Magnetic Effect of Electric Current & Mechanical Effct of Electric Current

8/14/2019 Magnetic Effect of Electric Current & Mechanical Effct of Electric Current

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SYNOPSIS

EXPERT TUTORIALS, DHARWAD.

MAGNETIC EFFECT OF ELECTRIC CURRENT &

MECHANICAL EFFCT OF ELECTRIC CURRENT

1. The magnetic effect of current was first discovered by Oersted.2. An electric charge in motion sets up a magnetic field around it.3. An electric charge moving in an external magnetic field experiences a force.4. A stationary electric charge produces only electric field. But a moving electric

charge Produces both an electric field and a magnectic field.5. Magnetic field due to a straight conductor carrying current:

i) The magnetic field is circle.ii) The magnetic line of force are concentric circles with their centers lying on

axis of the conductor.iii) The plane of circle is perpendicular to direction of the current.iv) If the current flows through the conductor in upward direction, magnetic

lines of force are formed in anticlock wise direction.v) If the current flows through the conductor in down ward direction,

magnetic lines of force are formed in anticlock wise direction.

Scan

Fig 8.1

6. The direction of the magnetic field with respect to the current can be found by theAmpere right hand rule or Maxwells cork screw rule.

7. Ampere right hand rule :

a) For linear currents: If the wire is grasped in the palm of the of the right handwith the stretched thump pointing in the direction of current, fingers curl in thedirection of magnetic field (i.e., the magnetic lines of force).

b) For circular currents : If the direction of current coincides with the direction of

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0 idl

sin 42


the curl of the fingers of the right hand, the stretched thumb points in thedirection of magnetic field at centre of the loop.

Conventionally the direction of the field perpendicular to the plane of page isrepresented by if into the page and byif out of the pag

8. Maxwells cork screw rule : If we imagine a right hand cork screw to be drivealong the direction of the current in the conductor, the direction in which thethumb rotates represents the direction of the magnitude field.

9. The magnitude of the field produced by an electric current can be determined byusing the Biot Savart law or Amperes law.

10. Biot Savarts law : If a current of I amp passes through a small element dl ofa conductor, the magnetic induction dB produced at a distance r form dl is

( Fig. 8.2)

dB = x( Fig. 8.2)

whereis the angle between I and r,0 is the permeability of air orvacuum and 0 = 4x 10-7 web or henry or newton

amp-m m amp2

(a) In vector form, dB =0 idl x r

4 r3(b) If = 00 or 1800, dB = 0

11. Magnetic induction due to straight condurctor iof finite length carrying current I at a distance 0 Pr is given byB = 0i - [ sin1 + sin 2]

4 r

B = 0i - [ cos 1 + cos 2] Fig. 8.34r

a) If the point P is at one end of the conductor then i

B =P

Fig. 8.412.Magnetic induction due to straight conductor of infinite length carrying current Iat


0 i sin4 r


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0 2i 2i4 r r


a distance r is given byi

B = =10-7P

(a) If the point P is at one end of the

conductor then B = Fig. 8.5r

13. The magnetic induction at centre of a square of current i inclockwise direction is

given byB =

a) The magnetic induction at corner of square is

B =

14. The magnetic induction at centre of equilateral triangle of side d due to of current Iin clockwise direction is given by

B =

a) At any vertex of triangle is

B =

15. The magnetic induction produced on the axial line of a circular coil ofradius r, containing n turns is

B =Fig. 8.6(a)

where x is the distance between the centre of the coil and the givenpoint on the axial line. If x > > rB = 0 nr2I = or B = 0 n r2 I

2x3 2 x3But r2 = A is the face area of the coil.

B = 0 nIA = 0M

2 x3 2x3where M = magnetic moment of loop.


0 8 2i4 l

0 2i4 2l

0 18i4 d

0 2 i4 3d

0 nr2 I2 (r2 + x2 ) 3/2


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M1 n2M2 n1

B1 r2B2 r1

Bnr22r12


SCAN

Circular coil Rectangular coil Circulating charge

Fig. 8.6 (b), (c), (d)

a) At centre of coil, B =

b) For half coil, the magnetic induction at centre B =

c) For quarter coil, the magnetic induction at centre B =

d) On surface of coil, B=

16. A wire of length L is bent in the form of a circular loop with n turns and carries

current i. Its magnetic moment M = l2

i4pn17. A wire carrying current i is first bent in the form of circular loop with n tums and

carries current i. Its magnetic moments ratio is given by =

18. If two coils are connected in series then ratio of magnetic inductions at theircenters is given by =

19. If two coil are connected in parallel then ratio of magnetic inductions at theircenters is given by

B1 = r22B2 r12

20. Two copper wires of lengths l1 and l2 have area of cross section A1 and A2. Theybent in circular loops with turns n1 and n2. If they are connected in parallel thenratio of magnetic inductions at their centers is given by

B1 = n22 A1 l22B2 r12 A2 l12

21. A wire carrying current i is first but in the circular loop with n1 turns and then withn2 turns B1 and B2 are magnetic inductions at centre of loop,

B1 = n12

B2 n22

22. A current carrying wire is bent in the form of a circular loop of radius r1. Itproduces a magnetic induction at centre is B If same wire is stretched so that itslength increased by n times and then the wire is in circular loop of radius r2 Themagnetic induction at its centre =

S.No

Current carrying inConductor

Position ofpoint ofobservation

Expression of B Direction for B


0 n i2 r

0 n i4 r

0 n i8 r

m0 ni25/2 r


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4.

5.

coils in which samecurrent is flowing inmutually oppositedirection.

