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Unformatted text preview: 31.2.1 Moving rod 8; Lenz law I D A metal rod AD is sliding to the right, along two N metal
railings, with a constant speed v. ADzL. The resistance is
R. Magnetic field B is constant. It points into the paper.
Determine the direction of BLnail (induced magnetic field).  1 2 3  4 
in ‘n'ght‘ out ‘ left ‘ direction of BM Hint: Lenz law says that Bind, the induced magnetic field
vector, opposes the change of the flux defined by the loop. Extra: Show that duMdt: BVL. Explanation: As the rod is sliding to the right, the
ingoing—flux, defined by the expanding loop ABCDA,
increases. To prevent the increase of this ingoing—flux, BM
must point outward. Extra: The growing ingoing—ﬂux defined by the loop is
given by ¢:BA:BxL. So the rate of change of the flux
dwdt= d(BxL)fdt 2 EV]... 31.2.3 Moving rod & power A metal rod AD is sliding to the right along the ] railings.
The resistance of the loop is R. B is into the paper.
Direction of induced current i is shown. Denote Fan to be
external force which keeps AD movin with constant v. direction of Fm _) [r <— J, Hint: Let the magnetic force on AD be FB=iLXB. To keep v constant, we must have Fm+ FB=0.
Extra: Show that the mechanical power applied to the rod is given by Pmech= Fm°v=izR. (Recall: smd=BvL)
Explanation: FBziLBsineziLBzFem. Direction of FB is: FB Direction of FExt is opposite to F3. Thus it is to the right.
Extra: Pmech = FmvziLBwiemd = 'ZR. This shows that
the mechanical energy, which is transferred to the
electrical energy, equals to the dissipative energy. 31.2.2 Moving rod & induced current I D A metal rod AD is sliding to the right along the H railings.
The resistance is R. The magnetic field B is into the paper.
L=AD. Determine the direction of the induced current. 3 direction of imd clockwise 0 counterclockwise Hint: Lenz law— Bind tends to maintain the original flux. dd 0’ , 5'
Extra: Based on slimf—E=—:II(M, [Ami (20’. Show that im: BVLIR.
Explanation: As the loop expands, the flux pointing into the paper increases. So Bind is pointing “ou ”. The light
hand rule #3 implies that it is counterclockwise. Extra: From the expressions given above, we get the
magnitude of the induced emf lewd: BVL. By the Ohm’s law, imam, IR 2 BvaR. 31.2.4 Rotating Bar cutting B—lines '\ Consider a metal bar 0A which is rotating in the plane L
to the ma tic field B. Which end has a hiher otential? Hint: The magnetic force due to B asserted on q, a
positive charge on the rod, is given by F=qva. This
force “pushes” q from a low potential point to a higher
potential point. Extra: Show IVDVﬁl, the magnitude of the potential
difference is (0BR212, where RzOA and 0) is the angular
frequency of rotation. m: When q is pushed by a
displacement Ar, the corresponding potential difference is given by AV: FAr/q = quArfqszrAr. Explanation: By inspection on the direction of the
vector—cross—product, the magnetic force on q, i.e. on a
positive charge on the rod, should be directed radially
inward. So a positive charge is being “pushed” by the
magnetic field from A to O. This implies that VD>VA.
Extra: From the hint, emf! dv=IwBrd1=wBR2/2, where the range of integration was taken from r:O to r:R. 31.2.5 A falling rod R
D C A metal rod AB is falling downward along two  vertical
metal railings. ABzL. The resistance is R. Magnetic field
B is constant. It points out of the paper. Determine the
direction of i11m (induced current). direction of iind Hint: Lenz law says that Bind, the induced magnetic field
vector, opposes the change of the flux defined by the loop.
Extra: If the rod weighs mg, how can one express its
terminal velocity in terms of R, m, g, B and L? Explanation: As the rod is falling, the outgoing—ﬂux,
defined by the expanding loop ABCDA, increases. To
oppose the increase of this flux, Bind must point inward.
The right hand rule implies that iiml is from D to C. Extra: When the terminal velocity is reached, the falling
speed is constant and the net force on the rod is zero. At
this point the magnetic force is balanced by the weight, so
imdLB=mg, where imd=BvL/R. These two expressions
enable one to solve for v. 31.4.1 Varying ﬂux & Induced emf /  \ [/"ﬂx ;‘\\
/ x x x l’x x x x \ XXXXX . 4‘ xxxxx/ \x x/ \z x x/ \‘_—o’ B1att1 Bzattz Given: 1=1m. AI; t1: 0 sec, B1=1T. At t2: 2 sec, B2=2T.
Find: Induced emf and, involts. 1 2  3 4
and 11: 11/2 11: 11/ 2
(in volts)
(ii: of clock clock counter counter
emd wise wise clockwise clockwise BzA—BIA Hint: g. dzlﬂz .Use Lenz law
m d! Extra: Show the induced electric ﬁeld Ema: (1/4)V/n1. Explanation: Based on the formula given in the hint, the
magnitude of induced emf 8de= (21)TE/(20) =Tt/2volts.
Direction: Bind opposes the increase of ﬂux Within the
circular loop. So Bind is out. RHR #3 implies that and is
counterclockwise. Extra: and = 27:1“ Emd = all or Ema =(l/4r)=(1/4)volts/m.
Verify the units: volts/m = N/C. 31.3.1 Levitation of a magnet Consider levitation of magnet by a superconducting
surface. The north pole is pointing downward. The
direction of induced current as viewed from above is: 1 2 3
clockwise 0 counterclockwise Hint: The loop of induced current should correspond to
an equivalent induced dipole which repels the magnet.
