Chapter - 4 Movine CHarces AND Macnetism 44 INTRODUCTION; OERSTED EXPERIMENT ‘The magnetic effect of current was discovered by Danish Physicist Hans Christians Oersted. He noticed that a current in a straight wire makes @ “deflection ina magnetic needle. The deflection in- creases on increasing current . He also found that reversing the direction of current reverses direction of needle. Oersted concluded that current produces ‘a magnetic field around it. 4,2 MAGNETIC FORCE 4,24 Sources and fields: The static charge is the source of electric field. The source of magnetic field is current or moving charge.Both the electric and mag- netic fields are vector fields and both obeys super- position principle. 4.2.3 Magnetic force oF n current carrying conduct Consider a rod of uniform cross: section ‘A’ and len, *p', Let ‘n’ be the number of electrons per u volume (number density). 'v, be the drift velociyt electrons for steady current v Total number of electrons inthe entire volume of tq =nAg Charge of total electrons = nA £.€ 'e’ is the charge of a single electron. The Lorentz force on electrons, F=nAte (VaxB) F =navgel? xB) .2.2 Lorentz Force : The force experienced by moving charge in electric and magnetic field is called Lorentz force. The Lorentz force experienced by charge ‘q’ moving with velocity ‘vis given by v3] +f, gnetic F=qE+q = Forewie The features of Lorentz Force ) The Lorentz force on positive charge is opposite to that on negative charge because it depends on charge’ il) The direction of Lorentz force is perpendicular to velocity and magnetic field, Its direction is given by screw rule or right hand rule. iil) Only moving charge experiences magnetic force. Forstatic charge (v=0), magnetic force is zero. Note : i) Acharge particle moving parallel or antiparallel to magnetic field will not experience magnetic force and moves undeviated il) The work done by magnetic force is zero. Because magnetic force is always perpendicular to direction of velocity. iil) A charged particle entering perpendicular mag- netic field (=807) will make a circular path. M The unit of B is Tesla MOTION IN MAGNETIC FIELD Case! The charged particle enters perpendicular { magnetic field.(V is perpendicular to B)> When charged particle moves perpendicular to netic field, it experiences a magnetic force of magq tude, qVB and the direction of the force is perpey dicularto both and \y. This perpendicular may netic field act as centripetal force and charged ticle follows a circular path. a Mathematical explanation : Let a charge 4 enters into a perpendicular magnetic field B vil velocity V. Let r be the radius of circular path. centripetal force for charged particle is providedd mannatin farne.ssn) Thus radius of circle described by charged particle depends on momentum, charge and magnetic fiekd, Ife Isthe angular frequency v eet ‘Thus from (1) we get QB m 2m m (o=2nv) 28 The frequency v= ‘Thus frequency of revolution of charge Is indepen. dent of velocity (and hence energy) : oe 2am ‘The time period T=—— ver QB Case Il ‘The charged particles enters at an angle ‘0’ with magnetic field. Since the charged particle enters at an angle ‘6’ with magnetic field, its velocity will have two components; ‘a component parallel to magnetic field, V, (Vcos®) ‘and a component perpendicular to the magnetic field, V, (vsin8). ‘The parallel component of velocity remains unaffected by magnetic field and it causes charged particle to move along the field. The perpendicular component makes the particle to move in circular path. ‘The effect of linear and circular movement produce helical motion. Pitch and Helix : The distance moved along mag- netic field in one rotation is called pitch ‘P’. P =VeosoT = 2™™ xvcoso 8 The radius of circular path of motion Is called helix. 4.4 MOTION IN COMBINED ELECTRIC AND MAG- NETIC FIELDS: [7 March 2007, SAY- 2000) Ma Eis March «2013, BAY AAA Velocity selector: A transverse electric and mag- otic field act as velocity selector. By adjusting value of E and 8, It 18 possible to select charges of particular velocity aut of a beam containing charges of different speed. Explanation : Consider two mutually perpendicular olectricand magnetic fields in a region. Acharged particle moving in this region, will exparience elac- tric and magnetic force, If net force on charge is zero, then it vill move undeflected .The mathematl- cal condition for this undeviation is Fsncront. = F magnate gE = qvB The charges with this velocity pass undeflected through the region of crossed fields. 