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Unformatted text preview: P202 Spring 2006
Ch 23: Magnetic Flux and Faraday’s Law Homework due at 11:59 pm Feb. 19, 2006
Error allowance on all numerical answers is 3%. 1. The magnetic ﬁeld produced by an MRI solenoid L=[2.0 — 2.5] In long and
D=[l .0 — 1.5] min diameter is B=[1.7 — 2.1] T. (a) What is the magnetic ﬂux through the solenoid? Wb
(b) What is the magnetic energy stored in the device? J
Solution: (a) Find the magnetic ﬁeld inside the solenoid using eqn (22—12) of Walker. Use
this ﬁeld to find the ﬂux through the solenoid using eqn (23—!) of Walker,
noting that the magnetic field of the solenoid is along the solenoid axis so cos 6:1' 5 33A LOSS l (1:359: l as: B we? (19) The magnetic energy density is given by eqn (23—20) of Walker. We already
calculated the magnetic ﬁeld within the solenoid in part (a) so all we need to
do is to plug this ﬁeld into eqn (23—20) to get the energy density. T he total
energy is given by the energy density times the volume of the solenoid which
we can easily ﬁnd from its diameter and length ac/VLL'raT alga/~33? u&: —L 5L x.) 2 Fa
‘l‘othﬁ «emu—c \ct‘a.3 Slﬁm_d 5 L46 xvoh" ""6
(J in
1.
‘3 rL ___J_. "T? ”‘2’ L__ d ﬁg <1.) :1 RLDLLXW:
.32 Solution: (a) = 0.7854B D2 (b) = 3.125 D2 L B2 105 2. A magnetic ﬁeld increases linearly from 0 to B=[0.1 — 0.2] T in t=[1.0 a 2.0] s.
How many turns of wire are needed in a circular coil D=[8.0 v 12.0] cm in
diameter to produce an induced EMF of v=[5.0 — 6.0] V. Assume the plane of the
coil is perpendicular to the ﬁeld. Solution: We are told that the magnetic ﬁeld changes linearly by a certain amount
(B) in a certain time (t). Thus we know the rate of change of B. Knowing this, we
can calculate the rate of change of magnetic ﬂux using eqn (231) of Walker. We
are told that the coil is perpendicular to the magnetic field so cos (9 = l in this
eqn. The area A in eqn (231) is just the area of the coil. Having calculated the
rate of change of ﬂux, we can plug this into eqn (23—4) of Walker to obtain an
expression for the EMF induced in the coil. The unknown in this can is the
number of turns for the coil, for which we can solve. 5; [3A 0,58 '1 c.5583; PF= A z  Til). 2
AC}: 35. (3 o) (2005 mama 2mg! Veréé r~N37T<2>1
At 3?: 200 5a M: vii than“
Tieso? Solution: =12732.0 (vi/BB2) . An airplane with a wing span ofL =[30.0 v 40.0] In ﬂies at a speed ofv:[600 —
850] km/h in a north westerly direction. _
(a) If the vertical component of the Earth’s magnetic ﬁeld is 5 x 10'6 T,
what is the induced EMF between the wing tips. InV
(b) If the plane turns to head due north does the induced EMF increase
[radio button], stay the same [radio button] or increase [radio buttou] Solution: (a) The trick here is to note that the horizontal component of the Earth’s magnetic
field is in the same plane at the velocity vector of the aircraft and the vector
across the wing span of the aircraft. Thus, this component of the Earth ’3 ﬁeld
produces no induced EMF between the wing tips (although it does produce
one between the top and bottom of the plane '5 fuselage — but that’s another
story). The EMF between the wing tips is induced by the vertical component of the Earth ’sfield which we are given. T 0 ﬁnd the EMF, we use can (235) of
Walker. E: one: E; yang to V‘VLB cow 3»; TL JE/khme‘ﬂ. 0:“?va 02 i ‘
rt; gmfbb énﬂci (b) D" the plane turns to ﬂy northwards nothing changes because the vertical
magnetic ﬁeld is still the same and the induced EMF is always perpendicular
to both the velocity of the conductor and the magnetic ﬁeld (in this case the
vertical component of the Earth 's ﬁeld) Solution: (a) 1.38889 10'3 Lv ; (b) stay the same . A circular coil of diameter D=[15 h 25] cm with N=[120 — 150] turns rotates with
an angular Speed of R=[1000 — 1500] rpm about a vertical axis that coincides with
a coil diameter. The only magnetic ﬁeld is the Earth’s which, at the location of the
coil, has a magnitude of 4.75 x 10'5 T and makes an angle of 36° with the
horizontal. What is the maximum EMF induced in the coil? mV Solution: Eqn (2311) of Walker tells us the EMF produced by a generator, which
is exactly what the problem is describing (ie a circular coil that is rotating about
its diameter in a magnetic field). We note that the question tells us that the axis of
rotation is vertical. Thus the vertical component of the Earth ’sﬁeld does not cut
the plane of the coil — only the horizontal component does. Thus the ﬁeld we use
in eqn (2311) is the horizontal component of the Earth ’s ﬁeld, or the total ﬁeld
(given) times cos (3 6 °). F urther, the question asks us for the maximum EMF
induced in the coil so we need to put sin at in eqn (23—11) to its maximum value
(ie. unity). T he diameter of the coil is given so we can calculate its area A. The
number of turns is also given. All we need to do is to convert the rotation rate from rpm to Hz and multiply it by 27rto get a).
