m33 knig9404 ism c33 - media.physicsisbeautiful.com

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Conceptual Questions

33.1. The induced current will be counterclockwise (ccw). As the bar moves upward through the constant-magnetic-field region, the area of the loop decreases, so the flux through the loop decreases. By Lenz’s Law, any induced current will tend to oppose this decrease. A ccw current will create a stronger magnetic field inside the loop, tending to increase the flux.

33.2. You need to push against a repulsive force. In its initial position, the loop encircles essentially no magnetic flux, whereas once the loop is positioned between the poles of the magnet, magnetic flux will pass through the loop. Lenz’s Law tells us that the current flows in the direction required to oppose a change in flux, so the current will flow in the direction required to create a repulsive force on the copper wire. The current induced in the loop is therefore in the direction shown (into the page on the left side of the copper loop). The magnetic poles of the induced current loop are also shown in the diagram. The resulting magnetic force on the loop is to the right. Note that if the magnetic poles were reversed, the induced current would reverse as well so that the resulting force would still be to the right.

33.3. Yes, there is a net upward force. Lenz’s Law tells us that a current will be induced in the loop to oppose the change in magnetic flux. The net force will be upward because the upper loop segment traverses the field B whereas the lower loop segment does not (the two side segments produce forces of equal magnitude but in opposite directions). We know the force must be upward because, by Lenz’s Law, it must oppose the force due to gravity that is causing the change in flux.

ELECTROMAGNETIC INDUCTION

33-2 Chapter 33

33.4. (a) Clockwise. The induced magnetic field is into the page and by the right-hand rule the induced current is clockwise. (b) No current. A changing magnetic flux is necessary to induce a current. (c) Counterclockwise. The induced magnetic field is out of the page.

33.5. (a) No current. No magnetic flux is changing, so no magnetic field is induced. (b) Clockwise. An increasing counterclockwise (from above) current in the lower loop creates an increasing upward-pointing magnetic field in the upper loop. Thus, a downward magnetic field is induced, resulting from a clockwise current in the upper loop. (c) No current. The current through the lower loop is constant, so there is no change in the flux through the upper loop. (d) Counterclockwise. The decreasing current in the lower loop causes the flux through the upper loop to decrease, inducing an upward magnetic field.

33.6. (a) Clockwise from above. The increasing flux through the loop is due to the upward-pointing field of the north pole of the magnet, so the induced magnetic field is downward. (b) Yes, upward. The induced downward field is like a magnet with its north pole on the bottom, which is repulsed by the bar magnet. (c) Yes, downward. By Newton’s third law, the bar magnet is pushed down.

33.7. There is no current induced in the loop. As the bar magnet is pushed upward, the net magnetic flux through the loop is zero. In other words, all flux lines that enter the loop will, by symmetry, also exit the loop. Note that the magnet must be under the center of the loop for this symmetry argument to hold.

33.8. (a) Left to right. The bar magnet’s field inside the coil points from right to left. The flux through the coil is increasing. inducedB will point left to right, and is caused by a current flowing left to right through the meter. (b) No current. The flux through the coil is not changing. (c) Right to left. Flux through the loop decreases as the magnet is withdrawn. Since originalB is right to left, inducedB is left to right. It is caused by a current flowing right to left through the meter by the right-hand rule.

33.9. B is increasing. For a clockwise induced electric field, the induced magnetic field is into the page (right-hand rule). Since the induced field is in the direction that opposes the change in flux, B must be increasing.

33.10. The energy stored in an inductor is 212 .U LI= To increase the energy to 2U, the current must increase by a

factor of 2. Thus the required current is 2(2.0 A) 2.8 A.=

33.11. a. No. Since /V LdI dtΔ = − and VΔ and L are both known, we can only find / ,dI dt not I. b. Yes, through the right-hand inductor, since / / .dI dt V L= Δ

c. Yes, it is possible to tell: The current is decreasing. If / 0dI dt < then L 0VΔ > so the input side is more negative and the potential from the induced current increases in the direction of the original current (see discussion accompanying Figures 33.40 and 33.41).

33.12. We have 1(2 ) 2000 Hz.f LCπ −= = Denoting the new circuit by primed quantities, with 4 ,L L=′

1 1 1 1 1 1000 Hz2 22 2 (4 ) 2

f fL C L C LCπ π π

⎛ ⎞′ = = = = =⎜ ⎟′ ⎝ ⎠

33.13 a / .L Rτ = For circuit b, note that the two inductors are in series. Recall that the inductance of an inductor is given by its turn density N/l. Because two identical inductor in series will have the same turn density but double the number of turns, the total inductance will be doubled. Another way to look at this is by considering the expression for inductance, which is proportional to 2 / .N l For two inductors in series we double N and l, so the total inductance increases by a factor of two. Thus,

Electromagnetic Induction 33-3

totalb a

2 2L LR R

τ τ= = =

The time constant for circuit c is

c atotal

L L LR R R

τ τ= = = =

Thus, the order from largest to smallest is c b aτ τ τ= >

33.14. (a) Zero. The potential across the resistor is zero, and there is a zero initial current because the current in the inductor cannot change instantaneously. (b) After the switch has been closed for a long time, the current in the inductor reaches the steady state ( / 0)dI dt = so there is no voltage drop across the inductor. Kirchhoff’s loop law then gives

10 V0 2 A5

− = ⇒ = = =Ω

Exercises and Problems

Section 33.2 Motional emf

33.1. Visualize:

To develop a motional emf the magnetic field needs to be perpendicular to both the velocity and the current, so let’s say its direction is into the page. Solve: This is a straightforward use of Equation 33.3. We have

1.0 V 2.0 10 m/s(1.0 m)(5.0 10 T)

vlB −= = = ×

Assess: This is about 45,000 mph, which is an unreasonable speed for a car. It’s unlikely you’ll ever develop a volt.

33.2. Solve: This is a straightforward use of Equation 33.3. We have 0.050 V 0.10 T

(0.10 m)(5.0 m/s)B

lv= = =E

33.3. Visualize:

33-4 Chapter 33

The wire is pulled at a constant speed in a magnetic field. This results in a motional emf and produces a current in the circuit. From energy conservation, the mechanical power provided by the puller must appear as electrical power in the circuit. Solve: (a) Using Equation 33.6,

pull pull4.0 W 1.0 N4.0 m/s

PP F v Fv

= ⇒ = = =

(b) Using Equation 33.6 again, 2 2 2

pull2 2

(0.20 )(1.0 N) 2.2 T(4.0 m/s)(0.10 m)

RFv l BP BR vl

Ω= ⇒ = = =

Assess: This is reasonable field for the circumstances given.

Section 33.3 Magnetic Flux

33.4. Model: Assume the field changes abruptly at the boundary between the two sections. Visualize: Please refer to Figure EX33.4. The directions of the fields are opposite, so some flux is positive and some negative. The total flux is the sum of the flux in the two regions. Solve: The field is constant in each region so we will use Equation 33.10. Take A to be into the page, which makes it parallel to the field in the left region so the flux there is positive, and antiparallel to the field in the right region, so the flux there is negative. The total flux is

L L L R R R2 2

cos cos

(0.20 m) (2.0 T) (0.20 m) (1.0 T) 0.040 Wb

A B A B A Bθ θΦ = ⋅ = +

= − =

Assess: The flux is positive because the areas are equal and the stronger field is parallel to the normal of the surface (as we have defined it).

33.5. Model: Consider the solenoid to be long so the field is constant inside and zero outside. Visualize: Please refer to Figure EX33.5. The field of a solenoid is along the axis. The flux through the loop is only nonzero inside the solenoid. Since the loop completely surrounds the solenoid, the total flux through the loop will be the same in both the perpendicular and tilted cases. Solve: The field is constant inside the solenoid so we will use Equation 33.10. Take A to be in the same direction as the field. The magnetic flux is

2 2 5loop loop sol sol sol sol cos (0.010 m) (0.20 T) 6.3 10 WbA B A B r Bπ θ π −Φ = ⋅ = ⋅ = = = ×

When the loop is tilted the component of B in the direction of A is less, but the effective area of the loop surface through which the magnetic field lines cross is increased by the same factor, so the flux through the loop remains unchanged, as can be expected on physical grounds (i.e., we still have the same numbe of magnetic field lines going through the loop).

33.6. Model: Assume the field is uniform inside the rectangular region and zero outside. Visualize: The flux measures how much of the field penetrates the chosen surface. We will break the surface up because the magnetic field has different strengths over different parts of the surface of the loop. Solve: For convenience we choose the normal to the loop to be into the page so it is in the same direction as the magnetic field. The total flux is

insiderectangleloop inside outside inside

rectangle rectangle rectangle

cosB dA B dA B dA BdA BA BabθΦ = ⋅ = ⋅ + ⋅ = = =∫ ∫ ∫ ∫

Assess: Only the region with nonzero magnetic field inside the loop contributes to the flux. Thus, the result is independent of the size of the loop, provided it encompasses the entire rectangular area.

33.7. Model: Assume the field strength is uniform over the loop. Visualize: Please refer to Figure EX33.7. According to Lenz’s Law, the induced current creates an induced field that opposes the change in flux.

Solve: The original field is into the page within the loop and is changing strength. The induced, clockwise current produces a field into the page within the loop that is opposing the change. This implies that the original field must be decreasing in strength so the magnetic field strength in the loop must be decreasing.

33.8. Model: Assume that the magnet is a bar magnet with field lines pointing away from the north end. Visualize: As the magnets move, if they create a change in the flux through the solenoid, there will be an induced current and corresponding field. According to Lenz’s Law, the induced current creates an induced field that opposes the change in flux. Solve: (a) When magnet 1 is close to the solenoid there is flux to the left through the solenoid. As magnet 1 moves away there is less flux to the left. The induced current will oppose this change and produce an induced current and a corresponding flux to the left. By the right-hand rule, this corresponds to a current in the wire into the page at the top of the solenoid and out of the page at the bottom of the solenoid. So, the current will be right to left in the resistor. (b) When magnet 2 is close to the solenoid, the diverging field lines of the bar magnet produce a flux to the left in the left half of the solenoid and a flux to the right in the right half. Since the flux depends on the orientation of the loop, the flux on the two halves have opposite signs and the net flux is zero. Moving the magnet away changes the strength of the field and flux, but the total flux is still zero. Thus there is no induced current.

33.9. Visualize: Please refer to Figure EX33.9. The changing current in the solenoid produces a changing flux in the loop. By Lenz’s Law there will be an induced current and field to oppose the change in flux. Solve: The current shown produces a field to the left inside the solenoid. So there is flux to the left through the surrounding loop. As the current in the solenoid increases there is more field and more flux to the left through the loop. There will be an induced current in the loop that will oppose the change by creating an induced field and flux to the right. By the right-hand rule, this requires a clockwise current.

