ELECTROMAGNETIC INDUCTION
1. MAGNETIC FLUX
(a) The magnetic nux ~ linked with any surface plncAd in a magnetic field 8 is measured by the total of
magnetic lines of force passing through It.
(b) Sti_ppose magnetic field B makes an angle o with the elemental area vector
d A , then magnetic flux through the area,
d4, := B.dA = B dAcosO
2
(c) It is a scalar quantity with S.I. unit waber = 1Nm/A and C.G.S. unit is gauss or Maxwell.
(d) The total magnetic flux qi linked with complete area Sis given by the algebraic sum of the magnetic
fluxes linked with all elementary areas of the surface, i.e.,
~= L d~ = IB.dA if dA ➔ 0 ~ = J B. dA
(e) Here, magnetic flux linked with a surface may be defined as the product of the normal component of surface
area and the magnetic field acting on the area. If may also be defined as the dot product of magnetic field
and surface area .
1.1 Special Cases
Flux through plane surface in unlfonn magnetic field:
Suppose a plane surface of area A is placed in a uniform magnetic field B when area
vector and field vector makes an angle e between them. The flux through plane
'A a
surface
f
~ = B. dA = Jscos 8.dA = BAcosS
<I> =BAcosS
Flux through non~oplanar surface in uniform magnetic field:
If can be calculated by the area of projection perpendicular to the direction of field, i.e., flux through equivalent
plane surface, then
f
<I>= B.dA = dAcosS J
Flux through a surface (plane or non-planer) in non uniform magnetic field:
It can be calculated by finding flux through each elemental area and then integrating it, i.e.,
<!> = J 13.dA
surface ... •' 11
1.2 Results:
7t
(i) When e = O or 1t, q> is maximum (ii) When 8 = - , q> is minimum = O
2
(iii) Magnetic flux through any closed surface is always zero.
4>~1 =(q>1n + q>out) with sings where q>in =(-) ve and q>out =(+) ve
(iv)
.
Net flux through a surface,
Illustration 1: A rectangular plane of area 100 cm 2 is placed in a uniform magnetic field of 100 Tesla such
that plane makes an angle 60° with field. Then find magnetic flux through the surface.
Solution: The angle between the area vector and field, A 13
e =2: - 60° = 30°
2
:. Magnetic flux, ~ = BA cos e
= 1oo x 1oo )( 1o""" ./3 = ✓3
2 2
·111ustratlon 2: The magnetic field In a certain region is given by B =40i- 18k G. How much flux passes
through a 6.0 - cm 2 - area loop In this region if the loop lies flat on the xy plane?
___
. ._,._,,, .. . '
tA these A = 6.0 x•10 - ◄ km 2 and B = 40i - 18k 10 -1 T. Then <I>= B.A = -900nWb .
·" ..,,.~ . .
,,-•--- -••• ••• ••• ••-- -~- :::; :=:;;;,~,
Illustration 3:
1; <.J~lei d Wfiifnlht ,,1111 i~ ;~,,~
enri'io;;/Jd.l.
wb Is produc ed In th• Iron ,or• o l•n<.,ld by th• t.lJm'I curt1,;r,r
A fl ux of 900µ ...-, , 't'lt,~:1"1._;
d In th• 61,rno io "' - ,~ 1;
flux (in air) of 0.5 µWb Is produce
-~
relative permeablllty of the Iron? , 00 .
11 9
Solution: <l> = BA inside the solenoid . Thus K.. 1
_ ~~ .. --··· :Ill' _____ .. 1e-oo
•
8 ;. t1>
;
O.5
f tides • & b place at •
Illustration 4: Find the magnetic flux through • rectJJngle Ocarryin g a ,urrent -
I •1
distance r from an Infinitely long co nductor
shown. . conduclOf, £0 tha1 ma~rlof!.x; f1.::1
Solution: Take a strip of width dx at a distance x from the
a t x is B = µol and the area of the 6trip is bdx ·
- 2nx
µI
As,~= JB.dA = J- 0
21tx
Jµ(ilb dx
.bdxcos0°:::: - -
21t X
= µalb [ enxr• 4> :: µ,)b
21t
tn(~
. r
J-
21t
Illustration 5: In the following figure there Is a + directed magnetic ::d y
of 0.2T. Find the magnetic flux through each face O e
box.
