electromagnets notes for a levels

8/2/2019 Electromagnets Notes For A Levels

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MAGNETIC EFFECTS OFA CURRENTA rr:!.agne tic field exists not __nly around a piece of magne t but is also produced by .1! current ina_cQ!ldl!fJOL ~ e s i d e s , there is the field of the earth. In this chapter we will examine how

m_gnetic fields can be produced by _ ___: t i . ~ o i l d u c t i ) i s a i - i o u : s shapes.C ~ f ! _ ' ! - s g y produc_d frqm c_uqent_._T h ~ _evers_ is also possible, Le.

electric currents can be produced using magnetism. This is called electromagnetism .

16.1 Magnetic Field PatternsI f a bar _ g 1 _ ~ g ~ e ! __s_s _ Q e c ! J I ~ e ! y . f ~ < ? . ~ - - ! l : l ! ~ j t __q_~ ~ ~ ~ _ i ~ ~ ~ . f _ i _ c ! s i n t i n o w a ~ __ ! . ! _ g f : l . __This ! " l ! a g f ~ as the ~ ~ k i p . g _ _ R o l ~ 9 J j nshort the north (N) po le. The _ s l __gf _ J n a g q _ e t _ __):cnQwn as _ the . soutl:L.(S.)._po le_(see

F l " g I . -- - . ---------- - ------

NTFig. 16.1 The poles of a magnet Fig. 16.2 The magnetic field around a bar magnet

The magnetic field pattern around a magnet is reviewed by placing the magnet below ~ i e e e _Qf_Q_

___r p . a g n . : c T : ? ~ J ' ! 0 ~ 9 ..?Y.__ - s : . 9 y .. ~ 1 ~ ~ ..i { e ' i i c ~ Q " v ~ ~ a f o n 1 l i 1 _ - : - r f 9 . f ~ ~ ~ f . ~ ~ _~ ~ ~ e c t i os h o ~ ~ a ~ 9 \ ! ' (see Fig. 16.3).


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16.2 Magnetic Fields Due to CurrentsW h ~ ~ ~ - l 1 1 ! e ~ f l . Q ~ ~ j ~ g ~ e : t field. _ o d u c ~ Q .. I . ~ ~ ~ ~ g ~ ~ P . ~ ~ ! ~ r n aroun9 __a long straight wire carrying a current is shown in Fig. 16.5. The direction of the field may bedetermifiecfuSiilg a- sffiair {Jiotting-coinpassplacec neii---iile ~ - - - - - - - - - - ..-- ........--------- -- - - - -

directionof field

I, current

t directionof current

right hand

cardboardplottingcompass

Fig. 16.5 The magnetic fi eld due to a current in a straight wire and the right hand ruleThe ! ! ! - ' ! g Q . ~ . ! i ~ .. l 9 : . ! I L . f . Q ! ! i __Qf...:g.n_reotric eire with p _ a r a t i o n ._be ween. th.eJin. ..s

of force increasing as the distance from the wire increases. This denotes that the field gets weaker- - - - - ~ .. - - - - - - - ..-....- -- __ .________---- - - - ~ - "" ..... . ..... - - - - - - - ~ - - - - - - - - - ---. - . .at a distance further away r e . In a vacuum, the m ~ g _ n . . i ! ! l d _f the ~ a g n e t i c flux density B at a point d i ~ t a n c e r from the wireis

(2I.._,where I is ' ? u q e n t ,

__ !fn


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16.3 Circular Coils

Fig. 16.6 The magnetic field due to a current in a circular coilig: ..!6.__ 9 ~ 2 . ~ ~ ~ g _ ! ? . ~ ~ _ . f i P . T q ~ 1 J C ~ < : L l?Y . s : u ~ e n ! in s : u l a r coiL }he direction of the

may be detennined using the right hand rule. Grasp the left s i(je of the coil with t h ~ _humbo"lnttiig-upwru=as-'"li1' .the -cfirectiori of" the 'current, a ~ d the f i ~ g e r s will curl in the j . of the_r:--slm1Iarly, grasp the iigh tsTcfeof the -co1f wi_h __the t h u O : : i b i i t cl9;;D; ard_ and the

