Answers to Courseook questions Chapter 56 Questions marke with a star (*) use the formula for the magnetic fiel create y a current μi ( = ) which is not on the syllaus an so is not eaminale See Figure 6 (page 343 in Physics for the I Diploma) We must apply the right-han rule for force a c e he magnetic fiel is into the page he force is into the page he magnetic fiel is out of the page he force is zero since the velocity is anti-parallel to the fiel he force is zero since the velocity is parallel to the fiel 3* he current of 4 A prouces a magnetic fiel into the page at P he other current prouces a magnetic fiel out of the page he magnitues are 4π 4 4 4π 3 = an = 6 π 7 π So the net fiel is 4 6 = 54 into the page 4π 5 3 4 4* At P, the fiel from the top wire is = 5 out of the page an π 4 4π 5 4 from the ottom wire it is 3 5 = also out of the page π 8 4 he net fiel at P is 375 out of the page At Q, the fiel is zero since the wires prouce magnetic fiels of equal magnitue an opposite irection 4π 5 4 3 = At R, the top wire prouces a fiel into the page an the π 4π 5 4 ottom wire prouces a fiel 3 67 = into the page π 6 4 he net fiel is 67 into the page Copyright Camrige University Press All rights reserve Page of 9
5 4 4 4 4 6 he magnetic fiel is irecte into the page In a the right-han rule (for a negative charge) gives a force ownwars away from the wire In it gives a force to the right 7 In a the fiel is to the right an so the force is into the page In the velocity is parallel to the fiel an the force is zero In c the force is towars the magnet (up the page) 8 No, ecause the fiel insie the solenoi is along the ais an so parallel to the velocity of the electron 9 We apply the right-han rule for force in each case to fin: fiel out of page, fiel out of page, fiel to the left, fiel to the left * a he A current attracts the current in DC an repels the current in A he forces on C an DA cancel out he magnetic fiel create at DC is 4π 6 8 4π = an at A it is = 4 π 5 π So the net force on the loop is upwars an has magnitue IL IL = (4 4 ) 5 5 = 3 N If the current in the loop is reverse the magnitue of the force will stay the same ut the force will reverse irection * a y the right-han rule for force, the force on A is into the page, on C it is zero (current an fiel parallel), on CD it is out of the page an zero on DA he magnitue of the force on A an CD is IL = 5 = N he net force is zero Copyright Camrige University Press All rights reserve Page of 9
* From NI μ 3 4π 5 N = we fin 6 = N 36 l 3 Hence the length of wire neee is L= πrn = π 36 7 m 3 y the right-han rule, at P the fiel is out of the page an at Q it is into the page 4* a he magnetic fiel can only e zero along a line in etween the wires an on the same plane as the wires his is ecause the fiel from the top wire is into the page, whereas that of the ottom wire is out of the page Let the istance from the 4π 4π 3 3 top wire e hen = giving = or π π ( ) ( ) 4 = 3 = 8 cm Now the fiels are opposite aove the top wire or elow the ottom wire Since the ottom current is the largest, the place where the fiel is zero has to e aove the top wire Let the istance from the top wire e hen 4π 4π 3 3 = giving = or π π ( + ) ( + ) 4 + = 3 = 4 cm 5 a he electric an magnetic forces on the electron must e equal an opposite he electric force is irecte upwars an so the magnetic force is ownwars Hence, y the right-han rule for force, the magnetic fiel must e irecte into the page V 3 he electric fiel is E = = = 4 N C 5 3 E 4 3 o fin the magnitue, use qe = qv = = = v 6 c he electric force an the magnetic force woul change irection ut keep the same magnitue So the net force will e zero for a proton with the same spee as that of the electron in a If the spee is oule the magnetic force woul oule ut the electric force woul e unchange Hence the forces will no longer alance an there will e a eflection ownwars 6 a here woul e equal an opposite forces at the ens of the ar magnet giving a net force of zero he forces woul, however, force the ar magnet to rotate 7 he force woul e F = IL θ = = 3 sin 5 3 3 sin 3 o 5 N Copyright Camrige University Press All rights reserve Page 3 of 9
8 Since the current is counter-clockwise in oth cases, the loops create magnetic fiels that can e approimate y small ar magnets as shown elow S N S N 9 Hence they will attract F F he raius of the circular path is foun from qv = R R = q π π π he electron takes a time R = = m to complete a revolution an so the v v q q q numer of revolutions per secon is f = A proton has a larger mass an so πm woul complete a smaller numer of revolutions per secon a he initial velocity of the proton may e ecompose into a component parallel to the positive - ais an a component parallel to the positive z-ais he 6 o 6 components are v = v cosθ = 5 cos 3 = 3 m s an 6 o 6 vz = vsinθ = 5 sin 3 = 75 m s Since the component along the z-ais is parallel to the magnetic fiel there will e no force in the vertical irection he -component of velocity is at right angles to the magnetic fiel an so the proton will eperience a force F = qv which will make the proton to perform a circle parallel to the y plane with spee v just as it also moves upwars with constant spee v z his means that the path of the proton is a heli (a spiral) Copyright Camrige University Press All rights reserve Page 4 of 9
