Chapter 32 - Inductance

P32.1 ( ) 0 0.500 A 1 V s2.00 H 100 V0.010 0 s 1 H A

ILt

ε ⎛ ⎞Δ − ⋅⎛ ⎞= − = − =⎜ ⎟⎜ ⎟Δ ⋅⎝ ⎠⎝ ⎠

P32.2 Treating the telephone cord as a solenoid, we have:

( )( ) ( )227 32

04 10 T m A 70.0 6.50 10 m

1.36 H0.600 m

N ALπ πμ

μ− −× ⋅ ×

= = =l

*P32.3 ( ) ( )( )( )3max maxsin cos 10.0 10 120 5.00 cosdI dL L I t L I t

dt dttω ω ω πε −= − = − = − = − × ω

)

( ) ( ) ( ) (6.00 cos 120 18.8 V cos 377t tπ πε = − = −

P32.4 From ILt

ε Δ⎛ ⎞= ⎜ ⎟Δ⎝ ⎠, we have

( )3

324.0 10 V 2.40 10 H10.0 A s

LI tε −

−×= = = ×

Δ Δ

From BNLIΦ

= , we have ( )( )3

22.40 10 H 4.00 A19.2 T m

500BLIN

μ−×

Φ = = = ⋅

P32.8 2

0 0

2 2B NI N AN NBA NAL

I I I R Rμ μπ π

Φ= = ≈ ⋅ =

FIG. P32.8

P32.9 0kt dIe L

dtε ε −= = −

0 ktdI e dtLε −= −

If we require as , the solution is 0I → t →∞ 0 kt dqI e

kL dtε −= =

0 02

0

ktQ Idt e dtkL k Lε ε∞

−= = = −∫ ∫ 02Q

k Lε

=

P32.10 Taking LR

τ = , tiI I e τ−= : 1t

idI I edt

τ

τ− ⎛ ⎞= −⎜ ⎟

⎝ ⎠

0dIIR Ldt

+ = will be true if ( ) 1 0t ti iI R e L I eτ τ

τ− − ⎛ ⎞+ − =⎜ ⎟

⎝ ⎠

Because LR

τ = , we have agreement with 0 0= .

P32.11 (a) At time t, ( )( )1 te

I tR

τε −−=

where 0.200 sLR

τ = =

After a long time, ( )

max

1 eI

R Rε ε−∞−

= =

At ( ) max0.500I t I= ( )( )0.200 s1

0.500te

R Rεε −−

=

1

0.5

00 0.2 0.4 0.6

t (s)

I (A)Imax

FIG. P32.11 so 0.200 s0.500 1 te−= − Isolating the constants on the right, ( ) ( )0.200 sln ln 0.500te− =

and solving for t, 0.6930.200 s

t− = −

or 0.139 st =

(b) Similarly, to reach 90% of , maxI 0.900 1 te τ−= − and ( )ln 1 0.900t τ= − −

Thus, ( ) ( )0.200 s ln 0.100 0.461 st = − =

P32.18 Name the currents as shown. By Kirchhoff’s laws:

(1) 1 2I I I= + 3

(2) 1 210.0 V 4.00 4.00 0I I+ − − =

( ) 31 310.0 V 4.00 8.00 1.00 0dII I

dt+ − − − = (3)

From (1) and (2), 1 1 310.0 4.00 4.00 4.00 0I I I+ − − + =

and 1 30.500 1.25 AI I= +

FIG. P32.18

Then (3) becomes ( ) ( ) 33 310.0 V 4.00 0.500 1.25 A 8.00 1.00 0dII I

dt− + − − =

( ) ( )331.00 H 10.0 5.00 VdI I

dt⎛ ⎞ + Ω =⎜ ⎟⎝ ⎠

We solve the differential equation using equations from the chapter text:

( ) ( ) ( )

( )

10.0 1.00 H 10 s3

10 s1 3

5.00 V 1 0.500 A 110.0

1.25 0.500 1.50 A 0.250 A

t t

t

I t e e

I I e

− Ω −

−

⎛ ⎞ ⎡ ⎤ ⎡ ⎤= − = −⎜ ⎟ ⎣ ⎦⎣ ⎦Ω⎝ ⎠

= + = −

FIG. P32.21

P32.21 0.140 28.6 ms4.90

LR

τ = = =

max6.00 V 1.22 A4.90

IRε

= = =Ω

(a) (max 1 tI I e )τ−= − so ( )0.220 1.22 1 te τ−= −

0.820te τ− = : ( )ln 0.820 5.66 mst τ= − =

(b) ( ) ( )( )10.0 0.028 6 350max 1 1.22 A 1 1.22 AI I e e− −= − = − =

(c) maxtI I e τ−= and 0.160 1.22 te τ−=

so ( )ln 0.131 58.1 mst τ= − =

P32.23 ( ) ( )22 2

2

0 0

68.0 0.600 108.21 H

0.080 0N AL

πμ μ μ

−⎡ ⎤×⎣ ⎦= = =l

( )( )22 61 1 8.21 10 H 0.770 A 2.44 J2 2

U LI μ−= = × =

P32.24 (a) The magnetic energy density is given by

( )( )

226 3

60

4.50 T8.06 10 J m

2 2 1.26 10 T m ABuμ −

= = = ×× ⋅

(b) The magnetic energy stored in the field equals u times the volume of the

solenoid (the volume in which B is non-zero).

( ) ( ) ( )26 38.06 10 J m 0.260 m 0.0310 m 6.32 kJU uV π⎡ ⎤= = × =⎣ ⎦

P32.28 From the equation derived in the text, ( )1 Rt LI eRε −= −

(a) The maximum current, after a long time t , is 2.00 AIRε

= =

At that time, the inductor is fully energized and

( ) ( )( )2.00 A 10.0 V 20.0 WI V= Δ = =P .

