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Physics 9 Fall 2011
Homework 8 - Solutions
Friday October 21, 2011
Make sure your name is on your homework, and please box your final answer. Because
we will be giving partial credit, be sure to attempt all the problems, even if you don’t finish
them. The homework is due at the beginning of class on Friday, October 28th. Because
the solutions will be posted immediately after class, no late homeworks can be accepted! You
are welcome to ask questions during the discussion session or during office hours.
1. When the current in an 8.00 H coil is equal to 3.00 A and is increasing at 200 A/s, find
(a) the magnetic flux through the coil, and
(b) the induced emf in the coil.
————————————————————————————————————
Solution
(a) The magnetic flux, Φ, is given in terms of the inductance and current by
Φ = LI.
With numbers, Φ = 8 × 3 = 24 T m2 , or 24 webers.
(b) The induced emf in the coil is
˙
E = Φ̇ = LI,
where, as always, a dot denotes the derivative with respect to time. We know the
rate of change of the current, and so
E = 8 × 200 = 1600 V.
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2. A long solenoid has n turns per unit length and carries a current that varies with time
according to I(t) = I0 cos ωt. The solenoid has a circular cross section of radius R.
Find the induced electric field, at points near the plane equidistant from the ends of
the solenoid, as a function of both tie time t and the perpendicular distance r from
the axis of the solenoid for
(a) r > R, and
(b) r < R.
(c) Do your answers match at the boundary, when r = R?
————————————————————————————————————
Solution
H
~ · d~s = − d Φm , where
(a) The induced electric field is given by Faraday’s law, E
dt
R
~ · d~a = BA is the magnetic flux, since the magnetic field inside the
Φm = B
solenoid points along the direction of the solenoid. Taking the integration loop
to be a circle of radius r > R, weH see that from
H the symmetry, the electric
~
field is constant along the circle, so E · d~s = E ds = E (2πr), while the net
flux through the surface area bounded by the circle is just B(t) (πR2 ), since the
2
magnetic field is zero outside the solenoid. Thus, E = − R2r dB
. Now, B(t) =
dt
dB
µ0 nI(t) = µ0 nI0 cos (ωt), and so dt = −µ0 nI0 ω sin (ωt). Thus,
µ0 nI0 ωR2
E=
sin (ωt) .
2r
(b) Here, the analysis proceeds exactly as before, except that the flux through a small
circle inside the solenoid is only Φm = B (πr2 ). Working things out as before, we
find
µ0 nI0 ωr
E=
sin (ωt) .
2
(c) The two solutions do agree at the surface of the solenoid, as we should expect.
So, we know that our solutions are trustworthy!
2
3.
(a) Compute the magnetic flux throug the rectangular loop shown in the figure to the right.
(b) Evaluate your answer for a = 5.0 cm, b = 10
cm, d = 2.0 cm, and I = 20 A.
————————————————————————————————————
Solution
(a) The magnetic field is not constant in the loop, so we can’t just write Φ = BA.
Instead, let’s look at the flux through a tiny little vertical strip of area dA = bdx. If
we take the strip small enough, then the magnetic field from the wire is essentially
constant on that strip. Thus, recalling that the magnetic field of a wire a distance
x away is
µ0 I
,
B=
2πx
then the small amount of flux through the tiny strip is
dΦ = BdA =
µ0 Ib dx
.
2π x
To get the total flux we just need to integrate this result from x = d to x = d + a,
Z
µ0 Ib d+a dx
µ0 Ib
d+a
Φ=
=
ln
.
2π d
x
2π
d
(b) With the numerical values we find
d+a
4π × 10−7 × 20 × 0.01
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µ0 Ib
Φ=
ln
=
ln
= 5 × 10−8 T m2 .
2π
d
2π
2
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4. A series RLC circuit that has an inductance of 10 mH, a capacitance of 2.0 µF, and a
resistance of 5.0 Ω is driven by an ideal ac voltage source that has a peak emf of 100
V. Find
(a) the resonant frequency, and
(b) the root-mean-square current at resonance.
(c) When the frequency is 8000 rad/s, find the capacitive and inductive reactances,
(d) the impedance,
(e) the root-mean-square current, and
(f) the phase angle.
————————————————————————————————————
Solution
(a) The resonant frequency is given by
ω0 = √
1
1
=√
= 7071 rad/s.
10 × 10−3 × 2 × 10−6
LC
(b) At resonance, when Z = R, the root-mean-square current is
Irms =
1 E
1 100
Erms
=√
=√
= 14.14 Amps.
Z
2R
2 5
(c) The capacitive and inductive reactances are XC =
So, when ω = 8000 we have
XC =
1
,
ωC
and XL = ωL, respectively.
1
1
=
= 62.5 Ω,
ωC
8000 × 2 × 10−6
and
XL = ωL = 8000 × 10 × 10−3 = 80 Ω.
(d) At this frequency, the impedance is
p
p
Z = R2 + (XC − XL )2 = 52 + (62.5 − 80)2 = 18.2 Ω.
(e) Finally, the phase angle, φ, is
−1 XL − XC
−1 80 − 62.5
φ = tan
= tan
= 74◦ .
R
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5. Show by direct substitution that
LQ̈ + RQ̇ +
1
Q=0
C
(as always, recall that a dot over a quantity denotes the time derivative) is solved by
Q(t) = Q0 e−t/τ cos (ωt) ,
where τ = 2L/R, ω =
p
ω02 − 1/τ 2 , and Q0 is the charge on the capacitor at t = 0.
————————————————————————————————————
Solution
Here we just need to plug in our trial solution to the differential equation, which we
can write
R
1
Q̈ + Q̇ +
Q = 0,
L
LC
or, writing R/L = 2/τ and LC = ω0−2 , we have
2
Q̈ + Q̇ + ω02 Q = 0.
τ
Taking the derivatives of our trial solution gives
Q0 −t/τ
−t/τ
−t/τ 1
cos(ωt) − ωQωe
sin(ωt) = −Q0 e
cos(ωt) + ω sin(ωt) ,
Q̇ = − e
τ
τ
and
Q0 −t/τ 1
1
−t/τ
Q̈ =
e
cos(ωt) + ω sin(ωt) − Q0 ωe
− sin(ωt) + ω cos(ωt) ,
τ
τ
τ
or
−t/τ
Q̈ = Q0 e
1
ω
2
− ω cos(ωt) + 2 sin(ωt)
τ2
τ
Plugging in these derivatives to our differential equation gives
1
ω
2 1
−t/τ
2
2
Q0 e
− ω cos(ωt) + 2 sin(ωt) −
cos(ωt) + ω sin(ωt) + ω0 cos(ωt) = 0
τ2
τ
τ τ
Canceling off the Q0 e−t/τ term we also see that the sine terms cancel, which also allows
us to cancel off the cosine terms. Thus we find
2
1
1
− ω 2 − 2 + ω02 = ω02 − 2 − ω 2 = 0.
2
τ
τ
τ
But, since ω =
equation.
p
ω02 − 1/τ 2 , we find that the solution does satisfy the differential
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