y F N +q + θ S v x z 4/4/6 Magnetc Felds ( F.obllard) 1
4/4/6 Magnetc Felds ( F.obllard)
Introducton: It has been known, snce antquty, that when certan peces of rock are hung by a thread, from ther centre of mass, they wll algn themselves n a North-South drecton. These rocks were naturally-occurrng magnets, composed of magnette (Fe 3 O 4 ). One end was called a North pole (N), and the other end, a South pole (S), dependng on the geographc drecton n whch they ponted. Lke poles (N & N, or S & S) were found to mutually repel each other. Unlke poles (N & S) were found to mutually attract. Ths force of attracton, or repulson, s the magnetc force. A regon n whch a magnetc force exsts s called a magnetc feld. Our frst task must be to defne, precsely, the magntude and drecton, of the magnetc feld. We do ths n terms of the magnetc force. (The NS algnment of magnets, was due to magnet forces produced by nteracton of the magnet, wth the earth s global NS magnetc feld.) 4/4/6 Magnetc Felds ( F.obllard) 3
Feld Defnton Usng a Test Magnet: The drecton of a magnetc feld can be defned as the drecton n whch the SN poles of a small test magnet would algn, when placed at a pont n the feld. N S S->N Drecton of the feld s n the S-to-N drecton of the test magnet. However, defnng the magntude (strength) of the feld n ths way s problematc. The net magnetc force on a small test magnet n a unform magnetc feld s zero the forces on N and S poles are equal and opposte. Use of an solated N pole would allevate ths dffculty, but sngle magnetc poles (monopoles) do not exst. 4/4/6 Magnetc Felds ( F.obllard) 4
Feld Defnton Usng a Charge: A better way to defne a magnetc feld s n terms of the magnetc force that acts on a movng electrcal charge When a charge s statonary n a magnetc feld, no magnetc force acts on t. ut when the charge moves, relatve to the feld, a magnetc force acts on the charge. Ths force depends on - - the charge, - the strength of the magnetc feld - the magntude and drecton of the velocty of the charge, relatve to the drecton of the feld. ecause the magnetc force depends on the magntude and drecton of the magnetc feld, t can be used to defne the feld. 4/4/6 Magnetc Felds ( F.obllard) 5
vsnθ z F +q + y v vcosθ θ Expermental Observatons: Say a unform magnetc feld s drected n the +x-drecton We move a partcle wth postve charge, +q, through the feld, wth a velocty, v, n the +xzplane, at an angle, θ, to the drecton of the feld. x Let s resolve the velocty, v, nto - a (v cos θ) component n the drecton of the feld, and - a (v sn θ) component perpendcular to the drecton of the feld. Experment shows, that a magnetc force, F, acts on the charge n the +ydrecton perpendcular to both the drecton of the feld, and the drecton of the velocty, v. The magntude, F, of ths force - s proportonal to the charge, q, - s proportonal to the (v sn θ) component of v - s not affected by the (v cos θ) component of v - depends on the strength of the feld ecause the magnetc force, F, depends on the magntude and drecton of the magnetc feld, t can be used to defne the feld. 4/4/6 Magnetc Felds ( F.obllard) 6
z F Defnton of Feld Vector : We represent the magntude and drecton of the magnetc feld, at the locaton of the charge, by a feld vector. y +q + θ v x Therefore the magnetc force, F, s perpendcular to both v, and drectons, as before. We defne the magntude (or strength) of the magnetc feld,, so that the magntude of the magnetc force, F, and the magntude,, of the feld are n drect proporton. Takng account of the expermental facts noted above, namely that the magntude, F, of the magnetc force s proportonal to the charge q, proportonal to the (v sn θ) component of the velocty of the charge (the component perpendcular to ), and followng our asserton, that F should be proportonal to the magntude of, we can wrte - F q v snθ Ths relatonshp defnes the magntude of the feld vector. The drecton of s defned accordng to the relatve drectons n the above fgure. 4/4/6 Magnetc Felds ( F.obllard) 7
Vector Defnton of : The defnton of feld vector can be expressed more elegantly usng the followng cross-product expresson, whch ncorporates magntudes and drectons of all vectors -- ( v ) q v F q Ths expresson should be thought of as a general mplct defnton of the magnetc feld vector,, n terms of the magnetc force, F that acts on a charge, q, that moves wth a velocty, v, at a partcular pont n a magnetc feld. Havng thus defned the feld vector,, we can use the above expresson to fnd both the magntude, and drecton, of the magnetc force, F, that would act on a movng charged partcle, n a magnetc feld. SI Unt for : F/(qv sn θ), therefore unts for wll be - N C -1 m -1 s Tesla T A charge of 1 C would experence a magnetc force of 1 N f t moved perpendcular to a magnetc feld of 1 T, at a velocty of 1 m/s. Examples: Earth s magnetc feld: ~5 µt; conventonal laboratory magnet: ~ T 4/4/6 Magnetc Felds ( F.obllard) 8
Along & Across the Feld: As the angle, θ, between v, and, changes, the magnetc force wll vary as (sn θ). y z N + v +q y F S x When the charge moves n the drecton of the feld (θ), the magnetc force s zero. F q v snθ z F + v +q x When a charge moves perpendcularly to the drecton of the magnetc feld (θ 9 deg), the magnetc force s maxmum. F q v sn θ q v The drecton of the magnetc force s gven by the ght Hand ule (H) for cross products. 4/4/6 Magnetc Felds ( F.obllard) 9
