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2. REASONING During its fall, both the length of the bar
and its velocity v are perpendicular to the horizontal
component Bh of the earth’s magnetic field (see the
drawing for an overhead view). Therefore, the emf ξ
induced across the length L of the rod is given by West
ξ = vBh L (Equation 22.1), where v is the speed of the
rod. We will use Equation 22.1 to determine the
magnitude Bh of the horizontal component of the earth’s
magnetic field, and RightHand Rule No.1 from Section
21.2 to determine which end of the rod is positive. North
Bh East
v F (velocity into page) South
Overhead view SOLUTION
a. Solving ξ = vBh L (Equation 22.1) for Bh, we find that
Bh = ξ
vL = 6.5 × 10 −4 V
= 3.7 × 10−5 T
( 22 m/s )( 0.80 m ) b. Consider a hypothetical positive charge that is free to move inside the falling rod. The bar
is falling downward, carrying the positive charge with it, so that the velocity v of the charge
is downward. In the drawing, which shows the situation as seen from above, downward is
into the page. Applying RightHand Rule No. 1 to the vectors v and Bh, the magnetic force
F on the charge points to the east. Therefore, positive charges in the rod would accelerate to
the east, and negative charges would accelerate to the west. As a result, the east end of the
rod acquires a positive charge. 3. REASONING AND SOLUTION The motional emf ξ generated by a conductor moving
perpendicular to a magnetic field is given by Equation 22.1 as ξ = vBL, where v and L are
the speed and length, respectively, of the conductor, and B is the magnitude of the magnetic
field. The emf would have been ( )( )( ) ξ = vBL = 7.6 × 103 m/s 5.1 × 10−5 T 2.0 × 104 m = 7800 V Chapter 22 Problems 1193 ______________________________________________________________________________
4. REASONING
a. The motional emf generated by the moving metal rod depends only on its speed, its
length, and the magnitude of the magnetic field (see Equation 22.1). The motional emf does
not depend on the resistance in the circuit. Therefore, the emfs for the circuits are the same.
b. According to Equation 20.2, the current I is equal to the emf divided by the resistance R
of the circuit. Since the emfs in the two circuits are the same, the circuit with the smaller
resistance has the larger current. Since circuit 1 has onehalf the resistance of circuit 2, the
current in circuit 1 is twice as large.
c. The power P is P = ξ 2 / R (Equation 20.6c), where ξ is the emf (or voltage) and R is the
resistance. The emf produced by the moving bar is directly proportional to its speed (see
Equation 22.1). Thus, the bar in circuit 1 produces twice the emf, since it’s moving twice as
fast. Moreover, the resistance in circuit 1 is half that in circuit 2. As a result, the power
delivered to the bulb in circuit 1 is 22 / ( 1 ) = 8 times greater than in circuit 2.
2 SOLUTION
a. The ratio of the emfs is, according to Equation 22.1 ξ1 vBL
=
=1
ξ 2 vBL
b. Equation 20.2 states that the current is equal to the emf divided by the resistance. The
ratio of the currents is
I1 ξ1 /R1 R2 110 Ω
=
=
=
=2
I 2 ξ 2 /R2 R1 55 Ω
c. The power, according to Equation 20.6c, is P = ξ 2/R. The motional emf is given by
Equation 22.1 as ξ = vBL . The ratio of the powers is ξ12 P
1 = R1 = ξ1
2 ξ P2 ξ 2 2 R2
2 2 R2 v1BL = R1 v2 BL 2 2 R2 R1 v R 2v R = 1 2 = 2 1 2 = 8
v R v R 2 1 2 2 2
______________________________________________________________________________ 1194 ELECTROMAGNETIC INDUCTION 5. SSM WWW REASONING AND SOLUTION For the three rods in the drawing in the
text, we have the following:
Rod A: The motional emf is zero , because the velocity of the rod is parallel to the
direction of the magnetic field, and the charges do not experience a magnetic force.
Rod B: The motional emf ξ is, according to Equation 22.1, ξ = vBL = (2.7 m/s)(0.45 T)(1.3 m) = 1.6 V
The positive end of Rod B is end 2 .
Rod C: The motional emf is zero , because the magnetic force F on each charge is directed perpendicular to the length of the rod. For the ends of the rod to become charged,
the magnetic force must be directed parallel to the length of the rod.
______________________________________________________________________________
6. REASONING The situation in the drawing given with the problem statement is analogous to
that in Figure 22.4b in the text. The blood flowing at a speed v corresponds to the moving
rod, and the diameter of the blood vessel corresponds to the length L of the rod in the figure.
The magnitude of the magnetic field is B, and the measured voltage is the emf ξ induced by
the motion. Thus, we can apply ξ = BvL (Equation 22.1).
SOLUTION Using Equation 22.1, we find that ( ) ξ = BvL = ( 0.60 T )( 0.30 m/s ) 5.6 ×10−3 m = 1.0 ×10−3 V 7. REASONING The average power P delivered by the hand is given by P = W / t , where W
is the work done by the hand and t is the time interval during which the work is done. The
work done by the hand is equal to the product of the magnitude Fha...
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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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