Download Spring 2005 Physics Problem Set Solutions: Magnetic Forces and Torques and more Assignments Physics in PDF only on Docsity!
Physics 1112. Spring 2005
Phvsics LtLz
Problem $et #LO Solutions
Multiple-Choice Qirestions
PS #10 Sohrtions
' PaAe I of.?
Question 1:^ The^ figure^ shorvs^ an end-on view^ of^ three^ parallel wires^ that^ are perpendicular^ to
the pfane^ of.the^ paper,^ with^ tuo^ currents^ pointing into^ the^ page^ and one^ pointing^ out.^ The two
outermost w'ies^ are^ fixed^ in^ place.^ Which^ vuay^ will the^ middle^ wire^ move?
The easy way to answer this question^ is to^ remember^ that^ currents flo*wing in^ the^ same^ direction willi^ attract,^ while^ currents^ in^ oppo-^ <\ n^1 ,)> gite (^) directionc will tepel. Thrrs the middle wire.will feel a force (^) \ | ,/ lJotfightl, because^ it's^ both attracted by the^ right^ wire^ and^ re-^
t',|.,t
FelfeA-6yln6 left^ wire.^ -
61 Aj" G,l
To answer in more detail, first^ rise^ the^ "gtab-the-wire'l right-hand^
-il' X
v)/
mle for the two outer wires: the right^ wire's magnetic field^ circulates^ "
'. clockwise around it, while the left ,wire's^ -.grr"ti" fleld^ circulates^ ---"/ (^) "----
counterclockwise. Thus at the.position of the middle^ wire (our^ "test
wire"), the magnetic field due to^ these^ other^ wires^ points^ upward.
Nexb, use the magnetic force right-hand mle to^ find the direction of the^ force^ that this^ field
causes on^ the current-ca,rrying wire^ in^ the middle: rightward.
Question 2:^ A^ square^ loop and^ a^ rectangular^ loop^ are each made^ from^ the^ same.length^ of
wire. Each loop contains a single^ turn,^ and^ carries^ the^ same^ amount^ of current.^ The^ long^ sides
of the^ rectangle'are tlr,o.times as long as^ its^ short sides. When^ these loops^ are^ placed^ in^ the
same uniform magrietic field, what is^ the ratio of the^ maximum torque on^ the^ square^ lbop^ to^ the maximum torque on the rectangular loop?
The ratio of the torques is going to be equal to the ratio of the loop axeasJ Everything else
in the torque
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r :^ NIABsind is^ the^ same^ for^ both^ loops. Let's^ say^ that^ the^ short side of the rectangle ha^s length c, so that the long side^is^ 2r and^ the^ total^ perimeter^ of the rectangle is 6r. Since the^ loops have^ the^ sa,me^ perimeter^ (same^ length of wire),^ the^ square
has"sides 3n/2. Tben the ratio is
rse
A"q : (grl2)2^ : f9 l
Tred. A,"a 2r2 (^) t L
f'
t
Physics 1112 Spring 2005
PS #10 Sohrtions
Page 2 of.
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Question 3:^ Trrro^ straight^ current-carrying^ wires are positioned^ at^ right^ angles^ to^ edch^ other,
as shown belor. What is^ the direction^ of^ the^ force^ on wire^1 due^ to^ wire^ 2?^ ,
It (^) is probably ea"siest to analyze this problem^ by looking^ at^ it
from above,^ so^ that^ current^ 2 is^ coming^ out^ of^ the^ page^ and^ r
curent 1 lies along the^ page.^ The^ diagram^ at right^ shows^ t^ l
the magnetic^ field^ that^ wire^2 produces^ at^ a few^ points^ along
wire i. [Ising the magnetic forcg^ right-hand-rule,^ we conchrde
that the force on wire 1 point-s^ out of the page^ at^ the^ topmost
point, and into the page at the bottom point.^ These two forces
are of equal magnitude (since^ these^ two^ points^ are^ the^ same
distance and angle from wire 2) but opposite direction, so^ they^ D
cancel. At the middle point,^ there is no force because the^ angle^ L' between E and the current^ is^ 180". [rhat this shows is^ that^ there^ is^ tnoGtfofrl on^ wire^1 due
to wire 2! (However, there 'is^ a torque that wants to'make
wire I rotate so that^ it^ is^ parallel^ to^ wire^ 2.)
