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Physics 9 Fall 2010
Midterm 2 Solutions
For the midterm, you may use one sheet of notes with whatever you want to put on it, front
and back. Please sit every other seat, and please don’t cheat! If something isn’t clear, please
ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR
FINAL ANSWERS! You have the full length of the class. If you attach any additional
scratch work, then make sure that your name is on every sheet of your work. Good luck!
1. A closed circuit consists of two semicircles of radius R and R/2 that are connected by straight
segments, as seen in the figure to the right. A
current I exists in this circuit and has a clockwise
direction.
(a) What is the direction of the magnetic field
at point P (i.e., into or out of the page), and
why?
(b) Calculate the magnitude of the magnetic
field at point P.
(c) Suppose that r = 40 cm, and I = 3 amps.
What’s the magnetic field?
————————————————————————————————————
Solution
(a) By the right-hand rule for currents, the magnetic field from the circuit points into
the page.
(b) The total magnetic field is just the sum of the two semicircles (the straight-line
segments contribute nothing, since they are inline with the point P). We can use
the Biot-Savart law to determine the magnetic field from each semicircle. From
the top, since the distance from the wire to the point is constant, r = R, and the
current is always tangent to the point, the Biot-Savart law says
R d~s×r̂
0I
Btop = µ4π
R r2
µ0 I
= 4πR
ds
2
µ0 I
= 4R ,
where we have recalled that the distance around the semicircle is s = πR. The
magnetic field from the bottom wire is done in exactly the same way, but now
0I
the radius is only r = R/2. So, Bbottom = µ2R
. Combining the two fields gives the
total,
3µ0 I
B=
.
4R
(c) Plugging in the numbers gives
3µ0 I
3 × 4π × 10−7 × 3
B=
=
= 7 µT.
4R
4 × 0.4
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2. An inductor and resistor is connected to a 60 Hz generator with a peak voltage of 100
V. At this frequency, the circuit has an impedance of 10 Ω and an inductive reactance
of 8 Ω.
(a) What is the resistance, R, of the resistor?
(b) What is the peak current in the coil?
(c) What is the phase angle (in degrees) between the current and the applied voltage?
(d) A capacitor is put in series with the inductor and the generator. What capacitance
is required so that the current is in phase with the generator voltage?
(e) What is the peak voltage measured across the capacitor for this in-phase condition?
————————————————————————————————————
Solution
p
(a) For a circuit with only a resistor and inductor, the impedance Z = R2 + XL2 ,
where R is the resistance, and XL = ωL is the inductive reactance. Since Z = 10
and XL = 8, then R = 6 Ω.
(b) The peak current in the circuit is I0 = E0 /Z = 100/10 = 10 amps.
(c) The phase angle is given by tan φ = XL /R = 8/6 ⇒ φ = 53◦ .
(d) When the current and voltage are in phase, then φ = 0◦ . When we include a
C
, and so the in-phase condition sets XL = XC =
capacitor, then tan φ = XL −X
R
8 Ω. Since XC = 1/ωC, then C = 1/ωXC = 1/(120π × 8) = 330 µF.
(e) Whenpthe current and voltage are in-phase, then XC = XL , and so the impedance
Z = R2 + (XL − XC )2 = R. Thus, the peak current is I = E0 /Z = E0 /R =
100/6 = 16.7 amps, and the peak voltage through the capacitor is ∆VC = I0 XC =
16.7 × 8 = 133 V.
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3. A circular loop is constructed from a flexible, conducting wire of resistance R. The
wire is being cooled such that it’s radius is shrinking at a rate
t
,
r(t) = r0 1 −
τ
where r0 is the initial, uncooled radius, τ is a constant, and t is the time. The loop is
perpendicular to a uniform, constant magnetic field B.
(a) Find an expression for the induced voltage in the loop as a function of time.
(b) What is the induced current in the loop?
————————————————————————————————————
Solution
(a) The induced voltage, E, is given by
E=
d
Φm ,
dt
where Φm is the magnetic flux. In this case, since the magnetic field is constant,
and perpendicular to the loop, then the flux is just Φm = BA = Bπr2 . Plugging
in for the radius we have
2
t
2
Φm (t) = πr0 B 1 −
.
τ
To get the induced current we just need to take the derivative. Doing so gives
t
2πr02 B
1−
.