Fig. 8.11

Electron revolvingaround the nucleolusin a circular path ofradius a

Fig. 8.12a. Two parallelcurrent CarryingconductorsFirst second

Fig. 8.13

b.

Fig 8.14

c.

Fig. 8.15

d.

commonCentre

At nucleus

At distance r form2

both conductors

At distance ( r+x) form firstand x formsecondconductors.

At distance r form2

both conductor

B = 0i n1 - n22 r1 r2

.if the number of

turnsin them is sameB = 0in 1 2

2 r1 r2

B = 0iqV4 a2

= 10-7 2 fqa

where f =frequency

B = 0

B = 0i 1 +1__2 x (r + x)

B = 20i r

Upwards.

i) If current isanticlockwise thendirection of B will normalto plane of paper upwards.If current is clock wisethen direcation of B will

be normal to plane of

paper downwards

Direction of B normal toplane of paper downward.

Direction of B will benormal to plane of paperdownwards.

Direction of B will be



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6.

Fig. 8.16

Straight wire andloop

Fig. 8.17Note : No contact at

Point O

At distance( r+x)form first and xform secondconductors.

At the centre ofthe loop and atdistance o formstraightconductor.

B = 0i 1 1__2 x

(r+x)

B = 0i 1_2a 1+

normal to plane of paperdownwards.

Normal to plane of paperupwards



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S.No

7.

8.

9.

10.

11.

Current carrying inConductor


Fig. 8.18


Fig. 8.19

Straight conductor

Fig. 8.20

Solenoida. Of finite length

Fig. 8.21

b. Of infinite length( = 0,= )

c. Of length 1 andradius a.

d. Long solenoid ( = 90o, = 180o)

Toroid

Position of pointof Observation

At the centre ofthe loop and at

distance a formstraightconductor.

At centre

At point P thewire

Along its axis

Along its axis

At its centre

At its end

Along its axis

Expression of B

B = 0i 1 - 1__2a

B = 0

B = 0

B =0Ni2l

( cos - cos )B = 0ni

2( cos - cos )

B = 0ni wheren = N

1B = 0ni___ 12 +4a2

B = 0ni 2

B = 0niWheren = N

2 RR = mean radius oftoroid

Direction for B

Normal to plane ofpaper upwards

Along the axis ofsolenoid

Direction as above

Along the axis oftoroid



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0 i2 BH


24. Amperes law: If a current of I amp passes through a long and straight conductor.the magnetic induction B Produced by the current at a perpendicular distance r fromthe conductor is IB = 0i x l

2 r B

The direction of B can be determined by the right hand thumb rule.

Fig. 8.3625) A Magnetic pole of strength m revolves around a straight current carrying

conductor in circular path of radius r. The work done by magnetic force onmagnetic pole for n rotations is given by W =0 mni

26) A straight conductor carrying current I in vertically upward direction then distance

of the null Point form conductor, r =

The null point is formed in west side of conductors.

27) Two infinite long thin insulated straight conductors charringcurrents i1 and i2Lie along x and y axes respectively as shown infig. 8.37 The location of thePoint where resultant magnetic field

of system is zero will be given byi1 = i2y x fig 8.37

if i1 = i2 then y = x

28) Flemings left hand rule : Stretch the fore finger, middle finger and the thumb ofthe eft hand mutually perpendicular to each other. If the fore- finger represents thedirection of the magnetic field andthe middle finger that of the force onthe conductor. (Fig. 8.38 a & b). The

above rule is true for only +vecharged particle. For ve charged particle. For ve charged particle,reverse the direction of force afterapplying the rule.

Fig. 8.38 (a) & (b)

29. Magnetic force on a charged particle: If a charged particle of positive charge qtravels with a velocity V an anglewith the direction of the mangnetic field ofinduction B, particle



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experience a force,F = Bq V sin

a) The direction of F can be determined by the relation F = q (V x B)

b) The direction of F can be determined by Flemings left hand rule.(This rule applies only for + ve charges. If it a ve charge, say an electron, its

direction motion should be reversed to apply Fleminss left hand rule to get theproper direction of B)c) The S. I unit of B is Tesla (T)d) 1T = 1 NA-1 m-1.e) The magnetic induction is said to be 1T if a charge of 1C moving with a speed of

1ms-1 at right angles to the field experiences a force of 1N.f) The C.G.S unit of B is gauss (G)

1G = 10-4 Tg) Magnetic induction B is a vector quantity.

Case (i) : When the charged particle is either at rest or moving parallel to magneticfield, thena. The magnetic force acting on it is zero Fm = 0b. The path of the particle will be a straight line i.e., particle will keep on the same path.c. The value of momentum (p) and kinetic energy remain constant.

Case (ii) : When the charged particle is moving at right angles to the magnetic filed :a. The magnetic force acting on the particle will be maximum i.e., F m = qVBb. The direction of Fm will be normal to the velocity of particle.c. The path of the particle will be circular.d. The momentum of the particle will remain constant magnitude but its direction

will constantly be change i. e.,p = qBre. The kinetic energy of the particle remains constant

Ek = mV2 = q2B2r22 2m

f. The magnetic force acting on the particle provides it necessary centripetal forcefor its circular motion.

qVB = mV2r _____

g. The radius of circular path of the particle r = 2m Ek

qBh. Angular velocity of charged particle = qBm

i. Time period of charged particle T

i) T = 2 = 2 m qB

ii) This does not depend on the speed of charged particle.j. Frequency of charged particle.

i) f = 1 = qB

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T 2 mii) this does not depend on the speed particle.iii) It depends on specific charges q /m or the nature of particle and on B.

k. The work done in revolving the particle in circular path is zero.

