Download 27.15. (a) Identify: Apply Eq.(27.2) to relate the magnetic force to the

27.15.
(a) IDENTIFY: Apply Eq.(27.2) to relate the magnetic force F to the directions of v and B. The electron has
negative charge so F is opposite to the direction of v  B. For motion in an arc of a circle the acceleration is
toward the center of the arc so F must be in this direction. a  v2 / R.
SET UP:
As the electron moves in the semicircle,
its velocity is tangent to the circular path.
The direction of v0  B at a point along
the path is shown in Figure 27.15.
Figure 27.15
EXECUTE:
For circular motion the acceleration of the electron arad is directed in toward the center of the
circle. Thus the force FB exerted by the magnetic field, since it is the only force on the electron, must be radially
inward. Since q is negative, FB is opposite to the direction given by the right-hand rule for v0  B. Thus B is
directed into the page. Apply Newton's 2nd law to calculate the magnitude of B :
 F  ma gives  F
rad
 ma
FB  m(v /R )
2
FB  q vB sin  q vB, so q vB  m(v2/R)
B
mv (9.109  1031 kg)(1.41 106 m/s)

 1.60  104 T
qR
(1.602  1019 C)(0.050 m)
(b) IDENTIFY and SET UP: The speed of the electron as it moves along the path is constant. ( FB changes the
direction of v but not its magnitude.) The time is given by the distance divided by v0 .
 (0.050 m)

 1.11  107 s
v0 1.41  106 m/s
EVALUATE: The magnetic field required increases when v increases or R decreases and also depends on the
mass to charge ratio of the particle.
EXECUTE: The distance along the semicircular path is  R, so t 
27.17.
R
IDENTIFY and SET UP: Use conservation of energy to find the speed of the ball when it reaches the bottom of
the shaft. The right-hand rule gives the direction of F and Eq.(27.1) gives its magnitude. The number of excess
electrons determines the charge of the ball.
EXECUTE: q   4.00  108  1.602  1019 C   6.408  1011 C
speed at bottom of shaft:
1
2
mv2  mgy; v  2gy  49.5 m/s
v is downward and B is west, so v  B is north. Since q  0, F is south.
F  q vB sin    6.408  1011 C   49.5 m/s  0.250 T  sin 90  7.93  10 10 N
27.33.
EVALUATE: Both the charge and speed of the ball are relatively small so the magnetic force is small, much less
than the gravity force of 1.5 N.
IDENTIFY: The magnetic force is F  IlB sin  . For the wire to be completely supported by the field requires
that F  mg and that F and w are in opposite directions.
SET UP: The magnetic force is maximum when   90°. The gravity force is downward.
mg (0.150 kg)(9.80 m/s 2 )

 1.34 104 A. This is a very large current and
lB (2.00 m)(0.55 104 T)
ohmic heating due to the resistance of the wire would be severe; such a current isn’t feasible.
(b) The magnetic force must be upward. The directions of I, B and F are shown in Figure 27.33, where we
have assumed that B is south to north. To produce an upward magnetic force, the current must be to the east.
The wire must be horizontal and perpendicular to the earth’s magnetic field.
EVALUATE: The magnetic force is perpendicular to both the direction of I and the direction of B.
EXECUTE: (a) IlB  mg. I 
Figure 27.33
IDENTIFY: F  IlB sin  .
SET UP: Since the field is perpendicular to the rod it is perpendicular to the current and   90° .
F
0.13 N
 9.7 A
EXECUTE: I  
lB (0.200 m)(0.067 T)
EVALUATE: The force and current are proportional. We have assumed that the entire 0.200 m length of the rod
is in the magnetic field.
28.1.IDENTIFY and SET UP: Use Eq.(28.2) to calculate B at each point.
 qv  rˆ 0 qv  r
r
B 0

, since rˆ  .
4 r 2
4 r 3
r
6
ˆ
v   8.00  10 m/s  j and r is the vector from the charge to the point where the field is calculated.
27.37.
EXECUTE: (a) r   0.500 m  iˆ, r  0.500 m
v  r  vrˆj  iˆ  vrkˆ
B
 6.00 106 C 8.00 106 m/s  kˆ
0 qv ˆ
7
k


1

10
T

m/A


2
4 r 2
 0.500 m 
B   1.92  105 T  kˆ
(b) r    0.500 m  ˆj, r  0.500 m
v  r  vrˆj  ˆj  0 and B  0.
(c) r   0.500 m  kˆ , r  0.500 m
v  r  vrˆj  kˆ  vriˆ
B  1  107 T  m/A 
 6.00 10
6
C  8.00  106 m/s 
 0.500 m 
(d) r    0.500 m ˆj   0.500 m kˆ, r 


