Download Φ21 Fall 2006 HW16 Solutions

Φ21 Fall 2006
1
HW16 Solutions
Problem K33.19
An FM radio station broadcasts at a frequency of 100 MHz. What inductance should be paired with a 5.00
pF capacitor to build a receiver circuit for this station?
Solution:
The resonant frequency of a capacitorinductor pair determines the inductance:
2
ω 2 = (2πf ) =
2
1
LC
⇒
L=
1
2
(2πf ) C
=
1
(2π (100 ×
106
2
Hz)) (5 × 10−12 F)
= 5.07 × 10−7 H = 0.507 µH
Problem K33.17
How much energy is stored in a 2.70 cm diameter, 14.0 cm long solenoid that has 160 turns of wire and
carries a current of 0.750 A?
Solution:
The inductance is
2
L = µ0 N 2 A/` = µ0 (160)
³ ¡¡
¢ ¢2 ´
π 2.70 × 10−2 /2 m
/ (0.14 m) = 1.316 × 10−4 H
The energy of an inductor is
UL =
3
¢
1 2
1¡
2
LI =
1.316 × 10−4 H (0.75 A) = 3.70 × 10−5 J
2
2
Problem OH 15-6-21
For this problem, you must determine the magnetic ux through a square
loop of side a if one side is parallel to, and a distance a from, a straight
wire that carries a current I .
~ eld curls around the wire, so it is directed perpendicThe B
ular to the loop. The ux is
Z
Z a Z 2a
µ0 I
~
~
dr dy
Φ =
B · dA =
2πr
a
0
µZ a ¶ µZ 2a
¶
µ0 I
µ0 Ia
1
=
dr =
ln 2
dy
2π
r
2π
0
a
Solution:
Notice that the integrals separated. The y integral just became a, the
height of the loop, and the r integral was:
Figure 1: A loop near a current.
Z 2a
1
2a
2a
dr = [ln r]r=a = ln 2a − ln a = ln
= ln 2
r
a
a
1
4
Induced Current in a Pair of Solenoids
For each of the actions depicted, determine the direction of the current induced to ow through the resistor in the circuit containing the secondary a)
coil.
~ eld (and hence the ux) will be increasing to the
Solution A: The B
left. The induced current must try to counter this increase by creating a
~ ind to the right, opposing B
~ . By the RHR, B
~ ind to the right is created
B
by a current going down the front of the solenoid, which would be going
to the right in the resistor.
~ eld (and hence the ux) will be decreasing to the
Solution B: The B
~ ind to b)
left. The induced current will try to boost the ux by creating a B
~ . By the RHR (see above), this means the current
the left, supporting B
in the resistor is to the left.
This is exactly the same as Part A. Swapping the positions
of the coils makes no dierence since they are still wound in the same
direction. The current in the resistor is to the right.
Solution C:
By moving the primary coil to the left, some of the magnetic
ux will escape between the coils, and the ux felt by the secondary coil
~ eld is still point toward the left. So this is the
will decraese. The B
same as Part B with a decreasing ux to the left, which leads to (see
above) a current to the left in the resistor.
Solution D:
5
c)
Induced EMF and Current in a Shrinking
Loop
d)
A circular loop of exible iron wire has an initial circumference of 160 cm,
but its circumference is decreasing at a constant rate of 15.0 cm/s due to
a tangential pull on the wire. The loop is in a constant uniform magnetic
eld of magnitude 0.700 T, which is oriented perpendicular to the plane
of the loop. Part A. Find the EMF E induced in the loop after exactly
time 9.00 s has passed since the circumference of the loop started to
decrease. Part B. Find the direction of the induced current in the loop,
as viewed along the direction of the magnetic eld.
Figure 2: Four scenarios for a pair
of solenoids for Problem 4.
The EMF is the time derivative of the ux. The ux is
¯ ¯
¯ ¯
¯ ¯
¯ ¯
¯ ¯
¯ ¯
~ A
~ = BA = πr2 B ⇒ |E| = ¯ dΦ ¯ = πB2r ¯ dr ¯ = BC(t) ¯ dr ¯
B·d
¯ dt ¯
¯ dt ¯
¯ dt ¯
Solution:
Z
Φ=
Note that the information given gives us an equation for the circumference of the circle C and the derivative of r:
C(t) = C0 +
dC
t = (1.6 m) − (0.15 m) t
dt
and
dr
1 dC
0.15
=
=−
t
dt
2π dt
2π
Plug this into the equation for the EMF, and the answer is
µ
¶¯ ¯
dC ¯¯ dr ¯¯
E (t = 9.0) = B C0 +
t ¯ ¯
dt
dt
=
(0.7 T) (1.6 m − 0.15 m/s (9.0 s))
2
µ
0.15 m
2π
¶
= 4.18 × 10−3 Tm2
~ eld). Therefore
For the direction, the ux is decreasing into the loop (remember we're looking along the B
~ ind is parallel to the eld. By the RHR, that
the induced current must try to boost the ux and hence B
means the current is clockwise.
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