W = zero

l. The ratios of quantities for two particles in magnetic field.i) r1 = q2 m1 when Ekand B are constant.

r2 q1 m2Ek1

ii) r1 = q2 m1 , when Ekand B are constant.r2 q1 m2 Ek2

iii) P1 = q2 r1, when B are constant.

P2 q1 r2

iv) 1 = q1 m2 , when B are constant. 2 m1 q2

v) T1 = m1 q2 , when B are constant.T2 q1 m2

vi) f1 = q1 m2 , when B are constant.f2 m1 q2

vii) Ek 1 , = q21 r21 m2 when B is constant.

Ek m1 q22 r22 2

Case (iii) : When the particle enters the magnetic filed at anangle of with B

a) The path of the particle will be helical.b) F = qVBc) Pitch of the path : The linear distance covered by

the particle in one time period in the direction of fig 8.39

magnetic field is defined as pitch.Pitch = TV cos = 2mV cos

qBCase (iv ) : When the charged particle enters crossed clectricmagnetic fileds.

a) If FB = FE then charged particle passed undeviated.b) The velocity of such a particle is given by

qVB = qE



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V = EB Fig. 8.40

c) The momentum of the particle p = qBr.

30. Difference in the behaviors of electric field and magnetic filed :

1) The electric field always acts whether particle is at rest (or) motion but magneticforce acts only when the particle in motion.

2) In electric field F = qE

In magnetic filed F = q ( V x B )

3) The speed of particle does not charge in magnetic filed.31. For constant velocity of charged particle, the region space may have

1) E = 0, B = 0

2) E = 0, B 03) E 0, B0 (if E B and v = E/B)

32. Force between two moving charges :i) If charges are stationary then there is only electric force between them(Fe= 9 x 109 q1 q2 )

d2

ii) If charges are in motion a magnetic force also acts between them in addition toelectric force.

iii) The magnetic force between two charges travel with velocities v1v2andseparated by distanced is F m =0 q1 q2 v1 v2

4 d2iv) The direction of magnetic force acting between two charges depends on (a)

nature of charges (b) direction of motion of charges.v) If q1 and q2 are similar nature. and moving in same direction then magnetic

force is attractive vi) If q1 and q2 are similar nature. and moving in oppositedirection then magnetic force is repulsive.

vii) If q1 and q2 are opposite in nature. and moving in same direction thenmagnetic force is repulsive.

viii) If q1 and q2 are opposite in nature. and moving in same direction thenmagnetic force is attractive

ix) Magnetic force < < electric between two moving charges.x) If V1 = V2 = V then Fm = V2Fe C 2

Where C = velocity of light.

33.Magnetic force on a current carrying conductor:If a current carrying of length l is placed making an angle with the direction of the magnetic field of induction B , theforce exerted by the magnetic field on the conductor (Fig.8.41),

F= BI i sin


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where I is the currenta. If = 00 then F = 0b. If = 900 then F = Bil Fig. 8.41

34. Force between two parallel conductorscarrying currents: If two straight

parallel and infinitely long conductorscurrying currents I1 and I2 amprespectively in the same direction andare separated by a distance r1 there is aforce of attraction between them.If the currents are in opposite direction [Fig 8.429b)]there is a force of repulsion.The magnetic induction B1 exerted by the conductorCarrying the current I1 at a distance r isB1 = 0. I1

2

r fig 8.42If l is the common length of the magnetic fore experienced by the secondconductor due toB1 isF = B1 I2 I or F = 0. I1 I2 l

2 rAn equal force is exerted by second conductor on first. The force of attraction or

repulsiorPer unit length of the conductor is

F = 0. I1 I2Nm-1

2 r

If I1 = I2 = I amp and r = 1m, thenF = 0. = 4 x 10-7 = 2 x 10-7 N.

2 2

Definition of ampere: An ampere is that steady current when flowing in each of twolong Straight parallel wires separated by a distance of 1m causes each wire to exert aforce of 2 x 10-7 N per unit length of the wire.Note: In case of non parallel currents

i) When two current approach a point (or) they flowaway from that point then force between them is

attractive (Fig. 8.43) Fig. 8.43ii) When one current approach a point and other currentthey flow away from that point then force between themis attractive (Fig. 8.44)

Fig. 8.4435. i)The work done when the distance between them increased to 2r,

= 0 i1 i2 l loge2 .2 Fig. 8.45

ii) A rod of length bcarrying a current i1 is placed in the field of along wire carrying current i2 as shown in fig. The force on rod is


9448301445

(a) (b)


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F = 0 i1 i2 I loge 1 + b2 a Fig. 8.46

The force F is directed vertically up.36. A horizontal wire carries current i1below which another wire

Carrying a current i2 is kept at d distance. If the wire kept

below hangs in air then,1) i1 and i2 must be flowing in same direction2) F = mg

0. i1 i2 l = mg2 d

37. If a magnetic pole of strength m is kept at distance r form a current carryingconductor then

force on it, F = mB = m x 0 i2 r

38. Torque on a current loop : A rectangular coil of area A containing n turns is place

in a uniform magnetic field of induction B.i) If is the angle between normal to the plane of coil and magnetic field then

orque on coil = BiAn sin = MB sin (or) = M x Bii) If = 00 (or) 1800 then = 0 (minimum)iii) If = 900 then = BiAn (maximum)iv) The work done to turn the coil form angle1 to2 is W = MB (cos 1 - cos

2)

Fig. 8.48

(a) If = 00 and = 0 then W = MB (1 - cos) v) P.E of the coil u = - MB cos = - M .B

(a) The change in P.E of coil when the coil is rotated form 0 to 1800 = 2 MBvi) If is the angle between plane of the coil magnetic field then Torque on coil = BiAn cos.

