Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Electrostatics wikipedia , lookup
Work (physics) wikipedia , lookup
Field (physics) wikipedia , lookup
Maxwell's equations wikipedia , lookup
Condensed matter physics wikipedia , lookup
Electromagnetism wikipedia , lookup
Magnetic field wikipedia , lookup
Neutron magnetic moment wikipedia , lookup
Magnetic monopole wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Superconductivity wikipedia , lookup
SF017 A group of phenomena associated with magnetic field. UNIT 6: MAGNETISM SF027 1 6.1 Magnetic Field Definition – is defined as the region around a magnet where a magnetic force can be experienced. { A stationary electric charge is surrounded by an electric field only. { When an electric charge moves, it is surrounded by an electric field and a magnetic field. The motion of the electric charge produces the magnetic field. field { Magnetic field has two poles, called north (N) and south (S). (S) This magnetic poles are always found in pairs whereas a single magnetic pole has never been found. Like poles (Nz (N-N or SS-S) repel each other. Opposite poles (Nz (N-S) attract each other. 6.1.1 Magnetic field lines { Magnetic field lines are used to represent a magnetic field. { By convention, magnetic field lines leave the north pole and enters the south pole of a magnet. z Magnetic field lines can be represented by straight lines or curves. The tangent to a curved field line at a point indicates the direction of the magnetic field at that point as shown in figure 6.1.a. { SF027 Fig. 6.1a P direction of magnetic field at point P. 2 SF017 z { Magnetic field can be represented by crosses of by dotted circles as shown in figures 6.1b and 6.1c. X X X X X X X X X X X X Fig. 6.1c : magnetic field lines Fig. 6.1b : magnetic field lines leave the page perpendicularly enter the page perpendicularly A uniform field is represented by parallel lines of force. This means that the number of lines passing perpendicularly through unit area at all cross-sections in a magnetic field are the same as shown in figure 6.1d. Fig. 6.1d { SF027 unit crosscross-sectional area A non-uniform field is represented by non-parallel lines. The number of magnetic field lines varies at different unit cross-sections as shown in figure 6.1e. 3 Fig. 6.1e A1 A2 weaker field in A2 stronger field in A1 The number of lines per unit crosscross-sectional area is proportional to the magnitude of the magnetic field. field { Magnetic field lines do not intersect one another. 6.1.2 Magnetic field lines Pattern { The pattern of the magnetic field lines can be determined by using two methods. Using compass needles (shown in figure 6.1f) z { SF027 Fig. 6.1f : plotting a magnetic field line of a bar magnet. 4 SF017 z { Using sprinkling iron filings on paper (shown in figure 6.1g). Fig. 6.1g : Thin iron filing indicate the magnetic field lines around a bar magnet. Figure below shows the various pattern of magnetic field lines around the magnets. a. Bar magnet SF027 5 b. Horseshoe or U magnet c. Two bar magnets (unlike pole) - attractive d. Two bar magnets (like poles) - repulsive SF027 Neutral point (point (point where the resultant magnetic 6 force is zero). zero). SF017 6.1.3 Magnetization of a Soft Iron { There are two methods to magnetized the soft iron. Using the permanent magnet. z { One permanent magnet z A permanent magnet is bring near to the soft iron and touching the surface of the soft iron by following the path in the figure 6.1h. N z z S Fig. 6.1h This method is called induced magnetization. magnetization The arrows in the soft iron represent the magnetization direction with the