. , - - - - - - , ; - - - - : : - ~ - - = - - = - : - = o - - - - -- - ---- - . --.- - - - -- . .. .. . . - .. .. . .. .ction _ f _ f i ~ l ~ _ ! the cl!!. of . f i J l g ~ r ~ -f r = radius of the coil, - ~ =:]Y__ ~ U ! ~ } ? e ~ _ i ~ ~ m S , and

I = cuiTent, /m a g n e i l c flux dens ity _ ? ! ~ t e n t r e of the coil is


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16.4 Long SolenoidsWhen a current flows in a solenoid, the magnetic field pattern produced is identical to that of abar magnet (see Fii 1 6 ~ 7) Hence. the solenoid haspoiesjus tl1ke . m a g n e t . To remember the_ole ~ c h e ~ = o f t . . - ~ } ~ ~ ~ J C f c s > . . ! l t - . ~ ~ d . J i : < ; > i l } ~ - ~ h ~ 9 i f ~ f 9 . L t b . ~ ..Qitrrintflow, we can d e d l ! _ th:_PC?._le. JJ

lf_jhe current flows in the a n t i c l o c k V : ~ ~ ~ ~ r ~ ~ : _ ~ ~ : _ ! ~ e end of the solenoid is a N-pole (_.1_:if the current flow is clockwise, then the end of the solenoid is a S-pole ( S see Fig. 16.7).__ ._______________ _____._________------------ ..------ -- - - -

I I

Fig. 16.7 The magnetic field due to a current in a solenoidNo tice that the lines of force inside the solenoid are parallel. Hence the magnetic field inside

the s o r ~ n . ~ i js uniforin. y f i ~ d u . - e c t i o n of the magnetic fleld inside . " 8 9 ~ J . i o . i ( j . _ 0 t t L " ~ : e d e d u c e d fr_Qm ~ h e po)_t_rity at t l ...C ! l c ! ~ of th.e ~ n o ~ d d i s c u s s ~ d . ~ b . ~ ~ ~ - ' : l s ~ _ g the righ t hand rule::.- Just grasp any one of the ~ l : g t_!_le e ~ ~ j i ~ h the_Iight hand t h a t ~ ! ~ _ ! ~ u m . b p o i n tin the direction of the current in the coil and the fingers curl into the solenoid . Then the direction

- ; r - t i e f i e c ( l n - n o i d follows the c l : [ i - ( o l r ~ e - f i f l . ~ s ~ 9 j : _ ~ T 6 ) 3 . -- --

/

right hand

I~ o f solenoid

direction of field

curl of fingers

Fig. I6.8


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..r n ~ g n e t i c flux density B within the solenoid depends on the current I?and n, the numberof tums I y n j ~ ? . ! j o l ~ n o i . d ~ is-gi\,eii -ilie e x : p r ~ s ~ i o n ---- - --- - ----- .. - --

:/'" IB = ,U0nl I--------- --' The magnetic_lLDes of - - ( ~ f u .:..J . ~ ! _ _ ~ h e _ ends of the solenoid spread out This showsthat the magnetic flux density at the ends is weaker. At the ends of the solenoid, the magneticriux C i e ~ s h Y . __i l ~ i i t " _ T l : l l ! ! f l . ~ b ~ ~ J ~ r i 0 1 c t , _i:e.:: ~ e t ; i : c f i u ~ ~ n s 1 i ) r auiieei1JS of thesolenoid

A

(a)1---- - - -+-X

(b)0

Fig. 16.9

III_-- - 1I\ B ' =flux density at the endsI'I

__ _ ~ ! ! : h ~ ~ __ e p < _ ? j _ ~ t w i ~ ~ _ _ ! ~ ] ~ ! __ t the ends because a long solenoid can beconsidered as being made up of two sections joined at A. The resultant flux density at A in16.9(a) is the vector the flux de nsities B' at the end of each of the two coils. The

directions of the flux densi ties are the-sarrie:Hetice the resulta-nt-fluxdensity - point within- - -a-- -- - - --------- -- - - --the solenoid, is B = B' + B' = 2B'.Fig . 16.9(b) shows the variation of the flux density along the length of a long solenoid.

