Download Physics 107 HOMEWORK ASSIGNMENT #18

Physics 107 HOMEWORK ASSIGNMENT #18
Cutnell & Johnson, 7th edition
Chapter 21: Problems 16, 20, 34, 42, 54
16 An -particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest
through a potential difference that has a value of 1.20 x 106 V and then enters a uniform magnetic
field whose magnitude is 2.20 T. The -particle moves perpendicular to the magnetic field at all
times. What is (a) the speed of the -particle, (b) the magnitude of the magnetic force on it, and
(c) the radius of its circular path?
*20 Review Conceptual Example 2 as background for this problem. A charged particle moves
through a velocity selector at a constant speed in a straight line. The electric field of the velocity
selector is 3.80 × 103 N/C, while the magnetic field is 0.360 T. When the electric field is turned
off, the charged particle travels on a circular path whose radius is 4.30 cm. Find the charge-tomass ratio of the particle.
*34 Consult Interactive Solution 21.34 to explore a model for solving
this problem. The drawing shows a thin, uniform rod, which has a length
of 0.45 m and a mass of 0.094 kg. This rod lies in the plane of the paper
and is attached to the floor by a hinge at point P. A uniform magnetic
field of 0.36 T is directed perpendicularly into the plane of the paper.
There is a current I = 4.1 A in the rod, which does not rotate clockwise or
counterclockwise. Find the angle . (Hint: The magnetic force may be
taken to act at the center of gravity.)
42 Two pieces of the same wire have the same length. From one piece, a square coil containing
a single loop is made. From the other, a circular coil containing a single loop is made. The coils
carry different currents. When placed in the same magnetic field with the same orientation, they
experience the same torque. What is the ratio Isquare/Icircle of the current in the square coil to that in
the circular coil?
*54 Multiple-Concept Example 8 and Conceptual Example 9
provide background that will help you understand the approach
needed in this problem. A rectangular current loop is located near a
long, straight wire that carries a current of 12 A (see the drawing).
The current in the loop is 25 A. Determine the magnitude of the net
magnetic force that acts on the loop.
16. REASONING The speed of the α-particle can be obtained by applying the principle of
conservation of energy, recognizing that the total energy is the sum of the particle’s kinetic
energy and electric potential energy, the gravitational potential energy being negligible in
comparison. Once the speed is known, Equation 21.1 can be used to obtain the magnitude
of the magnetic force that acts on the particle. Lastly, the radius of its circular path can be
obtained directly from Equation 21.2.
SOLUTION
a. Using A and B to denote the initial positions, respectively, the principle of conservation
of energy can be written as follows:
1
mvB2
2
Final kinetic
energy
+
1
EPE B
=
mvA2 + EPE A
2
Final electric
potential energy
(1)
Initial electric
potential energy
Initial kinetic
energy
Using Equation 19.3 to express the electric potential energy of the charge q0 as EPE = q0V,
where V is the electric potential, we find from Equation (1) that
1
mvB2
2
1
2
+ q0VB = mvA2 + q0VA
(2)
Since the particle starts from rest, we have that vA = 0 m/s, and Equation (2) indicates that
vB =
2q0 (VA − VB )
m
2  2 +1.60 × 10−19 C  1.20 ×106 V