2
iˆ =  1.92  105 T  iˆ
 0.500 m   0.500 m
2
2
 0.7071 m
v  r = v  0.500 m   ˆj  ˆj  ˆj  kˆ   4.00  106 m2 /s  iˆ
B  1107 T  m/A 
 6.00 10
6
C  4.00 106 m / s 
 0.7071 m 
3
iˆ    6.79 106 T  iˆ
EVALUATE: At each point B is perpendicular to both v and r . B = 0 along the direction of v .
28.3.
IDENTIFY: A moving charge creates a magnetic field.
0 qv sin 
.
4 r 2
EXECUTE: Substituting numbers into the above equation gives
 qv sin  4 107 T  m/A (1.6 1019 C)(3.0 107 m/s)sin30
(a) B  0

.
4 r 2
4
(2.00 106 m)2
SET UP: The magnetic field due to a moving charge is B 
B = 6.00  10–8 T, out of the paper, and it is the same at point B.
(b) B = (1.00  10–7 T  m/A)(1.60  10–19 C)(3.00  107 m/s)/(2.00  10–6 m)2
B = 1.20  10–7 T, out of the page.
(c) B = 0 T since sin(180°) = 0.
EVALUATE: Even at high speeds, these charges produce magnetic fields much less than the Earth’s magnetic
field.
28.9.IDENTIFY: A current segment creates a magnetic field.
 Idl sin 
SET UP: The law of Biot and Savart gives dB  0
.
4
r2
EXECUTE: Applying the law of Biot and Savart gives
4π 107 T  m/A (10.0 A)(0.00110 m) sin90°
(a) dB 
= 4.40  10–7 T, out of the paper.
4π
(0.0500 m)2
28.12.
(b) The same as above, except r  (5.00 cm)2  (14.0 cm)2 and  = arctan(5/14) = 19.65°, giving dB = 1.67  10–8
T, out of the page.
(c) dB = 0 since  = 0°.
EVALUATE: This is a very small field, but it comes from a very small segment of current.
IDENTIFY: A current segment creates a magnetic field.
 Idl sin 
SET UP: The law of Biot and Savart gives dB  0
.
4 r 2
Both fields are into the page, so their magnitudes add.
EXECUTE: Applying the law of Biot and Savart for the 12.0-A current gives
 2.50 cm 
(12.0 A)(0.00150 m) 

4π  10 T  m/A
 8.00 cm  = 8.79  10–8 T
dB 
2
4π
(0.0800 m)
7
The field from the 24.0-A segment is twice this value, so the total field is 2.64  10–7 T, into the page.
EVALUATE: The rest of each wire also produces field at P. We have calculated just the field from the two
segments that are indicated in the problem.
28.15.
IDENTIFY: We can model the lightning bolt and the household current as very long current-carrying wires.
I
SET UP: The magnetic field produced by a long wire is B  0 .
2 r
EXECUTE: Substituting the numerical values gives
(4π 107 T  m/A)(20,000 A)
(a) B =
= 8  10–4 T
2π(5.0 m)
(4π 107 T  m/A)(10 A)
= 4.0  10–5 T.
2π(0.050 m)
EVALUATE: The field from the lightning bolt is about 20 times as strong as the field from the household
current.
I
IDENTIFY: B  0 . The direction of B is given by the right-hand rule in Section 20.7.
2 r
SET UP: Call the wires a and b, as indicated in Figure 28.21. The magnetic fields of each wire at points P1 and
P2 are shown in Figure 28.21a. The fields at point 3 are shown in Figure 28.21b.
EXECUTE: (a) At P1 , Ba  Bb and the two fields are in opposite directions, so the net field is zero.
(b) B 
28.21.
(b) Ba 
I
0 I
. Bb  0 . Ba and Bb are in the same direction so
2 rb
2 ra
B  Ba  Bb 
0 I  1 1  (4  107 T  m/A)(4.00 A)  1
1

6

  

  6.67  10 T
2  ra rb 
2
 0.300 m 0.200 m 
B has magnitude 6.67  T and is directed toward the top of the page.
(c) In Figure 28.21b, Ba is perpendicular to ra and Bb is perpendicular to rb . tan 
5 cm
and   14.04° .
20 cm
ra  rb  (0.200 m)2  (0.050 m)2  0.206 m and Ba  Bb .
 I 
2(4 107 T  m/A)(4.0 A)cos14.04°
B  Ba cos  Bb cos  2 Ba cos  2  0  cos 
 7.54  T
2 (0.206 m)
 2 ra 
B has magnitude 7.53  T and is directed to the left.
EVALUATE: At points directly to the left of both wires the net field is directed toward the bottom of the page.
Figure 28.21