39. Moving coil galvanometer:


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a) Principle: When a current carrying coil is placed in a uniform magnetic filed itexperiences a torque.

b) It is used to measure low currents upto 10-9 A.c) It consists a powerful circular magnet with concave poles. So that the magnetic

field becomes radial.

d) Phosphor bronze wire has high value of. Y and low value of n (rigiditymodulus). So it is used as suspension wire.

e) By placing a soft Iron cylinder inside the coil magnetic flux of the filedincreases.

f) Deselecting coupe = BiAng) Restoring couple = C

where is angle of twist of wire in radian.h) In equilibrium position, BiAn = C

I = C BAn

I = KWhere K = C/BAn = galvanometer constanti) In M.C.G. ,I j) Current sensitivity of the moving coil galvanometer can be increased by

(i) increasing n (ii) increasing A (iii) increasing B and (iv) decreasing C.k) The reciprocal of current sensitivity is called figure of merit.i) Voltage sensitivity = = = BAn

V iR CRWhere R = galvanometer resistance.

40. Tangent galvanometer :i) It is based on Tangent law (B = BH tan )

ii) It is used to measure currents up to 10-6 Aiii) The plane of coil is must be in magnetic meridian.iv) I = K tan where K = 2rBH

0n

41. Comparison of M.C.G with T.G :

M. C. G T. G.


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G


1.2.

3.

4.

5.6.7.

In M.C.G, I It is a moving coil and fixed magnet typeGalvanometer.It can be measure the currents of order of10-9 A

The plane of coil need not be set inmagnetic meridianIt can be used in minesIt is not portable.The galvanometer constant does notdependOn BH.

1.2.

3.

4.

5.6.7.

In T.C.i, itanIt is a moving mangent and fixedcoil type galvanometer.It can be measure the currents oforder of

10-6 AThe plane of coil should be set inmagnetic meridian.It cant be used in mines.It is portable.The galvanometer constant dependOn BH.

42. Shunt: As the galvanometer is a low resistance instrument, it is damaged when a

large current is passed through it.To protect the galvanometer, a small resistance

known as shunt is sonnected in paralle with it. LetG be the resistance of the galvanometer and S bethe resistance of the shunt and let I,Ig and Is be thecurrent in the main circuit, the galvanometer andshunt repectively(Fig 8.48) then

i) Ig = IFig. 8.49

ii) The current through the shunt, Is = I

iii) The fraction of current passing through the galvanometer =

iv) The fraction of current passing through the shunt =

v) The total resistance R =

vi) The decrease in resistance of galvanometer due to shunt = G -

vii) The decrease in resistance of shunt = S -


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S S + G

S S + G

S S + G

GS G + S

GS G +

GS G + S


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G


43. Ammeter :

i) It is a direct current reading instrument.ii) An ammerter should always be connected in

series in a circuit.iii) The resistance of the ammeter should be as

small as small as possible.iv) The resistance of an ideal ammeter is zero.

Fig. 8.50A galvanometer can be converted into an ammeter by connecting a suitable shunt asshown inFig. 8.50I = total current to be measuredIg = current passing through the galvanometerIs = I Ig = current through the shuntG = resistance of the galvanometer.S = shunt resistance.

i) I = Ig + Isii) S = Ig GI Ig

iii) The range of resistance G can be increased from i1 to i2 by connecting a resistanceof

S = G__ in parallel where n = i2n - 1 i1

44. Voltmeter:i) It is instrument used to measure p.d. or

voltage or e.m.f.

ii) It should always be connected in parallelto a circuit.

iii) The resistance of the voltmeter should be aslarge as possible.

iv) The resistance of an ideal voltmeter is infinity. Fig : 8.51


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Ammeter

G

Voltmeter


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v) A galvanometer can be converted into a voltmeter connecting a suitable highresistance R in series with it shown in fig 8.51

vi) If V is the voltage range of the voltmeter, thenV = Ig (R + G)

Where Ig is the maximum current that can be passed through the voltmeter.Vii) The value of R can be calculated by using the formula

R = G.

viii) To increase the range of a voltmeter of resistance R from V1 toV2 ,the

resistance to be connected in series to the voltmeter isR1 = G (n 1)

Where n = =

45. i) If an ammeter of range I and resistance R is to be converted into voltmeter of

range V then high resistance R1 to be connected in series is R1 = -R

ii) If an voltmeter of range V and resistance R is to be connected into ammeter of

range i then low resistance s to be connected in paralle is S =


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VI

g

V2

new range

V1 old range

Vl

RV


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ELECTROMAGNETIC INDUCTION

1. Electromagnetic induction discovered by Faraday.

2. Magnetic Flux ( ) : The magnetic flux is the total number of magnetic lines of

force passing through a given area normally. It the closed area A is perpendicularto the magnetic induction, the magnetic flux is

= BA (1) (fig. 8.52 (a)

If the magnetic filed (B) makes an angle with the outward normal to the areaA, the Magnetic flux is,

= BA cos (1) (Fig. 8.52 (b)

SCAN

The S.I unit of magnetic flux is weber (b) [In C.G.S. system, the unit

of magnetic flux is Maxwell and 1 weber (wb) = 108 Maxwell (mx)]

3. Direction of induced e.m.f. and induced current:

The various position of relative motion between the magnet and

the coil are represented in

The following table .


