A charge q resides on a point mass of value m.The charge is
initially located BELOW the x-axis where B=0 and travels up along
the positive y-axis with velocity v. There is a uniform magnetic
field above the x-axis.
q=+4.50 mC
m=1.92×10-12 kg
v=7.50×105 m/s
^
j
B=-3.20 T
^
k
(a) [5 pts] As the charge moves through the origin, does it veer to the
left or right of the y-axis?
ANS: Using the right-hand-rule, the charge veers (bends) to the LEFT.
(b) [8 pts] Calculate the force on the charge as it just moves into the
magnetic field (magnitude and direction).
F
=
qv×B
=
(4.50×10-6 C)
æ è
7.50×105 m/s
^
j
ö ø
×
æ è
-3.20 T
^
k
ö ø
=
-10.8 N
^
i
(c) [8 pts] The charge will eventually leave the magnetic field region
(y ³ 0). Where, on the x-axis will this occur?
R
=
mv
qB
=
(1.92×10-12 kg)(7.50×105 m/s)
(4.50×10-6 C)(3.20 T)
=
0.100 m
The charge leaves the magnetic field at
x=-2R=-0.200 m
(d) [4 pts] With what speed does the charge leave the magnetic field?
The charge leaves the magnetic field traveling in the negative y-direction
with a speed of
v=7.50×105 m/s
A coil of N identical turns bounds the perimeter of a half-circular
area which lies in the x-z plane, as shown in the figure. Each turn
carries a constant current I. The radius of the semicircle is R. A uniform
magnetic field B is directed along the z-axis. For the given coil
position:
N=20 turns
I=12.0 A
R=4.00 cm
B=1.60 T
(a) [7 pts] calculate the magnetic force on the straight wire segment
[`OA]. Give both magnitude and direction.
F
=
NIL×B
=
(20 turns)(12.0 A/turn)
æ è
-8.00×10-2 m
^
i
ö ø
×(1.60 T
^
k
)
=
+30.7 N
^
j
(b) [7 pts] calculate the magnitude and direction of the magnetic
moment, m, of the coil.
m
=
NIA
=
(20 turns)(12.0 A/turn)p(0.0400 m)2/2
=
0.603 A·m2
m
=
+0.603 A·m2
^
j
(c) [7 pts] calculate the instantaneous torque on the coil (magnitude
and direction).
t
=
m×B
=
æ è
0.603 A·m2
^
j
ö ø
×
æ è
1.60 T
^
k
ö ø
=
+0.965 N·m
^
i
(d) [4 pts] determine the axis of rotation of the coil and the sense of
the rotation about the axis (CW or CCW).
ANS: The x-axis is the axis of rotation. Looking in along thex-axis
towards the origin, the rotation is CCW (thumb in the direction of the torque
and fingers in the direction of the rotation).
A semi-infinite straight wire is connected to a half-circular wire of
radius R as shown in the figure. The combination carries a constant current
I.
I=20.0 A
R=5.00 cm
(a) [7 pts] Calculate the magnetic field at the center of curvature due
to the half-circular wire. [Hint: What is the magnetic field at the center of
a full circular wire?]
Bc
=
moI
4R
=
(4p×10-7 T·m/A)(20.0 A)
4(0.0500 m)
=
1.26×10-4 T
Bc
=
-1.26×10-4 T
^
k
(b) [7 pts] Calculate the magnetic field at the center of curvature due
to the semi-infinite wire. [Hint: what is the magnetic field outside a full
infinite wire?]
Bw
=
moI
4pR
=
(4p×10-7 T·m/A)(20.0 A)
4p(0.0500 m)
=
4.00×10-5 T
Bw
=
-4.00×10-5 T
^
k
(c) [4 pts] What is the total magnetic field at the center of curvature
due to both wires?
Btot
=
Bc+Bw
=
-(1.26×10-4 T+4.00×10-5 T)
^
k
=
-1.66×10-4 T
^
k
(d) [7 pts] Suppose that a positive point charge q=8.00 mC moves
1.00 cm directly below the center of curvature to the right. How fast must
it be moving to cancel the field at the center of curvature due to the wires?
Neglect the magnetic field due to the wires on the moving charge.
Bq
=
moq
4p
v×
^
r
r2
=
+Bq
^
k
The equation reduces to
Bq
=
moqv
4py2
v
=
4py2Bq
moq
=
4p(0.0100 m)2(1.657×10-4 T)
(4p×10-7 T·m/A)(8.00× 10-6 C)
=
2.07×104 m/s
Two long parallel wires carry equal constant currents. The magnitude
of the magnetic field 1.00 cm from wire 2 is 3.60×10-4 T.
(a) [5 pts] In wire 2, using the figure, what is the direction of the
current?
ANS: According to the direction of the magnetic field along the path
surrounding wire 2, the direction of the current is DOWN.
(b) [5 pts] In wire 2, what is the magnitude of the current?
I
=
2prB2
mo
=
2p(0.0010 m)(3.60×10-4 T)
4p×10-7 T·m/A
=
18.0 A
(c) [5 pts] If the wires repel, what is the direction of the current in
wire 1?
ANS: If the wires repel, the currents must oppose. The current direction in
wire 1 is UP.
(d) [10 pts] If the force per unit length each wire exerts on the other
wire is 1.62×10-3 N/m, calculate the wire separation.
F
L
=
moI2
2pd
d
=
moI2
2p(F/L)
=
(4p×10-7 T·m/A)(18.0 A)2
2p(1.62×10-3 N/m)
=
0.0400 m ® 4.00 cm
FORMULAS
F=qv×B
F=NIL×B
R=mv^/qB
F=moI1I2L/2pd
B=moI/2pr
B=moI/2r
m = NIA
t = m×B
B=(moq/4p) (v×
^
r
)/r2
File translated from
TEX
by
TTH,
version 3.85. On 28 Mar 2016, 18:02.