27.3. Identify: The force
on the particle is in the direction of the deflection of the
particle. Apply the right-hand rule to the directions of 
and 
. See if your thumb is in the direction of , or opposite to that direction. Use 
with 
to calculate F.
Set Up: The
directions of , and are shown in Figure
27.3.
Execute: (a) When
you apply the right-hand rule to and , your thumb points east. is in this direction,
so the charge is positive.
(b) 
Evaluate: If
the particle had negative charge and and are unchanged, the particle would be deflected toward the
west.
Figure 27.3
27.5. Identify: Apply and solve for v.
Set Up: An
electron has 
.
Execute: 
Evaluate: Only
the component 
of the magnetic field
perpendicular to the velocity contributes to the force.
27.7. Identify: Apply 
.
Set Up: 
, with 
. 
and 
.
Execute: (a) 
.
which is consistent
with as given in the problem. There is no force component along
the direction of the velocity.
. 
.
(b) 
is not determined. No force due to this component of along ; measurement of the force tells us nothing about 
(c) 
. and are perpendicular (angle is 
.
Evaluate: The
force is perpendicular to both and , so 
is also zero.
27.9. Identify: Apply 
to the force on the proton and to the force on the electron. Solve
for the components of .
Set Up: is perpendicular to both and . Since the force on the proton is in the +y-direction, 
and 
. For the proton, 
.
Execute: (a) For the proton, 
so 
. The force on the proton is independent of 
. For the electron, 
. 
.
The magnitude of the force is 
. Since 
, 
. 
. The sign of is not determined by measuring
the magnitude of the force on the electron. 
. 
. 
. is in the xz-plane
and is either at 
from the +x-direction toward the 
or from the 
toward the .
(b) . 
. 
. 
. 
. 
. The force is in the xz-plane
and is directed at 
from the 
toward either the +z or 
axis, depending on the
sign of .
Evaluate: If the direction of the force on the first
electron were measured, then the sign of would be determined.
27.53. (a)
Identify: Use Eq.(27.2) to relate 
Set Up: The directions of 
are shown in Figure
27.53a.
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Figure 27.53a |
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The directions of 
are shown in Figure
27.53b.
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Figure 27.53b |
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Both pieces of information
taken together say that 
is in the y-direction;
Execute: Use the information given about 
to calculate 
(b) 
Evaluate: 
is perpendicular
to whereas only the
component of 
perpendicular to contributes to the
force, so it is expected that 
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 
and qvB – qE – mg = 0.
Execute: Solving for B gives
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.