E =−
τ
τ
(b) The induced current is just the voltage, divided by the resistance, I = E/R, or
t
2πr02 B
I(t) = −
1−
.
Rτ
τ
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4. The maximum electric field strength in air is 3.0 MV/m. Stronger electric fields ionize
the air and create a spark. What is the maximum power that can be delivered by a
1.0 cm diameter laser beam propagating through the air?
————————————————————————————————————
Solution
The power of a laser is related to its intensity by P = IA, where A is the area of
the beam. Furthermore, the intensity is the average of the Poynting vector, I =
hSi = 2µ1 0 E0 B0 , where E0 and B0 are the amplitude of the electric and magnetic fields,
respectively. But these amplitudes are related by E0 = cB0 , and so I = 2µ10 c E02 . Thus,
the power is
E2
P = IA = 0 A.
2µ0 c
So, plugging in the numbers gives
P = IA =
E02
(3 × 106 )2
A=
× π(0.5 × 10−2 )2 = 9.4 × 105 W.
2µ0 c
2(4π × 10−7 )(2.99 × 108 )
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Extra Credit Question!!
The following is worth 10 extra credit points!
Theorists have speculated about the existence of magnetic monopoles, and several
experimental searches for such monopoles have occurred. Suppose magnetic monopoles
were found and that the magnetic field at a distance r from a monopole of strength qm
is given by
µ0 qm
.
B=
4π r2
Modify the Gauss’s law for magnetism equation to be consistent with such a discovery.
————————————————————————————————————
Solution
Gauss’s law for magnetism says that
I
~ · dA
~ = 0,
B
which says that magnetic monopoles don’t exist; we want to adjust this expression to
include monopoles. We’ll do it by analogy with Gauss’s law for the electric field, which
says that
I
~ · dA
~ = Qencl .
E
0
For a single electric point charge of strength q, then a spherical Gaussian surface of
radius r gives the electric field as
E=
1 q
,
4π0 r2
as we’ve seen many times before. In analogy with Gauss’s law for the electric field, if
we had individual magnetic charges, qm , then we expect that we could write Gauss’s
law for these charges as
I
~ · dA
~ = µ0 Qm encl .
B
We can easily check that this gives the expected magnetic field of a monopole. Take
a spherical Gaussian surface of radius r, enclosing a magnetic point charge qm . Then,
the magnetic field is constant over the surface, and points along the direction of the
normal
to theH surface, and so the left-hand-side of the modified Gauss’s law gives
H
~
~
B · dA = B dA = B (4πr2 ), just as in the electric case. The right-hand-side reads
µ0 Qm encl = µ0 qm , and so
µ0 qm
B=
,
4π r2
which is the expected field. Thus we have the correct generalization.
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Some Possibly Useful Information
Some Useful Constants.
Coulomb’s Law constant k ≡
1
4π0
= 8.99 × 109
N m2
.
C2
The magnetic permeability constant µ0 = 4π × 10−7
N
.
A2
Speed of Light c = 2.99 × 108 m/s.
Newton’s Gravitational Constant G = 6.672 × 10−11
N m2
.
kg 2
The charge on the proton e = 1.602 × 10−19 C
The mass of the electron, me = 9.11 × 10−31 kg.
The mass of the proton, mp = 1.673 × 10−27 kg.
Boltzmann’s constant, kB = 1.381 × 10−23 J/K.
1 eV = 1.602 × 10−19 Joules ⇒ 1 MeV = 106 eV .
1 Å = 10−10 meters.
Planck’s constant, h = 6.63 × 10−34 J s = 4.14 × 10−15 eV s.
The reduced Planck’s constant, ~ ≡
h
2π
= 1.05 × 10−34 J s = 6.58 × 10−16 eV s.
Some Useful Mathematical Ideas.
( n+1
x
R n
n 6= −1,
x dx = n+1
ln (x) n = −1.
√
R dx
2 + x2 .
√
=
ln
x
+
a
2
2
a +x
R
√ x dx
a2 +x2
=
√
a2 + x 2 .
Other Useful Stuff.
The force on an object moving in a circle is F =
mv 2
.
r
Kinetmatic equations x(t) = x0 + v0x t + 21 ax t2 , y(t) = y0 + v0y t + 12 ay t2 .
The binomial expansion, (1 + x)n ≈ 1 + nx, if x 1.
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