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SYNOPSISSYNOPSIS


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Position of

Magnet

Directio

n of

Deflectio

n

Direction

of

current

in coil

BehaviourOf face of

coil

Type ofmagneti

cforce

opposed

Magnetic

field

1. When thenorth pole ofmagnetapproaches thecoil

Fig. 8.53

2. When thenorth pole ofmagnetrecedes awayform the coil

Fig. 8.54

3. When thesouth pole ofmagnetapproaches thecoil

Fig. 8.55

4. when thesouth pole ofthe magnet

Towardsleft

Towardsright

Towardsright

Towardsleft

In anit -clockwisedirection




As a northpole

As a southpole

As a southpole

As a northpole

RepulsiveForce

Attractiveforce

Repulsiveforce

Attractive

Increases

Decreases

Decreases

Increases


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4. If a bar magnet is dropped vertically into a metal ring then its acceleration is lessthan g. If the ring is cut (or we take broken ring) then acceleration of magnet is equalto g.

5. Faradays laws of electromagnetic induction:

a) Whenever there is a change in magnetic flux passing through a coil, an inducede.m.f is developed in it.

b) The induced e.m.f. lasts as long as the change in magnetic flux takes place.

c) The magnitude of the induced e.m.f. developed in a directly proportion al tothe rate of change of magnetic flux passing through the coil.

e = -

where e is the induced e.m.f. and d is the change of magnetic flux passing throughthe

dtcoil. The ve sign show that the induced e.m.f. opposes the growth of magnetic flux.

d) If a coil of area A containing n turns is placed perpendicular to a uniformmagnetic field of induction B, then,

= n AB.

Hence e = - d (nAB)dt

e) If the originalmagnetic flux is 1 and the final magnetic flux is 2 in a time t,e =

i) If a coil containing magnetic flux is turned quickly through 900, 1 = and 2= 0 in a time, t

e =

tii) If the rotates quickly through 1800, 1 = and 2 = - e = 2

tiii) If R is the resistance of the circuit in which the induced e.m.f. is developed. Theinduced current is,

I = e = - 1 dR R dt


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ddt

- 2 -1dtt


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iv) The induced electric change q developed is,q = I dt = - d = - (2 - 1)

R R6. Lenzs law:

i) The direction of the e.m.f. and the induced current can be found by Lenzs law.ii) Accounding to this law, the direction of the e.m.f. is such that is opposes the

motion which produces it.

iii) Lenzs law is a consequence of the law of conservation of energy.

7. Induced e.m.f. and determination of its direction:

A) Various formulae of induced e.m.f.:

a) The induced e.m.f. generated on account of rotation of conducting rod in aperpendicular magnetic field.

i) e = -

ii) e = -BAf where f = frequency of rotation and A = L2,

= angular velocity, L = length of conduction rod. Fig: 8.57b) Induced e.m.f. generated in a disc rotating with a constant angular velocity in a

perpendicularmagnetic field.

i) e = -Br2f =

ii) e = - BAf where

A = Area of disc

r = radius of disc = angular velocity of disc Fig. 8.58

iii) If the direction of is anticlockwise then the negative charge accumulatesat the rim and positive change accumulates at the centre.

iv) The e.m.f. is induced between the centre and the rim of the disc.


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B L22

B L22


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c) Induced e.m.f. generated due to linear motion of a rectangular coilin a uniform magnetic field. Fig. 8.59

i) = 0 e = 0

Here the coil is moving with a constant velocity in a uniform

magnetic field. and e will be zero.d) Induced e.m.f. generated on account of linear

motion of a coil in a uniform finite magnetic field

.

Fig. 8.60

i) On displacting the coil towards right, that area of the coil changes which emergesout of the magnetic field, i.e, A = - lvt

ii) Because = B A = - Blvt therefore e = - Bvl

e) Induced e.m.f. generated on account of linear motion of a conducting rod in aperpendicular uniform magnetic field.i) Induced e.m.f. generated in the rode = Blv

ii) If the rod moves making an angle

with thedirection of magnetic field. Thena) e = Blv sin

b) e = - ( v x B ) l

c) e = B ( l x v) Fig. 8.61

d) If the rod moves along the direction of magnetic field then no e.m.f. is induced.

iii) To determine direction of induced current:

a) The direction of induced current in the rod is given by Flemings right hand ruleRule : Stretch the index finger, middle fingerand the Thumb of right hand mutually at rightangle to each Other. If the index finger pointsin the direction of magnetic field, thumb

points in the direction of motion of the rod


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X X X X X

X X X X X

X X X V X

X X X X X

X X X X X


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then the middle finger represents the directionof induced current in the rod.

Fig. 8.62b) If the direction of magnetic field is at right angles to the plane of paper upwards,

then the direction of current in the rod the conducting rod is form Q to P.c) If the direction of magnetic field is normal to plane of paper downwards, then

the direction of induced current in the rod will be form P to Q.

iv) F out of l , v and B any two parameters are in same direction then e = 0

v) If the conducting rod is in horizontal position and moves in a horizontal directionthen an induced e.m.f. is generated in rod due to the vertical component of earth'magnetic field.

vi) If a horizontal rod in east - west direction is falling vertically downwards then apotential difference is generated due to horizontal component of earth' magneticfield.

f) Induced e.m.f. is generated in a conductor moving on two conducting rails in auniform magnetic field :

i) Induced e.m.f., e = Blva) Induced current, I = e

Rb) Force on conductor, F = B2l2vR

c) Power expended in moving the conductor p = Fv = B2l2v2R

g) Induced e.m.f. generated a rectangular loop moving in a non-uniform magneticfield .

i) e = - (B1 B2) vl

ii) e = - (B1 B2) x l t

iii) If the loop is moving in a non uniform magnetic field with gradient dB ,thendz

e = - vA dB = lv (B1 B2)dz

iv) If B changes with t and z, then


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e = AB + vAB = A B + vBt z t z

v) The induced e.m.f. opposes the change of current in the circuit.

