college physics 11th edition by raymond sm

college physics 11th edition by raymond sm

Topic 20
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1191
Topic 20
Induced Voltages and Inductance
QUICK QUIZZES
20.1 b, c, a. At each instant, the magnitude of the induced emf is proportional
to the magnitude of the rate of change of the magnetic field (hence,
proportional to the absolute value of the slope of the curve shown on the
graph).
20.2 Choice (c). Taking downward as the normal direction, the north pole
approaching from above produces an increasing positive flux through the
area enclosed by the loop. To oppose this, the induced current must
generate a negative flux through the interior of the loop, or the induced
magnetic field must point upward through the enclosed area of the loop.
Imagine gripping the wire of the loop with your right hand so the fingers
curl upward through the area enclosed by the loop. You will find that
your thumb, indicating the direction of the induced current, is directed
counterclockwise around the loop as viewed from above.
20.3 Choice (b). If the positive z-direction as chosen as the normal direction,
the increasing counterclockwise current in the left-hand loop produces an
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1192
increasing positive flux through the area enclosed by this loop. The
magnetic field lines due to this current curl around and pass through the
area of the xy-plane outside the left-hand loop in the negative direction.
Thus, the right-hand loop has an increasing negative flux through it. To
counteract this effect, the induced current must produce positive flux, or
generate a magnetic field in the positive z-direction, through the area
enclosed by the right-hand loop. Imagine gripping the wire of the righthand loop with the right hand so the fingers point in the positive zdirection as they penetrate the area enclosed by this loop. You should
find that the thumb is directed counterclockwise around the loop as
viewed from above the xy-plane.
20.4 Choice (a). All charged particles within the metal bar move straight
downward with the bar. According to right-hand rule number 1, positive
changes moving downward through a magnetic field that is directed
northward will experience magnetic forces toward the east. This means
that the free electrons (negative charges) within the metal will experience
westward forces and will drift toward the west end of the bar, leaving the
east end with a net positive charge.
20.5 Choice (b). According to Equation 20.3, when B and v are constant, the
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1193
emf depends only on the length of the wire cutting across the magnetic
field lines. Thus, you want the long dimension of the rectangular loop
perpendicular to the velocity vector. This means that the short dimension
is parallel to the velocity vector, and (b) is the correct choice. From a more
conceptual point of view, you want the rate of change of area in the
magnetic field to be the largest, which you do by thrusting the long
dimension into the field.
20.6 Choice (b). When the iron rod is inserted into the solenoid, the inductance
of the coil increases. As a result, more potential difference appears across
the coil than before. Consequently, less potential difference appears
across the bulb, and its brightness decreases.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
20.2 Consider the copper tube to be a large set of rings stacked one on top of
the other. As the magnet falls toward or falls away from each ring, a
current is induced in the ring. Thus, there is a current in the copper tube
around its circumference.
20.4 (a) The flux is calculated as ΦB = BA cos θ = B⊥A. The flux is therefore
maximum when the magnetic field vector is perpendicular to the
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1194
plane of the loop.
(b) The flux is zero when the magnetic field is parallel to the plane of the
loop (and hence, θ = 90°, making B⊥ = B cos θ = 0).
20.6 As water falls, it gains velocity and kinetic energy. It then pushes against
the blades of a turbine, transferring this energy to the rotor or coil of a
large generator. The rotor moves in a strong external magnetic field and a
voltage is induced in the coil. This induced emf is the voltage source for
the current in our electric power lines.
20.8 No. Once the bar is in motion and the charges are separated, no external
force is necessary to maintain the motion. During the initial acceleration
of the bar, an external applied force will be necessary to overcome both
the inertia of the bar and a retarding magnetic force exerted on the bar.
20.10 The current at time t is I = ε
R
1− e
−t/τ
( ) where τ = L/R.
(a) Doubling both ε and R has the net effect of reducing the time
constant by a factor of 2 (because ε /R = 2ε /2R) so that the current
will be greater than Iref.
(b) Doubling the inductance L doubles the time constant so that the
current will be less than Iref.