5’; NBA w 34. L/J't :2; SM“: NgAm) A; john?) 33: Lt.?§%\6r @536" we; 2,—__R_ 200 _— Zoo EM“.— M qux.5"h536‘ 1(3) )2: 27W 8  i a”; NRQZ 3,l(oi><uo V 62’ 2km? : NKDZ 3.!(3! xlgﬁq m\/ Solution: 3.161 N R D2 10'7 5. Detennine the inductance ofa solenoid with N=[500 ~ 700] turns in a length of
X=[20 i 30] cm. The circular cross section of the solenoid has a radius of R=[4.0
~ 5.0] cm. mH Solution: T he formula for the inductance of a solenoid is given by Walker eqn (23—14). We simply plug in the numbers the question gives us, paying attention to
units. Z ..
L=/‘°(‘l>l.'m)17? R 25: ﬂ°wHL£ = SauthiogNlﬂl
H30 ‘00 X K L: 3.QU?8(N1R2/x) l5?) m H
Solution: 3.9478 (N2R2/X)10'5 6. A resistance of R:[5.0—6.0] Ohms and an inductor are connected in series with a 9
Volt battery and a switch. When the switch is closed, the current in the circuit
increases from 0 to 0.25 A in T=[0.10 — 0.20] ms. (at) What is the value of the inductance? mH
(b) What is the maximum current that flows in the circuit? A Solution: (a) Eqn (23—16) of Walker tells us how the current through an inductor changes
with time after the switch is closed. We are given the current (0.25 A) at some
time tand we also know the EMF of the battery and the resistance. So the only
unknown in the equation is the inductance L. We solve the equation by
isolating the exponential term on one side of the equation and taking natural logs.
I: %_ (i—e—ERIL‘)
~3
0‘25: 10“ éTX\OtQ/1_)
K .‘
. _ 2 —e "7*mhle
(025 5%) RTE:
E qOZC. \\ _ BTW—KID” (/4...
a E " L: —Txuo’3e/€m(t— Care/q) H L: ﬁne/earlo— aura/q) wit (b) The maximum current in the circuit is the current that ﬂows at a long time
after the switch is closed (i. e. after the current has had time to ramp up to its
ﬁnal value). After a long time, the current is no longer changing so the
inductor has no effect. Thus, the current follows directly from Ohms law
applied to a circuit with a resistor and a battery. Solution: (a) L : — TR/natural_log(1 0.02778R) ; (c) 9lR . A stepup transformer has N1=[20 4 30] turns on the primary coil and N2=[400~
600] turns on the secondary coil. If the transformer produces an output of 4800 V
with a 12 mA current, calculate (a) the input voltage V (b) the input current mA Solution: We use the transformer equations given by Walker eqn (2322) to
calculate the current and voltage on the input side, given the current and voltage on the output side as well as the number of turns on both the primary and
secondary windings Isl: ~ VI  3.
N1. VL It Liam _ N1 is. Vt. biog git . Ti: gill: N?—
V; MI in. d N! NI Solution: (a) 4800 N1/N2 (b) 12 N2/N1 . You want to store j=[8 — 10] J of energy in the magnetic field of a solenoid with
N=[400—500] circular turns of diameter d=[6.08.0] cm distributed uniformly
along its L:[25—35] cm length. (a) How much current is needed? A
(b) What is the energy density inside the solenoid? J/rn3
Solution: (a) The energy stored in an inductance (eg. a solenoid) is given by eqn (23—19) of
Walker in terms of the inductance of the solenoid and the current ﬂowing
through it. The inductance ofa solenoid is given by eqn (23—14) of Walker in
terms of quantities we are given in the question (i.e. the length, cross sectional
area and length of the solenoid). Plugging this value for the inductance into
eqn (23—19), we see that the unknown is the current ﬂowing. We solve for this. obit. T1: LBZAQ t/MMZI‘AQ c: MAJ)
2’ Z—jvla 2 0‘
I: two/lilac“ 3177i 7: _ Iznta‘ : ‘°‘
2 KL. l RZOD>1CJD — “L: X” 0‘73 “O
a 7. ﬂ .— .
