Physics 272, Recitation #9 Spring 2023
Name (Last,First): _____________________, ______________ RUID: __ __ __ __ __ __ __ __ __
Name (Last,First): _____________________, ______________ RUID: __ __ __ __ __ __ __ __ __
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•
•
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In the fields above, enter your names and Rutgers 9-digit RUID in big, easily legible font.
This recitation has a maximum score of 18.
Ask questions and try to solve all the problems by the end of recitation.
Please ask questions!
1. (4 pts) A nonconducting thin ring of radius 𝑎𝑎 carries a static charge 𝑞𝑞. This ring is in a
magnetic field of strength 𝐵𝐵0, parallel to the ring’s axis, and is supported so that it is free to
rotate about that axis. If the field is switched off:
a) How much angular momentum will be added to the ring?
b) Supposing the mass of the ring to be 𝑚𝑚, show that the ring will acquire an angular velocity
𝜔𝜔 =
𝑞𝑞𝐵𝐵0
𝑚𝑚
, if initially at rest. Note that the result depends only on the initial and final values of
the field strength, and not on the rapidity of change.
GE.de
(a)
=
- TR2
-
xX
-
Torque
-
R
-
2
⑦
I S***t
2
=
-
line
xfE.dtxR
=
I
=
on
4
d
=
=
=
-
charge.
xR.d
=
=
TR3
x
Rx)**
R
some
If
-
=
Rx 0-B0) πR XBo
=
-o
(b)
hf πRBoX
=
=
a
=
Fw
mRw
=
2TRX
=w qB
=
Page 1 of 5
=
-
πXRdt
2. (5 pts) A metal crossbar of mass 𝑚𝑚 slides without friction on two long parallel conducting
rails a distance 𝑏𝑏 apart; as shown in the Figure below. A resistor 𝑅𝑅 is connected across the rails
at one end; compared with 𝑅𝑅, the resistance of bar and rails is negligible. There is a uniform
field 𝐵𝐵 perpendicular to the plane of the figure. At time 𝑡𝑡 = 0 the crossbar is given a velocity
𝑣𝑣0 toward the right.
a) Show that the velocity of the bar decreases exponentially with time. This is a result of Lenz's
law. Calculate the induced current and the corresponding applied force and use Newton's
laws.
b) How far does the bar travel before it comes to rest?
c) What is the total energy dissipated by the resistor?
-I
-
As
↑
the
with
-
bar
the
I
-
it
3
=
The
E
by
enclosed
slides, the flux
current
opposes
(Lenz's law)
=
the
change
(Clockwise
-
=
resulting
=
formed
-
flux
that
b* =
↳
Along
-
-
lorentz force
IB
in
current).
Y
E Ix
=
I(
=
=
loop
the
resistor yo
The induced
produced
↑
⑳
=
-
Page 2 of 5
-
b4)x(B2)
IbBY
newton's
Applying
m
=
F
=
law
-
r
(b
Mer ba
=
->
=(Inlm]=-t
velocity
decors
=
v
wet.
=
exponentially
(b)
-
Total
Woe
=
X
=>
travelled.
distance
t
re--
=
⑧
=
vo
(et].mr.
3. (3 pts) The figure below shows a solenoid of radius 𝑎𝑎1 and length 𝑏𝑏1 located inside a longer
solenoid of radius 𝑎𝑎2 and length 𝑏𝑏2 . The total number of turns is 𝑁𝑁1 on the inner coil, 𝑁𝑁2 on
the outer. Work out an approximate formula for the mutual inductance 𝑀𝑀.
-
If
solenoid
2
Br Monz
=
-Total
flux
Carries
Current
I,
the
magnetic field
MoI()
=
through
Coll
1:
Nex
Marx
en
# of
Area
turns
Ez
M1
=
M MoN, N2
=>
=
Page 3 of 5
Ia
Iz
B2.=
4. (3 pts) An electric generator contains a coil N turns of wire, each forming a rectangular loop
of area A. The coil is placed entirely in a uniform magnetic field with magnitude B and initially
perpendicular to the coil's plane. What is the maximum value of the emf produced when the
loop is spun at radial speed 𝜔𝜔 about an axis perpendicular to the magnetic field? (Use any
variable or symbol stated above as necessary.)
fit.
-⑦
-f
⑦ ct
=
-
Flux through the
B
-
N
BANCOSO
=
=
Induced
of
turns
EMF:E
=
-
loop
when
loop
is
a
angle
BANCos(wt)
44B
+
=
EMAX=BANw
->
Page 4 of 5
BANwsin(wt)
t:
5. (3 pts) A coil with resistance of 0.01 Ω and self-inductance 0.5 mH is connected across a large
12 V battery of negligible internal resistance.
a) Sketch the current in the resistor as a function of time.
b) How long after the switch is closed will the current reach 90 percent of its final value?
c) At that time, how much energy, in joules, is stored in the magnetic field? How much energy
has been withdrawn from the battery up to that time?
Equation
--
I
T
12V
=
E
L 0.5 mH
=
ICH
=>
=
2
E11-e
=
reach
current
90%
of
to
it's
final value,
1-e*t12
Et
=
-
-
0.9-
Rt/L
0.1
e
=
(/p(10.2)
=
1(u(10)
=
R
4.
=
-
Energy
stored
time:E
in
R+12)
a/m=12oo1
Anag,
the
IR.
--------
R 0.01
For
LR
+
=
-
Lum
-
by
arcult:
E
-
satisfied
the
2.3025
magnetic field
ILF2
=
Page 5 of 5
0.1155
=
-
& that
E
(0.90)
1
=
0.82.4L to2
=
(24(E)
(5x154)
0.82x0.3x()
=0.82.
265
=
=
-
Total
drawn from
energy
S
E.)
E VIIHdt=
=
battery
-
at that
time.
(l-et)
Instantaneous
power provided
by battery
-
It
=
-
Rt/L
e-
lo
-Iam
a.(z) [
=
12V
-
t
-
1200A
10085
-
0.1159-0.055
ne
I
0.05340.2
+
<
Rest of energy
dissipated
In
resistor.
at