he raius is foun from qv R q = R 7 6 67 3 Numerically this is R = = 7 m 9 6 5 c he proton takes a time the numer of revolutions per secon is f πr π πm = = to complete a revolution an so v v q q 9 6 5 6 7 76 q = = = πm π 67 v z 6 = 75 m s e he perio (time for one revolution) is the inverse of the frequency an so = = 3 s an so the istance covere vertically in this time is 6 76 6 z = v = 75 3 = 98 m z πm A charge particle performs a circle insie a magnetic fiel with perio = q In other wors the particle sweeps an angle of 36 o in a time equal to the perio In this case the particle only sweeps an angle of 3 o an so this will take a time 3 πm πm π 9 = = = = 596 s 9 q 6q 6 6 5 3* he 4 A current creates a fiel ownwars at P he other two create a fiel upwars he fiels are: 4π 4 6 53 4π = =, = = 4 an π 5 π 4π 3 = = 4 π 5 he net fiel is then + 3 = 7 upwars Copyright Camrige University Press All rights reserve Page 5 of 9
4* A iagram is the following: in A out θ C φ Using the cosine rule, the angles of the triangle forme y the currents an the origin of the aes are: + c a 5 + 8 4 A = cos = cos = 4 c 5 8 a + c 5 + 4 8 = cos = cos = 5 o ac 5 4 Hence the angles in the iagram are θ = 4 o an φ = 5 9 = 35 o, o o o he fiel from the in current is 4π 5 3 = = π 8 4π 5 5 = = out current is π 4 herefore, the components of the fiels are: an that from the o o = 3 cos(9 4 ) = o o y = 3 sin(9 4 ) = 74, an o = 5 cos 35 = 4 Hence the net fiel has components an o y = 5 sin 35 = 87 = + 4 = 88 an y = 74 + 88 = 4 So the net fiel is 4 + = at θ = arctan = 8 o to the horizontal 88 88 4 9 Copyright Camrige University Press All rights reserve Page 6 of 9
4π 3 5* he magnetic fiel of wire at wire is = = π 3 irecte to the left an the fiel from wire 3 at wire is 4π 4 3 = = irecte upwars he net magnetic fiel has π 4 magnitue + = 8 at 35 o to the positive -ais he force per unit F 5 length is then I = = 8 N = 8 μn L 6 he electron completes a full revolution in a time of = v hus a charge e moves past the circle in this time an so the current create y the Q e ev revolving electron is I = = = v ev Hence the fiel create y the electron at the nucleus is = μ 4πr 7* his is a flawe question as it stans μ he fiel insie a solenoi is given y NI V = he current is foun from I = l R L where R is the resistance of the wire use his resistance is R = ρ where L is the A μ length of the wire So = NVA Now the length of the solenoi is l = N( r) where lρl π π π r is the raius of the wire Hence μ NVA μ V r μ V r μ V r = = = = N r ρ L r ρ L ρ L We ρ L have a fie mass an hence volume of metal he volume is given y v= πr L Sustituting in the formula for the magnetic fiel we get μvπ r μvπ 3 = r ρ v = ρv πr So to maimize this fiel we nee a large wire raius his means that we shoul use all the metal into a single turn of wire his is why the question is flawe: with one single turn we o not have a solenoi! here must e aitional restrictions place on this prolem such as a given minimum numer of turns or solenoi raius 8* a oth currents prouce a magnetic fiel in the same irection an so the net fiel I I is just = μ + μ π( r ) π( r+ ) Copyright Camrige University Press All rights reserve Page 7 of 9
We may rewrite μ I π ( r ) ( r+ ) = + μ I π r ( ) ( + ) r r μ I ( ) ( ) use r r + μ I = + + + + + his is completely epecte From a large istance away (ie r >> ) we see just μ one current of magnitue I an so the fiel is epecte to e I = I I c he fiels are now opposite an so = μ μ π( r ) π( r+ ) We may rewrite μ I π ( r ) ( r+ ) = μ I π r ( ) ( + ) r r μ I ( ) ( ) use r r + μ I = + + + In this case, from a large istance away we see zero current to first approimation an so we o not see the familiar r in the magnetic fiel 9 a he raius of a circular path in a magnetic fiel is increases when the spee increases R = an so the raius q Copyright Camrige University Press All rights reserve Page 8 of 9
Acceleration takes place every time the particle crosses the gap At that instant, the potential on the other sie has to e of the right sign so as to accelerate the particle After half a revolution the particle will again e at the gap an again it must face a potential on the opposite sie that is again of the right sign for acceleration his sign must e opposite to what it was half a revolution efore Hence the frequency of the alternating source must e equal to the frequency of rotation c We know that R = an so q πr πm = v = q For an electron, f 9 q 6 5 3 4 = = = = πm π 9 e he kinetic energy increases y kev at every crossing After revolutions there have een crossings an so the energy increase 4 y = 4 kev = 4 MeV = 38 J 3* he magnetic fiel at the position of the A wire ue to the other wire is 4π 8 4 = an is irecte ownwars Hence the net fiel at the π 5 4 4 4 position of the A wire is 8 3 = 5 ownwars F 4 Hence the force per unit length is I = = 5 = 5 N m L y the right-han rule for force, the force is irecte to the left 3 a he comine magnetic fiel from the two wires at point R must point ownwars so as to cancel the uniform fiel Since R is closer to Q, the fiel of Q is larger than the fiel from P Hence the current in Q must go out of the page If the current increases, the net fiel from P an Q increases as well, so that the total fiel at R is no longer zero If we move closer to Q the fiel from Q will e much larger than the fiel from P an so their comine fiel will e ownwars an much larger than eternal fiel Hence the point has to move to the left 3 he raius of the circular path is given y R = an so the perio is q πr πm = v = q Since the particles are ientical, a = an E v 4 E = = v = = Copyright Camrige University Press All rights reserve Page 9 of 9