Read each question carefully. Write the SINGLE correct answer
in the grid given inside your blue book. No explanation is required, and no partial credit will be
given. (20 points total)
1. Consider the RL circuit below. What is the power supplied by the battery immediately
after the switch S is closed and a long time after the switch S is closed?
a)
b)
c)
d)
e)
immediately after S is closed : 375 Watts ; long time after S is closed : 562.5 Watts.
immediately after S is closed : 1125 Watts ; long time after S is closed : 750 Watts.
immediately after S is closed : 750 Watts ; long time after S is closed : 1125 Watts.
immediately after S is closed : 375 Watts ; long time after S is closed : 1125 Watts.
immediately after S is closed : 750 Watts ; long time after S is closed : 562.5 Watts.
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S
N
Back
Front
2. A bar magnet is dropped through a loop of copper wire, as shown. If the positive direction
of the induced current I is shown by the arrows on the loop, which of the following graphs
correctly illustrates the variation of I with time as the magnet falls. The time when the
midpoint of the magnet passes through the loop is indicated by C on each graph.
C
I
t
(A)
t
(B)
C
I
C
I
C
I
t
(C)
t
(D)
C
I
t
(E)
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3. A 300-turn coil with cross-sectional surface area of 100cm 2 carries a current of 1A. The
normal to the coil’s plane makes an angle of 60 0 with a magnetic field of 0.2T. The torque
on the coil is
a)
b)
c)
d)
e)
0.1Nm
0.2 Nm
0.3Nm
0.6 Nm
4.0 Nm
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4. In the figure shown below the rod has mass “m”, a resistance “R” and the horizontal
rails have negligible resistance. A battery of emf ε and negligible internal resistance is
connected between the points “a” and “b” such that the current in the rod is downward.
The rod is at rest at t=0. Which of the following is true?
a) The terminal velocity of the rod is ε l B ; the final current in the rod is 0.
b) The terminal velocity of the rod is ε l B ; the final current in the rod is ε / R .
c) The terminal velocity of the rod is ε/( l B) ; the final current in the rod is ε / R .
d) The terminal velocity of the rod is ε/( l B) ; the final current in the rod is 0.
e) None of the above
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5. A proton moves clockwise in a circle of radius r due to a uniform B field
perpendicular to the plane of motion. The proton is replaced with an electron. How must
the B field be changed in order that the electron moves in a circle with the same radius in
the same direction and with the same velocity .
a)
b)
c)
d)
e)
B reversed and reduced by factor me / mp
B same direction and reduced by factor me / mp
B reversed and increased by factor mp / me
B same direction and increase by factor mp / me
B reversed and increased by factor mp 2/ me2
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Problems: (20 points each): Write the complete solution to each problem in your blue book.
Remember that no partial credit will be given for an answer with no supporting work.
II. Motion in a Magnetic Field
A mass spectrometer is built to separate the two isotopes of Chlorine. The isotopes are
35
Cl, which we approximate as having a mass equal to 35 times the mass of a proton and
37
Cl, which we approximate as having a mass equal to 37 times the mass of a proton. The
atoms are singly ionized and have a charge Q = 1.6 x 10-19 C . The mass of the proton is
Mp = 1.67 x 10 –27 kg. The schematic for the mass spectrometer is shown below.
The spectrometer has two zones: an accelerating zone with electric field only and hence
a potential difference ∆V = 104 V between plates; and a magnetic field zone where there
is a uniform magnetic field of magnitude B=0.1 T only.
The two isotopes of Chlorine are accelerated through the full ∆V in the accelerating zone.
Then they pass through a slit and enter the B-field zone. There they experience the
Lorentz Force, follow a half-circular trajectory and strike a wall.
(a) Find the kinetic energy and the velocity of an arbitrary particle of charge Q and
mass M just as it reaches the slit. Do not plug in numbers. Simply obtain an
expression that relates the mass M and the velocity v to the charge Q and the
potential difference ∆V .