33.10. Model: Assume the plane of the loop is perpendicular to the field direction. Visualize: Please refer to Figure EX33.10. The flux is due to the field through the area of the triangle. Only the left half gives a contribution, as the field strength is zero on the right half. Solve: (a) The flux is .A BΦ = ⋅ Take A to be into the page, perpendicular to the triangle, and thus parallel to .B In this case ABΦ = where A is the area of half of the triangle. This smaller triangle has a base of 10 cm and height 20sin 60 cm 17.32 cm.° = Thus,

41 (0.10 m)(0.1732 m)(0.10 T) 8.7 10 Wb2

AB −Φ = = = ×

(b) The flux is directed into the loop. According to Lenz’s Law, the induced current will try to prevent the decrease of flux. To do this, the field of the induced current will have to point into this loop. This requires the induced current to flow clockwise.

Section 33.5 Faraday’s Law

33.11. Model: Assume the field is uniform. Visualize: Please refer to Figure EX33.11. If the changing field produces a changing flux in the loop there will be a corresponding induced emf and current. Solve: (a) The induced emf is /d dt= ΦE and the induced current is / .I R= E The field B is changing, but the area A

is not. Take A to be out of the page and parallel to ,B so .ABΦ = Thus,

2 2 3(0.050 m) (0.50 T/s) 3.9 10 VdB dBA rdt dt

π π −= = = = ×E

323.93 10 V 2.0 10 A 20 mA

−−×= = = × =

The field is increasing into the page. To prevent the increase, the induced field needs to point out of the page. Thus, the induced current must flow counterclockwise. (b) As in part (a), ( / ) 3.9 mVA dB dt= =E and 20 mA.I = Here the field is out of the page and decreasing. To prevent the decrease, the induced field needs to point out of the page. Thus the induced current must flow counterclockwise.

33-6 Chapter 33

(c) Now A (left or right) is perpendicular to B and so 0 Wb.A B⋅ = That is, the field does not penetrate the plane of the loop. If 0 Wb, then / 0 V/m and 0 A.d dt IΦ = = Φ = =E There is no induced current. Assess: Note that the induced field opposes the change.

33.12. Model: Assume the field is uniform. Visualize: Please refer to Figure EX33.12. The motion of the loop changes the flux through it. This results in an induced emf and current. Solve: The induced emf is /d dt= ΦE and the induced current is / .I R= E The area A is changing, but the field B is

not. Take A as being out of the page and parallel to ,B so and / ( / ).AB d dt B dA dtΦ = Φ = The flux is through that portion of the loop where there is a field, that is, .A lx= The emf and current are

( ) (0.20 T)(0.050 m)(50 m/s) 0.50 VdA d lxB B Blvdt dt

= = = = =E

0.50 V 5.0 A0.10

= = =Ω

The field is out of the page. As the loop moves, the flux increases because more of the loop area has field through it. To prevent the increase, the induced field needs to point into the page. Thus, the induced current flows clockwise. Assess: 5.0 A is a large current, but seems reasonable because the wire resistance is low and the loop is moving rapidly.

33.13. Model: Assume the field strength is changing at a constant rate. Visualize:

The changing field produces a changing flux in the coil, and there will be a corresponding induced emf and current. Solve: The induced emf of the coil is

2 3 23

( ) 0.20 T(10 ) (0.0050 m) 1.6 V10 10 s

d d A B dB BN N NA N rdt dt dt t

π π −⎛ ⎞Φ ⋅ Δ= = = = = =⎜ ⎟⎜ ⎟Δ ×⎝ ⎠

where we’ve used the fact that B is parallel to .A Assess: This seems to be a reasonable emf as there are many turns. Notice that the sign of the change in B does not enter the calculation.

33.14. Model: Assume the field is uniform across the loop. Visualize: Please refer to Figure EX33.14. There is a current in the loop so there must be an emf that is due to a changing flux. With the loop fixed the area is constant so the change in flux must be due to a changing field strength. Solve: The induced emf is /d dt= ΦE and the induced current is / .I R= E The B field is changing, but the area A is

not. Take A as being into the page and parallel to ,B so and / ( / ).AB d dt A dB dtΦ = Φ = This gives

2(150 10 A)(0.20 ) 4.7 T/s

(0.080 m)d dB dB IRAdt dt dt A

−Φ × Ω= = ⇒ = = =E

The original field and flux is into the page. The induced counterclockwise current produces an induced field and flux that is out of the page. Since the induced field opposes the change, the field must be increasing.

Section 33.6 Induced Fields

33.15. Model: A changing magnetic field creates an electric field. Visualize: Please refer to Figure EX33.15 in your textbook.

Solve: The magnetic field produced by a current I in a solenoid is

0solenoid

The magnitude of the induced electric field inside a solenoid is given by Equation 33.26. The direction can be found using Lenz’s Law, but the problem only asks for field strength, which is positive. Only the current is changing, thus

750 (4 10 T m/A)(0.010 m)(400) V s1.26 10

2 2 2(0.20 m) A mr dB r N dI dI dIE

dt L dt dt dtμ π −

−× ⎛ ⎞= = = = ×⎜ ⎟⎝ ⎠

In the interval 0 s to 0.1s,t t= =

5 A 50 A/s0.1s

In the interval 0.1 s to 0.2 s,t = the change in the current is zero. In the interval 0.2 s to 0.4 s, / 25 A/s.t dI dt= =

Thus in the interval 40 s to 0.1s, 6.3 10 V/m.t t E −= = = × In the interval 0.1s to 0.2 s, t E= is zero. In the interval 40.2 s to 0.4 s, 3.1 10 V/m.t E −= = × The E-versus-t graph is shown above, in the Visualize step.

33.16. Model: A changing magnetic field creates an electric field. Visualize:

33-8 Chapter 33

Solve: (a) Apply Equation 33.26. For a point on the axis, 0 m, so 0 V/m.r E= = (b) For a point 2.0 cm from the axis,

0.020 m (4.0 T/s) 0.040 V/m 40 mV/m2 2r dBE

dt⎛ ⎞= = = =⎜ ⎟⎝ ⎠

Assess: Notice that the solenoid diameter and the initial magnetic field strength do not enter into the calculation.

33.17. Model: Assume the magnetic field inside the circular region is uniform. The region looks exactly like a cross section of a solenoid of radius 2.0 cm. The proton accelerates due to the force of the induced electric field. Visualize: Please refer to Figure EX33.17. Solve: Equation 33.26 gives the strength of the induced electric field inside a solenoid with changing magnetic field. The magnitude at each point is thus 2 ,dBr

dtE = where r indicates the radial distance between each point and the

center of the circular region. The direction of E can be determined with Lenz’s Law. As B decreases, a clockwise electric field is induced that would create a current that opposes the decrease if a conducting loop were present. Using Newton’s second law,

2eE er dBF eE ma am m dt

= = ⇒ = =

The sign indicates that the acceleration is in the same direction as the induced electric field. At point a the acceleration is

19 24 2

27(1.6 10 C)(1.0 10 m) (0.10 T/s) (4.8 10 m/s , up)

2(1.6 10 kg)a a

− −

−− × ×= ⇒ = ×

At point b, 20 m so 0 m/s .r a= = At point c, 4 2(4.8 10 m/s , down).a = × At point d, 4 2(9.6 10 m/s , down).a = ×

Section 33.8 Inductors

33.18. Model: Assume that the current drops uniformly. Visualize:

Assume that the current is flowing left to right as in the figure. The changing current produces a changing flux in the inductor and a corresponding emf is developed across the inductor. Solve: For the inductor the induced potential difference is

6(50 A 150 A) 10(10 10 H) 100 V

10.0 10 sdI IV L Ldt t

−−

−Δ − × Δ = − = − = − × =Δ ×

As the current decreases, so does the magnetic flux in the inductor. The induced current will oppose this change and try to maintain the original flux. This means that the induced current will be in the same direction as the original current. This induced current carries positive charge carriers to the right, creating a higher potential at the right end of the inductor than at the left end. This induced potenial therefore increases in the direction of the original (i.e., non-induced) current in the inductor.

33.19. Model: Assume that the current changes uniformly. Visualize: We want to increase the current without exceeding the maximum potential difference. Solve: Since we want the minimum time, we will use the maximum potential difference:

3.0 A 1.0 A(200 10 H) 1.0 ms400 V

dI I IV L L t Ldt t V

−Δ Δ −Δ = − = − ⇒ Δ = = × =Δ Δ

Assess: If we change the current in over a shorter time the potential difference will exceed the limit.

33.20. Visualize: The battery will cause a current through the inductor. Hence, there will be stored energy. Solve: The energy stored in the inductor depends on the inductance and the current. According to Ohm’s law,

2 3 2bat 1 1L 2 2

12 V 2.0 A (100 10 H)(2.0 A) 0.20 J4.0 2.0

VI U LIR R r

−Δ= = = = ⇒ = = × =+ Ω + Ω

33.21. Visualize: The solenoid has inductance and when a current flows there is energy stored in the magnetic field. Solve: Using Equations 33.33 and 33.39, the inductance and energy of the solenoid are

2 7 2 240

2 4 2 51 1L 2 2

(4 10 H/m)(200) (0.015 m) 2.96 10 H0.12 m

(2.96 10 H)(0.80 A) 9.5 10 J

μ π π−−

− −

×= = = ×

= = × = ×

Section 33.9 LC Circuits

33.22. Model: A capacitor in combination with an inductor will have a resonant frequency determined by the inductance and capacitance. Solve: We know the frequency and the capacitance so we can find the inductance. We have

72 6 2 12

1 1 1 2.5 10 H 0.25 H2 (100 10 Hz) (10 10 F)

ω μω π

−−= ⇒ = = = × =

33.23. Model: Changing the variable capacitor in combination with a fixed inductor will change the resonant frequency of the LC circuit. Solve: Because the resonant frequency depends on the inverse square root of the capacitance, a lower capacitance will produce a higher frequency and vice versa. The maximum frequency is

6max 3 12

1 1 2.2 10 rad/s(2.0 10 H)(100 10 F)LC

ω − −= = = ×× ×

The corresponding minimum is 6

min 2.0 10 rad/s.ω = × These are angular frequencies so we can use /(2 )f ω π= to

find 5

min 2.5 10 Hzf = × and 5

max 3.6 10 Hz,f = × giving a range of 250 to 360 kHz.

