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Unformatted text preview: nd of the force exerted by
the hand, the magnitude x of the rod’s displacement, and the cosine of the angle between the
force and the displacement. Since the rod moves to the right at a constant speed, it has no acceleration and is, therefore,
in equilibrium. Thus, the force exerted by the hand must be equal to the magnitude F of the
magnetic force that the current exerts on the rod. The magnitude of the magnetic force is
given by F = I LB sin θ (Equation 21.3), where I is the current, L is the length of the moving
rod, B is the magnitude of the magnetic field, and θ is the angle between the direction of the
current and that of the magnetic field. Chapter 22 Problems 1195 SOLUTION The average power P delivered by the hand is
P= W
t (6.10a) The work W done by the hand in Figure 22.5 is given by W = Fhand x cos θ ′ (Equation 6.1). In
this equation Fhand is the magnitude of the force that the hand exerts on the rod, x is the
magnitude of the rod’s displacement, and θ ′ is the angle between the force and the
displacement. The force and displacement point in the same direction, so θ ′ = 0° . Since the
magnitude of the force exerted by the hand equals the magnitude F of the magnetic force,
Fhand = F. Substituting W = Fhand x cos 0° into Equation 6.10a and using the fact that
Fhand = F, we have that
W F x cos 0° F x cos 0°
P = = hand
=
(1)
t
t
t
The magnitude F of the magnetic force is given by F = I LB sin θ (Equation 21.3). In this
case, the current and magnetic field are perpendicular to each other, so θ = 90° (see Figure
22.5). Substituting this expression for F into Equation 1 gives
P= F x cos 0° ( I LB sin 90° ) x cos 0°
=
t
t (2) The term x/t in Equation 2 is the speed v of the rod. Thus, the average power delivered by
the hand is
P= ( I LB sin 90° ) x cos 0° =
t ( I LB sin 90° )v cos 0° = ( 0.040 A )( 0.90 m )(1.2 T ) sin 90° ( 3.5 m/s )cos 0° = 0.15 W
____________________________________________________________________________________________ 8. REASONING Once the switch is closed, there is a current in the rod. The magnetic field
applies a force to this current and accelerates the rod to the right. As the rod begins to move,
however, a motional emf appears between the ends of the rod. This motional emf depends
on the speed of the rod and increases as the speed increases. Equally important is the fact
that the motional emf opposes the emf of the battery. The net emf causing the current in the
rod is the algebraic sum of the two emf contributions. Thus, as the speed of the rod increases
and the motional emf increases, the net emf decreases. As the net emf decreases, the current
in the rod decreases and so does the force that field applies to the current. Eventually, the
speed reaches the point when the motional emf has the same magnitude as the battery emf,
and the net emf becomes zero. At this point, there is no longer a net force acting on the rod
and the speed remains constant from this point onward, according to Newton’s second law.
This maximum speed can be determined by using Equation 22.1 for the motional emf, with
a value of the motional emf that equals the battery emf. 1196 ELECTROMAGNETIC INDUCTION SOLUTION Using Equation 22.1 ( ξ = vBL) and a value of 3.0 V for the emf, we find that
the maximum speed of the rod is ξ 3.0 V
= 25 m/s
BL ( 0.60 T )( 0.20 m )
______________________________________________________________________________
v= 9. = SSM REASONING The minimum length d of the rails is the speed v of the rod times the
time t, or d = vt. We can obtain the speed from the expression for the motional emf given in
ξ
Equation 22.1. Solving this equation for the speed gives v =
, where ξ is the motional
BL
emf, B is the magnitude of the magnetic field, and L is the length of the rod. Thus, the length
ξ of the rails is d = vt = t . While we have no value for the motional emf, we do know BL that the bulb dissipates a power of P = 60.0 W, and has a resistance of R = 240 . Power is related to the emf and the resistance according to P = ξ2 (Equation 20.6c), which can be
R
solved to show that ξ = PR . Substituting this expression into the equation for d gives ξ PR d =
t
t = BL BL SOLUTION Using the above expression for the minimum necessary length of the rails, we
find that PR ( 60.0 W )( 240 Ω ) d = BL t = ( 0.40 T )( 0.60 m ) ( 0.50 s ) = 250 m 10. REASONING AND SOLUTION
a. Newton's second law gives the magnetic retarding force to be
F = mg = IBL Now the current, I, is
I= ξ
R = vBL
R So m= v ( BL )
Rg 2 ( 4.0 m/s )( 0.50 T ) (1.3 m )
=
( 0.75 Ω ) ( 9.80 m/s2 )
2 2 = 0.23 kg Chapter 22 Problems 1197 b. The change in height in a time ∆t is ∆h = – v∆t. The change in gravitational potential energy
is
∆PE = mg∆h = – mgv∆t = – (0.23 kg)(9.80 m/s2)(4.0 m/s)(0.20 s) = –1.8 J
c. The energy dissipated in the resistor is the amount by which the gravitational potential
energy decreases or 1.8 J .
______________________________________________________________________________
11. REASONING According to Equation 22.2, the magnetic flux Φ is the product of the
magnit...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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