Numeric Questions
Question 1:^ A^ long straight wire^ carries^ a^ current^ of^ fi - 15.0 A. Next to the wire is^ a^ square loop^ with^ sides^ I^ .10^ m^ in length, as shown in the figure. The distance D^ is^ 2.10^ m.^ The loop carries a current of 12.* (^) 4.10 A in the direction iridicated.
(You can ignore gravity^ for this problem.) (a)^ What is^ the^ di-
rection of the net force ocerted on the loop^ by the^ wire?^ (b)
Calculate the magnitude^ of^ the net^ force^ acting on^ the^ loop.
(a) The loop is our (^) "test wire,' experiencing the magnetic fieid that^ is^ generated^ by^ the long straight wire. First, the current right-hand nrle^ tells^ u.s^ that^ the^ magnetic field
of the long wire points^ into^ the^ page everywhere belovr^ the^ wire.^ Then, the^ magnetic
force right-hand mle tells r$ that^ each^ section^ of^ the^ curent-carrying^ loop^ experiencen
a force directed outward from the center of the loop. The forces on^ the left^ a,nd^ right
segments of the loop will cancel^ each^ other^ out,^ since^ both^ segments^ carry^ the^ sa,me
current, and are the same tength and distance away frcfrn^ the^ long^ wire.^ However,^ the
forces on the top and bottom segtnents will not cancel. (That's only true in^ a^ uniform
E netC.) Beca,rlce the top segment is^ closer to the^ long^ wire, the field^ it^ experiencm^ is
gre,q!S$bq4 for the bottom segment. Thus, the direction of the net foice on^ the^ loop
:r---:--l rs (^) lllpwa,ro. I
-/r-=--* *o l@ tl ll D il'l lr| I y *T
Physics 1112 Spr'"ng 2005
The angle I is given by
For the^ mrmbers given,
PS #10 Solutions
' (^) Page 4of 7 I
l^-a- Bz^
(w2!z\ (^) / (w?L\
ranar: B, :^ \z,. d ) / \2"-T ) -Iz- f,'
, Btot (^) = 2 x ro-? ofrl/EfiflT{.G
: 3.76'x 10-5 T
ll
_[__6,
tano:i*3 (^) =
; =+^
d: (^) b6.8'.
Since d is the angle from the^ -z axis,^ the^ arlswer we^ should quote^ is^ 180"^ - Q^ - W]
(However, since iSolve aetuaily wanted the angle relative^ to the^ -z axis, f, is^ what^ should
have been^ entered^ in^ on^ the^ Web.)
Question 3:^ Two^ long,^ straight^ wires are^ oriented^ perpendicular^ to^ the^ page, as^ shoun.^ The currents in the wires are (^) h :^ 2.7 A pointingout of^ the^ page,^ and^ Iz^ :^ 3.6^ A'pointing^ into the page. These tno wires are situated at turo corners of a square with^ side^ length^ L^ :^16 cm. (a)
Calculate the^ strength^ of the net^ magnetic^ field^ at the^ point^ P.^ (b)^ Determine^ the^ direction^ of
the net magnetic field'ai the point P. Express your result as an angle measured relative^ to^ the
positive x-axis.
Using the current right-hand rule, we can determine the
direction of the magaetic flelds generated at point^ P by each wire'individually;^ these are^ shown^ in^ the^ diagram. Because we're looking at a sqlla,re, point P is a distance
L away from wire 1 and -LJl away^ from wire^ 2.^ So the
magnitudes of the fields a.re
and
Bzr: B2cos45":BzlJ2: Bzu: B2sin45"Bz
The total field components are
Now we^ need to a.dd these vectors^ together. E1^ has only an ,r component, but E2 points^ up and to the^ right at a 45" angle. So we need to^ find the^ r^ and^ g^ components^ of
.B2 (which^ turn out to be equal):
ur:*+ and
",:#h
x\g,
I1 I
Fo Iz 2n 2L
B, :^ Bn * Bz, = y^ (-++^ 3) :^ *1.125^ x^ 1o-6^ T
2r\ L'2L/
##,:2.25x Bs (^) = 4ro- ro-G^ r.