33.24. Model: Solve: The resonant frequency depends on the inverse square root of the product of the inductance and the capacitance:

Detecting a frequency 450 MHzf = would therefore require a capacitance of

2 2 2 6 31 1 1 3.8 nF

2 2 4 4 (450 10 Hz)(15 10 H)f C

LC f Lωπ π π π −

1= = ⇒ = = =× ×

Section 33.10 LR Circuits

33.25. Visualize: Please refer to Figure EX33.25. This is a simple LR circuit if the resistors in parallel are treated as an equivalent resistor in series with the inductor. Solve: We can find the equivalent resistance from the time constant since we know the inductance. We have

eq 6eq

7.5 10 H 30025 10 s

−×= ⇒ = = = Ω×

33-10 Chapter 33

The equivalent resistance is the parallel addition of the unknown resistor R and 500 .Ω We have

1 1 1 (500 )(300 ) 750500 500 300

Ω Ω= + ⇒ = = ΩΩ Ω − Ω

33.26. Visualize: Please refer to Figure EX33.26. This is a simple LR circuit if the two resistors in series are treated as an equivalent resistor in series with the inductor. Solve: The time constant of the equivalent circuit is

eq 1 2

75 10 H 2.50 10 s300

L LR R R

−×= = = = ×+ Ω

If the current 0I is the maximum current, then it will decay exponentially with time according to /0( ) .tI t I e τ−= The

time 1/2t when the current drops to half the initial value is found as follows:

( )1/2/ 4 41 10 0 1/2 1/22 2ln / ln 2 (2.50 10 s)0.69 1.7 10 s 0.17 mstI I e t tτ τ τ− − −= ⇒ = − ⇒ = = × = × =

Assess: This is less than ,τ as expected, since the time constant is the amount of time it takes to decay to 0.37 of the initial current and we are only going to 1/2 of the initial current. This result is quite general and is characteristic of all exponential decay phenomena.

33.27. Visualize: Please refer to Figure P33.27. To calculate the flux we need to consider the orientation of the normal of the surface relative to the magnetic field direction. We will consider the flux through the surface in the two parts corresponding to the two different directions of the surface normals. Solve: The flux is

top left top left top left

cos45 cos45

2(0.050 m)(0.10 m)(0.050 T)cos45 3.5 10 Wb

A B A B A B A B−

Φ = Φ + Φ = ⋅ + ⋅ = ° + °

= ° = ×

33.28. Model: Assume the field is uniform in space although it is changing in time. Visualize: The changing magnetic field strength produces a changing flux through the coil, and a corresponding induced emf and current. Solve: (a) Since the field is perpendicular to the plane of the coil, A is parallel to B and .ABΦ = The emf is

( ) 20 (0.025 m) (0.020 0.020 ) T/s 7.9 10 (1 ) V

( ) 7.9 10 (1 ) V( ) 1.6 10 (1 ) A0.50

d dBt NA t tdt dt

t tI t tR

π −

−−

Φ= = = + = × +

× += = = × +Ω

(b) Using the expression for I(t) in part (a), 3 3 2( 5 s) 1.57 10 (1 5) A 9.4 10 A ( 10 s) 1.7 10 AI t I t− − −= = × + = × = = ×

Assess: The field is always increasing and is increasing more rapidly with time so the induced current is greater at 10 s than at 5 s.

33.29. Model: Assume the field is uniform in space although it is changing in time. Visualize: The changing magnetic field strength produces a changing flux through the loop, and a corresponding induced emf and current. Solve: Since the field is perpendicular to the plane of the loop, A is parallel to B and .ABΦ = The emf is

2(0.20 m) (4 4 )T/s 0.16(1 ) V 1.6(1 ) Ad dBA t t I tdt dt RΦ= = = − = − ⇒ = = −E

The magnetic field is increasing over the interval 0 s 1 st< < and is decreasing over the interval 1 s 2 s,t< < so the induced emf and current must have opposite signs in the second half of the time interval. We arbitrarily choose the sign to be positive during the first half. Inserting 0.0, 1.0, and 2.0 st = into the equation above gives a current of 1.6, 0.0, and −1.6 A, respectively.

33.30. Model: Assume that B changes uniformly with time. Visualize:

The magnetic strength is changing so the flux is changing, and this will create an induced emf. The magnetic field is at an angle 60θ = ° to the normal of the plane of the coils. Solve: The flux for a single loop of the coil is cos .A B BA θΦ = ⋅ = The radius and angle don’t change with time, but B does. According to Faraday’s law, the induced emf is

cos cos

1.50 T 0.50 T100 (0.010 m) (cos60 ) 2.6 10 V 26 mV0.60 s

d dB BN NA N rdt dt t

θ π θ

π −

Φ Δ= = =Δ

−= ° = × =

33.31. Model: Assume that the field is uniform in space over the coil. Visualize: We want an induced current so there must be an induced emf created by a changing flux. Solve: To relate the emf and the current we need to know the resistance, which may be found using Equation 30.22 and the resistivity values from Table 30.2:

8Cu wire Cu

2 3 2wire wire

2 2(100)(1.7 10 m)(0.040 m) 2.176(0.25 10 m)

l N rRA r

ρ ρ ππ

−× Ω= = = = Ω

The magnetic field is parallel to the axis of the coil so the flux for a single loop of the coil is ,A B BAΦ = ⋅ = if we take the normal to a coil loop to be in the same direction as the field. Using Faraday’s law,

(2.0 A)(2.176 ) 8.7 T/s100 (0.040 m)

dB dB dB IRIR NA N rdt dt dt N r

ππ π

Ω= = = ⇒ = = =E

33.32. Model: The loop remains perpendicular to the magnetic field as it shrinks. The magnetic field remains uniform and steady. Visualize: The changing area of the loop causes the magnetic flux through the loop to change, inducing an emf in the loop. Solve: The area of the loop at a time t is

2 2 20( ) .tA t r r e βπ π −= = Since the normal to the loop dA and the field

direction are parallel, .B dA BdA⋅ = The induced emf is m ,ddtΦ=E where

m( ) ( )t B dA BA tΦ = ⋅ =∫

Thus 2 2 2 2m

0 0( 2 ) 2t td dAB r e r Bedt dt

β βπ β βπ− −Φ= = = − =E

Assess: As the loop area shrinks to zero radius at t → ∞ the induced emf also approaches zero. The induced emf is proportional to the time rate of change of the area of the loop.

33.33. Model: The magnetic field changes with time, but not position. The loop does not move. Solve: (a) The changing magnetic field will induce a current in the loop. We can compute the magnetic flux through the loop, and then apply Faraday’s Law. The magnetic field B does not depend on the position variables x or y, so the flux .m B AΦ = ⋅ The area vector A points along the k direction.

33-12 Chapter 33

With 2 ˆˆ(0.030 0.50 ) TB t i t k= + and

2ˆ ˆ(10 cm 10 cm) 0.010 m ,A k k= × = 2 2 3 2 2(0.50 T)(0.010 m ) (5.0 10 ) Tm .B A t t−Φ = ⋅ = = ×

The induced emf is thus 3 2 3 2 32(5.0 10 T m ) (10.0 10 T m ) (10.0 10 V/s) .d t t t

dt− − −Φ= = × = × = ×E

At 30.5 s, 5.0 10 V 5.0 mV.t −= = × =E

(b) Similarly, at 310 s, 10.0 1.0 V 10.0 mV.t −= = × =E

Assess: Only the component of B along A contributes to the magnetic flux.

33.34. Visualize:

The bottom of the loop is on the x-axis. The field is changing with time so the changing flux will produce an induced emf and corresponding induced current in the loop. The field strength varies in space as well as time. Solve: Let’s take the normal of the surface of the loop to be in the z-direction so it is parallel to the magnetic field. The flux through the shaded strip is

2( ) (0.80 )d B dA BdA B bdy y t bdyΦ = ⋅ = = = To find the total flux we integrate over the area of the loop. We obtain

0.80(0.80 ) 0.803

b b tby t bdy tb y dyΦ = = =Ñ Ñ

To find the induced current we take the time derivative of the flux: 4 4 4

loop 4loop

1 1 0.80 0.80 0.80(0.20 m) 8.5 10 A 0.85 mA3 3 3(0.50 )

d d tb bIR R dt R dt R

−⎛ ⎞Φ= = = = = = × =⎜ ⎟⎜ ⎟ Ω⎝ ⎠

The induced current in the loop is a constant.

33.35. Model: Assume the wire is long enough so we can use the formula for the magnetic field of an “infinite” wire. Visualize:

The magnetic field in the vicinity of the loop is due to the current in the wire and is perpendicular to the loop. The current is changing so the field and the flux through the loop are changing. This will create an induced emf and induced current in the loop. Solve: The induced current depends on the induced emf and is

looploop

1 dIR R dt

Φ= =E

The flux through a rectangular loop due to a wire was found in Example 33.5:

Ib c ac

+⎛ ⎞Φ = ⎜ ⎟⎝ ⎠

We have 0.10 m and 0.20 m.c a b= = = The current through the loop is therefore 7

1 (4 10 T m/A)(0.020 m) 0.030 mln ln (100 A/s) 44 A2 (0.010 )2 0.010 m

b c a dIIR c dt

μ π μπ π

− ⎛ ⎞+ ×⎛ ⎞= = =⎜ ⎟⎜ ⎟ Ω⎝ ⎠ ⎝ ⎠

33.36. Model: Assume the wire is infinitely long. Visualize: See Figure P33.36. The current through the wire produces a steady magnetic field, but the magnetic field strength depends on the distance from the wire. The loop moves away from the wire at constant speed. The flux through the loop varies due to its motion. Faraday’s law gives the magnitude of the induced emf, and Ohm’s law will yield the current strength. The direction of the magnetic field is into the paper at the location of the loop. The magnetic field is perpendicular to the plane of the loop, so .B dA BdA⋅ = Solve: The flux through the loop is decreasing as it moves away from the wire. Lenz’s law implies that the induced current is clockwise in order to increase the flux. Using the results of Example 33.5, which treats the rectangular loop as a series of tall, infinitesimally thin strips,

(4.0 cm) 1.0 cmln2

+⎛ ⎞Φ = ⎜ ⎟⎝ ⎠

where x is the distance from the wire to the closer edge of the loop at the given instant. Note that mddtΦ =

m m ,d dx d vdx dt dxΦ Φ= where dxv

dt= is the speed of the loop.

The induced emf is thus

m 0 (4.0 cm) 1.0 cm2 ( 1.0 cm)

d dx I vdx dt x x

Φ= =+

At 2.0 ms,x = 7

5(4 10 T m/A)(15 A)(4.0 cm) 1.0 cm (10 m/s) 2.0 10 V2 (2.0 cm)(3.0 cm)

−−×= = ×E

Since the loop resistance is 0.020 ,R = Ω the induced current is thus 5

loop2.0 10 V 1.0 mA

0.020I

−×= = =Ω

33.37. Model: Since the solenoid is fairly long compared to its diameter and the loop is located near the center, assume the solenoid field is uniform inside and zero outside. Visualize: Please refer to Figure P33.37. The solenoid’s magnetic field is perpendicular to the loop and creates a flux through the loop. This flux changes as the solenoid’s current changes, causing an induced emf and corresponding induced current. Solve: (a) Using Faraday’s law, the induced current is

2 2loop sol sol sol 0 sol 0

loop1 d A dB r d NI r N dII

R R dt R dt R dt l Rl dtπ μ π μΦ ⎛ ⎞= = = = =⎜ ⎟

⎝ ⎠

33-14 Chapter 33

where we have used the fact that the field is approximately zero outside the solenoid, so the flux is confined to the area 2

sol solA rπ= of the solenoid, not the larger area of the loop itself. The current is constant from 0 s 1st≤ ≤ so / 0 A/sdI dt = and loop 0 AI = during this interval. Thus, (0.5s) 0.0 A.I =

(b) During the interval 1s 2 s,t≤ ≤ the current is changing at the rate / 40 A/s,dI dt = so the current during this interval is

loop(0.010 m) (4 10 T m/A)(100) (40 A/s) 1.58 10 A 158 A

(0.10 )(0.10 m)I π π μ

−−×= = × =

Thus (1.5 s) 0.16 mA.I = (c) The current is constant from 2 s 3 s,t≤ ≤ so again loop/ 0 A/s and 0 A so (2.5s) 0.0 A.dI dt I I= = = Assess: The induced current is proportional to the negative derivative of the solenoid current, and is zero when there is a constant current in the solenoid.