Example: An electron (q -1.6 x 1-19 C) passng, wth a velocty of 1 m/s, n a drecton perpendcular to the magnetc feld between the poles of a horseshoe magnet, experences a magnetc force of 3 an. What s the strength of the feld? N v - S F F q v thus F qv 3 x 1-18 ( -19 1.6 x 1 ) 1. T 4/4/6 Magnetc Felds ( F.obllard) 1
Magnetc Flux Lnes: Any pont, P, n a magnetc feld, s charactersed by a feld vector. Spaces whch have a characterstc vector defned at every pont, are called vector felds. Although the -vector s the quantty that fundamentally represents the magnetc feld, there s a more potent way to represent the feld, graphcally. As we have done wth electrc felds, magnetc felds can be represented graphcally by flux lnes. A magnetc flux lne s a lne drawn n a magnetc feld wth shape such that a tangent to the lne at any pont corresponds to the drecton of the magnetc feld vector at that pont. The strength of the feld s represented by the closeness of flux lnes drawn n a gven regon of the feld a strong feld has close-together flux lnes. 4/4/6 Magnetc Felds ( F.obllard) 11
Examples of Flux Lnes: Unform feld between the poles of a horseshoe magnet: Non-Unform magnetc feld near a bar magnet: N S N S flux lnes flux lnes Later we wll develop a quantty called flux, to represent the densty of flux lnes. 4/4/6 Magnetc Felds ( F.obllard) 1
F q v F ( ds ) 4/4/6 Magnetc Felds ( F.obllard) 13 Q P
z Moton of a Free Charged partcle n a Magnetc Feld: We turn our attenton frstly to the moton of charged partcles that move freely, n a magnetc feld. Say a free charged partcle, of mass, m, and charge, q, s projected nto a unform magnetc feld, of vector. How wll t move? v F q y Assume that v s perpendcular to the feld vector,,. x The magnetc force, F, on the partcle, s gven by F q (v x ) Drecton of F s always perpendcular to v. Therefore the magntude of v wll not be changed by the force. (Magntude of F) F q v sn 9 q v const., snce q, v, are all const. So we have a constant force actng on the partcle, always perpendcular to ts velocty, v. These are condtons for crcular moton the partcle wll move n a crcular path, whose spn axs s n the drecton of. 4/4/6 Magnetc Felds ( F.obllard) 14
v z m r y Fqv lookng n the ( x)-drecton adus of Crcular Path, r: Usng Newton : ΣF ma ( magnetc force) qv r m a v m r m v q snce path s crcular adus, r, of the crcular path s proportonal to v (the partcle velocty), nversely proportonal to (the strength of the feld), and proportonal to the rato of mass to charge, (m/q). Ths s an mportant relatonshp, n Physcs, and has been used to measure the mass-to-charge rato (whch s a fundamental quantum constant) for the electron, and other elementary partcles.( v,, and r are measured, allowng the computaton of (m/q). 4/4/6 Magnetc Felds ( F.obllard) 15
General Case: We have consdered the case where the partcle s njected nto a unform feld at rght angles to the feld. What happens f the partcle s njected at an angle θ to the feld vector,? Say s n the +x-drecton, and v s n the y xz-plane, at angle θ to. v snθ F q θ v v cosθ x Magnetc force, F, wll be n the +y-drecton, as before. v can be replaced by ts components v snθ and v cosθ, perpendcular to, and parallel to, respectvely. z The v snθ component wll produce the crcular path, as before the magnetc force, F, s due entrely to ths component. The v cosθ component wll smultaneously transport the partcle, at a constant speed, n the +x drecton. The partcle wll consequently follow a helcal path, spralng about the +x-drecton. 4/4/6 Magnetc Felds ( F.obllard) 16
The Lorentz Force: We have consdered the magnetc force on a charged partcle n a magnetc feld. If both magnetc and electrc felds are present, both electrc, and magnetc forces can act smultaneously on the partcle. In ths case, the overall electromagnetc force that acts s called the Lorentz force, and s the sum of the electrc and magnetc forces. F q E + q v x where: F the Lorentz force q charge of the partcle v velocty of the partcle E electrc feld vector magnetc feld vector The controllng of the movement of charged partcles, usng the Lorentz force, fnds some mportant nstrumental applcatons. We wll look at some of these now n partcular - 1. the velocty selector. the mass spectrometer 3. the cyclotron 4/4/6 Magnetc Felds ( F.obllard) 17
1.Velocty Selector: Our frst applcaton s the velocty selector. When produced, charged partcles generally have a range of veloctes. A velocty selector s used to select from charged partcles wth a range of veloctes, only those wth a specfc velocty. Hence, ths devce acts as a velocty flter by transmttng only those charged partcles wth a chosen velocty, and blockng all the rest. It s based on crossed electrc (E) and magnetc () felds. We arrange the E and felds such that the electrc and magnetc forces are opposte n drecton. For a partcular charged partcle, the electrc force depends only on the charge, whereas, the magnetc force depends on both the charge and the velocty of the partcle. Thus, the electrc and magnetc forces wll only balance each other, for partcles havng a specfc velocty. 4/4/6 Magnetc Felds ( F.obllard) 18
The Crossed Felds: Consder the unform magnetc feld between a N-pole and a S-pole. N q + E S v Cross ths feld wth the unform electrc feld between two oppostelycharged, flat, parallel, metal plates. Consder a partcle, wth charge q, movng between the pole faces, and the charged plates, wth a velocty, v, that s perpendcular to both feld drectons. ecause of the drectons of the felds, the electrc force, F E, and the magnetc force, F, wll be n opposte drectons, as shown. v F q + F E E 4/4/6 Magnetc Felds ( F.obllard) 19
