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Unformatted text preview: 30.1.1 Magnetic field at the center of a current
loop The loop has a current i and a radius r. Determine the
direction of the magnetic field at the center. H' l. U th B. t S V3111 AB “0 i—z—ALXf
ln : Se 9 10 — a. aw, = — .
43": 1' Extra: Based on the Biot—Savart law, show that the
magnetic field at the center is given by B2u0if2r. Explanation: Taking the cross product, one finds that
for any current segment along the circle, the corresponding AB at center always points into the paper.
So B due to the entire current loop should also point into
the paper. Extra: From the hint, the magnitude of the field at 0 due
to a current segment irAG is AB:(p.D/41t) irABsinQOolrz. For the circle, one replaces AG by 211. This gives
szuii2r. 30.1.2 Circular Arc A current loop consists of one circular arc subtended by an angle 0t, and two radial segments. There is a
clockwise current i and the radius is r. Determine the
direction of the magnetic field at the center, 0. H' U th B‘ S 1 AB H0 i Xf int: se e iot— avart aw, = — .
4:: 12 Extra: Based on the Biot—Savart law, show that the magnetic field at the center is given by B: uni 0c 1(4rtr). Explanation: Taking a cross product, one finds that for
any current element along the arc, the corresponding AB at center always points into the paper. So B due to the
entire arc should also point into the paper. Extra: From the hint, the magnitude of the field at 0 due
to a current segment irAB is AB=(.tU/4tt) irABsin900/r2. For the arc, one replaces AG by 0:. This gives B=poi 0t /(41tr). 30.1.4 Magnetic field due to a straight current
segment Consider a. current segment CD with current i. Let B be
the magnetic field vector at P due to this segment. Choose
one: 1 2 l 3 l 4
dir. of B into into I out I out
.3 Lei «fiuoi Lei «liuoi
4“ 4m. 4m 4m
A A0
Hint: Consider iAy at y where sine = E ,—g] = —. Its
l‘ r a
M ' A0
contribution at P is AB = "—U—Zysine = ﬂ —sin0.
4:: r 4a a Extra: Determine magnetic field vector at P due to CO. Explanation: By inspection iAyxr gives the direction of
AB to be into the paper. So B at P due to CD is into the
paper. Integrating from C to D, one obtains at P 3 4  
B: T [Er—esineQ/EPU‘
m4 41': a 411a Extra: Since iAyxr gives the direction of AB to be into the paper, at P the magnetic field vector due to the
segment CO should also be into the paper. For this case, the reader should verify that: B: ~/2p.Di/ (811a). 30.1.3 B due to a current loop C .
b 1
a
O A B
Determine the direction of B at 0 due to the curent loop.
1 2 3 dirofB out‘ into ‘ BisO _ M0 in.fo
Hint: = Extra: Find the magnitude of B. Explanation: B:BI+B2+Bs+B‘. At 0, 8228420.
B1=(J.o/4n )(iaJt/2)Ia2=.toi/(8a), into the paper.
B3:p.0if(8b), out of the paper. B1>B3. So resultant B is
into the paper. Extra: B=(.LUi/8)(1:’a1/b). 3 30.3.1 B field due to 2  wires 30.2.1 Magnetic force between "wire: ‘a O 3 i1 i2 ALP Two wires have currents i1 and i2. Dewmﬁne direcuon 0f the magnetic field BI! due to the Given 2  wires where both currents enter into the paper. currenti at P The distance of separation between them is 2a. Determine
the direction of resultant field at P, where OP=2a. left Extra: Sketch a vector diagram, showing that the
magnetic force vector, AFM, due to Bl on izAL is pointing
to the left. right ‘ Etra:Sh thd'ti thP"d dtfth
Explanation: By the RHrule foralong wire, RHR #1, vine of 01)::V e "ea one a ls m epen en 0 e the direction of B1 at P is into the paper. See sketch.
Extra: From the cross product rule, (RHR#2), the magnetic force AFM= iAL X B1 is pointing to the left. Explanation: Inspection on the sketch below shows that
A APP’2m’2 — I3, and A APP”: I3, so A P’PP”:90°. B is
along the negative x—axis. il
Bl iAL 30.3.2 r—dependent current density _~‘ P .__._'.«p
. a
I Front view Top view
Given: Current density szrz. Determine M of the Ampere’s law: M: [L B 'dS 2'10 iemir, where loop L is a circle with radius OP=a.
1 2 3 4 M ItaB/Z naB 2naB 41taB Extra: Based on iemi, =lJ21trdr, show that iemi, :
nba4/2. Explanation: M = des= 2JtaB. Extra: lenc = 30.3.4 Field due to 4 —wires Consider 4 parallel wires shown in a 3dview, and in a
projected view from the right. The latter indicates that the
wires are at four comers of a square, where each side has
a length “a”. Sketch the field vectors due to the four wires
at P, the center point, and the resultant vector B.
  1  2  3 4
direction of B u down ri ht left Hint: For a longwire at a distance r, Bzuol/(2nr).