33.38. Model: Since the solenoid is fairly long compared to its diameter and the loop is located near the coil center, assume the solenoid field is uniform inside and zero outside. Visualize: Please refer to Figure P33.38. The solenoid’s magnetic field is parallel to the loop’s axis and creates a flux through the loop. This flux changes as the solenoid’s current changes, causing an induced emf and corresponding induced current in the coil. Solve: The induced current from the induced emf is given by Faraday’s law. We have (see Example 33.8)

2 2loop loop loop loop loop 00 sol sol

loop looploop loop loop loop loop loop

1 1 ( )d A r r Nd dB d NI dII BA

R R dt R dt R dt R dt l R l dtπ π μμΦ ⎛ ⎞= = = = = =⎜ ⎟

⎝ ⎠

We used the fact that the coil flux is confined to the area 2loop loopA rπ= of the coil, not the larger area of the solenoid.

The current is changing uniformly over the interval 0 s 0.02 st≤ ≤ at the rate / 50 A/sdI dt = so the induced current during this interval is

loop(0.0050 m) (4 10 T m/A)(120) (50 A/s) 1.5 10 A

(0.50 )(0.080 m)I π π −

−×= = ×Ω

The current in the solenoid is changing steadily so the induced current in the coil is constant. The current is initially positive so the field is initially to the right and decreasing. The induced current will oppose this change and will therefore produce an induced field to the right. This requires an induced current in the coil coming out of the page at the top, so it is also positive. That is, it is clockwise when seen from the left.

33.39. Model: Assume that the magnetic field of coil 1 passes entirely through coil 2 and that we can use the magnetic field of a solenoid for coil 1. Visualize: Please refer to Figure P33.39. The field of coil 1 produces flux in coil 2. The changing current in coil 1 gives a changing flux in coil 2 and a corresponding induced emf and current in coil 2. Solve: (a) From 0 s to 0.1 s and 0.3 s to 0.4 s the current in coil 1 is constant, so its magnetic field is constant and the current in coil 2 must be zero. Thus (0.05 s) 0 A.I = (b) From 0.1 s to 0.3 s, the induced current from the induced emf is given by Faraday’s law. The current in coil 2 is

222 2 0 1 1 2 2 0 1 1

2 2 2 2 2 21 1

20 (0.010 m) (4 10 T m/A)(20) 20 A/s 7.95 10 A 79 A(2 )(20)(0.0010 m)

d dB d N I N r N dII N N A N rR R dt R dt R dt l Rl dt

μ π μπ

π π μ−

⎛ ⎞Φ= = = = =⎜ ⎟⎝ ⎠

×= = × =Ω

We used the facts that the field of coil 1 is constant inside the loops of coil 2 and the flux is confined to the area 2

2 2A rπ= of coil 2. Also, we used 1 1 20(1.0 mm)l N d= = and / (4 A)/(0.2 s) 20 A/s.dI dt = = From 0.1 s to 0.2 s the current in coil 1 is initially negative so the field is initially to the right and the flux is decreasing. The induced current

will oppose this change and will therefore produce a field to the right. This requires an induced current in coil 2 that comes out of the page at the top of the loops so it is negative. From 0.2 s to 0.3 s the current in coil 1 is positive so the field is to the left and the flux is increasing. The induced current will oppose this change and will therefore produce a field to the right. Again, this is a negative current. Hence (0.25 s) 79I μ= Α right to left through the resistor.

33.40. Model: Assume the field due to the solenoid is uniform inside and vanishes outside. Visualize: The changing current in the solenoid produces a changing field and flux through the coil. This changing flux creates an induced emf in the coil. Solve: The flux is only nonzero within the area of the solenoid, not the entire area of the coil. The flux is

A B BAΦ = ⋅ = and the induced current in the coil is

2coil sol 0 s scoil coil coil s coil s

2 2coil s 0 s s coil s 0 s

(2 ) cos(2 )

50 (0.010 m) (4 10 T m/A)(200) (2 )(60 Hz)(0.50 A)cos(2 )(0.50 )(0.20 m)

d dB d N II N N A N rR R dt R dt R dt l

N r N dI N r N f I ftRl dt Rl

π μ π μ π π

π π π π−

⎛ ⎞Φ= = = = ⎜ ⎟⎝ ⎠

7.4 mA)cos(2 )ftπ

Assess: The induced current oscillates because the current and field of the solenoid oscillate.

33.41. Model: Assume the magnetic field is uniform over the plane of the loop. Visualize: The oscillating magnetic field strength produces a changing flux through the loop and an induced emf in the loop. Solve: (a) The normal to the surface of the loop is in the same direction as the magnetic field so that .A B BAΦ = ⋅ = The induced emf is

2 20 cosd dB dBA r r B t

dt dt dtπ π ω ωΦ= = = =E

The cosine will oscillate between +1 and −1 so the maximum emf is 2 2 2 2 6 9

max 0 0(2 ) 2 (0.125) (150 10 Hz)(20 10 T) 0.93 Vr B r f Bπ ω π π π −= = = × × =E

(b) If the loop is rotated so that the plane is perpendicular to the electric field, then the normal to the surface will be perpendicular to the magnetic field. There is no magnetic flux through the loop and no induced emf.

33.42. Model: Assume the field due to the solenoid is uniform inside and vanishes outside. Visualize: The changing current in the solenoid produces a changing field and flux through the coil. This changing flux creates an induced emf in the coil. Solve: The flux is only nonzero within the area of the solenoid, not the entire area of the coil. The induced current in the coil is

2coil s 0 s scoil coil coil s coil s

2 2coil s 0 s s coil s 0 s

(2 ) cos(2 )

d dB d N II N N A N rR R dt R dt R dt l

N r N dI N r N f I ftRl dt Rl

π μ π μ π π

⎛ ⎞Φ= = = = ⎜ ⎟⎝ ⎠

The maximum coil current occurs when the cosine factor is .+1 The maximum current in the solenoid is

coil,max s0 2 2 7

coil s 0 s

(0.20 A)(0.40 )(0.20 m) 6.0 A(2 ) 40 (0.015 m) (4 10 T m/A)(200)2 (60 Hz)

N r N fπ μ π π π π−Ω= = =

Assess: This is a large current, but the induced current is large as well.

33-16 Chapter 33

33.43. Model: Assume an ideal transformer. Visualize: An ideal transformer changes the voltage, but not the power (energy conservation). Solve: (a) The primary and secondary voltages are related by Equation 33.30. We have

2 12 1 1 2

15,000100 12,500 turns120

N VV V N NN V

= ⇒ = = =

(b) The input power equals the output power and we recall that ,P I V= Δ so

2 2out in 1 1 2 2 1

(250 A)120 V 2.0 A15,000 V

I VP P I V I V IV

Δ= ⇒ Δ = Δ ⇒ = = =Δ

Assess: These values seem reasonable, because houses have low voltage and high current while transmission lines have high voltage and low current.

33.44. Model: The inner loop is small enough that the field due to the outer loop is essentially uniform over its area. Also, assume the current in the outer loop changes uniformly. Visualize: The current in the outer loop creates a field at the location of the inner loop. Changing the current in the outer loop results in a change in flux and a corresponding induced emf and current in the inner loop. Solve: The magnetic field due to the outer loop is perpendicular to the plane of the inner loop, so innerΦ =

inner outer inner outer ,A B A B⋅ = where outer 0 outer/2B I rμ= is the field of a current loop. The radius and angle don’t change with time, but outerB does. According to Faraday’s law, the induced emf is

2 2 2inner inner 0 outer inner 0 outer inner 0 outer

innerinner inner inner outer inner outer inner outer

12 2 2

(0.0010 m) (4 10 T m/A) 1.0 T 1.0 T 4.0 12(0.020 )(0.050 m) 0.10 s

d r d I r dI r IIR R dt R dt r R r dt R r t

π μ π μ π μ

π π −

⎛ ⎞Φ Δ= = = = =⎜ ⎟ Δ⎝ ⎠

× − −= = ×Ω

4.0 nA

33.45. Model: Assume the field changes abruptly at the boundary and is uniform. Visualize: Please refer to Figure P33.45. As the loop enters the field region the amount of flux will change as more area has field penetrating it. This change in flux will create an induced emf and corresponding current. While the loop is moving at constant speed, the rate of change of the area is not constant because of the orientation of the loop. The loop is moving along the x-axis. Solve: (a) If the edge of the loop enters the field region at 0 s,t = then the leading corner has moved a distance

0x v t= at time t. The area of the loop with flux through it is

( ) 2 21022 ( )A yx x v t= = =

where we have used the fact that y x= since the sides of the loop are oriented at 45° to the horizontal. Take the

surface-normal of the loop to be into the page so that .A B BAΦ = ⋅ = The current in the loop is

22 2 3

0 01 1 1 1 2(0.80 T)(10 m/s)( ) (2) (1.6 10 A)

0.10d dA dI B B v t B v t t t

R R dt R dt R dt R⎛ ⎞Φ= = = = = = = × ⎜ ⎟⎜ ⎟Ω⎝ ⎠

The current is increasing at a constant rate. This expression is good until the loop is halfway into the field region. The time for the loop to be halfway is found as follows:

10 cm (10 m/s) 7.1 10 s 7.1 ms2

v t t t −= = ⇒ = × =

At this time the current is 11 A. While the second half of the loop is moving into the field, the flux continues to increase, but at a slower rate. Therefore, the current will decrease at the same rate as it increased before, until the loop is completely in the field at 14 ms.t = After that the flux will not change and the current will be zero.

(b) The maximum current of 11 A occurs when the flux is changing the fastest, and this occurs when the loop is halfway into the region of the field.