(b) Without plugging in numbers, find the radius r of the circular trajectory of a
particle as it moves through the analyzing region. Express your answer for the
radius r in terms of the mass M, the velocity v , the charge Q and the magnetic
field B;
(c) Use the expressions from (a) and (b) to eliminate the velocity v .Without plugging
in numbers, obtain an expression for r in terms of ∆V, Q, M, and B.
(d) Find a numerical value for r35 and r37, the radii of the circular trajectories of 35Cl
and 37Cl respectively, as they pass through the spectrometer.
(e) What is the linear separation between the isotopes as they hit the lower horizontal
wall of the mass spectrometer?
wall
slit
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III. Ampere’s + Biot-Savart Law
An infinitely long, straight wire is bent, as shown in the figure below. It carries a current
I. The circular portion has a radius of R=10 cm with its center a distance r from the
straight part. Find r such that the magnetic field at the center of the circular portion of the
wire is zero, by using the steps given below:
(a) The circular part of the wire acts like a magnetic dipole. Is the magnetic dipole
moment associated with the circular part pointed into the paper or out of the paper? As a
result, does the magnetic field associated with the circular part point into the paper or out
of the paper?
(b) Use the Biot-Savart law to obtain the magnitude of the magnetic field in the center of
the circular portion due to the curved part of the wire. Write your answer in terms of the
current I, the radius R and µ0 .
(c) The straight part of the wire acts like an infinitely long wire that is carrying current to
the right. Does the magnetic field associated with the straight part point into the paper or
out of the paper?
(d) Use Amperes law to obtain the magnitude of the magnetic field at a distance r from
the straight part of the wire. Write your answer in terms of the current I, the radius r and
µ0 .
(e) Find the value of r for which the two magnetic fields obtained in parts (b) and (d) are
equal in magnitude. This is the condition for having a zero field in the center of the
curved portion of the wire.
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IV) Faraday’s Law
An electron, initially at rest, is located a distance r = 3 cm from the center of a circular region of
radius R = 1 cm where there is a uniform magnetic field. The field is directed out of the page,
which we define to be the ĵ direction. The magnetic field begins to change in time in the
G
following manner: dB / dt = −0.01 Tesla/sec k̂ , i.e., the field strength begins to decrease.
A) Use Faraday’s Law to find the magnitude of the electric field at the location of the electron.
B) Find the direction of the electric field at the location of the electron, using the coordinate
system given on the figure.
C) Find the magnitude and direction of the initial acceleration of the electron.
(use me = 9.1X10-31 Kg, e = 1.6X10-19 C)
B is out
of page
y
• • • •
R = 1 cm
• • • • •
• • • • •
• • • • •
• •
e−
r = 3 cm
z•
x
z is out
of page
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V) Ampere’s Law and Self-Inductance
This problem steps you through the process of evaluating the self-inductance of a toroidal
inductor, shown below. The inductor has a total number of turns given by “N”. “a” is the inner
radius and “b” is the outer radius. The height is given by “H”.
a) Send a current I through the wire and use Amperes law to write out an expression for the
magnetic field for a < r < b.
b) At a radius r such that a < r < b, consider an infinitesimal rectangular area having a radial
extent “dr”. Do this by writing out an expression for the infinitesimal magnetic flux dφm passing
through the infinitesimal area dA = H dr . Hint : see the hachured area in the figure below.
c) Find the total magnetic flux φm passing through the cross-sectional area of the toroid.
d) Find the total magnetic flux Φm associated with the “N” turns of the toroid.
e) Thus find the self-inductance of the toroidal inductor.
f) Say a current I flows through the inductor. What is the magnetic energy density um at a radius r
such that a < r < b ?
g) Use your answer to e) above to give an expression for the total magnetic energy Um stored in
the inductor when a current “I” flows through it.
Please look at the next page for the figure.
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