Selecton Condton: v F q + If the electrc and magnetc forces are equal n magntude, a charged partcle wll pass undeflected F E E ( Electrc force) ( Magnetc force) Eq q v v E...( 1) If a partcle s velocty satsfes the selecton condton, (1), t wll pass through the feld regon undeflected; all other charged partcles (wth other veloctes) wll be deflected. Hence ths system can be used to select charged partcles wth a chosen velocty, and act as a velocty flter. The undeflected partcles pass through an aperture n a plate; deflected partcles are blocked. Note: that the velocty selected, v, n (1), s ndependent of the charge, q, and the mass, m, of the charged partcles. It depends only on the rato of the strengths of the electrc and magnetc felds. 4/4/6 Magnetc Felds ( F.obllard)
.Mass Spectrometer: Our second applcaton s the Mass Spectrometer. The mass spectrometer s a wdely-used scentfc nstrument, that allows for the analyss of the composton of a sample t measures what atoms are present n the sample, and the percentage of each type of atom present. That s, t gves the dstrbuton (or spectrum) of atoms present. Ths devce uses electrc and magnetc felds to deflect charged atoms (ons). It conssts of three sectons an onzer, a velocty selector, and an analyser. We look frstly at the analyser secton. Consder a regon of unform magnetc feld, wth magntude. Say we nject a charged partcle, of mass, m, and charge, +q, nto the feld, perpendcularly to vector. As descrbed earler, ths partcle wll follow a crcular trajectory, wth radus, r, gven by - ( Magnetc force) ( Centrpetal force) v q v m r m r q v...() 4/4/6 Magnetc Felds ( F.obllard) 1 + r v
r m q v...() The Analyser: If we could fx the values of v and, equaton () shows that the radus, r, depends on the rato of mass-to-charge (m/q). The recprocal of ths, the charge-to-mass rato, (q/m) s tradtonally used. y measurng r, and knowng v and, we could compute (q/m), whch s characterstc of a partcular on, and can therefore be used to dentfy the on. Fxng v : We need to fx the velocty, v, of an on, ndependently of ts charge and mass. Ths can be acheved usng a the velocty selector, as dscussed earler, where the ons are drected through known, unform, crossed electrc, and magnetc felds. We can select the velocty of ons, v, accordng to the selecton condton, derved earler: + r v v E E S S...( 1) Where E S and S are the magntudes of the crossed electrc and magnetc felds n the Velocty Selector secton of the mass spectrometer. 4/4/6 Magnetc Felds ( F.obllard)
The System: We now put the three components of the mass spectrometer together. Frstly, we need to onze the atoms we wsh to analyse. Ths s typcally done by passng an electrc current through a vapor of the atoms, n a vacuum. Ths causes electrons to be knocked off the atoms, thereby onzng them. We then drect the ons through the perpendcularly crossed electrc (E S ) and magnetc ( S ) felds of the Velocty Selector. Only ons wth a selected velocty, wll be undeflected, and be able to pass through aperture A. Ionzer S E S + - Velocty Selector D + v A r v Analyser The ons then enter the Analyser secton, where they follow crcular paths, whose radus, r, depends on the charge-to-mass rato of the partcular on. A lnear detector, D, measures the numbers of ons, wth varous rad. Thus the spectrum of on (q/m) ratos, and hence atom masses can be measured. Note: the whole system must be under vacuum, so that ar partcles don t scatter the ons under analyss, or contamnate the results. 4/4/6 Magnetc Felds ( F.obllard) 3
v E E S S (1) () gves:...( 1) where: and r m v q (q/m) ato:...() m q E S S r radus of on path n analyser Es electrc feld n velocty selector s magnetc feld n velocty selector magnetc feld n analyser (q/m) the charge-to-mass rato of an on r E S, S and are fxed, so r wll be proportonal to (m/q). ecause only certan ons are present, wth characterstc (m/q) values, only correspondng dscrete values of r wll occur. We set E S, S and. Then by measurng the values of r that occur, we can determne the correspondng (m/q) values, and so dentfy the ons present. y countng the number of ons at each radus, we can determne the onc composton of the sample. A varaton of ths nstrument was used, n 1897, by J.J.Thompson, to measure for the frst tme, the charge-to-mass rato for the electron. 4/4/6 Magnetc Felds ( F.obllard) 4
Our thrd applcaton s the cyclotron. 3.The Cyclotron: Ths s a devce used to accelerate charged partcles to hgh veloctes. The large knetc energes of fast partcles allows them to collde and nteract at a fundamental level wth other partcles. Studes of such nteractons allow us to nvestgate physcs at a fundamental quantum level. The cyclotron uses electrc and magnetc felds to accelerate charged partcles n a vacuum. It conssts of two hollow, metal D-shaped S contaners, wth a gap between them. A unform magnetc feld,, cuts down perpendcularly through the plane of the D s. An oscllatng rectangular voltage, v, s connected across the D s, so that an electrc feld exsts n the gap between them. Ths feld perodcally s drected from D 1 to D, then reverses. V T Tperod of the voltage V 4/4/6 Magnetc Felds ( F.obllard) 5 t D 1 D vacuum v