Solution: "
Let n represent the outward unit normal vector
to a given face of the box and A the area of that
face. Then the outward flux through the face is
" A=
<I>m = 8. n BA cosa, where O= angle (B, n)
"
Clearly <I> m = o through the two sides faces ( n in ± z direction) and the bottom fa';}! { n r : - y
/•
direction). Through the front and back faces n is along + x and - x, res~:·1e!y.
fi'.:J
<I>1n,n1 =(0.2T)(40x10.... m2 ){1) = 0.8mVVb
<I>bad< =(0.2T)(90x10.... m2 ){-1) = -1.BmVVb (-ve sign shows inward flux )
For the top surface e = 60° and <I>iop =(0.2T)(100x 10.... m2 )(1/2) = 1.0mV'Jb
Exercise 1: Find the flux through butterfly net
,..
I
~ .. - .. ,, , .,
2. FARA DAY'S LAWS OF ELECTROMAGNETIC INDUCTION
When a current loop ts placed in a magnetic field it expenences a torque, which
makes it nt.a!e. tt a con::~:::n ;
loop of wire is rotated in a magnetic field, a al'l'ent flows in the loop.
This was first observed by Michael Faraday. On the basis of his experimental observa
tions he conclu~ d t:"Z:.
(i) Whenever there is relative motion between a magnet (souroe of magneti c field) and
a dosed concUQrg ~x~ cf
wire, an emf is induced in the loop: This Is due to the change in magnetic flux linked
by the loop.
(iij An emf causes a current to flow m the circuit. and so when loop and magnet are
in relative mctj~. a c.r.:;-~
flows in the loop. This implies that an emf is aet up in the loop. This emf
is known as induced emf a:-,1 r.s
magnitude is directly proportional to the rate of change of magnelc b With reapect
to time.
Magnetic flux associated with any area is given by lhe e,cp11111kln
q,8 =JB-ds .
Here B is the magnetic field and ds
is an area eln,1• ·
perpendicular to the surface.
Flux associated with a flat coil (having N turns) which ii
⇒ N(B.A)
kept••
⇒ <l>a = NBAco se
Here. N is the number of turns in the loop
, B = magne tic field strength, A = area of the loop
O = angle betwee n area vector and magne tic rield
atical form . Induced emf can be grven by
Now we come to Faraday's law of electromagnetic induction in mathem
the expres sion t: = - d4is
dt
Lenz's law is
Though Faraday's law, by itself, is sufficient to calculate the magnitude and polarity of induced emf.
directio n of the induce d current .
freque ntly used to determine the polarity of induced emf or the
t methods. For example, by changing N, by
Magne tic flux associated with a loop can be changed by differen
changing 8, by changing A or by changing the angle between
A
and 13 .
Hence Magnitude of the induced emf = Id$
dt
If~ represents the flux through a single tum, and the loop has
N turns, then Induced emf= Nld$
dt
magnit ude is given as
This induce d emf creates an induce d curren t in the circuit whose
. induce d emf 1 ld'I
1
= net resist ance of circuit = R dt
Is paralle l to the plane of
/1/ustration 6: A coil is placed in a consta nt magnetic field . The magnetic field
induce d In the coils 1 and 2.
the coil as shown in figure. Find the emf
oAre•=A =@:•
: . emf= 0
Soluti on: (i) ~ = 0 (always) since area is perpendicular to magnetic field.