(b) ( ) ( )22lost 2.00 A 5.00 20.0 WI R= = Ω =P

(c) ( )inductor drop 0I V= Δ =P

(d) ( )( )22 10.0 H 2.00 A20.0 J

2 2LIU = = =

P32.30 ( )1 max sintI t I e tα ω−= with max 5.00 AI = , 10.025 0 sα −= , and 377 rad sω =

( )1max sin costdI I e t t

dtα α ω ω ω−= − +

At 0. ,

800 st =

( ) ( ) ( )( ) ( )( )0.020 01 5.00 A s 0.025 0 sin 0.800 377 377cos 0.800 377dI edt

− ⎡ ⎤= − +⎣ ⎦

31 1.85 10 A sdIdt

= ×

Thus, 12

dIMdt

ε = − : 23

1

3.20 V 1.73 mH1.85 10 A s

MdI dtε− +

= = =×

P32.37 At different times, ( ) ( )max maxC LU U= so ( )⎡ ⎤ ⎛ ⎞Δ = ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠2 2

max max

1 12 2

C V LI

( ) ( )6

max 3max

1.00 10 F 40.0 V 0.400 A10.0 10 H

CI VL

−

−

×= Δ = =

×

P32.40 (a) ( )( )6

1 1 135 Hz2 2 0.082 0 H 17.0 10 F

fLCπ π −

= = =×

(b) ( ) ( )max cos 180 C cos 847 0.00100 119 CQ Q tω μ μ= = × =

(c) ( )( ) ( )max sin 847 180 sin 0.847 114 mAdQI Q tdt

ω ω= = − = − = −

P32.41 This radio is a radiotelephone on a ship, according to frequency assignments made by

international treaties, laws, and decisions of the National Telecommunications and Information Administration.

The resonance frequency is 01

2f

LCπ=

Thus,

( ) ( ) ( )2 26 60

1 1 608 pF2 2 6.30 10 Hz 1.05 10 H

Cf Lπ π −

= = =⎡ ⎤× ×⎣ ⎦

P32.44 (a)

( )( ) ( )

22

43 6 3

1 1 7.60 1.58 10 rad s2 2.20 10 1.80 10 2 2.20 10dR

LC Lω

− − −

⎛ ⎞⎛ ⎞ ⎜ ⎟= − = − = ×⎜ ⎟ ⎜ ⎟× × ×⎝ ⎠ ⎝ ⎠

Therefore, 2.51 kHz2

ddf

ωπ

= =

(b) 4 69.9 cLR

C= = Ω

P32.45 (a) ( )( )0 6

1 1 4.47 krad s0.500 0.100 10LC

ω−

= = =×

(b) 21 4.36 krad s

2dR

LC Lω ⎛ ⎞= − =⎜ ⎟

⎝ ⎠

(c) 0

2.53% lowerωωΔ

=

P32.49 (a) ( ) ( )20.01.00 mH 20.0 mVL

d tdILdt dt

ε = − = − = −

(b) ( ) 2

0 0

20.0 10.0t t

Q Idt t dt t= = =∫ ∫

( )2

2 26

10.0 10.0 MV s1.00 10 FC

Q tV tC −

− −Δ = = = −

×

(c) When 2

212 2Q LIC≥ , or

( )( ) ( )( )

2223

6

10.0 1 1.00 10 20.022 1.00 10

tt−

−

−≥ ×

×,

then . The earliest time this is true is at (4100 400 10t −≥ × )9 2t94.00 10 s 63.2 st μ−= × = .

P32.59 Left-hand loop: ( )2 1 2 2 0I I R I Rε − + − =

Outside loop: ( )2 1 0dII I R Ldt

ε − + − =

Eliminating gives 2I 0dIIR Ldt

ε ′− − =′

This is of the same form as the differential equation 32.6 in the chapter text for a simple RL circuit, so its solution is of the same form as the equation 32.7 for the current in the circuit:

FIG. P32.59

( ) ( )1 R t LI t eRε ′−= −′′

But 1 2

1 2

R RRR R

′ =+

and 2

1 2

RR R

εε =+

′ , so ( )( )

2 1 2

1 2 1 2 1

R R RR R R R R R

εε ε+= =

′ +′

Thus ( ) ( )1

1 R t LI t eRε ′−= −

P32.61 (a) While steady-state conditions exist, a 9.00 mA flows clockwise around the right

loop of the circuit. Immediately after the switch is opened, a 9.00 mA current will flow around the outer loop of the circuit. Applying Kirchhoff’s loop rule to this loop gives:

( ) ( )3 3

0

0

2.00 6.00 10 9.00 10 A 0

72.0 V with end at the higher potentialb

εε

−⎡ ⎤+ − + × Ω × =⎣ ⎦

+ =

(b)

FIG. P32.61(b)

(c) After the switch is opened, the current around the outer loop decays as

Rt LiI I e−= with max 9.00 mAI = , 8.00 kR = Ω , and 0.400 HL = .

Thus, when the current has reached a value 2.00 mAI = , the elapsed time is:

53

0.400 H 9.00ln ln 7.52 10 s 75.2 s8.00 10 2.00

iILtR I

μ−⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = × =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ × Ω⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

P32.63 When the switch is closed, as shown in

Figure (a), the current in the inductor is I :

12.0 7.50 10.0 0 0.267 AI I− − = → =

When the switch is opened, the initial current in the inductor remains at 0.267 A.

: IR V= Δ

( )0.267 A 80.0 VR ≤

300 R ≤ Ω

FIG. P32.63