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Unformatted text preview: 29.2.2 Direction of magnetic force ﬁfe 1% "External magnetic field" BEXt is along i.
Current I flows outward, along k . N .
i—‘ 1 Find direction of magnetic force based on intuitive
i 69 i . reasonin.
l
S k 29 .2 . 1 Magnetic force 1 2 3 4
Ill Givem B is downward H), Hint: Denote field due to I by BI . Current I is into the screen (k). ° 81 is C0111m?I‘CIOCkW‘iseIét
 Compare fields: B=Be + B1, at P1 and P2. Determine the direction of force. Choose one:
  1
Dir of ma . force ' Hint: Based on M =IA1xB  Force on the wire is pushing from the stronger field side to the weaker field side.
Explanation: Here we want to determine the direction of cross—product AF =1A1x3. From the directions given, we Explanation:
have (—k) X (—j) : —i. The natural tendency is to straighten up the wrinkle. So
the magnetic force is up, or in the direction of I X BeXt, i.e.kxi:i.Ans:3. 29.2.3 Force on current segments 2931 Magnetic diPOIe in B Consider a magnet placed in a uniform magnetic ﬁeld B.
Determine net force F on the magnet. Choose one:
Setup: A current ﬂows ﬁ'om Ato C and then to D. 1 4 3
AC= a, CD= b. B is constant, H to positive xaxis. F 2QMB QMB 0
Determine resultant force vector F, asserted by B on the Hint; Treat magnet as two magnetic charges (M: i QM,
segments separated by a distance d. On each m, F=qMB.
4 Extra: Recall torque: t=force><1ever_arm. Use a sketch to
i(a+b)B deﬁne the 1ever_arm, d i, which pivots about — QM. Show
In that the net torque is: diQMB.
Hint: Use A13”: mix]? Explanation: Forces on +QM & —QM are equal in magnitude and opposite in sign. Fnet=F+ + F = 0. Ans=3. Explanation: Since the current 1n the AC segment 1s Extra: Pivotﬁng about _ QM, the level—31m dL may be parallel to B, the magnetic force due to the segment AC is, F=iaB sin 0. The current in the CD segment is constructed as fOHOWSI
perpendicular to the magnetic force, so F=istin90°=ibB. + M righthandrule of crossproduct, direction of F is d F+=QMB
‘1n, see the sketch. d i . F_= — QMB ‘ Direction 0sz ix B . ........... _ — QM
g Applying the deﬁnition of 1:, 1:: QMB di.
B Digression: I=u><B, so I: diQMB=(dQM) iB=ui B 29.3.2 Magnet in nonuniform ﬁeld #2
I: #1’
X Consider force which dipole ﬁeld asserts on a small magnet. Direction ofxcomponent of the force on #1 is: 2 3
<——> Hint: F=QMBNQMBS =1?N + F5
Extra: Find direction of Xcomponent of the force on #2. Explanation: For #1, due to symmetry, we expect Fx=0. FS ‘ For case 2, since the Spole is closer, the ﬁeld is stronger. FS>FN. The net force is pointing to left. FN .—l:—' 29.3.3 Current loop in B Current I is counterclockwise. Loop area is axb.
B is along i. Find the direction of torque due to B.
1 2 3 4 direction of ‘I: j j k k
Hint: “F=ILx B. “For the direction of torque, you need 0 use the righthandrule for a rotation.
Extra: Sketch here is viewed from the bottom along +j. F 3 Here F1=F3=IaB. Show torque, I=IabB.
Explanation: Force on #1 is out from the paper and on #3 is opposite to it. This leads to a clockwise rotation as view
from the top. Righthandrule of rotation gives direction of r to be along j. Check that it agrees with: tszB. Extra: From the sketch, F1 and F3 are equal in magnitude
and opposite in direction. They are separated by a distance
b. These two forces form a “couple”. The corresponding torque is given by t=Flb=(IaB)b=IabB. 29.3.4 Flip an electric dipole d.
 +Q E
E j
: 5 _Q I II —i—, A dipole ofi Q, With a separation d is placed in a uniform
constant electric field E. Determine the potential energy
released in ﬂipping the dipole from I to 11, while pivotting
about its center. 1 2 3 4
mm QEd 2QEd 2QEd 4QEd Hint: The electric force on a q is F=q E.
Extra: What is the potential energy released if the ﬂip is
about the location of the Q in the diagram 1? Explanation: For +Q, the potential energy released
AU=Q (V1 —V11)= QEd. Thus the total potential energy
released is 2QEd. Extra: Since the net force due to a constant ﬁeld E, on the
dipole system is 0, no work is required in translating the
dipole horizontally. The total potential energy released
should again be 2QEd. 29.3.6 Torque on current loop A coil has N turns with radius r and current i. B is
directed in the manner as shown. Find the direction of torque, 1:. Hint: p1=p1_1°op=iAn. Use Righthand—rule (RHR) #3 for
n. Magnetic dipole moment for N—loop: plell. 12:}le.
Extra: Show the magnitude of the torque is iANBcosoc. Explanation: Right—hand—rule (RHR) #3 implies II. is
pointing upward. [LXB is pointing along k. \ 0:
Extra: The angle between B and II. is 0£+m’2. " ' So 1=J.Bsin(ot+1tf2)=(iAN)Bcos0t. 29.3.5 Current loop in a constant B
B The angle between the B ﬁeld and the loop is UL.
Determine direction of uloop and angle between uloop & 29.4.1 Magnetic force on
a moving charge B field is coming out of the paper. Determine the direction
of the magnetic force for each case. Case 1: Force on q], with q1>0. Case II: Force on  , with  <0. Extra: Show radius of the circular motion is r=mv/(qB).
Explanation: We leave it as an exercise for the reader.
Ans: 1. Extra: Equating centripetal force with the magnetic force gives mV’2/l=qVB. Solving for r gives: 1:mvf(qB). 29.4.2 Motion in constant B field
V A proton is moving in a. plane perpendicualr to constant B field.
Case 1: v1:10m/s. Case 2: v2:20m/s
Find the ratio of the two radii. Choose one: Hint: quzmv2/r. Exua: Show the two periods are the same. Useful relations: w=vli=qB/m, T=2nl w.
mv2 Explanation: ; = W = q =
qB Exua: TZIIT1= (Zn/(oz)! (Zn/w1)=1. For the last step, we used the fact that 031: w; = qB/m. So the period is
independent of speed. ...
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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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