343
—
Chapter 19
19.
Magnetism
(a) The direction of the current is the directi
on that a positive charge would have to move
to experience the given
force. Applying the right-hand force rule gives the
following current directions.
(i) to the [i] (ii) upwardl
(iii) unto the page
(iv) to the
(v) into or out of the page}; either would give zero force
(b) The force on a wire segment is F
F(5.5 A)(0.15 m)(l.0 x
ILB sin 0. In each case except (v), 0
T)(sin
900)
900,
so the force is
83 x l0 Ni
In case (v) the force is
20.
(a) The magnetic force is in the [(3)
(b) F
21.
F
22.
JLB sin 0= (5.0 A)(l.0 m)(0.30 T) sin 90°
ILB sin 0
(a) F
(20 A)(2.0 m)(0.050 T) sin 37°
JLB sin 0
(b)F=
d
=
=
jl.5
NI.
11.2 N perpendicular to the plane of
(15 A)(0.75 m)(l.0 x 10 T) sin 30°
=
5.6 x
B and
11.
NI
F,
ILBsinq
4- lLBsin3O°
sinço-
ojl4.5°
—
Also, sin(l80°- l4.5°)’
23.
+yj direction according to the right-hand force rule, (upward as +z).
-,
so 180°— 14.5°
(a) F/L
“
lB sin 0= jj since 0=0°
(b) F/L
=
(10 A)(0.40 T) sin 90°
[io N/mi
=
165.5°
wouldalsowork.
in the J+z directionj
(c) Using the same approach gives
F/L
4.0 N/mI in the
=
(d)F/L
—y directionj
14,ON/ml in the
(e)F/L= 14.0 N/mI in the
(f) F/L
24.
=
1—z
directioni
Hy directioni
(10 A)(0.40 T) sin 45°
12.8
N/mI in the l+z
direction 1
(a)FlLBsinO
0.050 N
1
=
1(0.25 m)(0.30 T) sin 90°
10.67 Al
(b) Using F
0.025 N
=
in the
direction 1
JLB sin 0, we have
1(0.25 m)(0.30 T) sin 90°
Copyright © 2010 Pearson Education, Inc. All rights
reserved. This material is protected under all
copyright laws as they currently exist. No
portion of this material may be reproduced, in any form
or by any means, without permission in writing
from the publisher.
College Physics Seventh Edition: Instructor Solutions Manual
1=10.33 Al in the
344
I+v directioni
(c) This is not possiblej because the force must be perpendicular to the magnetic field.
25.
(a) First find the angle 0 between the wire (the +x-axis) and the magnetic field.
tan 0 BJB, = (0.040 T)/(0.020 T)
—*
0= 63.43°
The magnitude of the magnetic field is
B
=
...JB
F/L
+
B
lB sin 0
=
\J(0.020 T)
2
+
(0.040 T)
2
0.04472 T
(10 A)(0.04472 T) sin 63 .43°
=
10.400 N/mJ in the
l+z
direction I
(b) Since B is parallel to the wire, changing it will have no effect on the force, so the answers are the
same as in part
(a)I.
(c) Using the same approach as in part (a), we have
tan 0= (0050 T)/(0.020 T)
B
26.
+
B
=
J(0.020 T)
2
+
(0.050 T)
2
(10 A)(0.05385 T) sin 68.20°
F/L
=
F
ILB sin 0
0= 68.20°
—*
=
10.500
=
0.053 85 T
N/mI in the
(1000 A)(15 m)(5.0 x i0 T) sin 45°
=
r—z directionl
10.53 NI.
The direction of the force is northward at an angle of 45° above the horizontall, according to the right-hand
rule.
27.
F= ILB sin 0= (5.0 A)(3 x 0.50 m)(1.0 T) sin 90°
28.
(a) For the magnetic torque to be at maximum, the plane of the coil should be [(1) parallelj to the magnetic
field.
=
17.5 N upward in the plane of the paper].
r= JAB sin Oand the angle 0 here is between the magnetic field and a direction perpendicular to the plane of the
coil. When 0=90°, ris maximum.
(b) m
=
1.4
(1.5 A)(0.20 m)(0.30 m)
r= JAB sin 0 mB sin 0(9.0 x
(c)sinO2O%0.20,
29.
(a) r
‘
=
19.0
x l02 Am21.
A.m)(1.6 T)sin 90°
=
[0.14
m.NI.
0Hl2°or16i].
NIAB sin 0= (l)(0.25 A)(0.20 m2)(0.30 T) sin 60°
=
10.013 mN I
(b) Since the torque is proportional to the magnetic field strength, doubling the fieldj would double the torque.
If
you are allowed to also change the direction of the field, any combination of B and 0 such that B sin Owas doubled
would also double the torque.
(c) Doubling the current[ will double the torque.
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all
copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from
the publisher.