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Unformatted text preview: Chapter 32 Solutions 1. Two coils have a mutual inductance of 2.0 H. If
current in the ﬁrst coil is changing at the rate of
60 A/s, what is the emf in the second coil? Solution Iiiom Equation 322, 82 = —M(dI;/dt) = —(2 H)x
(60 A/s) = —120 V. (The minus sign, Lenz‘s law.
signiﬁes that an induced emf opposes the process
which creates it.) 3. The current in one coil is given by I = I? sin 21rft,
where I], = 75 mA. f = 60 Hz, and t = time. Find
the peak emf in a second coil if the mutual
inductance between the coils is 440 mil. Solution Suppose 11 = 1? sin 27rft in Equation 322. Then 82 = —M dill/d: = —2'irfll{il'p cos 2rrft. and the peak
emf (when cos2nft = :l:1) is 21rfMIp = (2nx 60 Hz)(440 mH)(75 mA) = 12.4 V. 4. Two coils have a mutual inductance of 580 mH.
One coil is supplied with a current given by I = 3t2 — 2t + 4, where I is in amperes and t in seconds. What is the induced emf in the other coil at time
t. = 2.5 5? Solution Since dfl/dt = 6t — 2 (in A/s). Equation 322 gives, for
t : 2.5 s. 82 = —(580 mH}(6x2.5  2)(A/s) = —7.54 V
(see comment in solution to Problem 1). 9. A rectangular loop of length f and width at is
located a distance a from a long, straight, wire, as
shown in Fig. 3220. What is the mutual
inductance of this arrangement? t 0
—I—_ FIGURE 3220 Problem 9. Solution When current I1 flows to the left in the wire, the ﬂux
through the loop is (932 = (null/2n) f2+w€ drfr :
{pullf/21r)ln(1 + w/o) (see Example 312). Then
Equation 321 gives M = (bag/fl = (nag/2n) x 111(1 + w/a). (In calculating the ﬂux‘ the normal to the
loop area was taken into the page. so the positive sense
of circulation around the loop is CW. This determines
the direction of the induced emf 52 in Equation 322.) 11. What is the selfinductance of a solenoid 50 cm
long and 4.0 cm in diameter that contains
1,000 turns of wire? Solution Equation 324 gives L = ,ugNgA/f : (4nx10'7 H/m)x
(103)2rr(2 cm)2/(50 cm) 2 3.16 mil. (The long thin
solenoid approximation is valid here.) 13. A 2.0—A current is flowing in a 20H inductor. A
switch is opened. interrupting the current in
1.0 ms. What emf is induced in the inductor? Solution Assume that the current changes uniformly from 2 A
to zero in 1 [11.5 (or consider average values). Then
dI/dt = —2 Aims. and Equation 325 gives 8 = —(20 H)(—20 A/ms) = 40 kV. (The emf opposes the
decreasing current.) 22. A coaxial cable consists of an inner conductor of
radius a and outer conductor of radius b, as shown
in Fig. 3221. Current flows along one conductor
and back along the other. Show that the inductance per unit length of the cable is
$3 ln(b/u). Solution 'The magnetic ﬁeld strength between the conductors (a 5 r 5 b) is like that from a long straight wire. _B = pal/2n.“ {see Section 304). If We divide a
rectangular area of length E betWeen the conductors
into strips of width air. then the ﬂux linking this length of cable is $5 = f: Bf dr : (puff/2n) f: dr/r =
(puff/2r) ln(b‘i’u). Therefore, Equation 323 gives U5 = (#of'31rllnlb/al FIGURE 3221 Problem 22 Solution. Chapter 32 Solutions continued 25. The current in a series RL circuit rises to 20% of its ﬁnal value in 3.] us. If L = 1.8 mH. what is the
resistance R? Solution The buildup of current in an RL circuit with a battery
is given by Equation 328. In) = 100(1 — arm/L),
where 1m = fig/R is the ﬁnal current. Solving for R,
one ﬁnds R = ~(L/t)ln(1~ I/Im) = —(1.8 mH:— 3.1 ps)ln(1— 20%) = 130 Q. 27. A 10H inductor is wound of wire with resistance
2.0 Q. If the inductor is connected across an ideal 12V battery, how long will it take the current to
reach 95% of its ﬁnal value? Solution Reference to the solution to Problem 25 shonrs that
t: —(L/R) ln{1 — 1/130) = —(10 H/2 Q)1n(1 — 0.95} =
15.0 s. (The percentage I/Ix is independent of £0.) 39. What is the current in a 10an inductor when the
stored energy is 50 ,uJ? Solution
From Equation 3210. I = ‘/2L'/ =
«2(50 inn/10 mH = 0.1 A. 49. The Alcator fusion experiment at MIT has a 50T
magnetic ﬁeld. What is the magnetic energy
density in Alcator? Solution From Equation 3211. us = (50 T)2/(8nx
1n7 N/Az) = 995 MJ/ms. (This is about 2.8% of the
energy density content of gasoline; see Appendix C.) ...
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 Summer '08
 schuller

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