arrowhead being the north pole and arrow tail being the south pole. It is also known as domains ( the tiny magnetized region because of spin magnetic moment of the electron). SF027 7 In an unmagnetized piece of soft iron, these domains are arranged randomly but it is aligned in one direction when the soft iron becomes magnetized. z The soft iron becomes a temporary magnet with its south pole facing the north pole of the permanent magnet and vise versa as shown in figure 6.1h. Two permanent magnets z Bring and touch the first magnet to one end of the soft iron and another end with the second magnet as shown in figure 6.1i and 6.1j. z { Fig. 6.1i SF027 Fig. 6.1j N S S NN S 8 SF017 z Using Electrical circuit. { A soft iron is placed inside a solenoid (a long coil of wire consisting of many loops of wire) that is connected to the power supply as shown in figure 6.1k. N S I I Current anticlockwise Switch, S Current - clockwise Fig. 6.1k { { When the switch S is closed, the current I flows in the solenoid and produces magnetic field. The directions of the fields associated with the solenoid can be found by viewing the current flows in the solenoid from both end as shown if figure 6.1k or applying the right hand grip rule below. Important Thumb – north pole Other fingers – direction of current in solenoid. SF027 { Other examples: a. S I b. S I { 9 N I N I Note : If you drop a permanent magnet on the floor or strike it with a hammer, you may jar the domains into randomness. randomness The magnet can thus lose some or all of its magnetism. Heating a magnet too can cause a loss of magnetism. z z SF027 10 SF017 The permanent magnet also can be demagnetized by placing it inside a solenoid that connected to an alternating source. source Example 1 : (exercise) Sketch the magnetic field lines pattern around the bar magnets as shown in figures below. a. b. z { 6.2 Magnetic Flux Density and Magnetic Flux 6.2.1 Magnetic flux density, B { Definition – is defined as the magnetic flux per unit area across an area at right angles to the magnetic field. field Mathematically, Φ where B= SF027 B Φ B : magnetic flux A⊥ A⊥ : area at right angles to the magnetic field { It also known as magnetic induction (magnetic magnetic field intensity) intensity 11 { It is a vector quantity and its direction follows the direction of the magnetic field. Its unit is weber per metre squared (Wb m-2) or tesla (T). Unit conversion : -2 4 { { 1 T = 1 Wb m = 10 gauss(G ) 6.2.2 Magnetic flux, ΦB { Magnetic flux of a uniform magnetic field, field Definition – is defined as the scalar product between the magnetic flux density, B with the vector of the surface area, A. r r Mathematically, Φ B = B • A = BA cos θ where Φ B : magnetic flux θ : angle between the direction of B and A. { { It is a scalar quantity and its unit is weber (Wb). Consider a uniform magnetic field B passing through a surface area A as shown in figures 6.2a and 6.2b. area, A From the fig. 6.2a, θ =0°, thus r A Φ B = BA cos 0 o Φ B = BA SF027 Fig. 6.2a r B 12 SF017 r A area, A { r B θ { Note that the direction of vector A is always perpendicular (normal) to the surface area, A. Fig. 6.2b The magnetic flux is proportional to the number of field lines passing through the area. Let us consider an element of area dA on an arbitrarily shaped surface r as shown in figure 6.2c. { If the magnetic field at this element is B , ther magnetic r flux through the element r is B • dA dA θ r Where dA is a vector that is perpendicular to the surface and has a r B magnitude equal to the area dA. { Fig. 6.2c SF027 { between B and A is θ, thus Φ B = BA cos θ { { From the fig. 6.2b, the angle