(ii) ~ = BA (always) = const. :. emf= 0
a uniform magnetic field
Illustra tion 7: A condu cting circula r loop with variable radius r is placed in
of the loop is contra cting
B=0.020T with plane perpen dicula r to the field. While the radius
radius is 2 mm.
at a consta nt rate of 1.0 mm/s; find the induced emf in the loop when the
Soluti on: Radius (variab le)= r ⇒ 8 = 0.02 T
d dA dr dr 3
-=1. 0 x 10- n1 r=2 x 10·3 mt
lnduce demfE = -(BA )=B- = n8 x 2r- ⇒
dt dt dt dt
E = 0.02 x rrx 2 x 2 x 10·
3
x 1 x 10·
3
= 6 x 10-8 x 7t = 18.85 x 10"8 V.
EMF= - d4\
dt
=-~T µ 01 .bdx = -(
dt r 2nx
µ01 dy - ~ dy) = µ 0 lbva
2n(y + a) dt 2n(a) dt 2n(r + a)
3. LENZ'S LAW
cause due to which it •·=,s induced.
Accord ing to Lenz's law, induce d ;,,nf in a circuit opposes the
Consid er the following examp les.
towards a
(a) Suppo se that the north-p ole of a bar magnet is moved
conducting wire loop as shown in the figure. Due to a change
in the b I~
d. Due to
magne tic flux associated with the loop, a current is induce
magne tic field is induce d and this magne tic field
induced curren t, a
of bar magne t. The directio n of the induce d
oppose s the motion
pole is moving towards the loop;_therefore to
current can be deduce d by the following argument: the north
be induce d on that face of the loop which faces the
oppose the motion of the bar magne t only a north pole will
magnet.
field which X X X X X xo~ X
(b) A rectangular loop ABCD is being pulled out of the magnetic xA 1< 8
the plane of the XX XX XX XX XX X XX X
is directed into the plane of the paper. Perpen dicular to -
ted with the X X X X X X X X
paper. As the loop is dragge d out of the field, the flux associa c
in a sense so as X X X X X X X
loop decrea ses. The induced curren t flows in the loop
X
X X X X . X .P x. X
.
to oppose the decrea se in this flux. For
in the loop must be directed into the plane of the
this to happe n the magne tic field due to the induce d current
paper. Thus the current in the loop must flow be clockwise.
right, away from the long
Illustration 8: In figure the rectan gular loop of wire is being pulled to the
the curren t induce d in
straight wire throug h which a steady curren t i flows upward. Does
sense?
the loop flow In the clockw ise sense or in the counte r clockwise
, It,a ltfi!~lttr!lf. l1t:l,f lli411111 t;,1.1 ,,,,,,, jlJ l11J'Jl t1 ,,. ,1,,1:.,,ib,J ,nlfi 1J;41 Ji&(,t:I rto ti ~ l
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Ir/ti IM,(Jf,~J/, t l:11 ..
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1,l1, 1I ,,,t,, Ill's ,,,,.,," 1/f tJ,c, • •~ IJ((;
I,
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A,,,
w•,~~:~ -~.
_JI--:::,
II (
cp~p~ o C r,
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thJI tt,,,.,u~t, thF: f/,,f,
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ht lltt1 (,OIi will fl,,w .I) " ' In f1f()tj1Jf,ij 8 "'"' vmwrj 1114 lt:ft IJWJll!Jh ~t,,;lf 1-HilY th~ (i',Jh1--t,I11 /11
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tilt, tflb!Jhtil m,,v1:1i fJ,wwJ th'1 ri(Jhf, thtt tli11 inr:,ld~ th6 </Jil, &IVJ !J'1nE:rally to tt11: lf:ft.
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fr<,rn A tr, U 1111,J fttJfll I) I() C.
/l/11~ltblfo1110: 11,11 "'""'"'' ,,, ft,lfow/11u fiUllfO f"fJlal.tJtS H tillf)Wn (Jn II f'/VtJt thfQU(Jh It, ctmtre, At tJ,e
l11,tn11t •lmw11, w/1111 •ro tl,o 11/ruct/<Jns ,,, tllo lmJ11cod currents?