Extra: Find the magnitude of B at P. Explanation: The situation at P is shown below. The
vector sum gives the resultant field vector B, which is
pointing in the upward direction. Extra: The sum of vertical projections of four vectors, 1 2 1
BB+BD, & BA+BC gives B = 4 cos45D "—0 = "D .
“(7.5) 7“ 30.3.3 Flux through a square Consider the ﬂux due to currenti in a long wire. Find the
flux throu h the s uare area shown in the sketch. Hint: Att=BaAr, B=.L°i/(2nr) Extra: Show I" = q.” b(a + b) equally divides the flux through the square. . “oiaznﬂ_ El a : = +bBad=
xpanaonqﬁj'éz r2” b 1 b
Extra: Let 311% = E £n(a Z Solving for r gives 30.3.5 Field due to current sheet A sheet of current is in the yz—plane at x:0. Linear current
density L=dlzfdy=10 A/‘m. Determine the direction of B at P. 1 2 3 4
Dir of B J, _> 1~ (_ Extra: Consider the loop ABCDA, or L, where
' :la. lencir Show at P B=J.°l/2. CB__ 1 a
D E A"
Explanation: Righthandrule #1 together with the
superposition principle implies that B is up.
Extra: Ampere’s law says: the magnetomotive force over
the complete Amperian loop ABCDA is given by: MLzBa+O+Ba+02uDla. Or BzuDJU2 30.4.1 Field outside of solenoid Strictly speaking B outside of an infinitely long solenoid
is not zero. Find direction of B at oint P, OP=r. Hint: Sketch an Amperian loop through P. What is iemir? Extra: Show BP=u0i/(21tr). Explanation: Current from left to right is i. Applying
the righthand rule for a long wire, one finds that at P, B
is along k. Extra: Applying Ampere’s law for the Amperian loop,
which passes P, gives: 2an=tt0i, or BzuDi/(21tr). 30.5.1 B at one end of a long solenoid Given: A long solenoid which has a currenti and the
linear number density (turns per length), n. Find: B A, the magetic field at the point A, located on
the axis at the left end of the solenoid.
l 1 l 2 l 3 l 4 l
BA l pom  [Join  [Join/2 l [Join/2 l
dir of B A Extra: Determine BC, magnetic field at C (at right end). Explanation: Assume the solenoid is long. Near the
center BszR+BL=ZBR. By inspection, BAan, or BA:
Bin/2:1L0in/2. (It is a special case of B=,I,Din(sin11lz—
sintpl)/2 given in the text. Here $500, and 1111 =—90°.) Extra: Again in the long solenoid approximation, by
inspection, BC=BL. So BC is to the right, Bczpain/Z. 30.4.2 Ampere’s Law: Long Solenoid
AN, AL=AB ...., Bout
For a long solenoid with a current i, we assume the ﬁeld
inside B,,,, is uniform and constant and the ﬁeld outside
Bout, is 0. To evaluate the magnetic ﬁeld inside: B=Bm,
consider a rectangular Ampen'an loop of a height h and a
Width AL. Let the number of Wires enclosed by the loop be AN. Determine the magnetomotive force (mmf) M along
the loop ABCDA and the current enclosed: iencl. Hint: M involves an integral over the loop, Where the
integrand at each segment is given by BAs=BAs cos 0. Extra: Show the ﬁeld inside is B=B,n=p0iAN/AL.
Explanation: The total magnetomotive force is given by:
M=MAB+MBC+MCD+MDA. Since BJ_As, MBC=MDA=0,
Since the magnetic ﬁeld outside is 0, MCD=0. So
M=MAB=+BAL. By inspection imlq'AN. Ans=3. Extra: With the Ampere’s law: M=pﬂimb BAL=uﬂiAN. or
B=llgiANlAL. 30.8.1 Displacement Current Consider the set up shown in the sketch. Which of the
following quantities isfare numerically equal to i?
d dEA d¢E (WC 1. — , II. E —EE —, “LC—
dtme 0 dt 0 dt (it [only I & II
on] Extra: It is possible to have a situation where at some
instant i¢0, and to have Q:Vc:¢E=0? Explanation: me=cvc. For a parallel plate system,
EZQplate/(EUA), SO QplatE: Extra: Yes. We leave it as an exercise to the reader. 30.9.1 Magnetic dipole moment of an orbital
electron V Consider an atom, where an electron is orbiting along a
circular orbit with radius r, in a clockwise manner. The
orbital speed is v. The electron has a. mass m, and a
charge “— e”. Denote port, to be the corresponding
magnetic dipole moment vector. Choose one: Hint: uorb =iAﬁ. Current is i=e/T, with period T=2nrfv. Extra: Verify that port = [2i]L,where L=mvr is the
m orbital angular momentum. Explanation: By inspection, there is a counterclock
current. The loop—magnet identity implies that port, is out
of the paper. For a ciruclar orbit, the area is Aznrz, or new = iA= {%](nr2) = %. Extra: Since L=mvr, or VttL/m. Then [1.0m = {ZLJL
m ...
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 Fall '08
 Turner
 Physics, Magnetic Field

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