Therefore, the total magnetic flux through the surface is given by r r Φ B = ∫ B • dA = ∫ BdA cos θ 13 Example 2 : 30 o r B A Figure above shows a flat surface with an area of 3.0 cm2 is placed in the uniform magnetic field. The plane surface makes an angle 30 ° with the direction of the magnetic field. If the magnetic flux through the surface is 0.90 mWb, calculate the magnitude of the magnetic field. Solution: 30 o SF027 A=3.0x10-4 m2, ΦB=0.90x10-3 Wb r From the figure, the angle between B A r o o o B and A is θ = 90 − 30 = 60 θ 30 o By applying the equation of magnetic flux for uniform B hence Φ B = BA cos θ ΦB B= A cos θ B = 6.0 T 14 SF017 6.3 Biot-Savart Law { The law can be stated as : The magnitude of magnetic flux density dB at a point P which is a distance r from a very short length dl of a conductor carrying a current I is given by { dB ∝ Idl sin θ r2 (6.3a) where θ is the angle between the short length and the line joining it to point P. This law can be summarized by using the diagram shown in figure 6.3a. I θ r dl P r r̂ { I Fig. 6.3a r dB into the paper The direction of dB is given by the right hand grip rule (figure 6.3b). Fig. 6.3b Important Thumb – direction of current SF027 Other { fingers – direction of magnetic field. If the medium around the conductor is vacuum or air then the equation 6.3a can be written as dB = { 15 µ0 Idl sin θ 4π r 2 (6.3b) Magnitude form In vector form, using the unit vector r̂, we have dB = µ0 Idl × rˆ 4π r 2 (6.3c) Vector form where Idl : current element { µ0 : permeability of free space = 4 π x 10 −7 T m A−1 To find the total magnetic field B at any point in space due to the currentr in a complete circuit, eq. 6.3b needs to be integrated over all segment dl that carry current, symbolically B= µ0 Idl sin θ 4π ∫ r 2 (6.3d) Magnitude form or SF027 B= µ0 Idl × rˆ 4π ∫ r 2 (6.3e) Vector form 16 SF017 6.4 Ampere’s Law { { Ampere’s law is an alternative to the Biot-Savart law. For conductors in a vacuum or air the law states r r B ∫ • dl = µ0 I enc where ∫ Bdl cosθ = µ I or 0 enc integral of the flux density around ∫ Bdl cosθ : line a closed path It is the sum of the terms Bdl { (6.4a) cos θ for every very short length dl of the closed path. B : magnitude of the magnetic flux density I enc : current enclosed by the path r r θ : angle between the direction of B and dl This law can be summarized by using the diagram shown in figure 6.4a. r B Part of closed path r dl θ clockwise or anticlockwise Fig. 6.4a SF027 17 { Note : z Both Biot-Savart and Ampere’s law can be used to determine the magnetic field of a long straight conductor, a circular coil and a long thin solenoid. 6.5 Magnetic field produced by the electrical current When a current flows in a conductor wire or coil, the magnetic field will be produced. { The direction of magnetic field around the wire or coil can be determined by using the right hand grip rule as shown in figure 6.3b. 6.5.1 Magnetic field of a long straight conductor (wire) carrying current { The magnetic field lines pattern around a straight conductor carrying current is shown in figures 6.5a and 6.5b. { I r B or r B SF027 I Fig. 6.5a I r B r B Current out of the page 18 SF017 r B I r B r B I Fig. 6.5b a X Current into the page z Consider a straight conductor with length 2a carrying a current I as shown in figure 6.5c. z To find the field dB at point P caused by the element of the conductor of length 2 2 r dl θ r= x +y y π −θ P r X dB 0 x dl=dy shown in figure 6.5c, firstly we use Biot-Savart law. dB = SF027 -a r B The equation of magnetic flux density at any point from a long straight wire carrying current can be determined by using Biot-Savart