Cf?~3(?i° /, U
Cf ~ D
C D
t;o/11tio11: 'f ht, iitw111,111 Is ~111111.lr tn tltr,t In p1qvir1tJf5 probfc,m, lj)l<,;1:pt that thc, coil on the right is opp0s,t1;ly
wuul1J I ltirts, t,r,ltt G<JiftJ ,-,yr,nr1l-lnrA ~ (1,~c,1:as" In f11JX to the Jc,ft 8$ the magnr;t rota~ trc,m th1:
fJ1irullt,f pcJ1.111t,n 1:lirt1m ltJ tl,tt r,,,1p<,r1<llG1Jl,1r f)Q~ition. Ttte ln<11JG£:d current$ In both coils thus tend
tr, Uf!ulr, '2 f111Jd vmurr111111 Jt,ft ~o tile c,m,;nttJ fJ<JW from B to A &nd from C to o.
A r,11mJ11l1J1t1 ,,,r,a/tt• c,f 111>lvrJ(11,J rod"' tho lower ond of which there Is a metal ring. The
lm11llao tll11t ""' current 111 tllo rlnu. 1ho r,ond11/um 1wln9~ In tho piano of the ring. The ring
/1 ra/,o,J tmd rolooeod, At tho bottom of It• 1wlng, tho ring ent<m; a moynotic field normal
tfJ ti,., /,/1111<1 ,,, tho rl11u, In tho ur,p of a 1tror,y llororac..;jJo maanot. On entering this region,
It fl(JOII com110 ,,, n r,to1>, WIiy?
~(J/1111,,,,: A ,u,,,,,,t Iii i,1,Jw,,,,j In U1t1 rln~ M It ,111tfJ1~ tt11; ma~nr.tic fit:11 , By Lenz'a law, the magnetic force
c,n thti lndur.,t,tJ umtir,t c,J,fl'J~~«> tt,,,
molt<in wtilr;h ln<Ju<:<id tt,e current. It Is this force which halts
m11 tr1a,,~J.,1tnnuJ m<,Ur;n c,f tt,,; rhtQ T'tio ~lnfttlr, en<,r~y fJf the mar,roscopic motion is converted to '
tt,m,nhl tilltJl(JY by J,,,,Jc, htillltn() In !11~ rio~ tl<JIQ th:1t (<>ll1r;r things boing equal) the more highly
cx,r1tlu<Alt1!J Ill" rh,c, m111,,r1,,1 I!$, tt1,, m<Jrc; rapi<Jly the motion is btopped and the energy is
dl11tilr,1.itttd
lltu,tr,tJon 12:
lolul/ont
M•11not/c f/old 1, lncroHlrr(J ,,,,,, tho f)U(JO with limo whon a conducting
loop of dffflnlto ratllu,t I• filacod 011 tl,o pla110 <Jf t/111 J>npor. The find the
dlr•ctlon of c;urront In th11 1,,c,,,,
At, the nwc It 111crn1,l11g ln&i<J<; then tt1" cum,nt In t11,, IQop will be 5uc;;h that It
wtlf t,o C'f>PCitilng the, lnr.rotu+Q in moon,;tir, fit,l<J, l.tJ , tho lndur,ed
·o~
~
X
X
¥
X
k
•
X
X
X
X
ti 1
X
X
currtnt in th4 lotJp will crn11t~ ,u,;h o mo9nctic f11,ld which ia dh<:c;;te<.J out ward.
Thut b dlrtctJon of current wllt bo onticlnr.l<wl5.,
A '1111'9 loop 1, pulltd out of• uniform m11u11otlc tlald as ,t,own.
on of ourrtnt,
wlll be tuch that whlc~h will try to opp<,~c tho
field, I 1,, the current In th" Joop will try to
ftlld.