S Path of a Partcle: D 1 D P ecause they are statc hollow metal contaners, there wll be no electrc feld nsde the D s, only between them. E V Q v max Say a (+) partcle s released near P, n the mddle of the gap between the D s, when D 1 s (+) [ between t and tt/]. The partcle wll be accelerated across the gap, and nto D. V m When nsde D, because of the magnetc feld (and t zero electrc feld) the partcle wll follow a semcrcular path, that wll brng t back to the gap. T/ T 3T/ T 5T/ We tune the oscllatng voltage, V, so that D 1 s negatve, when the partcle returns to the gap, and, for a second tme, t wll be accelerated across the gap, and nto D 1. Insde D 1, the partcle wll agan follows a sem-crcular path. 4/4/6 Magnetc Felds ( F.obllard) 6
S Acceleraton of the Partcle: Thus the partcle wll traverse the gap numerous tmes, beng accelerated each tme, by the electrc feld n the gap. D 1 D P Each successve sem-crcular path nsde a D wll have a greater radus, because of the partcle s ncreasng velocty. V v Q vmax t Eventually, the radus of the partcle s path wll ncrease to the outer radus of a D, and the partcle wll emerge from the cyclotron, at hgh velocty, v max, (at Q). For ths to work, we need to get the correct perod, T, for the oscllatng voltage, v. T/ T 3T/ T 5T/ 4/4/6 Magnetc Felds ( F.obllard) 7
Partcle n a D Perod, T, of Oscllatng Gap Voltage: As we have seen before, the radus, r, of the sem-crcular path of a charged partcle n a unform magnetc feld ( such as nsde a D ) s gven by - ( Magnetc force) ( Centrpetal force) Transposng q v m r v m q v r q m v r...()... 4/4/6 Magnetc Felds ( F.obllard) 8 ( ) dstance velocty ( Tme for 1crcut) T...( 3) where: q partcle charge m partcle mass feld n the D v velocty of the partcle To ensure that the polarty s correct for an acceleraton of the partcle n the gap, the perod, T, of the oscllator needs to be synchronsed wth the tme for a complete crcut of the two D s. m q π π r v ( ) ( 3) gves T...( 4)
V v t T m q π...( 4) v q r... m ( ) T/ T 3T/ T 5T/ For a gven partcle (gven (q/m) rato), wth the magnetc feld set to a value,, the perod of the oscllatng voltage must be tuned to satsfy (4). Note: T s specfc to a partcular partcle, but s the same for all crcuts of that partcle wthn the D s ( whatever the v, and r). KE of Partcles Emtted by Cyclotron: The KE of emtted partcles wll be determned by ther ext velocty from the system. ( KE of emtted partcle) KE max 1 q m q m m 1 m v...( 5) max from ( ) 4/4/6 Magnetc Felds ( F.obllard) 9 v max partcle s ext velocty where largest radus wthn a D r correspondng to v max To maxmse KE, we need large and.
Example: A small cyclotron has D s of radus.5 m, and s used to accelerate protons. The magnetc feld nsde the D s s 1.7 T. Fnd the oscllator frequency requred between the D s, and the KE of protons emtted. Oscllator Frequency: Gven:.5 m 1.7 T For a proton, q 1.6x1-19 C & m 1.67x1-7 kg From (4): Proton KE: T f m q 1 T 1.67 1 1.6 1.6 1 7 Hz 7 19 π 1.7 6 MHz 3.8 1 8 s From (5): KE max q m ( 19 ) 1.6 1 ( 1.7) (.5) ( -7 1.67 1 ) 1.4 1 1 J 4/4/6 Magnetc Felds ( F.obllard) 3
Magnetc Force on Currents: So far, we have consdered the magnetc force on a freely movng charged partcle. In practce, charges often move through magnetc felds, as part of a current, whch s constraned to flow n a conductor. In ths case, the magnetc forces are transferred to the conductor. Ths s the prncple of operaton of the electrc motor. Clearly, the total magnetc force on the conductor, wll be the sum of all the component magnetc forces, on all the charges, that make up the current flow. Snce these component charges may be travelng n dfferent drectons, n dfferent parts of the conductor, relatve to the magnetc feld, the total force wll depend on the detaled shape of the conductor, and the geometry of the magnetc feld. To smplfy ths complex stuaton, we consder a tny dsplacement, at a gven pont along the conductor called a length element, ds, whch, because of ts nfntesmal length, must be straght. The drecton of the ds vector s taken to be the drecton of current flow, thus representng the drecton of the velocty of the charges, movng n the element. To fnd the total magnetc force on a real length of wre carryng a current n a magnetc feld (such as n an electrc motor), we need to do a vector addton of the forces on all the length elements that make up the conductor. 4/4/6 Magnetc Felds ( F.obllard) 31 ds v ds
Force on Dsplacement Element, ds: Say a conductor, carres a current,, from P to Q, n a magnetc feld, as below. y Consder a dsplacement element, ds, Q of the conductor. Say ds ponts n the +y-drecton df ds Say, at ds, the -vector of the feld ponts n the +x-drecton. Then (from F q v x ) the total x magnetc force, df, on all the movng P charges wthn the element, wll act n the ( z)-drecton. z ( Of course, ds could pont n any drecton, relatve to.) At any moment, there wll be some total charge dq flowng n the length element, ds, at an effectve velocty v. Let dt be the tme nterval for all the charge, dq, to travel through the length element, ds. ds dq df dq dt dt ( v ) dq ( ds ) ( ds ) ( s ) df d Magnetc force on current element (cross product). 4/4/6 Magnetc Felds ( F.obllard) 3
P z y ds df Force on a Length: Q x Say a conductor carres a current,, through a regon of magnetc feld, from P to Q. To get the total magnetc force, F, on the total length of conductor, we need to do a vector addton of all the elementary forces, df, on all the length elements, ds, that make up the conductor from P to Q. ( Total force) F df ( ds ) Q P Q P const. for all elements, over the summaton Ths s called a lne ntegral, snce we add up the elementary forces, df along the conductor. Dependng on the drecton of the partcular ds, and local, the drecton of the df s wll, n general, vary from pont to pont along the conductor. 4/4/6 Magnetc Felds ( F.obllard) 33