law. { 2a XI or I µ0 Idy sin θ 4π r2 19 Fig. 6.5c z From the figure 6.5c, r = x2 + y2 and sin θ = sin(π − θ ) = x x2 + y2 Therefore dB at point P produced by element dl is given by x dy 2 2 µ0 I x + y dB = 4π x2 + y2 µI xdy dB = 0 4π x 2 + y 2 3 / 2 ( z (6.5a) To find the magnitude of the total magnetic flux density B at point P supplied by the whole conductor, integrate eq. 6.5a from -a to a then µ0 I −a B= xdy a ∫ dB = 4π ∫ (x SF027 ) 2 + y2 µ0 I 2a 4π x x 2 + a 2 ) 3/ 2 (6.5b) 20 SF017 z When the length of the conductor is infinitely long then a is much x 2 + a 2 ≈ a. larger than x so that z Therefore the B at point P produced by whole conductor is where B= µ0 I 2πx x : distance of any point from the conductor 6.5.2 Magnetic field of a Circular Shaped Coil The magnetic field lines pattern around a circular shaped coil carrying current is shown in figures 6.5d. { I R N or S { X I S I I SF027 N I Fig. 6.5d 21 The equation of magnetic flux density at any point from a circular shaped coil carrying current can be determined by using Biot-Savart law. z Consider a circular shaped conductor with radius R that carries a current I as shown in figure 6.5e. y r dl R φ r = x2 + R2 O x I z I I Fig. 6.5e r dB dB y φ P dBx x z SF027 To find the field dB at a point P on the axis of the circular (loop) at distance x from the centre O, we need to apply Biot-Savart law. 22 SF017 z From the figure 6.5e, dl and r are perpendicular and the direction of z the field dB caused by the element dl lies in the yz-plane. Since r = x 2 + R 2, the magnitude dB of the field due to element z dl is given by µ Idl sin θ o and θ = 90 dB = 0 2 4π r µI dl dB = 0 2 4 π (x + R 2 ) The components of the vector dB are x-component : dBx = dB cos φ and cos φ = (x R + R2 2 dBx = µ0 I dl R 2 2 4 π (x + R ) (x 2 + R 2 )1 / 2 dBx = µ0 IR dl 4π x 2 + R 2 ( ) 1/ 2 ) 3/ 2 SF027 23 y-component : dB y = dB sin φ and sin φ = y z I X I Fig. 6.5f SF027 + R2 ) 1/ 2 elements around the loop, the resultant component is zero. zero It because the current in any element on the one side of the loop sets up a dBy that cancels the dBy set up by the current through the element diametrically opposite it. it (figure 6.5f) Therefore the resultant field at point P must along the x axis and is given by r dB P O 2 µ0 Ix dl 2 4 π (x + R 2 )3 / 2 By symmetry, when the component dBy are summed over all dB y = z (x x r dB ∫ dB = ∫ x x Bx = Bx = µ0 IR dl 4π x 2 + R 2 µ0 IR ( ( 4π x 2 + R 2 ( µ0 IR 2 2 x2 + R2 ) 3/ 2 dl and ∫ dl = 2πR ) ∫ Circumference of 3/ 2 the circular coil ) 3/ 2 24 SF017 z In general , B= µ0 IR 2 ( 2 x +R 2 On the axis of a circular coil ) 2 3/ 2 or B= µ0 NIR 2 ( 2 x +R 2 On the axis of N circular coils ) 2 3/ 2 The magnitude of magnetic field B at point O (centre centre of the circular coil or loop) loop , x=0 is given by z B= where ( µ0 IR 2 20+R ) 2 3/ 2 B= µ0 NI 2R At the centre of N circular coils R : radius of the circular coil N : number of coils (loops) µ0 : permeability of free space = 4 π x 10 −7 T m A−1 I : current x : distance between a point with a centre of the coil SF027 25 6.5.3 Magnetic field of a Solenoid { The magnetic field lines pattern around a solenoid carrying current is shown in figures 6.5g. N S I I or Fig. 6.5g XI XI XI XI N S I SF027 I I I 26 SF017 { The equation of magnetic flux density around a solenoid carrying current I can be determined by using Ampere’s law. z Consider a very long solenoid with closely packed coils, the field is nearly uniform and parallel to the solenoid axes within the entire cross section, as shown in figure 6.5h. Fig. 6.5h r r c B dl r r dl dl b r l d r dl a B(outside) = 0(very small ) z