I: l_____:1_ 1 i0 0.2024 .:> _\_ : Lhrg'gyxlon/La (A)
Nd (b) We are told what the total energy stored in the solenoid is to be. The energy
density is just the total energy divided by the volume of the solenoid. ‘  —— ~— 3
W51 castmtz.k : s/Ae = 4/(T<§iae)%> J/m
a : _Zi_‘_. leﬁ‘Bx [0
Lil Solution: (a) 1.424 sqrt(Lj) 104/(Nd) (b) 1.273 106 j/(Ldz) 9. A car with a vertical radio antenna L=[75 — 100] cm long drives due east at v=[20
— 30] mls. The Earth’s magnetic ﬁeld at this location has a magnitude 0f5.9 x 10'5
T and points northwards, 72° below horizontal.
(a) Is the top [radio button] or bottom [bottom] of the antenna at higher potential?
(b) Find the induced EMF between the ends of the antenna mV Solution: (a) We have a conductor (the antenna) pointing vertically upwards moving
perpendicular (ie eastwards) to the horizontal component of the Earth ’sﬁeld
(which points northwards). We know that movement in a magnetic ﬁeld
produces an electric field that is perpendicular to both v (velocity) and B, so
we know that the electric ﬁeld produced by the motion is along the antenna.
The question is m ”is the ﬁeld upwards or downwards? " One way to get the
answer is simply to remember that the force on a charge in a magnetic ﬁeld is
q v x B, which tells us that the force is upwards and hence that the induced
electric ﬁeld is also upwards. Another method is to draw a completed circuit,
such as ﬁgure 2313 of Walker, and apply Lenz ’3 law to ﬁnd the direction of
the induced motional EMF. Note that this ﬁgure has the magnetic field
coming out of the plane of the paper whereas the problem has the magnetic
ﬁeld going into the plane of the paper if the motion of the antenna is
eastwards (to the right). Thus the current in the antenna ﬂows up the antenna
(rather than down as in does in Fig 23—3) ifwe complete the circuit as done
in Fig 2313. Since the currentflows up the antenna, the bottom must be at a
higher potential. (b) The EMF is given by eqn 23—5 of Walker. We simply have to remember that the eﬂective magnetic ﬁeld is the horizontal component of the Earth ’5 field i.e.
the totalﬁeld (given) times cos 72°. _ q» (E: ﬁat = {.qmsltmh“ :_ u— = L223Luxm V \00 (El: 1.9123 LU’X to”+ W‘V Solution: (a) bottom (b) 1.823 L.v. 10'4 10. A magnetic ﬁeld with the time dependence shown in the ﬁgure below is at right
angles to a N=[150 — 200] turn circular coil with a diameter of d=[3.5 w 5.0] cm.
What is the induced EMF in the coil at (a) t z 2.50 ms mV
(b) t = 7.50 ms mV
(0) t = 25.0 ms mV Solution: The induced EMF in a coil is given by eqn 233 of Walker. Theﬂux in
this eqn is deﬁned by eqn 23—]. The area ofthe coil A can befoundfrorn its
diameter, which the question tells us. Note that the question tells us that the
magnetic field is perpendicular to the coil so cos (9 = l in eqn 23—] . The ﬁgure
above tells us that the rate of change of B is zero at i = 2.5 ms (the curve is
horizontal), so the induced EMF is also zero. The rate of change of B at t = 7.5
ms is —(0.02 — (—0.01))/5 171115. The rate ofchange att = 25 ms is + (0.02 — (—
0.0]))/5 T /ms. Note the diﬂerence in the sign of the rate of change in parts ([9) and
(c). Note also the minus Sign in the deﬁnition eqn 233. The upshot is that the
answer to (b) is a positive number while that to (c) is a negative number. Solution: (a) 0; (b) 0.4712 Ndz; (c) — 0.157 Nd2 ...
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 Spring '09
 WARREN
 Physics

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