z P Straght Length n Unform Feld: y ds df L Q x Say the length of wre n the feld, PQ, s straght, and of dsplacement L. (L s the sum of all the ds s from P to Q). ( Total force) F df ( ds ) F (ecause all ds s are parallel, and all s are parallel, all the df s are n the same drecton, whch consequently s the drecton of ther sum, F.) Say the feld between P and Q s unform ( const). ds ( snce const) ( L ) snce ds L P 4/4/6 Magnetc Felds ( F.obllard) 34 Q P Q P Q P Q L vector PQ. (L s n the drecton of current ). Magntude of F L sn θ, and s n a drecton gven by the rght-hand-rule for cross products
Closed Loop n Unform Feld: y Here we work out the total force, F, on a closed loop, that carres a loop current,, n a unform magnetc feld.( the loop s entrely wthn the feld, but can have any shape ). z ds df x Note: the elementary forces wll be n opposte drectons, on opposte sdes of a symmetrcal loop. df df ( Total force on loop) ds snce ( ds) sum of elementary forces around loop snce constant for unform feld ( ds) s zero round a closed loop F df The total magnetc force on a closed, current-carryng loop n a unform magnetc feld s zero. However, the total magnetc torque on the loop s not zero. 4/4/6 Magnetc Felds ( F.obllard) 35
Q z a θ a P ectangular Loop n a Unform Feld: y b θ A A S x The total magnetc torque on a rectangular loop n a unform magnetc feld s the bass for the electrc motor, and therefore has partcular practcal mportance. We consder such a loop, wth sdes of lengths a and b, and area A ab, that carres a loop current of. If we take dsplacement vectors along sdes of the loop, n the drectons of current flow, then we can defne a PQ & b Q Vector area of the loop A a x b (cross product) b (A H conventon works for the drecton of A. If the fngers of the rght hand curl about the loop n the drecton of the loop current, the thumb ponts n the drecton of area vector A.) A unform magnetc feld,, passes through the loop at angle θ to area vector A. 4/4/6 Magnetc Felds ( F.obllard) 36
(-F) S θ z a P f y b θ A (-f) F Q Forces on the Loop: x The magnetc force on a straght current-carryng conductor, n a unform feld s gven by -, F L x Hence, the magntude and drecton of the force on each sde of the loop s gven by - (- F) a...( 1) f b...( ) (The H gves the drectons of these forces F s n the ( y)-drecton; f s n the +z-drecton. ) We have prevously shown that the total force on any closed loop current s zero. Ths can be confrmed from the Fgure, snce (Total force) +F F +f -f The total magnetc torque on the loop, however, s not zero. We wll now fnd ths torque. 4/4/6 Magnetc Felds ( F.obllard) 37
S (-F) z θ (-F) a f b a P f ½ a y b C θ A (-f) F (-f) F Torque on the Loop: Q x We take torques about the symmetrcal centre pont of the loop, C. We use the defnton of torque, τ, namely τ r x F where r s the perpendcular dsplacement from C to the sde of the loop on whch the magnetc force acts. (Total torque due to f, -f, F, -F) τ ( ½ a x f) + (( - ½ a ) x ( -f)) + (½ b x F) + (( - ½ b ) x ( -F)) (½ a x f) + (½ b x F) + (b x F) Therefore: τ (b x F)...(3) ½ b snce a s parallel to f, & therefore a x f a f sn ( f & ( -f ) try to stretch the loop n the drecton of the z-axs, but exert no torque upon t; F & (-F) exert all the torque.) 4/4/6 Magnetc Felds ( F.obllard) 38
y (-F) P A a S θ θ b f z Magntude of τ: Drecton of τ: Total Torque on Loop: (-f) F Q x F a x...(1) τ (b x F)...(3) τ τ b x F b F sn θ [snce angle from to F s θ...see fgure]. b ( a ) sn θ from (1), snce a s perpendcular to ] (a b) Β sn θ Α Β sn θ [snce A ab] From the H, (b x F) s n the (-z)-drecton. b θ θ θ x F Vew the loop, edge-on, lookng n the ( z)-drecton oth magntude, and drecton of τ are consstent wth the followng expresson for the magnetc torque on the rectangular loop. τ (A x ) (cross product) 4/4/6 Magnetc Felds ( F.obllard) 39 S y A
Generalsaton : τ (A x ) (cross product) We have derved ths result for the magnetc torque, τ, on a rectangular loop, of vector area, A, n a unform magnetc feld, of vector. However, t s true for a flat loop of any shape, that carres a current,, n a unform feld. If the loop s replaced by a col wth N turns, then the torque wll be gven by τ N (A x ) (The torques for ndvdual loops of the col are addtve.) Ths formula s the bass for the electrc motor, and also of the movng-col galvanometer, from whch the movng-col ammeter and voltmeter are derved. 4/4/6 Magnetc Felds ( F.obllard) 4
S 3m z θ y P a b 4m θ I A m Q x Example: The loop PQS, shown, carres a current I 1 A, n a unform magnetc feld of T. The feld s drected n the (+x)-drecton. Fnd the magnetc torque, τ, on the loop. (T) Area vector of the loop A (A cos θ ) + (A sn θ ) j (ab cos θ ) + (ab sn θ ) j Torque τ I (A x ) I [(ab cos θ ) + (ab sn θ ) j ] x [ ] I (ab sn θ) () (-k) 1 ( x 5 x 4/5) () (-k) - 16 k (N m) Torque s 16 N m, drected n the (-z)-drecton. ( If the loop starts from rest, the z-axs wll be the spn axs of the loop.) 4/4/6 Magnetc Felds ( F.obllard) 41
ds Σ 4/4/6 Magnetc Felds ( F.obllard) 4
