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Unformatted text preview: m =
= = = ( R0 A / ρ )carbon ( Rcarbon )0 ρ tungsten 9 5.6 × 10 –8 Ω ⋅ m 70 where we have used resistivity values from Table 20.1 and the fact that the two sections
have the same crosssectional areas.
______________________________________________________________________________
122. REASONING AND SOLUTION The voltage VCu between the ends of the copper rod is
given by Ohm’s law as VCu = IRCu, where RCu is the resistance of the copper rod. The
current I in the circuit is equal to the voltage V of the battery that is connected across the
free ends of the copperiron rod divided by the equivalent resistance of the rod. The copper
and iron rods are joined endtoend, so the same current passes through each. Thus, they are
connected in series, so the equivalent resistance RS is RS = RCu + RFe. Thus, the current is I= V
V
=
RS RCu + RFe The voltage across the copper rod is VCu = IRCu = V
R
RCu + RFe Cu The resistance of the copper and iron rods is given by RCu = ρCuL/A and RFe = ρFeL/A,
where the length L and crosssectional area A are the same for both rods and ρCu and ρFe 1132 ELECTRIC CIRCUITS denote the resistivites. Substituting these expressions for the resistances into the equation
above and using resistivities from Table 20.1 yield V
VCu = ρ ρ + ρ Cu Fe Cu 12 V
−8
VCu = (1.72 × 10 Ω ⋅ m) = 1.8 V
−8
−8 1.72 × 10 Ω ⋅ m + 9.7 × 10 Ω ⋅ m ______________________________________________________________________________
123. SSM When two or more capacitors are in series, the equivalent
1
1
1
1
capacitance of the combination can be obtained from Equation 20.19,
=
+
+
....
Cs C1 C2 C3
Equation 20.18 gives the equivalent capacitance for two or more capacitors in parallel:
1
Cp = C1 + C2 + C3 + ... . The energy stored in a capacitor is given by 2 CV 2 , according to
REASONING Equation 19.11. Thus, the energy stored in the series combination is
1
1
1
–1
=
+
= 0.476 ( µ F )
Cs 7.0 µ F 3.0 µ F or Cs = 1
2 CsVs2 , where 1
0.476 ( µ F ) –1 = 2.10 µ F 2
Similarly, the energy stored in the parallel combination is 2 Cp Vp where
1 Cp = 7.0 µ F + 3.0 µ F = 10.0 µ F The voltage required to charge the parallel combination of the two capacitors to the same
total energy as the series combination can be found by equating the two energy expressions
and solving for Vp .
SOLUTION Equating the two expressions for the energy, we have
1
2 2
CsVs2 = 2 CpVp
1 Solving for Vp , we obtain the result
Vp = Vs Cs
Cp = ( 24 V ) 2.10 µ F
= 11V
10.0 µ F ______________________________________________________________________________ Chapter 20 Problems 1133 124. REASONING AND SOLUTION The resistance of the thermistor decreases by 15%
relative to its normal value of 37.0 °C. That is, ∆R R − R0
=
= −0.15
R0
R0
According to Equation 20.5, we have
R = R0[1 + α (T – T0)] or (R – R0) = α R0(T – T0) or R − R0
R0 = α (T − T0 ) = −0.15 Rearranging this result gives
T = T0 + −0.15 −0.15 = 39.5 °C
−1
−0.060 ( C° )
______________________________________________________________________________ α = 37.0 °C + CHAPTER 21 MAGNETIC FORCES AND MAGNETIC FIELDS
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
____________________________________________________________________________________________ 1. (d) RightHand Rule No. 1 gives the direction of the magnetic force as −x for both drawings
A and B. In drawing C, the velocity is parallel to the magnetic field, so the magnetic force is
zero. 2. (b) Using RightHand Rule No. 1 (see Section 21.2), we find that the direction of the
magnetic force on a positively charged particle is to the west. Reversing this direction
because the particle is a negative electron, we see that the magnetic force acting on it points
to the east. 3. (a) Using RightHand Rule No. 1 (see Section 21.2), we find that the direction of the
magnetic force on a positively charged particle is straight down toward the bottom of the
screen. 4. B = 1.1 × 10−1 T, south 5. (c) The electric force points out of the screen, in the direction of the electric field. An
application of RightHand Rule No. 1 shows that the magnetic force also points out of the
screen, parallel to the electric force. When two forces have the same direction, the
magnitude of their sum has the largest possible value. 6. (e) In this situation, the centripetal force, Fc = mv /r (Equation 5.3), is provided by the 2 2 magnetic force, F = qvB sin 90.0° (Equation 21.1), so mv /r = qvB sin 90.0°. Thus,
q = mv / ( rB ) , and the charge magnitude q is inversely proportional to the radius r. Since the
radius of curve 1 is smaller than that of curve 2, and the radius of curve 2 is smaller than
that of curve 3, we conclude that q1 is larger than q2, which is larger than q3.
7. 8. (a) The magnetic force that acts on the electron in regions 1 and 2 is always perpendicular
to its path, so the force does no work. According to the workenergy theorem, Equation 6.3,
the kinetic energy, and hence speed, of the electron does not change when no work is done.
(d) According to Equation 21.2, the radius r...
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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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