Note: If a rectangular coil is entering or coming out of a magnetic field then also allabove formula of e will be valid.

8. Self induction: The phenomenon of production ofe.m.f. in a coli due to change of current in same coli

is called self induction.Consider a coli through which a current of a I ampereflow. The magnetic flex produced in the coil is

proportional to the current I, when the permeability Fig. 8.65of the medium remains constant.

or, l

or, = Ll

where L is a constant of proportionality called the co-efficient of self induction o

self inductance of the coil.Form Faradays law the induced e.m.f. in the coil is

e = - d = - d (Ll)dt dt

e = - L dIdt

if dI = I; L= e (numerically).

Thus the self inductance of coil can be defined as numerically equal to the inducede.m.f. (in volt) in the coil when the current in the coil change at the rate of oneampere per second.

The S.I unit of L is Henry (or) Volt secAmp

9. A straight conductor has no self inductance.


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If a rod (r) is kept inside the solenoid then L =o rn2A1

18. Mutual Induction: The Phenomenon of the production of e.m.f in one coil whenthe current in another coil changes is called mutual induction.

i) If the current in primary coil (p) is increased, negative e.m.f.is induced insecondary coil (s).

ii) If the current in primary coil is decreased positive e.m.f. is induced in secondarycoil.

iii) The flux linked with secondary coil ( ) is directly proportional to the currentfloming in primary coil.

i = Mi

where M is a constant of proportionality called the mutual inductance or co efficient of mutual induction of the two coils.

According to Faradays law of induction

E = d = - d (MI)dt dt

e = - M dI = 1 : M = e = ( numerically).Dt

Thus the mutual inductance of two coils may be defined as numerically equal tothe e.m.f induced one coil when the current in the second coil changes at the rateof one ampere per second. The S.I Unit of M is Henry.

19. M depends upon

i) number of turns in the coilii) distance between coils

iii) Geometrical shape of coils

iv) The angle between the axes of the coils.v) The permeability of the core material inside the coil

20. If L1 and L2 are the slef inductance of two coils then M = K L1 L2where K = coefficient of couplingK = 1 for tight coupling, 0 < K < 1for loose coupling.


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21. Mutual induction of two solenoids: If Np is the number of turns in the primaryof a solenoid and Ns the number of turns of the secondary solenoid linked with

primary then the co-efficient of mutual induction of the two solenoids is,

M = 0NpNsA1

where A is the common cross sectional area of the two solenoids, l is the length ofthe Primary solenoid.

22. Transformer:

1) It is based on the principle of mutual induction.2) Transformer works on ac and not on dc.3) It can increase or decrease either voltage or current but not both

simultaneously.4) The transformation ratio (or) turns ratio, K= Ns = Es = s = Ip

Np Ep p Is(a) For step up transformer, K > 1

(b) For step down transformer, K > 1

(c) For step transformer,

Fig. 8.66

5) Step up transformer:a) It converts low voltage at high current into high voltage at low current.b) Primary coil is made of thick wire and secondary coil made of thin wire.c) The number of turns in secondary coil greater then that of primary coil.

6)Step down transformer:a) It converts high voltage at high current into high voltage at high current.b) Primary coil is made of thin wire and secondary coil made of thick wire.c) The number of turns in primary coil greater then that of secondary coil.

7) Efficiency of a transformer:There are some losses of energy due to primary coil resistance, hystersis in the

core, eddy Currents in the core, ect. Ordinary transformer. The efficiency of atransformer is defined as the ratio of output power to input power.

= output power = Esisinput power Epip

Percentage of efficiency = output power X 100


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input powerFor an ideal transformer, = 100% Due to energy losses, the efficiency of

ransformer varies Between 90% and 99%.

23. Growth and decay of current in L R circuit:1) The current in the circuit at any time t during growth of current, I = I0

-Rt1 e L

where I0 = maximum value of currentL = inductance of inductorR = resistance of resistor

a) It t = L then I = 0.632 I0.R

b) The value of (L/r) is known as inductive time constant and is expressed by (1/ ) I = I0 [1 e -v ]

c) Greater in the value of time constant, smaller will be the rate of growth of current.

2) The current in in circuit any time during decay of current. -Rt

I = I0e L = I0e -v

(a) If t = L then I = 0.368 I0

R(b) For small value of L the rate of decay of current will be large.

24. Growth and decay of charge in C R circuit:

1) The charge on capacitor at any time T during charging.

q = q0 [ 1 e -vCR]

where q0 = maximum value of charge

C = capacity of capacitor

R = resistance of resistor

(a) If t = CR then q = 0.632qo

(b) The value of (CR) is known as capacitive time constant.


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(c) Smaller in the value of time constant (CR), more rapid is the growth of chargeon capacitor.

2) The charge on capacitor at any time t during distance discharging, q = q0 e -VCR

(a) If T = CR then q = 0.368 q0(b) Smaller the time constant, the quicker in the discharge of capacitor.

3) The voltage during charging of capacitor, V = E0 [1e-VCR ]

4) The voltage during discharging of capacitor, V = E0 e-VCR

5) The voltage during discharging of capacitor, I = I 0 e-VCR

ALTERNATING CURRENT (AC)

1. An alternating current or voltage is one magnitude and directions vary periodicallywith time. Alternating current shows only heating effect. Therefore meters used for acare based on heating effect and are called hot wire meters.