What Creates a Magnetc Feld?: In our dscusson of the magnetc feld, we frstly defned the feld n terms of a vector,. We then looked n detal at the propertes, and applcatons, of the magnetc force. We now ask the fundamental queston - what created the magnetc feld n the frst place?. At a fundamental physcal level, a magnetc feld s observed, when a charge moves, relatve to an observer. If the charge s statonary, only an electrc feld s detected, when the charge moves, an addtonal magnetc feld s detected, by the observer. The feld s due to the relatve velocty between a charge, and an observer. ut movng charges consttute a current - so Magnetc felds are produced by electrc currents. ecause a current flows along a lne n space (that s, the current s essentally one dmensonal), the feld produced by the current must have axal symmetry t must be symmetrcal about the lne. So magnetc felds curl about the current that produces them. 4/4/6 Magnetc Felds ( F.obllard) 43
Drecton of the Magnetc Feld due to a Current: Consder a long straght current,. eye The magnetc feld (flux lnes) curl around the current drecton. From the eye s vew - current,, upward The magnetc flux lnes are crcles, coaxal wth the current. The drecton of the flux lnes s gven by the ght Hand (Grp) ule (H.) eye s vew Clasp the current n the ght Hand, wth the thumb pontng n the drecton of current flow. The fngers wll then curl about the current drecton n the drecton of the magnetc flux. 4/4/6 Magnetc Felds ( F.obllard) 44
P ot-savart Law - Setup: In 18 J.. ot & F. Savart measured the expermental relatonshp between a current and ts magnetc feld - the ot-savart Law. ecause the feld produced by a current-carryng conductor, at a pont, P, depends both on the length, and on the shape, of the conductor, we consder a short, straght dsplacement element, vector ds, of the conductor. ( ds s taken n the drecton of current flow.) Say a conductor carres a current ds cosθ ds snθ θ r ds d We want to fnd the elemental magnetc feld, d, at a pont P, produced by the currentcarryng element ds, of the whole conductor. Usng the H for magnetc feld drecton the d-vector wll be downward, at P. Say that the poston vector of pont P from the dsplacement element, ds, s r. Vector ds can be resolved nto components (ds cos θ), n the drecton of r (e. the drecton of P), and (ds sn θ), perpendcular to r Experment shows, that the component n the drecton of P, (ds cos θ), contrbutes zero feld at P; d s due entrely to the perpendcular component, (ds sn θ). 4/4/6 Magnetc Felds ( F.obllard) 45
ds θ r d P So, the magntude of d can be expressed by - ot-savart Law : ot & Savart found that the magntude of d s proportonal to - - the current - the nverse square of the magntude of r - the (ds sn θ) component of the element ds d where: (µ /4π) s the constant of proportonalty for the SI unt system. µ s a magnetc constant called the permeablty of vacuum. (µ 4 π x 1-7 N A - :exact SI value ) 4 ds r ds sn r sn θ s the angle from ds to r The drecton of d the same drecton as that of the cross product ˆ s gven by the H, and turns out to be ( ) 4/4/6 Magnetc Felds ( F.obllard) 46
ds θ r d P d Vector Form : Takng account of the expermental facts of the prevous slde, both magntude and drecton of d, at P, can be ncorporated nto a sngle vector expresson, whch s the general form for the ot-savart Law: 4 ds rˆ r cross product where: µ s a magnetc constant called the permeablty of vacuum. (µ 4 π x 1-7 N A - :exact SI value ) and rˆ s the unt vector n the drecton of vector r. The ot-savart law plays an analogous role n magnetc felds, to that played by Coulomb s law n electrc felds both gve the lnk between the feld, and the source of the feld. 4/4/6 Magnetc Felds ( F.obllard) 47
ot-savart Law For an Entre Current : ds r ot-savart gves the feld due to one element, ds, of a current path. To fnd the total feld,, due to the entre current path, we need to sum all the feld contrbutons, d, from all of the elements, ds, along the path. d 4 d P ds rˆ r entre current path d 4 entre current path ds rˆ r (snce s constant for all path elements) Ths ntegral s a vector summaton of all the d s produced at a gven pont P, by all of the segments of a lne (a lne ntegral along the current path). We wll now apply the ot-savart law to fnd the magnetc feld produced by two smple current geometres. 4/4/6 Magnetc Felds ( F.obllard) 48
d Current Loop: As a frst applcaton of the ot-savart law, we wll fnd the magnetc feld at the centre of a flat, crcular, current loop that carres a constant currant. d ds Consder a crcular current loop, of radus, that carres a constant current. Let ds be a typcal dsplacement element of the loop. Let d be the element of feld produced at the centre of the loop by the current n ds. y symmetry, and the H, d must be perpendcular to the plane of the loop, as shown. Ths drecton s consstent wth the drecton of the cross product (ds x r). The full d vector s gven by the ot-savart law: d ds rˆ ds 4 r 4...( 1) d ds rˆ r Here, ds s perpendcular to r, and r const. Therefore the magntude of d d s gven by: ds rˆ 4/4/6 Magnetc Felds ( F.obllard) 49 4 ds 1 sn 9 o ds
d ds 4 entre current path 4 d ds ( π ) Total at Centre: around crcular current loop crcumference of crcular loop ds d 4...( 1) The total magnetc feld vector,, at the centre, s the vector sum of all the ndvdual d s, due to each of the length elements, ds, composng the current loop. ut, for ths case, by symmetry, and by ot-savart, all the ndvdual d s wll be n the same drecton. Therefore we can add the ndvdual magntudes, d, to get the total magntude, knowng that the vector wll have the same drecton as any of the ndvdual d vectors. Summary: The total feld,, at the centre of the loop, has magntude and s perpendcular to the plane of the loop accordng to the H. 4/4/6 Magnetc Felds ( F.obllard) 5