In clockwlH dtrect1on
1. MAGNETIC FLUX
(a) The magnetic nux ~ linked with any surface plncAd in a magnetic field 8 is measured by the total of
magnetic lines of force passing through It.
(b) Sti_ppose magnetic field B makes an angle o with the elemental area vector
d A , then magnetic flux through the area,
d4, := B.dA = B dAcosO
2
(c) It is a scalar quantity with S.I. unit waber = 1Nm/A and C.G.S. unit is gauss or Maxwell.
(d) The total magnetic flux qi linked with complete area Sis given by the algebraic sum of the magnetic
fluxes linked with all elementary areas of the surface, i.e.,
~= L d~ = IB.dA if dA ➔ 0 ~ = J B. dA
(e) Here, magnetic flux linked with a surface may be defined as the product of the normal component of surface
area and the magnetic field acting on the area. If may also be defined as the dot product of magnetic field
and surface area .
1.1 Special Cases
Flux through plane surface in unlfonn magnetic field:
Suppose a plane surface of area A is placed in a uniform magnetic field B when area
vector and field vector makes an angle e between them. The flux through plane
'A a
surface
f
~ = B. dA = Jscos 8.dA = BAcosS
<I> =BAcosS
Flux through non~oplanar surface in uniform magnetic field:
If can be calculated by the area of projection perpendicular to the direction of field, i.e., flux through equivalent
plane surface, then
f
<I>= B.dA = dAcosS J
Flux through a surface (plane or non-planer) in non uniform magnetic field:
It can be calculated by finding flux through each elemental area and then integrating it, i.e.,
<!> = J 13.dA
surface ... •' 11
1.2 Results:
7t
(i) When e = O or 1t, q> is maximum (ii) When 8 = - , q> is minimum = O
2
(iii) Magnetic flux through any closed surface is always zero.
4>~1 =(q>1n + q>out) with sings where q>in =(-) ve and q>out =(+) ve
(iv)
.
Net flux through a surface,
Illustration 1: A rectangular plane of area 100 cm 2 is placed in a uniform magnetic field of 100 Tesla such
that plane makes an angle 60° with field. Then find magnetic flux through the surface.
Solution: The angle between the area vector and field, A 13
e =2: - 60° = 30°
2
:. Magnetic flux, ~ = BA cos e
= 1oo x 1oo )( 1o""" ./3 = ✓3
2 2
·111ustratlon 2: The magnetic field In a certain region is given by B =40i- 18k G. How much flux passes
through a 6.0 - cm 2 - area loop In this region if the loop lies flat on the xy plane?
___
. ._,._,,, .. . '
tA these A = 6.0 x•10 - ◄ km 2 and B = 40i - 18k 10 -1 T. Then <I>= B.A = -900nWb .
·" ..,,.~ . .
,,-•--- -••• ••• ••• ••-- -~- :::; :=:;;;,~,
Illustration 3:
1; <.J~lei d Wfiifnlht ,,1111 i~ ;~,,~
enri'io;;/Jd.l.
wb Is produc ed In th• Iron ,or• o l•n<.,ld by th• t.lJm'I curt1,;r,r
A fl ux of 900µ ...-, , 't'lt,~:1"1._;
d In th• 61,rno io "' - ,~ 1;
flux (in air) of 0.5 µWb Is produce
-~
relative permeablllty of the Iron? , 00 .
11 9
Solution: <l> = BA inside the solenoid . Thus K.. 1
_ ~~ .. --··· :Ill' _____ .. 1e-oo
•
8 ;. t1>
;
O.5
f tides • & b place at •
Illustration 4: Find the magnetic flux through • rectJJngle Ocarryin g a ,urrent -
I •1
distance r from an Infinitely long co nductor
shown. . conduclOf, £0 tha1 ma~rlof!.x; f1.::1
Solution: Take a strip of width dx at a distance x from the
a t x is B = µol and the area of the 6trip is bdx ·
- 2nx
µI
As,~= JB.dA = J- 0
21tx
Jµ(ilb dx
.bdxcos0°:::: - -
21t X
= µalb [ enxr• 4> :: µ,)b
21t
tn(~
. r
J-
21t
Illustration 5: In the following figure there Is a + directed magnetic ::d y
of 0.2T. Find the magnetic flux through each face O e
box.
