Homework for the week of November 3. 6th week of classes.
Ch. 27: 6, 8, 16, 20, 23, 33, 35, 36
6.
The magnetic force must be equal in magnitude to the force of gravity on the wire. The
maximum magnetic force is applicable since the wire is perpendicular to the magnetic field.
The mass of the wire is the density of copper times the volume of the wire.
FB = mg → I l B = ρπ ( 12 d ) l g →
2
I=
ρπ d 2 g
4B
(8.9 × 10 kg
=
3
) (
) ( 9.80 m s ) = 1400 A
4 ( 5.0 × 10 T )
m3 π 1.00 × 10−3 m
2
2
−5
This answer does not seem feasible. The current is very large, and the resistive heating in
the thin copper wire would probably melt it.
8.
We find the force per unit length from Eq. 27-3. Note that while the length is not known, the
r
direction is given, and so l = l ˆi.
r r
r
r
FB = I l × B = I l ˆi × B →
r
FB
l
r
= I ˆi × B = ( 3.0 A )
ˆi
ˆj
kˆ
1
0
0
(
0.20T −0.36T 0.25T
(
 1m 
= −0.75ˆj − 1.08kˆ N m 

 100cm 
)
)
= − 7.5 ˆj + 11 kˆ × 10−3 N cm
16.
Since the charge is negative, the answer is the OPPOSITE of the result given from the
right hand rule
applied to the velocity and magnetic field.
(a) left
(b) left
(c) upward
(d) inward into the paper
(e) no force
(f) downward
20.
The velocity of each charged particle can be found using energy conservation. The
electrical
potential energy of the particle becomes kinetic energy as it is accelerated. Then, since the
particle is moving perpendicularly to the magnetic field, the magnetic force will be a
maximum. That force will cause the ion to move in a circular path, and the radius can be
determined in terms of the mass and charge of the particle.
Einitial = Efinal → qV = 12 mv 2 → v =
Fmax = qvB = m
v
2
r
→ r=
mv
qB
m
=
2qV
m
2qV
1
m
=
qB
B
2mV
q
rd
rp
rα
rp
=
1
B
2mdV
qd
=
md
mp
1 2mpV
B
qp
qd
qp
1 2mαV
B
qα
mα
mp
=
1
B
=
2mpV
qα
=
2
= 2 → rd = 2rp
1
=
4
2
= 2 → rα = 2rp
qp
qp
23. The kinetic energy of the proton can be used to find its velocity. The magnetic force
produces centripetal acceleration, and from this the radius can be determined.
K = 12 mv 2 → v =
r=
mv
qB
m
=
2K
m =
qB
2K
qvB =
m
2Km
qB
=
(
mv 2
r
→ r=
mv
qB
)(
)(
2 6.0 × 106 eV 1.60 × 10−19 J eV 1.67 × 10−27 kg
(1.60 × 10
−19
)
C ( 0.20T )
)
= 1.8 m
θ8
33. (a) In the magnetic field, the proton will move along an arc of a
circle. The distance x in the diagram is a chord of that circle,
and so the center of the circular path lies on the perpendicular
bisector of the chord. That perpendicular bisector bisects the
central angle of the circle which subtends the chord. Also recall
that a radius is perpendicular to a tangent. In the diagram,
θ1 = θ 2 because they are vertical angles. Then θ2 = θ 4 , because
θ5
θ4
θ6 θ7
r θ3 θ
2
they are both complements of θ3 , so θ1 = θ 4 . We have θ4 = θ5
θ1
since the central angle is bisected by the perpendicular bisector
of the chord. θ5 = θ7 because they are both complements of θ6 , and θ7 = θ8 because
they are vertical angles. Thus θ1 = θ 2 = θ4 = θ5 = θ7 = θ8 , and so in the textbook
diagram, the angle at which the proton leaves is θ = 45o .
mv
The radius of curvature is given by r =
, and the distance x is twice the value
( b)
qB
of r cos θ .
(1.67 × 10 kg )(1.3 × 10 m s ) cos 45
x = 2r cos θ = 2
cos θ = 2
qB
(1.60 × 10 C) ( 0.850T )
mv
−27
5
o
−19
= 2.3 × 10−3 m
35. The work required by an external agent is equal to the change in potential energy. The
r r
potential energy is given by Eq. 27-12, U = −µ B.
(a)
r r
r r
r r
r r
W = ∆U = ( −µ B )final − ( −µ B )initial = ( µ B )initial − ( µ B )final = NIAB ( cos θinitial − cos θfinal )
= NIAB ( cos 0° − cos180° ) = 2 NIAB
(b)
W = NIAB ( cos θinitial − cos θfinal ) = NIAB ( cos 90° − cos ( −90° ) ) = 0
36. With the plane of the loop parallel to the magnetic field, the torque will be a maximum. We
use Eq. 27-9.
τ
0.185 m N
τ = NIAB sin θ → B =
=
= 3.32 T
2
NIAB sin θ (1)( 4.20 A ) π ( 0.0650 m ) sin 90°