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SYNOPSISSYNOPSIS


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2. If the current or voltage varies periodically as sin or cos function of time thenthe current or voltage is said to be sinusoidal.

3. The value of current at any time t (instantaneous value) is given by I = I 0 sint(or) I = I 0 cos t

4. The value of voltage at any time t is given by V = V0 sin t where V0 is the peakvalue of alternating voltage.

5. The frequency of in India is 50Hz, i.e., f= 50 Hz so = 2 = 100 rad/ s andtime T = 0.002sec

6. The average (or ) mean value of ac is defined for positive or negative half cycle

1 T/2 1 T/2I = T/2 Idt = T/2 I0 sin t dt

0 0

/

= / I0 sin t dt = 2I0 T = 20

I (or) Imean = 2l0 = 0.637 I0

a) The mean value of alternating current during, positive half cycle is 0.637 times or63.7% of its peak value.

b) V or Vmean = 2V0 = 0.637 V0

7. The effective (or) virtual (or) root mean square (r.m.s) value of alternating current

is defined as the square root of the average of I2

during a complete cycle55 T I mean = 1/T I2 dt = I0 = 0.7071I0 0 2(a) The r.m.s. value of alternating current over a complete cycle is 0.707 times.

(Or) 70.7% of its peak value.(b) Vms = V0 = 0.707 V0

2(c ) All insterments measure virtual values of ac.8. Form factor of alternating current, F = Irms =

I 22

9. Peak to peak value of alternating current , I pp = 2I0 = 22Ims = 2.828 ImsPeak to peak value of alternating voltage, Vpp = 2 V0 = 22 Vms = 2.828Vms

10. The opposition for the flow of ac through resistor (or) inductor (or) capacitor (or)combination of them is called Impedance (z)(a) Its unit is ohm. Its value depends on frequency of alternating current or voltage(b) Formalae for Impendance


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i) Z =

ii) Z =

iii) if is the phase difference between alternating voltage and alternatingcurrent then

Z = =

11. The opposition for the flow of ac due to induced current is called Reactance (x).(a) Its Unit is ohm. Its value depends on frequency of alternating source,

inductance and capacitance.(b) There are two types of reactance :

i) Inductive reactance (xL) is the resistance offered by an inductor and is given byXL= L = 2fLIf = 0 then xL = and if = then xL = . Therefore the opposition of

inductor todc is Zero and it ac as a conducting wire. The inductor is called low pass filter.

(ii) capacitive reactance (xc) is the resistance offered by a opposition of capacitor to dc

is infinite and it acts as open circuit. The capacitor is called high pass filter.

12. The reciprocal of Impedance is called Admittance. Its Unit is mpo.

13.The reciprocal of Impedance is called susceptance. Its unit is same as that of admittance.

14. The close path through which the ac flows is called ac circuit. When ac flowsthrough the circuit, the current and voltage need not attain their maximum valueat the same instant. If the current reaches maximum first then current is said tolead the voltage or voltage is said to lag current.

15. A resistor in an ac circuit :The instantaneous current through the circuit, I = V

R= V0 sint = I0 sin t Fig.

8.69


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Vms

Ims

Vpp

Ipp

Vms

Ims

Vpp

Ipp


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RThe frequency of current in circuit is same as applied voltage (v). There fore thevoltage is in phase of current in circuit.

16. An Inductor in an ac circuit:

According to Kirchoffs loop rule, V L dI = 0 (or) dI = V0sin tdt dt L

I = V0 (-cos )= Vo sin t - Lw 2

I = I0 sin t - Fig.8.70

2The instantaneous current through the circuit,

I = I0 sint -. The current through the inductor lags (due to

2Negative sing in front of /2 the applied voltage by rad (or) voltage

2 Fig. 8.71leads the the current by The phase different between current and

2voltage is shown graphically in Fig.Peak value of current, I0 = V0 = V0

xL L

17. A capacitor is an ac circuit:According to Kirchoffs loop rule, V q = = (or) q = cv0 sint

CI + dq = c V0cos t = I0 cos t = I0 sin t +

2

The Instantaneous current through the circuit, I = I0 sin t + Fig.8.72

2

The current though the capacitor leads the applied voltage (V) by/2 rad. The phase difference between current and voltage is showngraphically in fig.

Peak value of current, I0 = V0 = V0c


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V_______ R2 + 12

c


18. Analysis of series ac circuits :

To calculate I and without dealing the theory the following procedure can beadopted.

(a) Specify the total resistance R of the given circuit which is independent offrequency

(b) Calculate the reactance in circuit, x = xL xC

(c) Calculate the Impedance in circuit, Z = R2 + x2

(d) The peak value of current, I0 = V0Z Fig. 8.74

(e) The phase difference () between I and V is given by tan = x

R(f) In case of ac circuits V VL + Vx but V2 = VR2 + Vx2 andV VL + Vx but Vx = VLVc but

19. ac applied to C-R series circuit:

1) Reactance of circuit, X = xL xC__=_ 1 c

2) Impedance of circ uit, Z = R2 + x2 = ( R2 +[1/ c]

3) Current in circuit, I = V =Z

4) Peak current, I0 = V0Z5) Phase difference, = tan-1 x = tan-1 1__

R cR6) The instatneous current through the circuit, I = I0 sin( t - )