Solution: "
Let n represent the outward unit normal vector
to a given face of the box and A the area of that
face. Then the outward flux through the face is
" A=
<I>m = 8. n BA cosa, where O= angle (B, n)
"
Clearly <I> m = o through the two sides faces ( n in ± z direction) and the bottom fa';}! { n r : - y
/•
direction). Through the front and back faces n is along + x and - x, res~:·1e!y.
fi'.:J
<I>1n,n1 =(0.2T)(40x10.... m2 ){1) = 0.8mVVb
<I>bad< =(0.2T)(90x10.... m2 ){-1) = -1.BmVVb (-ve sign shows inward flux )
For the top surface e = 60° and <I>iop =(0.2T)(100x 10.... m2 )(1/2) = 1.0mV'Jb
Exercise 1: Find the flux through butterfly net
,..
I
~ .. - .. ,, , .,
2. FARA DAY'S LAWS OF ELECTROMAGNETIC INDUCTION
When a current loop ts placed in a magnetic field it expenences a torque, which
makes it nt.a!e. tt a con::~:::n ;
loop of wire is rotated in a magnetic field, a al'l'ent flows in the loop.
This was first observed by Michael Faraday. On the basis of his experimental observa
tions he conclu~ d t:"Z:.
(i) Whenever there is relative motion between a magnet (souroe of magneti c field) and
a dosed concUQrg ~x~ cf
wire, an emf is induced in the loop: This Is due to the change in magnetic flux linked
by the loop.
(iij An emf causes a current to flow m the circuit. and so when loop and magnet are
in relative mctj~. a c.r.:;-~
flows in the loop. This implies that an emf is aet up in the loop. This emf
is known as induced emf a:-,1 r.s
magnitude is directly proportional to the rate of change of magnelc b With reapect
to time.
Magnetic flux associated with any area is given by lhe e,cp11111kln
q,8 =JB-ds .
Here B is the magnetic field and ds
is an area eln,1• ·
perpendicular to the surface.
Flux associated with a flat coil (having N turns) which ii
⇒ N(B.A)
kept••
⇒ <l>a = NBAco se
Here. N is the number of turns in the loop
, B = magne tic field strength, A = area of the loop
O = angle betwee n area vector and magne tic rield
atical form . Induced emf can be grven by
Now we come to Faraday's law of electromagnetic induction in mathem
the expres sion t: = - d4is
dt
Lenz's law is
Though Faraday's law, by itself, is sufficient to calculate the magnitude and polarity of induced emf.
directio n of the induce d current .
freque ntly used to determine the polarity of induced emf or the
t methods. For example, by changing N, by
Magne tic flux associated with a loop can be changed by differen
changing 8, by changing A or by changing the angle between
A
and 13 .
Hence Magnitude of the induced emf = Id$
dt
If~ represents the flux through a single tum, and the loop has
N turns, then Induced emf= Nld$
dt
magnit ude is given as
This induce d emf creates an induce d curren t in the circuit whose
. induce d emf 1 ld'I
1
= net resist ance of circuit = R dt
Is paralle l to the plane of
/1/ustration 6: A coil is placed in a consta nt magnetic field . The magnetic field
induce d In the coils 1 and 2.
the coil as shown in figure. Find the emf
oAre•=A =@:•
: . emf= 0
Soluti on: (i) ~ = 0 (always) since area is perpendicular to magnetic field.