If- a-so_t_i ~ o ~ piece is nset1eCi in-to the solenoid, the flux density"" within the solenoid is givenby ffie expression -

I B = J.lrJlonl Iwhere J.1 r is a constant known as the relative permeability of soft iron. The value of J.1 r for softlrOI1IS 37'0- bi.ii some alloys containing 1i:-on-;TiickeT-and cobaltlnayhave the value of Pras -highas 1 Hence phicing- an ..rron-bar-in-Side a olelioidTnc i-eases._ihe inagneilcflux-d-ensity to ai?ili __ ~ ~ ~ ~ 2 : : a n:aterjal ~ ~ _ ? ~ l ' f i ~ a h 1 e j J I ~ ~ s ~ r u n 5 n e , ~ ~ ~ 2 : ~ ~ y - ' 0 ' : 9 9 9 " " 9 9 .


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.,... Fig 15.4 Magnetic forceacting on current-carrying

conductor

current-carrying conductor in aMagnetic Field Wl1.en a conductor carrying a current I is placed in a magnetic

field B, a magnetic force F that is perpendicular to both thecurrent and the field will be exerted on the conductor (Fig 15 .4).

r-F ---- ' ( J I ~

"2 r. te /.'C / ;; .J.r o ./ ..wA/ tvfJ I

The magnitude of the magnetic force on the conductor inFig 15.4 is

F = BH'sin ewhere B is the magnetic flux density, ; _ _ -

I is t h e conventional current,l is the length of conductor a n ~ e is the angle between the direction of the c ~ t andm 9 - ~ e t i c flux density.

(a) When e= 0 or e= 180,F = BIZ sin oo = BIZ sin 180 = 0.No force acts on the conductor whenever the currentflowing through it is oriented parallel to the magnetic field(Fig 15.5) .

.,.. Fig 15.5 No force is exertedon conductor

(b) Wl!en e= 90, F = BIZ sin 90 = BIZ (Fig 15.6a).The magnitude of the magnetic force is

F= BIZWl!en e= 2 ~ 0 , F =BIZ sin 270 = - ~ B I Z (Fig 15.7a).Maximum magnetic force is exerted on the conductorwhenever the current flowing is oriented perpendicularly tothe magnetic field.


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...,. Fig 15.6 Maximum force isexerted at () = 90

Fig 15.7 Maximum force isexerted at () = 270

Maximum force F

(a)

Maximu m force F(a)

Thumb( Force or Motion )

(b)

First finger(Magneticfield B )

Second finger( Convent ionalcurren t I)

Second finger(C on ventional current 1)

Thumb( Force or Motion )(b)

Fleming's Left Hand Rule is a convenient way to predict thedirection of th e magnetic force exerted on a current-carryingconductor placed in a magnetic field. Hold the first three fingersof the left hand perpendicularly to each other (Fig 15.6b andFig 15.7b).(a) The first finger po in ts in the direc tion of the magnetic

f ield B.(b) The second finger points in the direction of the conventional

current I.(c) The thumb points in the direction of the magnetic force Fexerted on the curren t-carrying wire if the first and second

finger is oriented in the same direction as the magnetic fieldand conventional current in Fig 15.6b and Fig 15.7b.

Magnetic Flux Density and Tesla The magnetic flux density B is used to measure the strength and

direction of a magnetic field. It is a vector quantity.pFrom F = Bll, B - -II


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..,. Fig 15.12 Current balance

WeightmgPivot

Magnetic force F.t. Fig 15.13 Side view of th e

current balance

Measurement of Magnetic Flux DensityUsing a current Balance The magnetic force acting on a current-carrying conductor can

be used to measure the flux density of the magnetic field byusing a current balance (Fig 15.12).

/1/1/1/1/11 111111 /I IConductoP 1 1 .., 1 1/ I /

I- ; ~ . ; . . . . ; . . ~ ~ ~ .........B

The side view of the current balance is shown in Fig 15.13 /-:where d1 is the perpendicular distance from the pivot to theweight and d2 is the perpendicular distance from the pivot tothe magnetic force.The wire frame XABY is a conductor while the frame XDCY ismade of an insulating material. Both sections are mergedtogether and pivoted at XY to form a rectangular frame ABCD.Before a current is sent in to the wireframe, rigers are placed onCD and AB to balance the frame horizontally.The current-carrying solenoid provides a horizontal magneticfield that cuts AB perpendicularly. When a current is sent intothe wire-frame along XABY, a downward magnetic force F actsonAB.