33.46. Visualize: Please refer to Figure P33.46. The area within the loop is changing so the flux will change. This will produce an induced emf and a corresponding current. The area of the loop is the area of a square. Solve: (a) The field is out of the page, so the flux is increasing outward as the loop expands. According to Lenz’s Law, the induced current tries to prevent the increase of outward flux. It does so by generating a field into the page. This requires a clockwise current flow. (b) Take A parallel to ,B so and / ( / ).AB d dt B dA dtΦ = = Φ =E The field is constant but the loop area is changing. Let the length of each edge of the loop be x at time t. This length is increasing linearly with time as the corner of the loop moves outward with speed 10 m/s.v = We have

0 0 [(10 m/s)cos45 ] 7.07 mxx x v t t t= + = + ° = The loop’s area at time t is

2 2 2 2 2(7.07 m) 50 m 100 m /sdAA x t t tdt

= = = ⇒ =

Consequently, the induced emf at time t is

2(0.10 T)(100 m /s) 10 VdAB t tdt

= = =E

The induced current is / ,I R= E where R is the resistance of the loop. The wire has resistance 0.010 Ω per meter, so (0.010 /m) where 4R l l x= Ω = is the perimeter of the square at time t. That is,

4(0.010 /m)(7.07 m) 0.283R t t= Ω = Ω Thus, the induced current is

10 V 35 A0.283

= = =Ω

(c) At 0.10 s, 1.0 V and 35 A.t I= = =E Assess: The induced emf depends on time, but so does the resistance. This means the induced current is constant.

33.47. Visualize: Please refer to Figure 33.26. The moving wire in a magnetic field results in a motional emf and a current in the loop. Solve: (a) The induced emf is

loop (100 m/s)(0.040 m)(1.0 T) 4.0 VvlB= = =E (b) The total resistance of the square loop is 4(0.010 ) 0.040 .R = Ω = Ω The induced current is

loop/ (4.0 V)/(0.040 ) 100 AI R= = Ω =E

(c) The slide wire generates the emf, loop.E However, the slide wire also has an “internal resistance” due to the resistance of the slide wire itself—namely 0.01 .r = Ω Thus, the slide wire acts like a battery with an internal resistance. The potential difference between the ends is

loop 4.0 V (100 A)(0.01 ) 3.0 VV IrΔ = − = − Ω =E

It is this 3.0 V that drives the 100 A current through the remaining 0.03 Ω resistance of the “external circuit.” Assess: The slide wire system acts as an emf just like a battery does.

33-18 Chapter 33

33.48. Model: Assume that the field is uniform in space over the loop. Visualize:

The moving slide wire in a magnetic field develops a motional emf and a corresponding current in the wire and the rails. The current through the resistor will cause energy dissipation and subsequent warming. Solve: (a) The induced emf depends on the changing flux, not on where the resistance in the loop is located. We have

loop / (10 m/s)(0.20 m)(0.10 T) 0.20 Vv B= = =E

The total resistance around the loop is due entirely to the carbon resistor, so the induced current is

loop/ (0.20 V)/(1.0 ) 0.20 AI R= = Ω =E

(b) To move with a constant velocity the acceleration must be zero, so the pulling force must balance the retarding magnetic force on the induced current in the slide wire. Thus,

3pull mag (0.20 A)(0.20 m)(0.10 T) 4.0 10 NF F IlB −= = = = ×

(c) The current in the resistor will result in power being dissipated. The power is 2 2(0.20 A) (1.0 ) 0.040 W 0.040 J/sP I R= = Ω = =

During a 10 second period 0.40 JQ = of energy is dissipated by the current, increasing the internal energy of the carbon resistor and raising its temperature. This is, effectively, a heat source. The heat is related to the temperature rise by ,Q mc T= Δ where c is the specific heat of carbon. Thus,

50.40 J 11 K

(5.0 10 kg)(710 J/kg K)QTmc −Δ = = =

Assess: This is a significant change in the temperature of the resistor.

33.49. Model: Assume that the magnetic field is uniform in the region of the loop. Visualize: Please refer to Figure P33.49. The rotating semicircle will change the area of the loop and therefore the flux through the loop. This changing flux will produce an induced emf and corresponding current in the bulb. Solve: (a) The spinning semicircle has a normal to the surface that changes in time, so while the magnetic field is constant, the area is changing. The flux through in the lower portion of the circuit does not change and will not contribute to the emf. Only the flux in the part of the loop containing the rotating semicircle will change. The flux associated with the semicircle is

cos cos(2 )A B BA BA BA ftθ πΦ = ⋅ = = =

where 2 ftθ π= is the angle between the normal of the rotating semicircle and the magnetic field and A is the area of the semicircle. The induced current from the induced emf is given by Faraday’s law. We have

1 1 cos(2 ) 2 sin (2 )2

2(0.20 T) (0.050 m) sin (2 ) (4.9 10 ) sin (2 ) A2(1.0 )

d d B rI BA ft f ftR R dt R dt R

f ft f ft

ππ π π

π π π−

Φ= = = =

= = ×Ω

where the frequency f is in Hz.

(b) We can now solve for the frequency necessary to achieve a certain current. From our study of DC circuits we know how power relates to resistance:

2 / (4.0 W)/(1.0 ) 2.0 AP I R I P R= ⇒ = = Ω =

The maximum of the sine function is 1,+ so the maximum current is

3 2max 3

2.0 A4.9 10 A s 2.0 A 4.1 10 Hz4.9 10 A s

I f f−−= × = ⇒ = = ×

Assess: This is not a reasonable frequency to obtain by hand.

33.50. Model: Assume there is no resistance in the rails. If there is any resistance, it is accounted for by the resistor. Visualize: Please refer to Figure P33.50. The moving wire will have a motional emf that produces a current in the loop. Solve: (a) At constant velocity the external pushing force is balanced by the magnetic force, so

2 2 2 24 4

push mag(0.50 T) (0.10 m) (0.50 m/s) 6.25 10 N 6.3 10 N

2.0Blv B l vF F IlB lB lB

R R R− −⎛ ⎞= = = = = = = × ≈ ×⎜ ⎟ Ω⎝ ⎠

(b) The power is 4 4(6.25 10 N)(0.50 m/s) 3.1 10 WP Fv − −= = × = ×

(c) The flux is out of the page and decreasing and the induced current/field will oppose the change. The induced field must have a flux that is out of the page so the current will be counterclockwise. The magnitude of the current is

2 2(0.50 T)(0.10 m)(0.50 m/s) 1.25 10 A 1.3 10 A2.0Ω

− −⎛ ⎞= = = × ≈ ×⎜ ⎟⎝ ⎠

(d) The power is 2 2 2 4(1.25 10 A) (2.0 ) 3.1 10 WP I R − −= = × Ω = ×

Assess: From energy conservation we see that the mechanical energy put in by the pushing force shows up as electrical energy in the resistor.

33.51. Model: Assume that the magnetic field is constant in the area in which the coil rotates. Solve: From Equation 33.29, we know that the emf produced in a coil rotating in a constant magnetic field is

sin ( )ABN tε ω ω= −

where 2A rπ= is the cross-sectional area of the coil, 100N = is the number of turns in the coil, and 2 .fω π= The rms voltage, which is the voltage measured by standard voltmeters, is

2 2 2 2 2ms 2 2 (0.010 m) (100) (0.14 m )

2ABN r fBN Bf Bfωε π π= = = =

Thus, if we plot the rms voltage as a function of frequency, the slope s is 2(0.14 m ) .s B= Note that we can also add the datum rms( , ) (0, 0)f ε = to the data, because a coil that does not rotate will have no induced voltage. The plot is shown below.

The result is 20.039489 T m , so 0.28 T.s B= =

33-20 Chapter 33

33.52. Model: Assume the magnetic field is uniform over the region where the bar is sliding and that friction between the bar and the rails is zero. Visualize: Please refer to Figure P33.52. The battery will produce a current in the rails and bar and the bar will experience a force. With the battery connected as shown in the figure, the current in the bar will be down and, by the right-hand rule, the force on the bar will be to the right. The motion of the bar will change the flux through the loop and there will be an induced emf that opposes the change. Solve: (a) As the bar speeds up the induced emf will get larger until finally it equals the battery emf. At that point, the current will go to zero and the bar will continue to move at a constant velocity. We have

batterm bat termBlv v

Bl= = ⇒ = EE E

(b) The terminal speed is

term1.0 V 33 m/s

(0.50 T)(0.060 m)v = =

Assess: This is pretty fast, about 70 mph.

33.53. Model: Assume the magnetic field is uniform in the region of the loop. Visualize:

The moving wire creates a changing area and corresponding change in flux. This produces an induced emf and induced current. The flux through the loop depends on the size and orientation of the loop. Solve: (a) The normal to the surface is perpendicular to the loop and the flux is inner cos .A B AB θΦ = ⋅ = We can get the current from Faraday’s law. Since the loop area is ,A lx= We have

1 1 cos coscosd d Bl dx BlvI lxBR R dt R dt R dt R

θ θθΦ= = = = =E

(b) Using the free-body diagram shown in the figure, we can apply Newton’s second law. The magnetic force on a straight, current-carrying wire is mF IlB= and is horizontal. Using the current I from part (a) gives

mcoscos sin sinx x

B l vF F mg mg maR

θθ θ θ= − + = − + =∑

Terminal speed is reached when xa drops to zero. In this case, the two terms are equal and we have

term 2 2tancos

mgRvl B

33.54. Model: Assume the magnetic field is uniform over the loop. Visualize: Please refer to Figure P33.54. As the wire falls, the flux into the page will increase. This will induce a current to oppose the increase, so the induced current will flow counterclockwise. As this current passes through the slide wire, it experiences an upward magnetic force. So there is an upward force—a retarding force—on the wire as it falls in the field. As the wire speeds up, the retarding force will become larger until it balances the weight.

Solve: (a) The force on the current-carrying slide wire is m .F IlB= The induced current is

1 1d d B d BlvI AB lxR R dt R dt R dt R

Φ= = = = =E

Consequently, the retarding magnetic force is 2 2 2 2

ml B v l BF v

R R⎛ ⎞

= = ⎜ ⎟⎜ ⎟⎝ ⎠

The important point is that mF is proportional to the speed v. As the wire begins to fall and its speed increases, so does the retarding force. Within a very short time, mF will increase in size to where it matches the weight .w mg= At that point, there is no net force on the loop, so it will continue to fall at a constant speed. The condition that the magnetic force equals the weight is

term term 2 2l B mgRv mg v

⎛ ⎞= ⇒ =⎜ ⎟⎜ ⎟

⎝ ⎠

(b) The terminal speed is 2

term 2 2(0.010 kg)(9.8 m/s )(0.10 ) 0.98 m/s

(0.20 m) (0.50 T)v Ω= =

33.55. Model: Assume the magnetic field is uniform over the loop. Visualize: The motion of the eye will change the orientation of the loop relative to the fixed field direction, resulting in an induced emf. Solve: We are just interested in the emf due to the motion of the eye, so we can ignore the details of the time dependence of the change. From Faraday’s law,

cos coscos

20 (3.0 10 m) (1.0 T) cos(85 ) cos(90 )2.5 10 V

0.20 s

N r BdN N NABdt t t t

π θ θθ

π −−

−Φ ΔΦ Δ= = = =Δ Δ Δ

× ° − °= = ×

Assess: This is a reasonable emf to measure, although you might need some amplification.