4.4.2 CYCLOTRON Uses : It is a device used to accelerate to high energy. Principles : Cyclotron is based on 4) An electric field can acceler particle. 2) A perpendicular magnetic fj circular path. Constructional Details : semicircular dees D, and D, harged the iona C. This chamber is plac sen two magnets. An alternating vottage i in between D, and D,. Anion is kept ina chamber. Working : At certal let D, be positive and D, be negative. I ll be accelerated towards icircular path (inside it). aches the gap, D, becomes “omes positive. So ion is accet fand undergoes a circular motion is. This process repeats again and \es near the edge of the dee with high ion can be directed towards the target by ing plate.REQeitiacr 4.920%? ~ Io 7 ke So * ‘Thus tho kinetic onorny that can bo gained depong, ° (a3 ‘on mass of particle chargo of particle, magnetic fey and radius of cyclotron. [So Magnet Limitations 4) As tho particle gains extromoly high volocity, K ee ee tho mass of particlo will bo changed from constant valuo, This wil affoct the normal works of cyclotron as frequency depends of mass particlo. Mathematical expression : Let 'v' bo the volocity of ion, q the charge of the lon and B the magnetic flux density. Ifthe ion moves along a semicircular path of radius 2) Very small particios like olactron can not ‘r, then we can write accolorated using cyclotron. This Is bocause mv? eB tho mass of-electron Is vory small tho eyclotr Re aq(ix8) frequency required becomes oxtremely high mm? which Is practically difficult, eee 3, Neutron can't be accelerated [Since 6 =90°, Bis perpendicularto v] reli g007, SAV 2010) M aBr or v=SE or m (1) 4.5 MAGNETIC FIELD DUE TO CURRENT ELEMENT, Time taken by the ion to complete a semicircular BIOT SAVART LAW path. distance_xr velocity v zr were Ifrom eq(1)] m 8 a “The magnetic field at any point due to an elemental Eq. (2) shows that time is independent of radius and current carrying conductor is velocity. 4) Directly proportional to the strength of thy Resonance frequency (cyclotron frequency) : The current (|) 4, condition for resonance is half the period of the ac- 2) Directly proportional to the length of the element celerating potential of the oscillator should be Gh. (ie., T/2=t or T= 21). Hence period of AC 3) Directly proportional to the sine of the angle between the element and the line joining the midpoint of the element to the point. 4) Inversely proportional to the square of th distance of the point from the element Tr ap od dlsin® F , Resonace frequency, v=-22- °F acm [ap — Ho Halsind K.E of positive ion an? 1 coe KE=prm? Invector for a = te dxe 2 In vector form a _1(qBr)? The direction of magnetic field is perpendicular to =m plane containing d/ and rand is given by right ha screw rule.46 He in the above expression 7 is the constant of proportionality and |4, is called the permeability of yacuum. Its value Is 47x 10-7 Tmas, note :Amagnetic field acting perpendicularty into tne plane of the paper Is represented by tho symbol and a magnetic field acting perpendicularly out ofthe paper is represented by the symbol , comparison between Biot-Savart Law and Coulomb's law similarities : ')_ The two laws are based on inverse square of dis« tance and hence they are long range, i Both electrostatic and magnetic fields oboy su- perposition principle, ji) The source of magnetic field is linear; (the cur= rent element Idi). The source of electrostatic force is also linear; (the electric charge). Difference: Blot-Savart Law The source of magnetic force is vector: (current e- ement Idl)- i) The scurce of electro static fore is scalar; (elect charge) i) The direction of etecto- static free is along dis- placement vector join- ing source and field point ci) The electrostatic force is independent of angie. The direction of magnetic. field is perpendicular to plane containing dis- placement vector. ¢ and curent element 13) ‘There is an angle depen- dence in Bict-Savart Law. March -2014, SAY- 2014 MAGNETIC FIELD ON THE AXIS OF ACIRCULAR CURRENTLOOP: Consider a circular loop of radius ‘a’ and carrying current “I, Let P be a point on the axis of the coil, at distance x from A and r from ‘O'. Consider a small length di at A. The magnetic field at ‘p' due to this small element di, _ tigldlsin90 oe dB 4n x’ _ tld! B=, ) [since sin 90°- 1} ‘Tho dB can be resolved into dB cosb (along Py) and dBsinp (along Px). Similarly considor a small oloment at B, which pro- duces a magnotic flold ‘dB at P, If we resolve this magnotle