=
= 1.08 × 107 m/s
−27
6.64 ×10
kg
(
)(
)
b. According to Equation 21.1, the magnitude of the magnetic force that acts on the particle
is
(
)(
)
F = q0 vB B sin θ = 2 1.60 ×10−19 C 1.08 ×107 m/s ( 2.20 T ) sin 90.0° = 7.60 × 10−12 N
where θ = 90.0°, since the particle travels perpendicular to the field at all times.
c. According to Equation 21.2, the radius of the circular path on which the particle travels is
6.64 ×10 −27 kg )(1.08 × 107 m/s )
(
r=
=
= 0.102 m
q0 B
2 (1.60 × 10 −19 C ) ( 2.20 T )
mvB
20. REASONING AND SOLUTION The magnitudes of the magnetic and electric forces must
be equal. Therefore,
FB = FE or
q vB = q E
This relation can be solved to give the speed of the particle, v = E/B. We also know that
when the electric field is turned off, the particle travels in a circular path of radius
r = mv/( q B). Substituting v = E/B into this equation and solving for q /m gives
q
m
=
E
3.80 × 103 N/C
=
= 6.8 × 105 C/kg
2
2
−
2
rB
4.30 × 10 m ( 0.360 T )
(
)
34. REASONING Since the rod does not rotate about the axis at P, the net torque relative to
that axis must be zero; Στ = 0 (Equation 9.2). There are two torques that must be considered,
one due to the magnetic force and another due to the weight of the rod. We consider both of
these to act at the rod's center of gravity, which is at the geometrical center of the rod
(length = L), because the rod is uniform. According to Right-Hand Rule No. 1, the magnetic
force acts perpendicular to the rod and is directed up and to the left in the drawing.
Therefore, the magnetic torque is a counterclockwise (positive) torque. Equation 21.3 gives
the magnitude F of the magnetic force as F = ILB sin 90.0°, since the current is
perpendicular to the magnetic field. The weight is mg and acts downward, producing a
clockwise (negative) torque. The magnitude of each torque is the magnitude of the force
times the lever arm (Equation 9.1). Thus, we have for the torques:
τ magnetic = + ( ILB ) ( L / 2 )
τ weight = − ( mg ) ( L / 2 ) cos θ 
and
force lever arm
force
lever arm
Setting the sum of these torques equal to zero will enable us to find the angle θ that the rod
makes with the ground.
SOLUTION Setting the sum of the torques equal to zero gives Στ = τmagnetic + τweight = 0, and
we have
ILB
+ ( ILB )( L / 2 ) − ( mg ) ( L / 2 ) cos θ  = 0
or
cos θ =
mg
 4.1 A 0.45 m 0.36 T 
(
)(
)(
) =
θ = cos
2
 ( 0.094 kg ) 9.80 m/s 


–1 
(
)
44°
42. REASONING The magnitude τ of the torque that acts on a current-carrying coil placed in a
magnetic field is given by τ = NIAB sinφ (Equation 21.4), where N is the number of loops
in the coil (N = 1 in this problem), I is the current, A is the area of one loop, B is the
magnitude of the magnetic field (the same for each coil), and φ is the angle (the same for
each coil) between the normal to the coil and the magnetic field. Since we are given that the
torque for the square coil is the same as that for the circular coil, we can write
(1) Isquare Asquare B sin φ
τ square
= (1) I circle Acircle B sin φ
τ circle
This relation can be used directly to find the ratio of the currents.
SOLUTION Solving the equation above for the ratio of the currents yields
Isquare
I circle
=
Acircle
Asquare
If the length of each wire is L, the length of each side of the square is
the square coil is Asquare =
1
4
L , and the area of
( 14 L )( 14 L ) = 161 L2 . The area of the circular coil is
Acircle = π r 2 ,
where r is the radius of the coil. Since the circumference (2π r) of the circular coil is equal to
the length L of the wire, we have 2π r = L, or r = L / ( 2π ) . Substituting this value for r into
2
the expression for the area of the circular coil gives Acircle = π [ L / ( 2π )] . Thus, the ratio of
the currents is
2
Isquare
I circle
=
Acircle
Asquare
 L 
π 
4
2π
=  2 = = 1.27
1 L
π
16
54. REASONING AND SOLUTION The net force on the wire loop is a sum of the forces on
each segment of the loop. The forces on the two segments perpendicular to the long straight
wire cancel each other out. The net force on the loop is therefore the sum of the forces on
the parallel segments (near and far). These are
–4
Fnear = µ oI1I2L/(2π dnear) = µ 0(12 A)(25 A)(0.50 m)/[2π (0.11 m)] = 2.7 × 10 N
–4
Ffar = µ oI1I2L/(2π dfar) = µ 0(12 A)(25 A)(0.50 m)/[2π (0.26 m)] = 1.2 × 10 N
Note: Fnear is a force of attraction, while Ffar is a repulsive one. The magnitude of the net
force is, therefore,
–4
–4
F = Fnear − Ffar = 2.7 × 10 N − 1.2 × 10 N = 1.5 × 10 −4 N