Download 27.3. Identify: The force on the particle is in the direction of the

27.3.
IDENTIFY: The force F on the particle is in the direction of the deflection of the particle. Apply the right-hand
rule to the directions of v and B . See if your thumb is in the direction of F , or opposite to that direction. Use
F  q vB sin  with   90° to calculate F.
SET UP: The directions of v , B and F are shown in Figure 27.3.
EXECUTE: (a) When you apply the right-hand rule to v and B , your thumb points east. F is in this direction,
so the charge is positive.
(b) F  q vB sin  (8.50 106 C)(4.75 103 m/s)(1.25 T)sin90°  0.0505 N
EVALUATE: If the particle had negative charge and v and B are unchanged, the particle would be deflected
toward the west.
Figure 27.3
27.5.
IDENTIFY:
Apply F  q vB sin  and solve for v.
SET UP: An electron has q  1.60  1019 C .
EXECUTE:
v
F
4.60  1015 N

 9.49  106 m s
19
q B sin  (1.6  10 C)(3.5  103 T)sin 60
EVALUATE: Only the component B sin  of the magnetic field perpendicular to the velocity contributes to the
force.
27.7.
IDENTIFY: Apply F = qv  B .
SET UP:
v = vy ˆj , with vy   3.80  103 m s . Fx   7.60  103 N, Fy  0, and Fz   5.20  103 N .
EXECUTE: (a) Fx  q(v y Bz  vz By )  qv y Bz .
Bz  Fx qvy  (7.60  103 N) ([7.80  106 C)(  3.80  103 m s )]   0.256 T
Fy  q(vz Bx  vx Bz )  0, which is consistent with F as given in the problem. There is no force component along
the direction of the velocity.
Fz  q(vx By  v y Bx )   qv y Bx . Bx   Fz qv y   0.175 T .
(b) By is not determined. No force due to this component of B along v ; measurement of the force tells us
nothing about B y .
(c) B  F  Bx Fx  By Fy  Bz Fz  (0.175 T)(+7.60 103 N)  (0.256 T)(5.20 103 N)
B  F  0 . B and F are perpendicular (angle is 90) .
EVALUATE: The force is perpendicular to both v and B , so v  F is also zero.
27.9.
IDENTIFY: Apply F  qv  B to the force on the proton and to the force on the electron. Solve for the
components of B .
SET UP:
F is perpendicular to both v and B . Since the force on the proton is in the +y-direction, By  0 and
B = Bx iˆ  Bz kˆ . For the proton, v = (1.50 km/s)iˆ .
EXECUTE: (a) For the proton, F = q(1.50 103 m/s)iˆ  (Bx iˆ  Bz kˆ)  q(1.50 103 m/s)Bz ( ˆj).
2.25 1016 N
 0.938 T . The force on the proton is
(1.60 1019 C)(1.50 103 m/s)
independent of Bx . For the electron, v = (4.75 km/s)(  kˆ ) .
F = (2.251016 N) ˆj , so Bz  
F  qv  B  (e)(4.75 103 m/s)(kˆ)  ( Bxiˆ  Bz kˆ ) = e(4.75 103 m/s) Bx ˆj .
The magnitude of the force is F  e(4.75 103 m/s) Bx . Since F  8.50 1016 N ,
Bx 
8.50 1016 N
 1.12 T . Bx  1.12 T . The sign of Bx is not determined by measuring
(1.60 1019 C)(4.75 103 m/s)
the magnitude of the force on the electron. B  Bx2  Bz2  (1.12 T)  (0.938 T) 2  1.46 T .
tan  
Bz 0.938 T

.   40° . B is in the xz-plane and is either at 40° from the +x-direction toward the
Bx
1.12 T
 z-direction or 40° from the  x-direction toward the  z-direction .
(b) B = Bx iˆ  Bz kˆ . v = (3.2 km/s)( ˆj ) .
F  qv  B  (e)(3.2 km/s)( ˆj )  (Bx iˆ  Bz kˆ )  e(3.2  103 m/s)(Bx (kˆ )  Bz iˆ ) .
F = e(3.2 103 m/s)([1.12 T]kˆ  [0.938 T]iˆ )   (4.80 1016 N)iˆ  (5.73 1016 N)kˆ
Fz 5.73 1016 N
.   50.0° . The force is in the xz-plane and is

Fx 4.80 1016 N
directed at 50.0° from the  x-axis toward either the +z or  z axis, depending on the sign of Bx .
F  Fx2  Fz2  7.47  1016 N . tan  
27.53.
EVALUATE: If the direction of the force on the first electron were measured, then the sign of Bx would be
determined.
(a) IDENTIFY: Use Eq.(27.2) to relate v , B, and F .
SET UP: The directions of v1 and F1 are shown in Figure 27.53a.
F  qv  B says that F is perpendicular
to v and B. The information given here
means that B can have no z-component.
Figure 27.53a
The directions of v2 and F2 are shown in Figure 27.53b.
F is perpendicular to v and B, so B can
have no x-component.
Figure 27.53b
Both pieces of information taken together say that B is in the y-direction; B  By ˆj.
EXECUTE: Use the information given about F2 to calculate Fy : F2  F2iˆ, v2  v2kˆ, B  By ˆj.
F  qv  B says F iˆ  qv B kˆ  ˆj  qv B (iˆ) and F  qv B
2
2
2
2
y
2
y
2
2
y
By  F2 /(qv2 )  F2/(qv1 ). B has the maginitude F2 /(qv1 ) and is in the  y-direction.
(b) F1  qvB sin   qv1 By / 2  F2 / 2
EVALUATE:
v1  v2 . v2 is perpendicular to B whereas only the component of v1 perpendicular to B
contributes to the force, so it is expected that F2  F1 , as we found.
27.55.
IDENTIFY: The sum of the magnetic, electrical, and gravitational forces must be zero to aim at and hit the
target.
SET UP: The magnetic field must point to the left when viewed in the direction of the target for no net force.
The net force is zero, so  F  FB  FE  mg  0 and qvB – qE – mg = 0.
EXECUTE: Solving for B gives
B
qE  mg (2500 106 C)(27.5 N/C) + (0.0050 kg)(9.80 m/s2 )

 3.7 T
qv
(2500106 C)(12.8 m/s)
The direction should be perpendicular to the initial velocity of the coin.
EVALUATE: This is a very strong magnetic field, but achievable in some labs.