flold wo get. dB sind (along px) and dB cosb (along py’) dB cos) compononts cancel each other, because thoy aro in opposite diraction, So only dB sinh com pononts aro found at P, 60 total filed at P is Be faving ig lal 2 sing 4n x’ but from AAOP we get, sing =a/x , We get, Ho Ia 3 B= dl 4n x Ho Ta dl an? vf B= [85 ona [since f st=2-3) 4x? Be SS 1? from AAOP we get x=(P +a2)!2 Hg fa? 2? + a2)? Let there be N turns in the loop then, ie, B= Hig Nl? We + aye Point at the centre of the loop : When the point is at the centre of the loop, (r= 0) Then, 2a HR’ 208 +R?)4.8(1) Long straight conductor : 4.6(1) Magnetic field at the centre of loop : The magnetic field at a distance x from centre of loop is given by = HOR"! 267 +R?) At the center (x=0) at The direction of magnetic field due to current camry- ing circular loop is given by right hand thumb rule. Thumb Rule : Curl of palm of right hand around circular coil with fingers pointing in the direction of current. Then extended thumb gives the direction of magnetic field. Note : }) An anticlockwise current gives a magnetic field out of the coil and a clockwise current gives a magnetic field into the coil. li) The current carrying loop is equivalent to mag- netic dipole of dipole moment SIA 4.7 AMPERE'S CIRCUITAL LAW According 0 ampere’s law the line integral of ‘Magnotic field along any closed path is equal toy, times the current passing through the surface. Surface i * Closed path "March = 2010 4.8 APPLICATIONS OF AMPERE'S CIRCUITAL LAW Consider a long Straight conductor carying ‘T ampere current. To find magnetic field at ‘P’, we construct a circle of radius (passing through P). 4.8(2) ‘According to Ampere's circuital law we can write fadi-,1 fBal=u,t {B and d/are parallel} Bfde=pol B2nr=p,l = Bol 2m Magnetic field due to long solenoid Consider a solenoid having radius ‘r. Let ‘n' be number of tums per unit length and I be the curr flowing through it. In order to find the magnetic field (inside the so} Noid ) consider an Amperian loop PARS. Let ',' the length and 'b’ the breadth Applying Amperes law, we can write (Bis 1 to QR and SP} JBai-o (since RS is completely out side the so Noid, for which B = 0) Substituting the above values in eq (1),we get B r= Hol, (2) But te al > = where 'nl"is the total number of tums that carr Current I (inside the loop PARS) -. eq (2) can be written as Br = yale B=y nl‘4.8 (2)The toroid : Consider a toroid of if core of solenoid s filled with a medium of relative permittivity w, then B= nya "r.Let ‘n’be the number of turns p ‘the total number of tums of the sole- es current I (inside the Amperian loop) Integi 1@ €q(1) we get of the solenoid Is filled with a medium of permeability 11, then the above equation is led as Note: The magnetic field due to toroid is same as that due to solenoid. ‘THEAMPERE FORCE BETWEEN TWO PARALLEL, CONDUCTORS: P and Q are two infinitely long conductors placed parallel to each other and separated by a distance f, Let the current through P and Q be |, and |, respectively. Magnetic field at a distance ‘r from P is peti, "ORT Conductor ‘Q's placed in this magnetic field. {tListhe length of the conductor’, the Lorentz force on ‘O's a, ([eh) = 1B, [Since 0 = 90} bel, bol ahh since Byer] .!. Force per unit length can be written as, lp = Holle F Where {=~ a Note: 4) When currents are in the same direction, the force isattractive 2) Ifthe currents are in the opposite direction, the force is repulsive. Definition of ampere : An ampere is defined as that constant current which if maintained in two straight parallel conductors of infinite lengths placed one meter apart in vacuum will produce between a force of 2x107 Newton per meter length. 4.10. FORCE ON A CURRENT LOOP, MAGNETIC DIPOLE 4.10.1 Torque on a rectangular current loop in unl- form magnetic filed Consider a rectangular coil PARS of N tums which is suspended in a magnetic field, so that it can ro- tate (about yy1). Let’/"be the length (PQ) and ‘b' be the breadth (QR). ‘When a current / flows in the coil, each side pro- duces a force. The forces on the QR and PS will not produce torque. But the forces on PQ and RS will produce a Torque. Which can be written asBul, foro = BI 2) |sinco0 = 90") And trom agit, wont 4 distonoo (QT) bsind ‘Suiting the vales of en (2) and 9 (8) nea) WO et reblilbsind =BIA sind [since Ib = A (aren)) 7 =1AB sind + =mBsin@— {sineem=1A] n= ix Bl {there are N tums inthe coll, then 7= NIAB sind 4.10.2 Circular ee asa magnetic dipole Current loop of any shape act as magnetic