Download Recitation Week 8

Recitation Week 8
Chapter 27
Problem 27.39. A thin, 50.0 cm long metal bar with mass 750 g rests on, but is not attached to, two metallic supports in
a uniform 0.450 T magnetic field, as shown in Fig. 27.51. A battery and a 25.0 Ω resistor in series are connected to the
supports. (a) What is the highest voltage the battery can have without breaking the circuit at the supports? (b) The battery
voltage has the maximum value caculated in (a). If the resistor suddenly gets partially short-circuited, decreasing its resistance
to 2.0 Ω, find the initial acceleration of the bar.
V
R
B
(a) The current flowing to the right in the bar feels a lifting force from the magnetic field FB = Il × B which balances the
gravitational force Fg = mg. Because the current and magnetic field are perpendicular to each other, we can focus on the
magnitudes
FB = IlB = Fg = mg
(1)
2
0.750 kg · 9.8 m/s
mg
=
= 32.7 A .
lB
0.500 m · 0.450 T
This maximum current would occur when the voltage V from the battery balanced an IR drop across the resistor, so
(2)
I=
V = IR = 32.7 A · 25.0 Ω = 817 V .
(3)
(b) When the resistor shorts, the current jumps to
V
,
(4)
R0
because the resistor voltage still has to match the battery voltage. This creates a net lifting force and acceleration on the
bar.
V lB
F = FB − Fg = I 0 lB − mg =
− mg = ma
(5)
R0
V lB
817 V · 0.500 m · 0.450 T
a= 0 −g =
− 9.8 m/s2 = 113 m/s2 .
(6)
Rm
2.00 Ω · 0.750 kg
I0 =
Problem 27.64. A particle of charge q > 0 is moving at speed v in the +z-direction through a region of uniform magnetic
field B. The magnetic force on the particle is F = F0 (3î + 4ĵ), where F0 is a positive constant. (a) Determine the components
Bx , By , and Bz , or at least as many of the three components as is possible from the information given. (b) If it is given in
addition that the magnetic field has magnitude 6F0 /qv, determine as much as you can about the remaining components of B.
(a) Writing FB = qv · B in terms of components
FB = q[î(vy Bz − vz By ) − ĵ(vx Bz − vz Bx ) + k̂(vx By − vy Bx )] = q(−vz By î + vz Bx ĵ) = F0 (3î + 4ĵ) .
(7)
Matching components and writing vz = v we have
−qvBy = 3F0
−3F0
By =
<0
qv
qvBx = 4F0
4F0
Bx =
>0.
qv
(8)
(9)
We can’t find Bz because it does not contribute to the force felt by the charge, which is currently our only handle on B.
(b) With |B|, we can solve for |Bz |.
|B|2 = Bx2 + By2 + Bz2
Bz2 = |B2 | − Bx2 − By2 =
(10)
F2
36 2 0 2
q v
√ F0
.
Bz = ± 11
qv
However, we still cannot find the direction of Bz .
−
F2
16 2 0 2
q v
−
F2
9 202
q v
= (36 − 16 −
F2
9) 2 0 2
q v
=
F2
11 2 0 2
q v
(11)
(12)
Problem 27.68. A 3.00 N metal bar, 1.50 m long and having a resistance of 10.0 Ω, rests horizontally on conducting wires
connecting it to the circuit shown in Fig. 27.62. The bar is in a uniform, horizontal, 1.60 T magnetic field and is not attached
to the wires in the circuit. What is the acceleration of the bar just after the switch S is closed?
B
10.0 Ω
120.0 V
25.0 Ω S
This problem is extremely similar to Prob. 27.39, but with a more complicated circuit driving current through the bar, and
a top-down view in the figure.
When S is closed, current begins to flow, creating a magnetic force Fb = Ib l × B on the bar, where Ib is the current through
the bar. Using our cross product right-hand rule and magnitude formula, we see that the force on the bar is out of the page
with a magnitude Fb = Ib lB. We just need to find Ib .
Because the bar and vertical resistor are in parallel, the equivalent resistance for the vertical resistor and bar is
−1
1
1
= 5.00 Ω ,
+
R0 =
10.0 Ω 10.0 Ω
(13)
and the resistance for the entire circuit is
R = 25.0 Ω + 5.00 Ω = 30.0 Ω .
(14)
This effective resistance must drop all the voltage generated by the battery, so we can manipulate Ohm’s law V = IR to
yield
V
120.0 V
I=
=
= 4.00 A .
(15)
R
30 Ω
This is the total current through the battery. Now we need to find the current through the bar. The voltage across the
vertical-resistor/bar portion is
Vb = IR0 = 4.00 A · 5.00 Ω = 20.0 V ,
(16)
and current through the bar is
Ib =
Vb
20.0 V
=
= 2.00 A
Rb
10.0 Ω
(17)
The lifting force and acceleration are then
F = Fb − Fg = Ib lB − mg = ma
2.00 A · 1.50 m · 1.60 T
Ib lB
− 1)g =
− 1 · 9.8 m/s2 = 5.88 m/s2 .
a=(
mg
3.00 N
(18)
(19)
Note that I factored out g to mg in the denominator, because they give the weight of the bar (mg = 3.00 N), not the mass
m. Obviously, you could get a number for m if you want, but keeping mg together allows us to avoid the trouble.
Problem 27.73. A long wire carrying a 6.00 A current reverses direction by means of two right-angle bends, as shown in
Fig. 27.64. The part of the wire where the bend occurs is in a magnetic field of 0.666 T confined to the circular region of
diameter 75 cm, as shown. Find the magnitude and direction of the net force that the magnetic field exerts on this wire.
6.00 A
45.0 cm
75.0 cm
The horizontal portions of wire feel equal and opposite magnetic force, because FB = Il · B and the only differences are in
the sign of l. Therefore we only need worry about the vertical bit of wire, where the force is to the left from the right hand
rule. The magnitude of the force is
FB = IlB = 6.00 A · 0.450 m · 0.666 T = 1.80 N .
(20)