Extra: 1f the polarity of susupending magnet is reversed,
what is the direction of induced current? Will the magnet
fall? Explanation: The flux at the superconducting surface
due to the ﬂoating magnet is pointing downward. The
induced currrent loop prevents the penetration of the
downward flux, so the induced dipole points upward, or
the current is counterclockwise. Extra: When the polarity of suspending magnet is
reversed, Lenz law now implies that the induced current
should also reverse its direction. Again no penetration of
the magnetic flux lines is possible, or the magnet remains
being levitated. 31.4.3 Thee light bulbs with a pair in parallel A solenoid is producing the same steadily increasing
magnetic flux through two circular circuits shown above.
Case A: Two identical bulbs 1 and 2 are in series. Each
has a resistance R. Their brightnesses (or their electric
powers) are the same, i.e. P1=P2. Case B: Bulbs 2 and 3 are in parallel, and the bulbs 2 and
3 is in series with the bulb 1. Each bulb has a resistance
R. We label the respective electric powers of the bulbs by
Pl’, P2’ and P3’. Compare the power of bulb #1 for two
cases. _ Extra: Show Pz’f P2249. Explanation: For case A, denote the current i, the loop
equation is: s2iR=0, or i=E/(2R). For case B, label
currents through the bulbs by il’, i2’ and i3’ respectively.
By symmetry i2’2i3’. Since il’:iz’+i3’, then il’:2i2’. This
leads to the loop equation: Eil’RiZ’R : E(3f2)i1’R:0.
Solving for il’ gives: il’=(2/3)£/‘R =(2/3)(2i) =(4/3)i. Here
il’>i, so P1’> P1. Ans=3. Extra: Notice that i2: i: s/(2R) and i2’=i1’/2=213R. I {TiltM=[(%R>/(%R)12=3 31.4.2 Light bulbs in series Case A Case B A solenoid is producing a same steadily increasing
magnetic flux through two circular circuits shown above.
Case A: Two identical bulbs 1 and 2 are in series. Each
has a resistance R. Their brightness (or their electric
power) is the same, i.e. Plsz. Case B: Three bulbs are in series, each with a resistance
R. #3 is close to #2. Electric powers are labelled by P1 ’,
Pf’ and P3’. Compare the powers of bulbs #1 and #2. Extra: Show PZ’: P224/9. Explanation: For case A, denote the current i. The loop
equation gives £2iR=0, or i=E/(2R). Correspondineg for
case B, £=3i’R, or i’=E/(3R)=(2/3)i. Since i’<i, P1’< P1.
Bulb 1 and bulb 2 are in series, so Pl’z Pz’. Ans=3. . i.2
Em%=ﬁ=<r2wrag/(amig 31 .4.4 Two light bulbs and a short wire A solenoid is producing the same steadily increasing
magnetic flux through two circular circuits shown above.
Case A: Two identical bulbs 1 and 2 are in series. Each
has a resistance R. Their brightnesses (or their electric
powers) are the same, i.e. P1=P2. Case B: There is a short CD which is to the right of the
center line We label the respective electric powers of the
bulbs by Pl’, and PZ’. Compare the power of bulb #1 for
the two cases. Extra: If CD coincides with the center line compare P1’
with PI for this new situation. Explanation: For case A, denote i to be the loop
current, the loop equation is: £—2iR=0, or i=8/(2R). For
case B, label currents through the bulbs by i1’ and iz’, and
through the short by i’. The loop equations for C1 DC and C2DC are: E,’— il’R=0 and 82* iz’R=0 respectively. The
emf is proportional to area of the enclosed ﬂux. By
inspection E,’> sl2> 82’. Thus i1’>i> iz’, or P1’> P1. Extra: For the case where CD is the center line, the
identity of two areas of the enclosed flux implies that 81’: 52’. In other words, i1’= iz’:[ £I2]/‘R:i. So Pl’: P1. 31.4.5 Current along the short Case A Case B A solenoid is producing the same steadily increasing
magnetic ﬂux through two circular circuits shown above.
Case A: Two identical bulbs 1 and 2 are in series. Each has
a resistance R. The area of magnetic region is A. The induced emf is 8=AdB/dt. The loop equation s2iUR=0,
implies that the induced current is ngS/(QR). Case B: There is a short CD. Given the area of the
magnetic region to the left of CD be AL=2A/3 and to the right of CD be AR=A/3. The direction of the current ﬂow
along the short is (choose one):
1 2
frothoC fromCtoD Extra: Show the magnitude of the current ﬂow along the
Shell is Explanation: For case B, by inspection current through bulb #1 is larger than the current through #2, i.e. il’ >i2’ . At C, steady ﬂow requires the current to go from D to C.
Extra: For case B, the leﬁ loop (CIDC) has the induced emf 8L=(2A/3)dB/dt =28/3, giving the current
i1’=(28/3)/R =4ig/3. The right loop (i.e. D2CD) has the
induced emf 8R=81 3, giving i2’=(8./ 3)/R=2i0/ 3. The node equation at C reads i1’= i2’+ iDC, or iDc= i1’ if: Zia/3. 31 .6.1 Damped Pendulum Consider the setup of a daiﬁpEdpendulumThe magnetic
field B is into the paper. Find the direction of the force due to B asserted on metal plate, as it leaves the region. 1 2
Dir of PB / / Extra: Find the direction of F3 as it reenter the region.
Explanation: As the plate is leaving the rectangular region, Bind opposes the flux change in the plate. Bind
has the same direction as that of B, iind is clockwise,
FﬁziimAL XB, which leads to the pulling back force. XXXXX
XXXXXX /'swing up Extra: As the plate is swinging back, to maintain the
noﬂux status, Bind should be opposite to B, i.e. Bind is
out of the paper. Thus the direction of magnetic force
vector asserted on the plate is also reversed. ...
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This note was uploaded on 02/04/2012 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics

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