dipole, Current loop acts as magnetic dipole : The mag- netic field due to circular loop of radius R carrying current | at a distance ‘x'from the centre of loop (on the axis of loop) is given by, gl? axe +R ‘The magnetic field at large distance (>R) on axis, ofloopis oR? a Dividing and multiplying by x B= ‘A=aR®, area of loop 2a z= Holl L oer m=iA, magnetic moment of loop to 2a Pp) oO Comparison of magnetic dipole and electric dipole : The equation (1) is similar to electric fleld due to electric dipole at a distance 'x’ from the centre of dipole on its axial line. oe Company en) nd (2), 0 Cot ] _ | oe wera | moe | bot ) rom this comparison It Is dloat thal 4 nog ‘currant loop adie as A magnatio, oe Maree 6 ee alia 4.40,3 Tho magnotle dipolo moment of a rave) Ing oloctron : According to Bohr's modal of aloq lootrons aro rovolving around nucleus in is orty ‘Tho oloctron rovolving ints orbit can fs crcular curront loop, {orm circular motion of elect 1FTisthe period of revolution Applying Quantum Theory, Bohr has proposed hd Angular Momentum of electron can take only 43, crete values given by,(Bohr's quantization condition where n=1,2,3, 2n a 610) Where h is Plank’s constant, Thus ‘The orbital magnetic moment of electron is given by c a) ‘dam Bohr Magneton : We get the minimum valuo of magnetic moment, when n=1 4xm ts value is 9.27x10Am?, Thisis called Bohr mag- neton. . Gyromagnetic Ratio : The orbital magnetic mo- ment of electron is related to orbital angular mo- mentum ‘Tas leh (when n= 1) e xe Mom Me | £ 2m The ratio of orbital magnetic moment to orbital an- gular momentum is constant. This constant is called gyromagnetic ratio. Its value is 8.8x10" c/kg for an electron. 4.11 THE MOVING COIL GALVANOMETER Itis an instrument used to measure small current. Principle : A conductor canying current when placed in a magnetic field experiences a force, (given by Fleming's left hand rule). Construction coil Radial magnetic field Amoving coil galvanometer consists of rectangular, coil of wire having area ‘A’ and number of turns ‘n’ which is wound on metallic frame and is placed be- tween two magnets. The magnets are concave in ‘shape, which produces radial field. Working : Let ‘Ibe the current flowing the coil, Then the torque acting on the coil 1 = NIAB Where Alsthe area of coll and B is the magnetic field. This torque produces a rotation on coll, thus fiber is twisted and angio (4). Duo to this twisting a restoring torquo (+ = Ke) Is produced in spring, Undor oquilibrlum, we can write ‘Torque on tho coll = restoring torque on the spring Or = NIAB = Kb Or Tho quantity insido the bracket is constant for a gal- vanometer. bal The above equation shows that the deflection de- pends on current passing through galvanometer. ‘March = 2010, SAY -2013 4.14.4 Ammeter and voltmeter For measuring large current, the galvanometer can be converted in to ammeter and voltmeter. Ammeter ‘Ammeter is an instrument used to measure current inthe circuit, T (Io) Agalvanometer can be converted into an ammeter by a low resistance (shunt) connected parallel to it. Theory : Let G be the resistance of the galvanom- eter, giving full deflection for a current Ig To convert it into an ammeter, a suitable shunt resis- tance 'S' is connected in parallel. In this arrange- ment Ig current flows through Galvanometer and re- ‘maining (L-g) current flows through shunt resistance. Since G and S are parallel PdAcrossG = pd across S IgxG (gs is = 19S} (Ig)| Connecting this shunt resistance across galvanom- eter we can convert a galvanometer into ammeter. “March -2010,SAY-2013 4.11.2 Conversion of galvanometer into voltmeter To convert a galvanometer into a voltmeter, a high resistance is connected in series with it.Theory : Let Ig be the current flowing through the galvanometer of resisiance G Let R be the high re- | sistance 60 Connected in series with G Ig R : _) L From figure we can wete Vo = laR+1eG V-laG = lo Using this resistance we can covert galvanometer in to voltmeter. Current sensitivity : The current sensitivity of gal- vanometer is the deflection produced by unit cur- rent The current sensitivity can be increased by increas- ing number of tums. The voltage sensitivity : The voltage sensitivity of galvanometer is the deflection produced by unit volt- ‘The increase in number of turns will not change volt- age sensitivity. When number of turns double (N —> 2N), the resis- tance of the wire will be double (ie. R > 2R). Hence the voltage sensitivity does not change. March-2010 00