Unformatted text preview: e magnitude of the electron’s charge is 1.60 × 10
C, and the
−31
electron’s mass is 9.11 × 10 kg (see the inside of the front cover), so a= q0 vB sin θ
m (1.60 × 10−19 C )( 2.1 × 106 m/s )(1.6 × 10−5 T ) sin 90.0° = 5.9 × 1012 m/s2
=
9.11 × 10−31 kg 1138 MAGNETIC FORCES AND MAGNETIC FIELDS 3. SSM REASONING According to Equation 21.1, the magnitude of the magnetic force on
a moving charge is F = q0 vB sin θ . Since the magnetic field points due north and the proton moves eastward, θ = 90.0°. Furthermore, since the magnetic force on the moving
proton balances its weight, we have mg = q0 vB sin θ , where m is the mass of the proton.
This expression can be solved for the speed v.
SOLUTION Solving for the speed v, we have
v= 4. mg
(1.67 × 10 –27 kg)(9.80 m/s 2 )
=
= 4.1×10 –3 m/s
q0 B sin θ (1.6 ×10 –19 C)(2.5 ×10 –5 T) sin 90.0° REASONING The magnitude q0 of the electric
charge of the bullet is related to the magnitude F
of the magnetic force exerted on the bullet
F
(Equation 21.1).
according to B =
South
q0 ( v sin θ ) Up
11° v
North 58° θ
Here, θ is the angle between the direction of the
bullet’s velocity v and the earth’s magnetic field
B
B. (See the drawing, which depicts the situation
as seen from the side.) We will find the
Down
magnitude of the bullet’s charge from Equation
21.1, and RightHand Rule No. 1 will allow us to determine the algebraic sign of the charge. SOLUTION Solving B = F
(Equation 21.1) for q0 , we obtain
q0 ( v sin θ )
q0 = F
B ( v sin θ ) (1) The angle θ is the angle between the directions of the bullet’s velocity v and the direction of
earth’s magnetic field B. Therefore, θ is the sum of the angles that these vectors make with
the horizontal (see the drawing): θ = 58o + 11o = 69o
Applying RightHand Rule No. 1 to the vectors v and B (see the drawing), we see that the
force on a positively charged bullet would point into the page, which is west. Because the
force on this bullet points to the east, out of the page, the bullet’s charge must be negative. Chapter 21 Problems 1139 When removing the absolute value brackets from Equation (1), therefore, we must insert a
minus sign on the right hand side. Making this change, we find that
q0 = − 5. F
2.8 × 10−10 N
=−
= −8.3 × 10−9 C
−5
o
B ( v sin θ )
5.4 × 10 T ( 670 m/s ) sin 69 ( ) REASONING According to Equation 21.1, the magnetic force has a magnitude of
F = q vB sin θ, where q is the magnitude of the charge, B is the magnitude of the
magnetic field, v is the speed, and θ is the angle of the velocity with respect to the field. As
θ increases from 0° to 90°, the force increases, so the angle must lie between 25° and 90°.
SOLUTION Letting θ1 = 25° and θ2 be the desired angle, we can apply Equation 21.1 to
both situations as follows: F = q vB sin θ1
144
244
3 and Situation 1 2 F = q vB sin θ 2
1442443
Situation 2 Dividing the equation for situation 2 by the equation for situation 1 gives
2 F q vB sin θ 2
=
F
q vB sin θ1 or sin θ 2 =2sin θ1 =2sin 25° = 0.85 θ 2 = sin −1 ( 0.85 ) = 58° 6. REASONING A moving charge experiences no magnetic force when its velocity points in
the direction of the magnetic field or in the direction opposite to the magnetic field. Thus,
the magnetic field must point either in the direction of the +x axis or in the direction of the
– x axis. If a moving charge experiences the maximum possible magnetic force when
moving in a magnetic field, then the velocity must be perpendicular to the field. In other
words, the angle θ that the charge’s velocity makes with respect to the magnetic field is
θ = 90°.
SOLUTION The magnitude B of the magnetic field can be determined using Equation 21.1:
B= F
0.48 N
=
= 0.12 T
q v sin θ
8.2 × 10−6 C 5.0 × 105 m/s sin 90° ( )( ) In this calculation we use θ = 90°, because the 0.48N force is the maximum possible force.
Since the particle experiences no magnetic force when it moves along the +x axis, we can
conclude that the magnetic field points 1140 MAGNETIC FORCES AND MAGNETIC FIELDS either in the direction of the +x axis or in the direction of the –x axis . 7. REASONING When a charge q0 travels at a speed v and its velocity makes an angle θ with
respect to a magnetic field of magnitude B, the magnetic force acting on the charge has a
magnitude F that is given by F = q0 vB sin θ (Equation 21.1). We will solve this problem by applying this expression twice, first to the motion of the charge when it moves
perpendicular to the field so that θ = 90.0° and then to the motion when θ = 38°.
SOLUTION When the charge moves perpendicular to the field so that θ = 90.0°,
Equation 21.1 indicates that
F90.0° = q0 vB sin 90.0° When the charge moves so that θ = 38°, Equation 21.1 shows that
F38 = q0 vB sin 38°
Dividing the second expression by the first expression gives
F38°
F90.0° = q0 vB sin 38°
q0 vB sin 90.0° sin 38° sin 38° −3
−3
F38° = F90.0° = 2.7 × 10 N = 1.7 × 10 N sin 90.0°...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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