33.56. Model: Assume the coil is moved to a location where the magnetic field is zero. Visualize:

The removal of the coil from the field will change the flux and produce an induced emf and corresponding induced current. The current will charge the capacitor. Solve: The induced current is

coilcoil

N dIR R dt

Φ= =E

The definition of current is / .I dq dt= Consequently, the charge flow through the coil and onto the capacitor is given by

f i( )dq N d q N d N Nqdt R dt t R dt R R

Φ Δ Φ= ⇒ = ⇒ Δ = ΔΦ = Φ − ΦΔ

33-22 Chapter 33

We are only interested in the total charge that flows due to the change in flux and not the details of the time dependence. In this case, the flux is changed by physically pulling the coil out of the field. Since the coil is oriented for maximum flux, the initial flux through the coil is

2 2 8i (0.0050 m) (0.0010 T) 7.85 10 Wbr Bπ π −Φ = = = ×

After being pulled from the field, the final flux is f 0 Wb.Φ = The charge that flows onto the capacitor is

(10) 0 Wb 7.85 10 Wb 3.93 10 C3.93 10 C 3.9 V0.20 1.0 10 F

− −−

− × Δ ×Δ = = × ⇒ Δ = = =Ω ×

33.57. Model: Assume the field is uniform in the region of the coil. Visualize:

The rotation of the coil in the field will change the flux and produce an induced emf and a corresponding induced current. The current will charge the capacitor. Solve: The induced current is

coilcoil

N dIR R dt

Φ= =E

The definition of current is / .I dq dt= Consequently, the charge flow through the coil and onto the capacitor is given by

f i( )dq N d q N N Nqdt R dt t R t R R

Φ Δ ΔΦ= ⇒ = ⇒ Δ = ΔΦ = Φ − ΦΔ Δ

We are only interested in the total charge that flows due to the change in flux and not the details of the time dependence. In this case, the flux is changed by physically rotating the coil in the field. The flux is A BΦ = ⋅ =

cos .AB θ The change in flux is 2 6 7

f i(cos cos ) (0.020 m) (55 10 T)(cos30 cos210 ) 1.2 10 WbAB θ θ π − −ΔΦ = − = × ° − ° = ×

Note that the field is 60° from the horizontal and the normal to the plane of the loop is vertical. The final angle, when A points down, is f 30 ,θ = ° so the initial angle is i f 180 210 .θ θ= + ° = ° The charge that flows onto the capacitor is

f i 6200(1.2 10 Wb) 1.2 10 C( ) 1.2 10 C 12 V

2.0 1.0 10 FCN qq VR C

− −−

−× Δ ×Δ = Φ − Φ = = × ⇒ Δ = = =

33.58. Model: Use Faraday’s law of electric induction and assume that the magnetic field inside the solenoid is uniform.

Visualize:

Solve: (a) With 10.0 T (2.0 T)sin[2 (10 Hz) ],B tπ= + Equation 33.26 gives

0.015 m (2.0 T)[2 (10 Hz)] cos[2 (10 Hz) ] (0.94 V/m) cos[2 (10 Hz) ]2 2r dBE t t

dtπ π π= = =

The field strength is maximum when the cosine function is equal to 1 or −1. Hence max 0.94 V/mE = (b) E is maximum when cos[2 (10 Hz) ] 1 or 1tπ = − which means when sin[2 (10 Hz) ] 0.tπ = Under this condition,

(10.0 T) (2.0 T) sin [2 (10 Hz) ] 10.0 TB tπ= + =

That is, 10.0 TB = at the instant E has a maximum value of 0.94 V/m.

33.59. Model: Use Faraday’s law of electric induction. Visualize:

Equation 33.23, / ,E d s A dB dt⋅ = describes the relationship between an induced electric field and the flux through

a fixed area A. To solve this equation, choose a clockwise direction around a circle of radius r as the closed curve. The electric field vectors, as can be seen from the figure, are everywhere tangent to the curve. The line integral of E then is

(2 )E d s Eπ⋅ =

The direction of the induced electric field is chosen to agree with the right-hand rule applied to the flux calculation for magnetic field. Note that B = 0 outside the solenoid.

33-24 Chapter 33

Solve: Equation 33.23 now simplifies to 2

2(2 )2

dB R dBE r R Edt r dt

π π= ⇒ =

Assess: At 12, ( ),r R E R dB/dt= = which is the same result as Equation 33.26.

33.60. Visualize: The changing current produces an emf across the inductor. Solve: When the current is changing we know the induced emf, so we can get the inductance. We have

0.20 V 20 mH/ 10.0 A/s

dI VV L Ldt dI dt

ΔΔ = − ⇒ = = =

We know the flux per turn for a given current so we can connect the inductance to the magnetic field and the flux per turn. The flux through one turn of the solenoid is

(1) 0 0sol

NI NAB A A Il l

μ μ⎛ ⎞ ⎛ ⎞Φ = = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

But the inductance is 2

0 0 0 solsol

N A NA NA LL Nl l l N

μ μ μ⎛ ⎞ ⎛ ⎞= = ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Substituting into the previous expression we get 3

(1) sol sol(1) 6

(20 10 H)(0.10 A) 400 turns5.0 10 Wb/turn

L LI N IN

−×⎛ ⎞Φ = ⇒ = = =⎜ ⎟ Φ ×⎝ ⎠

33.61. Model: Assume the straight wire in part (b) is “infinite.” Visualize: The energy density depends on the field strength in a particular region of space. Solve: (a) We will use the formula for the magnetic field at the center of a current loop and then find the energy density. We have

5 22 4 3

(4 10 T m/A)(1.0 A) 3.14 10 T2 2(0.020 m)

1 (3.14 10 T) 3.9 10 J/m2 2(4 10 T m/A)

−−

×= = = ×

×= = = ××

(b) We will use the standard formula for the magnetic field of a long straight wire. We have 2 2 2 2 2 4 3

20 0B 7

1 1 8 8 (0.020 m) (3.9 10 J/m ) 3.1 A2 2 2 2 (4 10 T m/A)

BI I r uB u B Ir r

μ μ π ππ μ μ π μ π

−×⎛ ⎞= ⇒ = = ⇒ = = =⎜ ⎟ ×⎝ ⎠

33.62. Model: Assume the solenoid is long enough that we can approximate the field as being constant inside and zero outside. Visualize: The energy density depends on the field strength in a particular region of space. Solve: (a) We can use the given field strength to find the energy density and then determine the volume of a cylinder. We have

22 6 3

B 6 3 3 670 tot B sol

2 2 3sol sol

1 (5.0 T) 9.95 10 J/m2 (9.95 10 J/m )(0.126 m ) 1.3 102(4 10 T m/A)

(0.20 m) (1.0 m) 0.126 m

u BU u V

⎫= = = × ⎪

= = × = ×× ⎬⎪= = = ⎭

(b) Using the magnetic field of the solenoid, the number of turns is calculated as follows:

(1.0 m)(5.0 T) 4.0 10(4 10 T m/A)(100 A)

NI lBB Nl I

μμ π −= ⇒ = = = ×

Assess: This is a reasonable number for a solenoid that is a meter long.

33.63. Model: Assume the field is constant over the cross section of the patient. Visualize: The changing field results in a changing flux and an induced emf. Solve: To determine the largest emf, assume the normal to the surface is parallel to the field. From Faraday’s law, the emf is

d d dB BAB A Adt dt dt tΦ Δ= = = =

The time interval acceptable for the change is 2(0.060 m ) 5.0 T 0 T

3.0 s0.10 V

Δ −Δ = = =

Assess: This is a reasonable amount of time in which to change the field.

33.64. Model: Assume we can ignore the sharp corners when the current changes abruptly. Visualize: The changing current produces a changing flux, an induced emf, and a corresponding potential difference. Solve: Break the current into time intervals over which the current is changing linearly or not at all. For the intervals 2 to 3 ms and 5 to 6 ms, the current does not change, so the potential difference is zero. For the interval 0 to 2 ms, the current goes from 0 to 2 A, so the potential difference is

3L L 3

2 A 0 A(10 10 H) 10 V(2 s 0 s) 10

dI IV L L Vdt t

−−

Δ −Δ = − = − ⇒ Δ = − × = −Δ − ×

Similarly for the interval 3 ms to 5 ms, the potential difference is +20 V.

Assess: The potential difference is proportional to the negative slope of the current versus time graph.

33.65. Model: Assume we can ignore the edges when the potential difference changes abruptly. Visualize: The existence of a potential difference indicates there is a changing current. If the potential difference is zero, the current is constant. Solve: Break the potential difference into time intervals and determine the corresponding rate of change of the current for each time interval. Knowing the starting current for an interval, we can determine the current at the end of the interval by finding the change. For the intervals 10 to 20 ms and 30 to 40 ms, the potential difference is zero, so the current remains the same as at the start of the interval. For the interval 0 to 10 ms, the potential difference is −1 V. The rate of change of the current is calculated as follows:

L 0 10 3( 1 V)(10 s 0 s) 10 s 0.20 A

50 10 HdI I V tV L L Idt t L

→ −Δ −Δ Δ − − ×Δ = − = − ⇒ Δ = = − =Δ ×

To find the current at 10 ms we need the current at the start of the interval. We have

0 10(10 ms) (0 ms) 0.20 A (0.20 A) 0.40 AI I I →= + Δ = + =

A constant potential difference implies that the current is changing linearly. So, the current starts at 0.2 A and goes linearly to 0.40 A over the first 10 ms. For the interval 10 to 20 ms, the current remains at 0.40 A. For the interval 20 to 30 ms, the current goes linearly from 0.40 to 0 A. For the interval 30 to 40 ms, the current remains at 0 A.

33-26 Chapter 33

33.66. Visualize: The potential difference across the inductor depends on the rate of change of the current with respect to time. Solve: (a) We can find the potential difference by taking a derivative. We have

L 0 0sin cosdI dV L L I t L I tdt dt

ω ω ωΔ = − = − = −

(b) The cosine function oscillates between +1 and −1 so the maximum potential difference is

L, maxL, max 0 0 3 6

0.20 V 1.3 mA2 (500 10 Hz)(50 10 H)

VV LI I

ω π −

ΔΔ = ⇒ = = =

33.67. Visualize: The current through the inductor is changing with time, which leads to a changing potential difference across the inductor. Solve: (a) To find the potential difference we differentiate the current. We have

/ /0L 0( )t tdI d LIV L L I e e

dt dtτ τ

τ− −⎛ ⎞Δ = − = − = ⎜ ⎟

⎝ ⎠

(b) For t = 1 ms, the potential difference is 3 33 3

(1.0 10 s)/(1.0 10 s) 1L 3

(20 10 H)(50 10 A) (1.0 V) 0.37 V1.0 10 s

V e e− −− −

− × × −−

× ×Δ = = =×

Similar calculations give 1.0 V, 0.13 V, and 0.05 V for t = 0 ms, 2 ms, and 3 ms.