(ii) ~ = BA (always) = const. :. emf= 0
a uniform magnetic field
Illustra tion 7: A condu cting circula r loop with variable radius r is placed in
of the loop is contra cting
B=0.020T with plane perpen dicula r to the field. While the radius
radius is 2 mm.
at a consta nt rate of 1.0 mm/s; find the induced emf in the loop when the
Soluti on: Radius (variab le)= r ⇒ 8 = 0.02 T
d dA dr dr 3
-=1. 0 x 10- n1 r=2 x 10·3 mt
lnduce demfE = -(BA )=B- = n8 x 2r- ⇒
dt dt dt dt
E = 0.02 x rrx 2 x 2 x 10·
3
x 1 x 10·
3
= 6 x 10-8 x 7t = 18.85 x 10"8 V.
EMF= - d4\
dt
=-~T µ 01 .bdx = -(
dt r 2nx
µ01 dy - ~ dy) = µ 0 lbva
2n(y + a) dt 2n(a) dt 2n(r + a)
3. LENZ'S LAW
cause due to which it •·=,s induced.
Accord ing to Lenz's law, induce d ;,,nf in a circuit opposes the
Consid er the following examp les.
towards a
(a) Suppo se that the north-p ole of a bar magnet is moved
conducting wire loop as shown in the figure. Due to a change
in the b I~
d. Due to
magne tic flux associated with the loop, a current is induce
magne tic field is induce d and this magne tic field
induced curren t, a
of bar magne t. The directio n of the induce d
oppose s the motion
pole is moving towards the loop;_therefore to
current can be deduce d by the following argument: the north
be induce d on that face of the loop which faces the
oppose the motion of the bar magne t only a north pole will
magnet.
field which X X X X X xo~ X
(b) A rectangular loop ABCD is being pulled out of the magnetic xA 1< 8
the plane of the XX XX XX XX XX X XX X
is directed into the plane of the paper. Perpen dicular to -
ted with the X X X X X X X X
paper. As the loop is dragge d out of the field, the flux associa c
in a sense so as X X X X X X X
loop decrea ses. The induced curren t flows in the loop
X
X X X X . X .P x. X
.
to oppose the decrea se in this flux. For
in the loop must be directed into the plane of the
this to happe n the magne tic field due to the induce d current
paper. Thus the current in the loop must flow be clockwise.
right, away from the long
Illustration 8: In figure the rectan gular loop of wire is being pulled to the
the curren t induce d in
straight wire throug h which a steady curren t i flows upward. Does
sense?
the loop flow In the clockw ise sense or in the counte r clockwise
, It,a ltfi!~lttr!lf. l1t:l,f lli411111 t;,1.1 ,,,,,,, jlJ l11J'Jl t1 ,,. ,1,,1:.,,ib,J ,nlfi 1J;41 Ji&(,t:I rto ti ~ l
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I,
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1Jlrt.t,ti1,11• ,,f
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A,,,
w•,~~:~ -~.
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cp~p~ o C r,
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thJI tt,,,.,u~t, thF: f/,,f,
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ht lltt1 (,OIi will fl,,w .I) " ' In f1f()tj1Jf,ij 8 "'"' vmwrj 1114 lt:ft IJWJll!Jh ~t,,;lf 1-HilY th~ (i',Jh1--t,I11 /11
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tilt, tflb!Jhtil m,,v1:1i fJ,wwJ th'1 ri(Jhf, thtt tli11 inr:,ld~ th6 </Jil, &IVJ !J'1nE:rally to tt11: lf:ft.
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(J"i11tir .,,,.,'+ fh,1, II> lltt1 riylit lnl:,UJi, ftf;.l;lf if It 111 m,,,mt flQw~ from C tr.> 0 thffMJh tt~ resi~t◊r.
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fr<,rn A tr, U 1111,J fttJfll I) I() C.
/l/11~ltblfo1110: 11,11 "'""'"'' ,,, ft,lfow/11u fiUllfO f"fJlal.tJtS H tillf)Wn (Jn II f'/VtJt thfQU(Jh It, ctmtre, At tJ,e
l11,tn11t •lmw11, w/1111 •ro tl,o 11/ruct/<Jns ,,, tllo lmJ11cod currents?