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1195
(c) Doubling L, R, andε has no net effect on the circuit (ε /R = 2ε /2R
and τ = L/R = 2L/2R ) and the current will therefore equal Iref.
20.12 The increasing counterclockwise current in the solenoid coil produces an
upward magnetic field that increases rapidly. The increasing upward flux
of this field through the ring induces an emf to produce a clockwise
current in the ring. At each point on the ring, the field of the solenoid has
a radially outward component as well as an upward component. This
radial field component exerts an upward force on the current at each
point in the ring. The resultant magnetic force on the ring is upward and
exceeds the weight of the ring. Thus, the ring accelerates upward off of
the solenoid.
20.14 As the magnet moves at high speed past the fixed coil, the magnetic flux
through the coil changes very rapidly, increasing as the magnet
approaches the coil and decreasing as the magnet moves away. The rapid
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1196
change in flux through the coil induces a large emf, large enough to cause
a spark across the gap in the spark plug.
ANSWERS TO EVEN NUMBERED PROBLEMS
20.2 (a) 1.00 × 10−7 T ⋅ m2 (b) 8.66 × 10–8 T ⋅ m2
(c) 0
20.4 zero
20.6 (a) 1.69 m (b) 10.8 Wb
20.8 0.10 mV
20.10 34 mV
20.12 4.74 V
20.14 into the page
20.16 (a) left to right (b) no induced current
(c) right to left
20.18 (a) 1.88 × 10–7 T ⋅ m2 (b) 6.27 × 10–8 V
20.20 8.8 A
20.22 (a) |ε | = NB0πr
2
/t (b) clockwise
(c) I = B0πr
2
/tR
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1197
20.24 1.20 mV, west end is positive
20.26 2.36 × 103 V
20.28 (a) 6.0 µT
(b) The magnitude and direction of the Earth’s field varies from one
location to another, so the induced voltage in the wire will change.
20.30 (a) 2.00 m/s (b) 6.00 W (c) 3.00 N
(d) 6.00 W
20.32 13 mV
20.34 (a) 7.5 kV
(b) when the plane of the coil is parallel to the magnetic field
20.36 (a) 8.0 A (b) 3.2 A (c) 60 V
20.38 (a) 1.32 × 10−8 H (b) 6.34 × 10−6 V
20.40 See Solution.
20.42 1.92 × 10–5 T ⋅ m2
20.44 (a) 1.00 kΩ (b) 3.00 ms
20.46 (a) 0 (b) 3.8 V
(c) 6.0 V (d) 2.2 V
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1198
20.48 (a) 2.00 ms (b) 0.176 A
(c) 1.50 A (d) 3.22 ms
20.50 (a) 4.44 mH (b) 0.555 mJ
20.52 (a) 1.3 Ω (b) 4.8 × 102 turns (c) 0.48 m
(d) 0.76 mH (e) 0.46 ms (f) 3.6 A
(g) 3.2 ms (h) 4.9 mJ
20.54 (a) increasing (b) 62.2 mT/s
20.56 (a) 0.73 m/s, counterclockwise
(b) 0.65 mW
(c) Work is being done on the bar by an external force to maintain
constant speed.
20.58 (a) 2.1 × 106 m/s (b) from side to side
(c) 1.7 × 1010 V
(d) The very large induced emf would lead to powerful spontaneous
electrical discharges. The strong electric and magnetic fields would
disrupt the flow of ions in their bodies.
20.60 0.158 mV
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1199
20.62 0.120 A, clockwise
20.64 1.60 A
20.66 (a) ε = NBℓv (b) I = NBv/R
(c) P = N2
B2
ℓ2
v2 R (d) F = N2
B2
2
v R
(e) clockwise (f) toward the left
PROBLEM SOLUTIONS
20.1 The angle between the direction of the constant field and the normal to
the plane of the loop is θ = 0°, so
ΦB = BA cos θ = (0.50 T)[(8.0 × 10–2 m)(12 × 10–2 m)]cos 0° = 4.8 × 10–3 T ⋅ m2
20.2 The magnetic flux through the loop is given by ΦB = BA cos θ, where B is
the magnitude of the magnetic field, A is the area enclosed by the loop,
and θ is the angle the magnetic field makes with the normal to the plane
of the loop. Thus,
ΦB = BAcos θ = 5.00 × 10−5 ( T) 20.0 cm2 10−2 m
1 cm
⎛
⎝
⎜
⎞
⎠
⎟
⎡
⎣
⎢ ⎤
⎦
⎥cos θ = 1.00 × 10−7 T⋅m2 ( )cos θ
(a) When 
B is perpendicular to the plane of the loop, θ = 0° and ΦB =
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1200
1.00 × 10–7 T ⋅ m2
(b) If θ = 30.0°, then ΦB = (1.00 × 10–7 T ⋅ m2
) cos 30.0° = 8.66 × 10–8 T ⋅ m2
.
