Unformatted text preview: rms emf, which
is known: ξ 0 = 2 ξ rms (Equation 20.13). We are also not given the angular speed ω, but we
know that it is related to the frequency f in hertz according to ω = 2π f (Equation 10.6).
SOLUTION According to Equation 20.13, the peak emf is ξ 0 = 2 ξ rms = 2 (120 V ) = 170 V
Substituting ω = 2π f and the value for the peak emf into Equation (1) gives Total length = 4 Nξ 0
Bω =4 Nξ 0 B 2π f =4 100 (170 V ) ( 0.50 T ) 2π ( 60.0 Hz ) = 38 m 1220 ELECTROMAGNETIC INDUCTION 46. REASONING When the motor has just started turning the fan blade, there is no back emf,
and the voltage across the resistance R of the motor is equal to the voltage V0 of the outlet.
Under this condition the resulting current in the motor is I0 = V0 (1) R according to Ohm’s law. When the fan blade is turning at its normal operating speed, the
back emf ξ in the motor reduces the voltage across the resistance R of the motor to V0 − ξ,
so that, the current drawn by the motor is I= V0 − ξ
R (22.5) The current I drawn at the normal operating speed is only 15.0% of the current I0 drawn
when the fan blade just begins to turn, so we have that I = 0.150 I0 (2) SOLUTION Substituting Equation (2) into Equation (22.5) yields 0.150 I 0 = V0 − ξ
R (3) Substituting Equation (1) into Equation (3) and solving for ξ, we find that V V −ξ
0.150 0 = 0
R
R 47. or ξ = V0 − 0.150V0 = 0.850V0 = 0.850 (120.0 V ) = 102 V SSM REASONING The peak emf ξ 0 produced by a generator is related to the number
N of turns in the coil, the area A of the coil, the magnitude B of the magnetic field, and the
angular speed ωcoil of the coil by ξ0 = NABωcoil (see Equation 22.4). We are given that the
angular speed ωcoil of the coil is 38 times as great as the angular speed ωtire of the tire.
Since the tires roll without slipping, the angular speed of a tire is related to the linear speed
v of the bike by ωtire = v / r (see Section 8.6), where r is the radius of a tire. The speed of
the bike after 5.1 s can be found from its acceleration and the fact that the bike starts from
rest. Chapter 22 Problems 1221 SOLUTION The peak emf produced by the generator is ξ0 = NABωcoil (22.4) Since the angular speed of the coil is 38 times as great as the angular speed of the tire,
ωcoil = 38ωtire .
Substituting
this
expression
into
Equation
22.4
gives ξ0 = NABωcoil = NAB ( 38ω tire ) . Since the tire rolls without slipping, the angular speed of
the tire is related to the linear speed v (the speed at which its axle is moving forward) by
ωtire = v / r (Equation 8.12), where r is the radius of the tire. Substituting this result into the expression for ξ 0 yields v
r ξ0 = NAB ( 38ωtire ) = NAB 38 (1) The velocity of the car is given by v = v0 + at (Equation 2.4), where v0 is the initial velocity,
a is the acceleration and t is the time. Substituting this relation into Equation (1), and noting
that v0 = 0 m/s since the bike starts from rest, we find that v
r ( v0 + at ) r ξ0 = NAB 38 = NAB 38 0 m/s + ( 0.550 m/s 2 ) ( 5.10 s ) = (125 ) ( 3.86 × 10−3 m 2 ) ( 0.0900 T )( 38 ) = 15.4 V
0.300 m ______________________________________________________________________________
48. REASONING The generator coil rotates in an external magnetic field B, inducing a
varying emf with a peak value ξ0 and a varying current I with a peak value I0. Once the
current is flowing, the external magnetic field B exerts a torque τ on the coil, according to τ = NIAB sin φ (21.4) In Equation 21.4, N is the number of turns in the generator coil, A is its crosssectional area,
and φ is the angle between the external magnetic field and the normal to the plane of the
coil. Following Lenz’s law, this torque opposes the rotation that induces the emf and the
current, hence it is labeled a countertorque. The induced emf is given by ξ = NABω sin ω t = ξ 0 sin ω t
where ω is the angular frequency of the rotation of the coil. (22.4) 1222 ELECTROMAGNETIC INDUCTION In terms of the peak current I0 and peak voltage V0 across the light bulb, the average power
P that the generator delivers to the light bulb is P = 1 I 0V0 (Equation 20.10). The peak
2 voltage V0 across the bulb is equal to the peak emf ξ0 of the generator, and we have that P = 1 I 0V0 = 1 I 0ξ0
2
2 (20.10) SOLUTION The peak value ξ0 of the induced emf occurs when sin ωt = 1 in Equation
22.4, which is the instant when the plane of the coil is parallel to the external magnetic field.
At this instant, φ = 90° in Equation 21.4. Because the peak value I0 of the current occurs at the same instant as the peak value ξ 0 = NABω of the emf, the maximum value τ0 of the
countertorque is τ 0 = NI 0 AB sin 90o = NI 0 AB (1) Solving Equation (20.10) for I0, we obtain I0 = 2P ξ0 (2) Substituting Equation (2) into Equation (1) yields 2 P 2 PNAB
= ξ0 ξ0 τ 0 = NAB (3) Substituting ξ 0 = NABω from Equation 22.4 into Equation (3), we obtain τ0 = 2 P NAB 2 P
=
ω
NAB ω (4) We are given the frequency f = 60.0 Hz, which is related to the angular frequency by
ω = 2π f (Equation 10.6). Making this substitution into Equation (4) gives the maximum
countertorque: τ0 = 2P ω = 2P
P
75 W
=
=...
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 Spring '13
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 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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