## 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 = NBv/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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publicly accessible website, in whole or in part.

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

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