sn( -) r...( 1) Long Straght Current - Setup: As a second applcaton of the ot-savart law, we wll fnd the total magnetc feld vector,, at a radal dstance of, from a long, straght, constant current of. Y sn tan P d r θ -tan x ( -)...( ) x dx Take the x-axs along the current drecton. Let P be a pont on the y-axs, at a radal dstance of from the current. Consder a dsplacement element, dx, of the current, at coordnate x. Let r be the dsplacement vector from dx to P. θ s the angle from the (+x)-drecton to the drecton of vector r. Let the element of magnetc feld produced by dx, at pont P, be vector d. From the dagram - X 4/4/6 Magnetc Felds ( F.obllard) 51
P Y d r x θ dx Fnd d: X sn To fnd the total feld,, at pont P, we frstly fnd the element of feld, d, produced at P, by the current n the length element dx y the H, the drecton of d s perpendcularly upward from the plane of the fgure. The magntude of feld element d s gven by the ot-savart law - tan r x...... ( 1) ( ) d d 4 4 dx rˆ 4 r dx 1 sn r sn r dx...( 3) Equaton (3) gves d n terms of three varables, θ, r, and x. ut θ, r, and x are dependent varables as one of them vares, the other two vary, as functons of the frst. We need to express two of θ, r, and x, as functons of the other varable. It s easest to change the varable n (3) to θ. 4/4/6 Magnetc Felds ( F.obllard) 5
sn d 4 r Fnd r as a functon of θ: From (1): r sn Fnd d n terms of θ: dx...( 3)...( 4) Fnd dx as a functon of θ: From (): x tan dx sec d tan cos 1 sn cos sn dx d...(5) sn (4),(5) --> (3): 4 4 sn sn... r x 4/4/6 Magnetc Felds ( F.obllard) 53 d sn 1 tan sn d sn...( 6)... ( 1) d Ths gves the magntude of the feld element produced, at P, by the dsplacement element, ds, of the current path, for whch the bearng of pont P s θ. ( )
x+ x - d 4 4 4 [ -cos] P 4 Y d [ cos -cos] r x sn snce 4 d Fnd : const. [ ( 1) -( + 1) ]...(7) θ dx d 4 sn...(6) To compute the total feld,, at pont P, we need to add all the elementary feld components d, from all segments dx along the total length of the current: All d s pont n the same drecton, and therefore ther magntudes can be added, to get the total. 4/4/6 Magnetc Felds ( F.obllard) 54 X d When the dx s at the left end of the lne of current, θ, when at the rght end, θ π...( 7)
P Y d Summary: The magnetc feld produced by a long, straght, constant current, curls symmetrcally about the current drecton. Its drecton at any pont s gven by the H. Its magntude, at any pont, s proportonal to the magntude of the current,, and nversely proportonal to the radal dstance,, from the current. const 4/4/6 Magnetc Felds ( F.obllard) 55
Ampere s Law: ot-savart gves the magnetc feld vector, n terms of the current that produces t. It s the magnetc analogue to Coulomb s law, whch gves the electrc feld vector, n terms of the electrc charge producng t. However, ot-savart, lke Coulomb, becomes mathematcally dffcult when appled to many practcal cases. Ampere s law s a powerful reformulaton of ot-savart, that allows many otherwse dffcult cases, to be easly solved. Ampere s law s the magnetc analogue to Gauss law for electrc felds. It explots the essental symmetry present n many stuatons, to facltate ther soluton. We wll derve Ampere s law for the partcular case of the feld due to a long straght current. Our result, however, has general applcablty. We start, by makng an observaton about our result for the long, straght current. 4/4/6 Magnetc Felds ( F.obllard) 56
Observaton: Consder the magnetc flux lnes about a long straght current. eye These flux lnes are crcular. Consder one wth radus. Our prevous dervaton, for ths case, equaton (7), gave ( 7) : ( ) eye s vew where (π) s the length around the crcumference of a crcular flux lne, and s the (const.) magntude of the feld vector, along that flux lne. Concluson: Multplyng by the dsplacement around a flux lne, s proportonal to the current enclosed by the flux lne. 4/4/6 Magnetc Felds ( F.obllard) 57
ds µ π More Generally:...7 ( ) Let ds be an element of path around the flux lne. Let be the feld vector at that element of path. eye s vew ds Lets generalse our observaton, by summng the dot product.ds around the flux lne (that s, the lne ntegral of around the flux lne). s a sum of around a (.ds ) closed for all path ponts,.ds snce s parallel to ds at all ponts around the flux lne,.ds. ds. ds µ π ( π ) snce const. at all ponts on the flux lne, by symmetry. sum of all the ds segments s the crcumference of the crcular flux lne ( π ). From (7). 4/4/6 Magnetc Felds ( F.obllard) 58
Generalzaton: For a long, straght current, we have shown that ds The lne ntegral of (.ds) around a flux lne s proportonal to the current enclosed by that flux lne (that s, the current that produces the flux lne and the assocated feld vector ). µ s the permeablty constant 4π x 1-7 N A - Ths turns out to be vald for any closed path that encloses any constant currents. In General: ds Σ...( 8) The lne ntegral of.ds around any closed path s equal to µ Σ, where Σ s the algebrac sum of all the currents that cut through a surface that s bounded by the closed path, and µ s the permeablty constant Such a closed path s called an Amperan Path (AP). The AP s composed of dsplacements elements, ds, of path, and at a partcular element ds, the feld vector s. 4/4/6 Magnetc Felds ( F.obllard) 59 Equaton (8) s a general statement of Ampere s law. Ampere s Law