(c) If θ = 90.0°, then ΦB = (1.00 × 10–7 T⋅m2
)cos 90.0° = 0.
20.3 The magnetic flux is ΦB = BAcosθ where θ is the angle between the
magnetic field and the normal to the plane of the loop. The loop area is A
= ℓ2 = (0.250 m)2 = 0.0625 m2
.
(a) Here θ = 0° so that ΦB = BAcos0° = (2.00 T) 0.062 5 m2 ( ) = 0.125 Wb .
(b) When the plane of the loop makes an angle of 60.0° with the magnetic
field, the loop’s normal makes an angle θ = 30.0° so that
ΦB = BAcos30° = (2.00 T) 0.0625 m2 ( )cos(30.0°) = 0.108 Wb .
(c) Here the loop’s normal is perpendicular to the magnetic field so that
θ = 90° and ΦB = 0 (because cos(90°) = 0).
20.4 The magnetic field lines are everywhere parallel to the surface of the
cylinder, so no magnetic field lines penetrate the cylindrical surface. The
total flux through the cylinder is zero.
20.5 (a) Every field line that comes up through the area A on one side of the
wire goes back down through area A on the other side of the wire.
Thus, the net flux through the coil is zero.
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1201
(b) The magnetic field is parallel to the plane of the coil, so θ = 90.0°.
Therefore, ΦB = BA cos θ = BA cos 90.0° = 0.
20.6 (a) The angle between the magnetic field and the loop normal is θ = 0° so
that, with a loop area of A = π R2
,
ΦB = BA = B πR2 ( ) → R = 2.70 Wb
π (0.300 T) = 1.69 m
(b) When the loop radius is doubled, the area increases by a factor of 4.
Because the flux is proportional to the area, the flux also increases by
a factor of 4 so that ΦB,new = 4ΦB = 4(2.70 Wb) = 10.8 Wb .
20.7 (a) The magnetic flux through an area A may be written as
ΦB = (B cos θ)A
= (component of B perpendicular to A).A
Thus, the flux through the shaded side of the cube is
ΦB = Bx ⋅ A = (5.0 T) ⋅ (2.5 × 10–2 m)2
= 3.1 × 10–3 T ⋅ m2
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1202
(b) Unlike electric field lines, magnetic field lines always form closed
loops, without beginning or end. Therefore, no magnetic field lines
originate or terminate within the cube and any line entering the cube
at one point must emerge from the cube at some other point. The net
flux through the cube, and indeed through any closed surface, is zero.
20.8
ε = ΔΦ
B
Δt = (ΔB) Acos θ
Δt =
(1.5 T − 0) π 1.6 × 10−3
( m)
2 ⎡
⎣
⎢ ⎤
⎦
⎥cos 0°
120 × 10−3 s
= 1.0 × 10−4 V = 0.10 mV
20.9 (a) As loop A moves parallel to the long straight wire, the magnetic flux
through loop A does not change. Hence, there is no induced current
in this loop.
(b) As loop B moves to the left away from the straight wire, the magnetic
flux through this loop is directed out of the page, and is decreasing in
magnitude. To oppose this change in flux, the induced current flows
counterclockwise around loop B producing a magnetic flux directed
out of the page through the area enclosed by loop B.
(c) As loop C moves to the right away from the straight wire, the
magnetic flux through this loop is directed into the page and is
decreasing in magnitude. In order to oppose this change in flux, the