Physics 1112 Spring 2005
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So the magnitude of the net field^ is
B_
The direction with respect.to^ the^ -r axis^ is
PSr#10 Solutions ' Page 5 of
lFl I
tan6: !,aY',^ :2.0^ :==+ lb'l
(^6) = 63-4".^
\
This.means that the direction with respect to the^ *z axis^ is^ 180"^ *^ O (^) - FT71l (Well,^ 120"
to two signiflcant digits.)
Question 4:^ Consider a^ system consisting^ of^ two^ long, concentric^ solenoids; as^ shown^ belornr^ in cross section. The current in the'outer solenoid is (^) [ :^ 1.87 A, while th'e^ current in^ the^ inner sofenoid is (^) Iz- 2.56 A flowing in the opposite direction. The outer plenoid^ has^ 117 turns^ per
centimeier, and the inner solenoid has 138 turns per^ centimeter. (a)^ Find^ thd^ magnetic^ field
strength in^ the region^ betureen^ the two^ solenoids. Enter your^ answer as^ a positive^ number^ if^ the
field points^ up. (b)^ Find the magnetic field strength inside^ the^ inner^ solenoid.^ Enter your^ anslrver
as a positive number^ if^ the field^ points^ up.
(a) The magnetic field between the solenoids is due only to thd outer
solenoid (since we^ a^ssume^ for^ ideal^ solenoids^ that^ they^ create mag-
netic flelds only inside themselves and not orrtside.) The^ crtrrent
right-hand rule can tre u.sed to show that the field has to point^ up-
ward. Remembering that 117 turns per centimeter is 11700 turns'
per meter, the strength of the field is
81: panl:^ (4n x 10-7)(11700)(1.87 A) :2.74g^ x^ 10-2^ T.
So orr answer is 2.75 x 10-2 T.
(b) The inner solenoid creates a field that points^ down, with^ magnitudd
82: psn2l2:^ (4o^ x 10-?Xis800)(2.b6 A)^ = 4.489^ x^ 10-2^ T.
However, this is not the t'otal freId, inside the inner solenoid, because the outer solenoid's
field also makes a contribrrtion. Since the^ fields^ point^ in^ opposite^ dirrhtions, the^ net
field will be their differenlce, z.i+g x 10-2 T (^) - 4:439 x' 10-2^ T^ = (^) -1.69 x^ 10-2^ T.
B3+ B?:
Physics 1112 $pring 2005
.NIR^ I:B(o:Foo,lt
Now we can^ plug^ in^ the^ numbers:
PS #10 $olrrtions
Page 7 of (^7)
==+ r=w.
J= (^) = 1.25^ A.
Sine the rod^ is moving^ so as^ to^ increa^se^ the^ area^ of^ the^ loop, the^ magtetic flux^ will be increasing. The magnetic field created^ by the^ induced current^ will^ need^ to^ point
into the loop in order to counteract the^ increasingjgr_lg^ lh3,lgreans that^ the current
needs to^ florn' cloclrwise. Thrrs^ our finai^ answer^ is^ l.[ (^) = (^) -1.25 A.^ I
(b) (^) The electrical power dissipated is
P :^ 12R= (r.zb^ 4)2(4.00^ o)^ = 16.34 lvl
while the^ mechanical power^ is"
P :^ Fospu (^) = (4.80 N)(1.30 m/s)^ :^ {624 W.l
The two^ calculations^ yield^ identical^ answers, as^ enpected:^ the^ po$/er generated^ by
pulling this rod through a ma,gnetic fieid is dissipated as heat through the resistor.
(4.80 NX1.30 m/s)