To balance the rectangular frame horizontally, weight mg isattached to CD. At the equilibrium position,clockwise mome11t about pivot= anticlockwise moment about pivotf_d4 mgd1.Bild2 = mgd 1where B is the mapetic flux density of the solenoid,

I is the current flowing along AB andI is the length of AB.

.. B = mgdiIld2

or B = mg if d1 = d2Il


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f!it1 ~ ~ : . ; ~ ~ . . .. ' ,. ' .. ' .I : t ~ ~ g ~ . . . . , "

Fig 15 .15 Applying Fleming'sLeft Hand Rule on the motion ofpositive charge

The direction of the magnetic force on a charge moving . n amagnetic field can be determined with Fleming's Left HandRule. By convention, the direction of cOnventional current isgiven by the direction of flow of positive charges.(a) When the charge is positive, the middle finger points in thedirection of movement of the positive charge.The magnetic

force acting on the charge is in the upward direction and isrepresented by the thumb. The first finger points in thedirection of the magnetic field. All the three fingers must beperpendicular to each other (Fig 15.15).

(Thumb)Force F

(Second finger)Direction of movement orvelocity of positive charge v

F

B

v


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.,.. Fig 15.16 Applying Fleming'sLeft Hand Rule on the motion of

negative charge

F

.A Fig 15.17 Positive chargemoves at an angle e o themagnetic field

(b) When the charge is negative, the directioh of conventionalcurrent is in the opposite direction to that of the movementof the negative charge; The middle .finger points in thedirection of conventional current. Hence the direction offorce is downwards as shown in Fig 15.16.

F

~ o n v e n t i o n a l ' current'

F

B

Electronflowv

Expression for Force on a Moving Charge Consider a positive charge Q moving with a constant speed v at

right angle to a magnetic field B. If it travels a distance l in

time t, the current I is r -.f'->) Q=J .I= Qt

and the constant speed v byv =-

Fleming's Left Hand Rule ca n be applied to the conventionalcurrent du e to the motion of a single positive charge. Themagnetic force on this current isf.= BIZ .

~ B ( 7 } =BQill

i.e. F= BQp In general, the magnitude of the magnetic force acting on a

charge Q, moving with velocity v .at an angle e o a magneticfield B is


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In the case where the charge is an electron, the 1nagnitude of themagnetic .force is

. .It should be noted .that if a charge is stationary (i.e. v . 0) andthe magnetic field is switched on, the charge will experience nomagnetic force. It will remain stationary. A magnetic force willact on a charge only when it moves.


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X X

X X

X X

Deflection of Charges in a Magnetic Field . When a positive charge (proton) enters a uniform magnetic

field perpendicularly, the magnetic force acting on ~ h e proton isalways at 90 to the direction of motion (Fig 15.19).

The subsequent trajectory of the proton is along a path that isan arc of a circle. The magnetic force provides the centripetalforce for circular motion.

F= mv 2r BQv = mv2r

The magnetic force does no work on the proton as it isperpendicular to v. Hence the speed and kinetic energy of theproton remains the same as it moves along the circular path.

If T is the time taken to complete one revolution in the circularmotion and v is the constant speed of the proton,

v = f(!J

The period T is depen0-ent on the mass, magnetic flux densityand charge of the particle. It is independent of the speed.

'


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'1) 'The elec ~ T i ~ field can exert an electric forceon a stationary or n1oving charged particle.

2) The electric force acts in the direction of theelectric field. Deduced from the la1v ofelectrostatics (i.e. Like charges repel; unlikecharges a.thact).v ' .

3) Electric force is not dependent on the speedand direction of the charged particle.

4) Parabolic motion is obtained when a chargedparticle enters an electric fieldperpendicularly.

_ ' , ,' .. ,, _,gv ._ _ _ , . c ~ n e x e ~ : t r i ~ ~ ~ ~ t i ( = ' ... for.se o:nly on a _ 1oving charged particle. Statipn.aiy charged particles experience noforce."__: . : - - : ' _ ::: ... --: _ -2) The magnetic force is perpendicular to the

1nagnetic' field and the direction of n1otionof the cl1arged particle. Deduced fromFlem.ing's Left Hcmd Rule.

3) l\1agnetic force is dependent on the speedand direcbon of n:1otion of the chargedparticle.

4) Circular motion is obtained 1-vhen a chargedparticle enters a ma311etic fieldperpendicularly.

electromagnets notes for a levels

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