33.68. Model: Assume any resistance is negligible. Visualize: Please refer to Figures 33.43 and 33.44. Solve: (a) Equation 33.50 describes the dynamics of the charge on the capacitor. The first time the charge goes through zero is when the argument of the cosine is π /2. From this criterion, we can find the time t at which this occurs:

3 12 7(20 10 H)(8.0 10 F) 6.3 10 s2 2 2 2

t t LCπ π π πωω

− − −= ⇒ = = = × × = ×

(b) The initial charge on the capacitor is 0 C 0( )Q C V= Δ

From Equation 33.52, the inductor current at /(2 )t π ω= is

0 0 0 38.0 10 F( ) sin ( ) sin ( ) (25 V) 0.50 mA

2 20 10 HC CCI t Q t C V VL

πω ω ω ωω

−×⎛ ⎞= = = Δ = Δ = =⎜ ⎟ ×⎝ ⎠

Assess: The inductor current is maximum when the capacitor charge is zero. This seems reasonable from considerations of conservation of energy. When the capacitor charge is maximum, all the circuit’s energy is stored in the capacitor. When the (magnitude of the) capacitor charge is minimum, all the energy must be stored in the inductor, so its current must be at a maximum.

33.69. Model: Assume any resistance is negligible. Visualize: The potential difference across the inductor and capacitor oscillate. The period of oscillation depends on the resistance and capacitance, and the potential difference across the capacitor depends on the charge.

Solve: The current is 0( ) cos (0.60 A)cos .I t I t tω ω= = We can relate the extreme values of the current and the

capacitor potential difference. Using C1/ and ,LC Q C Vω = = Δ we find 2 2 30

0 0 C, max C, max 2 2C, max

1 (0.60 A) (10 10 H) 1.0 F(60 V)

I LI Q C V C V CLC V

ω ω μ−×= = Δ = Δ ⇒ = = =

Assess: This is a reasonable capacitance.

33.70. Visualize: The rate at which the charge leaves the capacitor will determine the current through the inductor. Solve: The charge on the capacitor is 0 cosQ Q tω= where 0Q C V= Δ is the maximum charge at t = 0 s. The current is given by 0/ sin .I dQ dt Q tω ω= − = Since the sine function oscillates between +1 and −1, the maximum current is

max 0 31 0.10 10 F(5.0 V) 5.0 10 A 50 mA

1.0 10 HCI Q C V VLLC

−−

×⎛ ⎞= = Δ = Δ = = × =⎜ ⎟ ×⎝ ⎠

33.71. Visualize: The oscillation frequency depends on the capacitance and the inductance. The inductance will depend on the number of turns, the length, and the cross-sectional area of the solenoid. Solve: First we’ll find the inductance necessary to give the proper frequency, then we can design our inductor. For an LC circuit,

22 2 2 2 2

1 1 12 2.53 10 H4 4 (1.0 Hz) (1.0 F)

f LLC f C

ω ππ π

−= = ⇒ = = = ×

The inductor length is ,l Nd= where N is the number of turns and d is the diameter of the wire used. We find that 2 2 2 3 2

0 02 7 2

(0.25 10 m)(2.53 10 H) 4000(4 10 T m/A) (0.02 m)

N A N r dLL Nl Nd r

μ μ πμ π π π

− −

−× ×= = ⇒ = = =

If we have one layer, the length will be 34000(0.25 10 m) 1.0 m.l Nd −= = × = Because we use two layers of closely spaced turns, the inductor length will be 0.50 m.

33.72. Model: Assume there is no resistance in the LC circuit. Visualize: The LC circuit oscillates at a constant rate. The energy in the circuit alternates between the inductor and capacitor. Solve: The maximum energy stored in the capacitor is the same as the total energy of the circuit, which is the same as the maximum energy stored in the inductor. Maximum energy in the inductor occurs when the circuit has maximum current, max 0.10 A.I = The maximum energy in the inductor is 2 51

L max2 1.0 10 J.U LI −= = × Solving yields

2.0 mH.L = The circuit oscillates with angular frequency 2 21 (2 10 kHz) ,LC

ω π= = ⋅ thus 0.13 F.C μ=

Assess: These are reasonable values for L and C.

33.73. Model: Assume negligible resistance in the LC part of the circuit. Visualize: With the switch in position 1 for a long time the capacitor is fully charged. After moving the switch to position 2, there will be oscillations in the LC part of the circuit. Solve: (a) After a long time the potential across the capacitor will be that of the battery and 0 batt .Q C V= Δ When the switch is moved, the capacitor will discharge through the inductor and LC oscillations will begin. The maximum current is

62batt

0 0 batt batt 3(2.0 10 F) (12 V) 7.6 10 A 76 mA(50 10 H)

C V CI Q C V VLLC

ω ω−

−−

Δ ×= = Δ = = Δ = = × =×

(b) The current will be a maximum one-quarter cycle after the maximum charge. The period is

3 62 2 2 (50 10 H)(2.0 10 F) 2.0 msT LCπ π πω

− −= = = × × =

So the current is first maximum at 1max 4 0.50 ms.t T= =

33-28 Chapter 33

33.74. Model: Assume negligible resistance in the circuit. Visualize: Energy is conserved. The maximum voltage on the right capacitor will occur when all of the energy from the left capacitor is transferred to the right one. Solve: (a) The voltage is calculated as follows:

2 2 11 11 C1 2 C2 C2 C12 2

300 (100 V) 50 V1200

CC V C V V VC

Δ = Δ ⇒ Δ = Δ = =

(b) Closing 1S causes the charge and current of the left LC circuit to oscillate with period L.T After one-quarter of a period, the 300 Fμ capacitor is completely discharged and the current through the inductor is maximum. At that instant we’ll open 1S and close 2S . Then the right LC circuit will start to oscillate with period RT and the inductor current will charge the 1200 Fμ capacitor. The capacitor will be fully charged after one-quarter of a period, so we will open 2S at that time to keep the charge on the 1200 Fμ capacitor. The periods are

6 1L 1 L4

6 1R 2 R4

2 2 2 (5.3 H)(300 10 F) 0.25 s 0.063 s

2 2 2 (5.3 H)(1200 10 F) 0.50 s 0.125 s

T LC T

π π πω

= = = × = ⇒ =

So the procedure is to close 1S at 0 s,t = open 1S and close 2S at 0.0625 s,t = then open 2S at 0.0625 st = + 0.1250 s 0.1875 s.=

33.75. Model: Assume an ideal inductor and an ideal battery (i.e., zero internal resistance). Visualize: Please refer to Figure P33.75. The current through the battery is the sum of the currents through the left and right branches of the circuit. Solve: (a) Because the switch has been open a long time, no current is flowing the instant before the switch is closed. A basic property of an ideal inductor is that the current through it cannot change instantaneously. This is because the potential difference L ( / )V L dI dtΔ = − would become infinite for an instantaneous change of current, and that is not physically possible. Because the current through the inductor was zero before the switch was closed, it must still be zero (or very close to it) immediately after the switch is closed. Conservation of current requires the current through the entire right branch to be the same as that through the inductor, so it is also zero immediately after the switch is closed. The only current is through the left 20 Ω resistor, which sees the full battery potential of the battery. Thus bat left bat / (10 V)/(20 ) 0.50 A.I I V R= = Δ = Ω = (b) The current through the inductor increases as time passes. Once the current rightI reaches a steady value and is no

longer changing, the potential difference across the inductor is L ( / ) 0 V.V L dI dtΔ = − = An ideal inductor has no resistance, so the inductor simply acts like a wire and has no effect on the circuit. The circuit is that of two 20 Ω resistors in parallel. The equivalent resistance is 10 Ω and the battery current is bat (10 V)/(10 ) 1.0 A.I = Ω =

33.76. Model: Assume an ideal inductor and an ideal battery (i.e., zero internal resistance). Visualize: Please refer to Figure P33.76. Solve: (a) Because the switch has been open a long time, no current is flowing the instant before the switch is closed. A basic property of an ideal inductor is that the current through it cannot change instantaneously. This is because the potential difference L ( / )V L dI dtΔ = − would become infinite for an instantaneous change of current, and that is not physically possible. Because the current through the inductor was zero before the switch was closed, it must still be zero (or very close to it) immediately after the switch is closed. Consequently, the inductor has no effect on the circuit. It is simply a 10 Ω resistor and 20 Ω resistor in series with the battery. The equivalent resistance is 30 ,Ω so the current through the circuit (including through the 20 Ω resistor) is bat eq/ (30 V)/(30 ) 1.0 A.I V R= Δ = Ω = (b) After a long time, the currents in the circuit will reach steady values and no longer change. With steady currents, the potential difference across the inductor is L ( / ) 0 V.V L dI dtΔ = − = An ideal inductor has no resistance ( 0 ),R = Ω so the inductor simply acts like a wire. In this case, the inductor “shorts out” the 20 Ω resistor. All current from the 10 Ω resistor flows through the resistanceless inductor, so the current through the 20 Ω resistor is 0.0 A.

(c) When the switch has been closed a long time, and the inductor is shorting out the 20 Ω resistor, the current passing through the 10 Ω resistor and through the inductor is (30 V) (10 ) 3.0 A.I = / Ω = Because the current through an inductor cannot change instantaneously, the current must remain 3.0 A immediately after the switch reopens. This current must go somewhere (conservation of current), but now the open switch prevents the current from going back to the battery. Instead, it must flow upward through the 20 Ω resistor. That is, the current flows around the LR circuit consisting of the 20 Ω resistor and the inductor. This current will decay with time, with time constant

/ ,L Rτ = but immediately after the switch reopens the current is 3.0 A.

33.77. Visualize: Please refer to Figure P33.77. Solve: (a) After a long time has passed the current will no longer be changing. With steady currents, the potential difference across the inductor is L ( / ) 0 V.V L dI dtΔ = − = An ideal inductor has no resistance ( 0 ),R = Ω so the inductor simply acts like a wire. The circuit is simply that of a battery and resistor R, so the current is 0 bat/ .I V R= Δ (b) In general, we need to apply Kirchhoff’s loop law to the circuit. Starting with the battery and going clockwise, the loop law is

bat R L bat

dIV V V V IR Ldt

dI V IR R V RI I Idt L L L R L

Δ + Δ + Δ = Δ − − =

Δ Δ⎛ ⎞= − = − = −⎜ ⎟⎝ ⎠

This is a differential equation that we can solve by direct integration. The current is 0 A at 0 s,I t= = so separate the current and time variables and then integrate from 0 A at 0 s to current I at time t:

0 00 0

ln ( ) ln/

t tl tdI R R I I tdt I I t

I I L L I L R⎛ ⎞−= ⇒ − − = ⇒ − =⎜ ⎟− ⎝ ⎠

Taking the exponential of both sides gives

/( / ) /( / )00 0

t L R t L RI I e I I I eI

− −− = ⇒ − =

Finally, solving for I gives /( / )

0(1 )t L RI I e−= −

The current is 0 A at t = 0 s, as expected, and exponentially approaches 0 as .I t → ∞ (c)

Assess: The current is zero at the start and approaches the steady final value. The behavior is similar to the charging of a capacitor.