Cf?~3(?i° /, U
Cf ~ D
C D
t;o/11tio11: 'f ht, iitw111,111 Is ~111111.lr tn tltr,t In p1qvir1tJf5 probfc,m, lj)l<,;1:pt that thc, coil on the right is opp0s,t1;ly
wuul1J I ltirts, t,r,ltt G<JiftJ ,-,yr,nr1l-lnrA ~ (1,~c,1:as" In f11JX to the Jc,ft 8$ the magnr;t rota~ trc,m th1:
fJ1irullt,f pcJ1.111t,n 1:lirt1m ltJ tl,tt r,,,1p<,r1<llG1Jl,1r f)Q~ition. Ttte ln<11JG£:d current$ In both coils thus tend
tr, Uf!ulr, '2 f111Jd vmurr111111 Jt,ft ~o tile c,m,;nttJ fJ<JW from B to A &nd from C to o.
A r,11mJ11l1J1t1 ,,,r,a/tt• c,f 111>lvrJ(11,J rod"' tho lower ond of which there Is a metal ring. The
lm11llao tll11t ""' current 111 tllo rlnu. 1ho r,ond11/um 1wln9~ In tho piano of the ring. The ring
/1 ra/,o,J tmd rolooeod, At tho bottom of It• 1wlng, tho ring ent<m; a moynotic field normal
tfJ ti,., /,/1111<1 ,,, tho rl11u, In tho ur,p of a 1tror,y llororac..;jJo maanot. On entering this region,
It fl(JOII com110 ,,, n r,to1>, WIiy?
~(J/1111,,,,: A ,u,,,,,,t Iii i,1,Jw,,,,j In U1t1 rln~ M It ,111tfJ1~ tt11; ma~nr.tic fit:11 , By Lenz'a law, the magnetic force
c,n thti lndur.,t,tJ umtir,t c,J,fl'J~~«> tt,,,
molt<in wtilr;h ln<Ju<:<id tt,e current. It Is this force which halts
m11 tr1a,,~J.,1tnnuJ m<,Ur;n c,f tt,,; rhtQ T'tio ~lnfttlr, en<,r~y fJf the mar,roscopic motion is converted to '
tt,m,nhl tilltJl(JY by J,,,,Jc, htillltn() In !11~ rio~ tl<JIQ th:1t (<>ll1r;r things boing equal) the more highly
cx,r1tlu<Alt1!J Ill" rh,c, m111,,r1,,1 I!$, tt1,, m<Jrc; rapi<Jly the motion is btopped and the energy is
dl11tilr,1.itttd
lltu,tr,tJon 12:
lolul/ont
M•11not/c f/old 1, lncroHlrr(J ,,,,,, tho f)U(JO with limo whon a conducting
loop of dffflnlto ratllu,t I• filacod 011 tl,o pla110 <Jf t/111 J>npor. The find the
dlr•ctlon of c;urront In th11 1,,c,,,,
At, the nwc It 111crn1,l11g ln&i<J<; then tt1" cum,nt In t11,, IQop will be 5uc;;h that It
wtlf t,o C'f>PCitilng the, lnr.rotu+Q in moon,;tir, fit,l<J, l.tJ , tho lndur,ed
·o~
~
X
X
¥
X
k
•
X
X
X
X
ti 1
X
X
currtnt in th4 lotJp will crn11t~ ,u,;h o mo9nctic f11,ld which ia dh<:c;;te<.J out ward.
Thut b dlrtctJon of current wllt bo onticlnr.l<wl5.,
A '1111'9 loop 1, pulltd out of• uniform m11u11otlc tlald as ,t,own.
on of ourrtnt,
wlll be tuch that whlc~h will try to opp<,~c tho
field, I 1,, the current In th" Joop will try to
ftlld.
In clockwlH dtrect1on