Σ : Notes: Σ s the algebrac sum of the currents that cut a surface bounded by the AP (an Amperan Surface). AP I I AP I AP A Σ I + I I I Σ I - I 5A Σ + - 5-3A The postve current drecton s taken to be consstent wth the H. If we curl the fngers around the AP n the drecton of travel for the lne ntegral, then the thumb wll pont n the postve current drecton. Note: that Ampere s law assumes all currents to be constant. 4/4/6 Magnetc Felds ( F.obllard) 6
Notes: AP: In prncple, an AP of any shape can be used,. However, to facltate a soluton, t s mportant to choose a shape such that the evaluaton of.ds s as smple as possble, for all ponts around the AP. The evaluaton of.ds wll be smple f () s parallel to ds (for example, f we follow a flux lne) () s perpendcular to ds ( f we cut perpendcularly, across flux lnes ) () (the magntude of s zero, or the feld s very weak) To acheve these smplfyng condtons, we need to choose an AP whose shape corresponds to the symmetry of the feld. Steps n usng Ampere s law: 1. Draw a dagram, showng the magnetc flux lnes.. Choose an optmally shaped AP. 3. Wrte down Ampere s law. 4. Integrate Ampere s law around the AP, carefully justfyng all steps, to fnd n terms of currents,. 4/4/6 Magnetc Felds ( F.obllard) 61
The Long Straght Current evsted: As our frst applcaton of Amperes law, we wll recompute our expresson for the magnetc feld vector,, at a radal dstance of, from a long, straght, constant current,. eye AP ds Draw the AP to follow the flux lne of radus. The AP s a crcle of radus, coaxal wth the current. eye s vew We wll ntegrate around the AP n the drecton of the flux. We show a typcal dsplacement element, ds, of the AP, and the correspondng vector. wll be parallel to ds at all ponts around the AP, snce we are followng a flux lne. Also, around the AP, the magntude of wll be constant, even though ts drecton changes. Ths follows from symmetry, snce at all ponts around the AP, the current has the same relatve orentaton, and s the same dstance away. 4/4/6 Magnetc Felds ( F.obllard) 6
We now nvoke Ampere s law. Fnd : AP ds.ds ds µ µ µ Σ Ampere s law s the total current. eye s vew snce parallel to ds along a flux lne, and thus.ds. ds ds ( π ) µ snce const., by symmetry. µ π AP s a crcle, of radus, and the sum of all scalar segments s the crcumference (π). Ths s the same result that we derved earler, usng the ot-savart law, but the dervaton here, has been much smpler. 4/4/6 Magnetc Felds ( F.obllard) 63
Feld Insde a Solenod: A solenod s a col of wre, wound on a cylndrcal former. For our second applcaton of Ampere s law we wll derve an expresson for the magnetc feld nsde such a solenod. Ths dervaton would be dffcult usng ot-savart, but s greatly smplfed by usng Ampere. Say the col has a total of N turns, unformly wound over a total col length of L. L Let the number of turns per unt length n N/L Say the current n the solenod s. Consder the solenod s magnetc feld. N turns The flux loops of the magnetc feld pass through the centre of the solenod, where they are confned, and are close together, producng a strong feld wthn. Outsde of the solenod, the flux loops can spread out, resultng n a weaker feld outsde. 4/4/6 Magnetc Felds ( F.obllard) 64
Drecton of the Internal Feld: L The drecton of the feld wthn the solenod s gven by a verson of the ght Hand ule (H) Hold the solenod so that the fngers of the ght hand, curl about the solenod n the drecton of the current n the turns. The thumb wll then pont n the drecton of the nternal feld through the solenod. N turns y symmetry, the flux lnes wthn the solenod wll be parallel to each other. Ths rule for the drecton of the nternal magnetc feld of a solenod, s a consequence of the H for a sngle conductor. The felds for the ndvdual adjacent conductng turns of the col, merge together nto the overall solenod feld. 4/4/6 Magnetc Felds ( F.obllard) 65
The AP: L N turns For the AP we choose a rectangle, of length d, nserted nto the solenod as shown the top sde s nsde the solenod; the bottom sde s outsde. Let the feld vector along the top sde of the rectangle be. Integrate around the rectangle n the clockwse drecton. d ds ds ds d ds Drectons of at each sde are shown. The bottom sde of the rectangle s outsde the solenod, where the feld s weak ( ) 4/4/6 Magnetc Felds ( F.obllard) 66
ds ds ds d ds Fnd : Invoke Ampere s law. top top ds +.ds + snce ds along top. Ampere's law.sde ds + + bottom along bottom ds + + L.sde ds µ Σ ds µ Σ top ds d µ µ µ µ Σ ( nd) ( nd) n perpendcular to ds along L. and. sdes Integratng along each sde of the rectangle const. along top by symmetry. there are (nd) turns cuttng the Amperan plane, each carryng current. top sde length d 4/4/6 Magnetc Felds ( F.obllard) 67
Conclusons: n The magntude,, of the nternal feld nsde a solenod, s proportonal to the current,, n the wndngs, and to the number of turns per unt length, n. ecause s ndependent of the locaton of the top sde of the rectangular AP, wthn the solenod, the nternal feld must be unform. Note: s ndependent of the dameter of the solenod. Example: Fnd the nternal feld of a solenod, that has 5 turns wound over a length of 1 cm, when t carres a current of A. µ n (4π x 1 7 ) x (5/.1) x.16 (T) 4/4/6 Magnetc Felds ( F.obllard) 68
Conclusons: In our consderatons of the magnetc feld, we frstly quantfed the feld, by defnng a magnetc feld vector,, usng the magnetc force on a movng charge. We then consdered varous stuatons, and applcatons, nvolvng the magnetc force, on both movng charges, and current-carryng conductors. Fnally, we looked at current as the source of the magnetc feld, and the law connectng the current to the feld. 4/4/6 Magnetc Felds ( F.obllard) 69
N S 4/4/6 Magnetc Felds ( F.obllard) 7