33.78. Visualize: Please refer to Figure P33.78. Solve: With the switch in position b, the current will decay following Equation 33.60,

/( / )0( ) t L RI t I e−=

33-30 Chapter 33

Using ( ) ( )/ ,RI t V t R= where ( )RV t is the voltage across the resistor, this may be recast as

/( / )0

( )( ) ( ) ln( )

t L R RR R

V t RV t V e tV L

− ⎡ ⎤= ⇒ = −⎢ ⎥

⎣ ⎦

Thus if we plot 0ln[ ( )/( ) ]R RV t V as a function of time t, the slope will give us the ratio R/L, from which we can find L. The plot is shown below and gives a slope of −33008 s−1, so

1 1100 3.0 mH

33008 s 33008 sRL − −

Ω= = =

33.79. Model: Assume the straight wire is very long so we can use the formula for the magnetic field produced by an “infinite” wire. Visualize: Please refer to Figure CP33.79. As the short wire moves in the field of the long wire, the charges in the short wire will experience a magnetic force. They will move inside the wire until the electric field created by the charge separation has a corresponding force that balances the magnetic force. The potential difference between the ends of the wire is due to this electric field. Solve: To find the potential difference between the ends of the short wire we need to know the electric field along it. The magnetic field of the long wire is into the page near the short wire, so the magnetic force on positive charges will be to the left. Positive charges will move to the left, leaving negative charges on the right. This creates an electric field and corresponding force to the right on the positive charges. Consider the charges at a distance x from the long wire. In equilibrium, the magnetic and electric forces balance. Using the magnetic field for an “infinite” wire,

IqvB qE E vB vx

= ⇒ = =

Because the electric field points to the right (positive x-direction), it takes the vector form

0 ˆ2v IE i

Integrating the electric field from the negative end of the wire (left end) to the positive end (right end), the potential difference is

0 0 0ˆ ˆ( ) ln2 2 2

d l d l

d l d l d

v I v I dx v I d lV E dl i idxx x d

μ μ μπ π π

+⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ = − ⋅ = − ⋅ = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠Ñ Ñ Ñ

where we have used ˆ .dl idx= Assess: This seems reasonable, because the potential difference increases with current (stronger magnetic field) and velocity (larger motional emf). In addition, for a moving wire of length 0,l = the potential difference is zero, as expected.

33.80. Model: Assume the RC circuit has wires with zero resistance. Assume that only the bottom horizontal wire in the RC circuit closest to the metal loop contributes to the magnetic flux through the loop. Assume that the wire appears infinite in length compared to the metal loop. Visualize: See Figure CP33.80. The current in the RC circuit decreases as the capacitor is discharged. There is a changing magnetic flux through the metal loop due to the changing current through the RC circuit, inducing an emf through the loop.

Solve: (a) Current /( )0( ) t RCI t I e−= flows from right to left through the wire at the bottom of the RC circuit,

causing a magnetic field out of the paper through the metal loop. Since the RC circuit current is decreasing, the induced current through the loop is counterclockwise in order to increase the magnetic flux. (b) Using the result of Example 33.5, the flux through the metal loop due to the wire is

7 /( )0m 0

( )(2.0 cm) 0.5 cm 1.0 cmln (4 39 10 )2 0.5 cm

t RCI t I eμπ

− −⎛ ⎞+Φ = = . ×⎜ ⎟⎝ ⎠

The initial current through the RC circuit 0 (20 V)/(2.0 ) 10 AI = Ω = The magnitude of the induced emf in the metal loop at 5.0 st μ= is

7 67 /( )0

(4.39 10 )(10 A)exp (5.0 10 s)/[(2.0 )(5.0 F)](4.39 10 ) 0.725 V(2.0 )(5.0 F)

t RCd I edt RC

− −− − × × ΩΦ ×= = = =Ω

The current through the loop

looploop

0.725 V 15 A0.050

= = =Ω

Assess: As the current through the RC circuit decreases to a constant value of zero at large times, the induced emf also approaches zero.

33.81. Model: Assume that the sides remain straight during the collapse. Also, assume the magnetic field is constant in the region of interest. Visualize: As the loop collapses the area will change, there will be a corresponding change in flux and, therefore, an induced emf and current. The loop is made up of four right triangles of equal size with width x, height y, and hypotenuse 10 cm.b = The corner of the loop is moving in the x direction.

Solve: (a) For simplicity we will focus on the upper-right triangle. The width of the upper-right triangle is given by

0 ,x x vt= + with 0 / 2 (0.10 m)/ 2 0.0707 mx b= = = and 0.293 m/s.v = The height y of the triangle is 2 2 .y b x= −

The loop will be a straight line when x − b. The time when this occurs is

0collapse

0.10 m 0.0707 m 0.10 s0.293 m/s

− −= = =

(b) The induced current is due to the changing flux as a result of the changing area. We have

1 d B dAIR R dt R dt

Φ= = =E

The total area at any time is four times the area of the upper-right triangle: 2 21

24 2A xy x b x= × = −

33-32 Chapter 33

Using the chain rule,

( ) 2 22 2

2(0.293 m/s)(0.50 T) (0.10 m) 2[0.0707 m (0.293 m/s) ]0.10 (0.10 m) [0.0707 m (0.293 m/s) ]

(0.10) 2[0.0707 (0.293) ]2.93(0.10) [0.0707 (0.293)

B dA dx B d Bv b xI v x b xR dx dt R dx R b x

⎛ ⎞−= = − = ⎜ ⎟⎜ ⎟−⎝ ⎠⎛ ⎞− +⎜ ⎟=⎜ ⎟Ω − +⎝ ⎠

− +=− + 2

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

where t is in seconds. (c)

t (s) I (A) 0.000 0.000 0.020 0.078 0.040 0.186 0.060 0.348 0.080 0.671 0.090 1.078 0.100 diverges

33.82. Model: Assume the field changes abruptly at the boundary and is constant. Visualize:

The loop moving through the field will have a changing flux. This will produce an induced emf and corresponding current. The induced current will oppose the change in flux and result in a retarding force on the loop.

Solve: (a) The loop moving into the field will have an increasing outward flux. The induced current will oppose this change and create a field that is into the page. This requires that the induced current flow clockwise. The current-carrying wires of the loop will experience a force. The force on the leading edge is to the left, retarding the motion. The forces on the top and bottom edges cancel and the trailing edge experiences no force since the field is zero there. The induced current is

1 d B dA Bl dx BlvIR R dt R dt R dt R

Φ= = = = =E

With this current we can find the force on the leading edge. Remembering that it opposes the motion, we have 2 2Blv B lF IlB lB v

R R⎛ ⎞⎛ ⎞= = = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Newton’s second law is 2 2 2 2

0btdv B l dv B lma m v v v v e

dt R dt mR−⎛ ⎞ ⎛ ⎞

= = ⇒ = ⇒ = ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

where 2 2/( )b l B mR= and the initial velocity is 0.v (b) The value of b is

2 21(0.1 m) (0.1 T) 100 s

(0.001 kg)(0.001 )b −= =

Assess: This exponential decay of the velocity is called eddy-current braking.

33.83. Model: Assume the field changes uniformly when turned on and that the loop is initially at rest. Visualize:

The changing magnetic field produces a changing flux that will generate an induced emf and current. The current-carrying wire will experience a force that opposes the change. Solve: (a) The increasing field strength produces an increasing outward flux. The induced current will oppose this change and create a field that is into the page. This requires that the induced current flow clockwise. The current-carrying

33-34 Chapter 33

wires of the loop will experience a force. Take A parallel to B so the flux is .A B ABΦ = ⋅ = Because the field strength is changing and the loop is moving, the emf has two terms. We have

1d dB dAA Bdt R dt dtΦ ⎛ ⎞= = +⎜ ⎟

⎝ ⎠E

We will assume that the loop does not move much and ignore the dA/dt term (see below). The current is 2 21 (0.080 m) (1.0 T) 32 A

2 2(0.010 )(0.010 s)dB l BI A

R R dt R tΔ= = = = =Δ Ω

(b) For a clockwise current, the force on the leading edge is to the left, so it is away from the magnetic field. The forces on the top and bottom edges cancel, and the trailing edge experiences no force since the field is zero there. From our assumption of a uniformly changing magnetic field, we know that the average magnetic field during the interval is [ ( ) (0)]/2 0.50 T.B B t B= − = The average force on the loop is

(32 A)(0.080 m)(0.50 T) 1.28 NF IlB= = =

To find the speed after being “kicked” we calculate the impulse. Using the average force found above gives

aveave f i f

(1.28 N)(0.010 s) 1.3 m/s0.010 kg

F tF t p mv mv vm

ΔΔ = Δ = − ⇒ = = =

Assess: This shows that the loop would be shot out of the field region.

33.84. Model: Assume the solenoid is an ideal solenoid, with 0 ,R = Ω so that the magnetic field at any instant in time is the same everywhere within the solenoid. Assume dI/dt is constant. Visualize: An electric field is induced by the changing magnetic field. Solve: Using Equation 33.26, the electric field strength and rate of change of the magnetic field in a solenoid are

related by ,2r dBE

dt= where r is the distance from the solenoid’s central axis at which E is measured. Thus

0 01( )

2 2 2r dB r d dIE nI nr

dt dt dtμ μ= = =

With 4 25.0 10 V/m at 0.50 cm, 1.6 10 A/s.dIE rdt

−= × = = ×

Assess: This current change is occurring in the solenoid coils. At first glance this seems like a rapid rate of change, but is really quite acceptable. For example, if a circuit carrying 1.0 A decreases to zero current in a millisecond (the time for a switch to disconnect) the rate of change of the current is 1000 A/s.

33.85. Model: Assume the conductor is long enough to use the formula for the magnetic field of an “infinite” wire. Visualize: Please refer to Figure CP33.85.The current in the inner conductor creates a field that circles it in the space between the inner and outer conductor. Magnetic flux exists in that space. Solve: (a) We will consider a rectangle between the conductors of length l that is perpendicular to the magnetic field produced by the inner wire. Take A parallel to B so the flux through a small piece of the rectangle of width dr is

.d B dA BldrΦ = ⋅ = The total flux is 2 2

0 0 0 2

I Il dr Il rd ldrr r r

μ μ μπ π π

⎛ ⎞⎛ ⎞Φ = Φ = = = ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

∫ Ñ Ñ

where we have used the formula for the field of an infinite wire. We want the inductance per unit length, so we take the inductance and divide by the length. We have

I Il rμπ

⎛ ⎞Φ Φ= ⇒ = = ⎜ ⎟⎝ ⎠

(b) The inductance per unit length is 7

7(4 10 T m/A) 3.0ln 3.6 10 H/m 0.36 H/m2 0.50

π μπ

−−× ⎛ ⎞= = × =⎜ ⎟

⎝ ⎠

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