7) The current leads the voltage (V) by angle

8) p.d across R, VR = IR


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Or resonant frequency

I = V = V_______ L 1 2

cI will be maximum if L = 1 = 1

c LCAt resonance, = 0 1__

LC Resonant frequency, 0 = 1__

2 LC22.ac applied to C-L series circuit:

1) The current in circuit at any time,I= I0 sin (t )(a) If xL > xC then the current lags the voltage by (b) If xL < xC then the current leads the voltage by

(c) If xL = xC then the current is in phase of voltage. Fig. 8.79

2) p.d across R, VR = IRp.d across L, VL = IXC = I

cp.d across L, VL = IXL IL ___________p.d across R and L, VRL = VR2 + VL2

___________p.d across R and C, VRC = VR2 + VC2 Fig. 8.80p.d across al and C, VLC = VL VC

= VLVC if VL > VC= VC VL if VC > VL= 0 if VCVL. Fig. 8.81

________________Total p.d of R.L and C,V =VR2 +( VLVC)2

3) Reactance of circuit, x = XL XC = L 1 C

4) Impendance of circuit, Z = R2 + X2

5) Current in circuit, I = VZ6) Peak current, I0 V0

Z

7) Phase difference, = tan-1 xR

8) Average electric field energy stored in capacitor = 1 CV2

29) Average magnetic field energy stored in inductor = 1 CV2


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2

23. Resonance in L C R circuit:Resonance takes place in circuity when Impedence (Z) of circuit is minimum (o)

the current (I) is maximum.

I = V = V_______Z R2 + L 1 2

CI will be maximum ifL = 1

c = 1

LCAt resonance, = 0 = 1_

LC Resonant frequency,0 = 1__

2 LCResonant frequency is independent of R.At resonancei) The reactance of circuit, x = xL xC = 0ii) = tan-1 x = tan-1 O = 0

R RSo current in the phase of applied voltage (V)

iii) Power factor, cos = R = 1 (maximum)iv) Current through L and C are same but 1800 out of phase with respect to each

other so that

V = VR because VLC = 0v) Impendance of circuit is minimum and is equal to Rvi) The current depends on R and not on L and C. I0 = V0

Rvii) VL = I0 xL = V0 ( 0 L) = QV0 Q = 0L

R RWhere Q = quality factor (or) figure of merit of resonant circuit. It is an

Indicator of sharpness of current peak [i.e., higher value of Q, sharper is peak] The p.d across L is Q times of applied voltage. This characteristic is calledvoltage magnification,

viii) If f1 is lower half power frequency and f2 is upper half power frequency then(a) Band width = f2 f1(b) quality factor, Q = _f0___

f2 f1where f0 = resonant frequency

ix) The frequency of source at which power of circuit is half of the power at resonantfrequency = f0

2where R

2L


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24. The series resonant circuit is called an acceptor circuit , because if number offrequencies are given to it then the circuit accepts one frequency f0 and rejects allother frequencies. The current will be maximum for that frequency. The seriesresonance is called voltage resonance.

25. A metal detector works on the principle of resonance in ac circuits.26. ac applied to L-C-R parallel circuit :

The voltage V across each element is same. The currents are related as

I = IR2 + (IC IL) 2

Fig. 8.83= V 2 + V C V 2 = V 1 C 1 2

R L R2 L

This shows that at resonance, i.e., when L = 1 , the current is minimum. C27. Power in ac circuits:

The power in an electric circuit is the rate at which electrical energy is consumedin the Circuit. If an ac voltage E = E0 sint is applied to the circuit then currentin circuit, I = sin ( t - ). The Instantaneous power consumed by circuit, P = EI= E0I0 sin t sin ( t - )The power consumed varies with time. Hence in ac circuit we find the average

power consumed by circuit,

T P dt

0_____ = 1 E0I0 cos = Erms Irms cos PCV = T 2

dt0

Erms Irmss is called the virtual (or) apparent power. Cos is called power factor.

cos = True power__

Apparent PowerA wattmeter measures the true power.

cos = RZ

Special cases :


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(a) In case of a pure resistance ac circuit, cos = R = R = 1and Pav = Erms Irms= E0I0(max)

Z R 2

(b) In case of a pure inresistance (or) capacitance ac circuit, cos R = O = 0 and Pav =

0Z Z

If the power consumed by the circuit is zero then the circuit is called ,watt less Thecurrent through the pure inductance or capacitance is called wattles current or Idealcurrent In practice power consumed by Inductor or capacitor is not zero but very small.

(c) In case of L-C-R series ac circuit, cos = R R = R______________Z R2 + x2 R2 + L 1 2

C

and Pav = Erms I

rms cos

= E0I0 cos

2

At resonance of L-C-R series ac circuit, cos = R = R = 1Z R

and PCV = Ems Ims = E0I0 _ = I(rms)2 R(max)2

28. Parameters of various A.C circuits:

S.

No.Parameter C R Circuit C L Circuit L- C- R Circuit

1. Circuit diagram

Fig. 8.85 Fig. 8.86 Fig. 8.87

2. Reactance x = xC = 1

C

X = XLXC= L 1

L

X = XLXC= L 1

c


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3. Impendance Z = R2 + xC 2 Z = R2 + x 2 =x

Z = R2 + (xL -xC )2

4. Peak current I0 = V0__ R2 + xC 2 I0 = V0

x I0 = V0_____R2 + (xL -xC )2

5. Average power Consumed by theCircuit by

Erms Irms R R2 + xC 2

(In practice, it isnot equat to zero

but very small)

Ems Ims R____________R2 + (xL -xC )2

6. Power factor (cos )

I0 = R_________

R

2

+ xC2

0R .

R2 + (xL -xC )2

Magnetic Effect of Electric Current & Mechanical Effct of Electric Current

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