z SF027 To find the magnetic field inside the solenoid and at the centre, we choose and draw the rectangle closed path abcd as shown in figure 6.5h (clockwise) for applying Ampere’s law. By considering this path consists of four segment : ab, bc, cd and 27 da, then Ampere’s law becomes r r B ∫ • dl = µ0 I enc and I enc = NI r cr r d r r ar r b r B • d l + ∫ B • dl + ∫ B • dl + ∫ B • dl = µ0 NI ∫ a b c d : number of coils (loops) our path encircles c d a o o o ∫ Bdl cos 90 + ∫ Bdl cos 0 + ∫ Bdl cos 90 = µ0 NI b where N d c B ∫ dl = µ0 NI c B= µ0 NI l and B = µ0 nI z c dl = l N =n l At the centre/ midmidpoint/ inside of N turn solenoid where n : number of coils per unit length The magnetic flux density at the end of the solenoid is given by B= SF027 ∫ and or d d 1 ( µ0 nI ) 2 28 SF017 { Example 3 : Two long straight wires are placed parallel to each other and carrying the same current I. Sketch the magnetic field lines pattern around both wires a. when the currents are in the same direction. b. when the currents are in opposite direction. Solution: I I a. I I or I I SF027 29 b. I I I I or I SF027 XI 30 SF017 { Example 4 : A long wire (X) carrying a current of 30 A is placed parallel to and 3.0 cm away from a similar wire (Y) carrying a current of 6.0 A. a. Find the magnitude and direction of the magnetic flux density midway between the wires : i. when the current are in the same direction. ii. when they are in opposite direction. b. When the currents are in the same direction there is a point somewhere between X and Y at which the magnetic flux density is zero. How far from X is this point ? (Given µ0 = 4π x 10-7 H m-1) Solution: IX=30 a. i. A, IY=6.0 A, d=3.0x10-2 m d r BX rX A rY or r BY IX SF027 IY rX IX r BX A rY IY r B d Y rX = rY = = 1.5 x10 − 2 m 31 2 By using the equation of magnetic field at any point near the straight wire, then at point A µI Magnitude of BX : B = 0 X B = 4.0 x10 −4 T 2πrX µI BY = 0 Y 2πrY X Magnitude of BY : X Direction : into the page (upwards) BY = 8.0 x10 −5 T Direction : out of page (downwards) Therefore the total r magnetic r flux r density at point A is Sign convention of B BA = BX + BY Into the page ⇒ + (positive) Out of page ⇒ - (negative) a.ii. BA = BX − BY BA = 3.2 x10 −4 T BX r BY d rX r B r X BY A Direction : into the page (upwards) r rY or IX rX rX = rY = SF027 IX IY A rY X IY d = 1.5 x10 − 2 m 2 32 SF017 By using the equation of magnetic field at any point near the straight wire, then at point A µI Magnitude of BX : B X = 0 X B = 4.0 x10 −4 T 2πrX Magnitude of BY : BY = X Direction : into the page (upwards) µ0 I Y 2πrY BY = 8.0 x10 −5 T Direction : into the page (upwards) Therefore the total r magnetic r flux r density at point A is BA = BX + BY BA = BX + BY BA = 4.8 x10 −4 T d b. r BX r BX A r Y rX or r BY SF027 Direction : into the page (upwards) IX IY IX rX A rX = r ry = d − r rY r BY IY 33 By using the equation of magnetic field at any point near the straight wire, then at point A µI Magnitude of BX : B = 0 X Direction : into the page X (upwards) 2πr Magnitude of BY : BY = µ0 I Y 2π(d − r) Direction : out of page (downwards) Since the total magnetic at point A is zero, hence r r flux density r BA = BX + BY 0 = BX − BY µ0 I X µ0 I Y = 2πr 2π(d − r) dI X r= I X + IY r = 2.5 x10 −2 m SF027 34 SF017 { Example 5 : a. A 2000 turns solenoid of length 40 cm and resistance 16 Ω is connected to a 20 V supply. Find the magnetic flux density at the mid point of the axis of the solenoid. b. A solenoid 1.30 m long and 2.60 cm in diameter carries a current of 18.0 A. The magnetic field inside the solenoid is 23.0 mT. Find the length of the wire forming the solenoid. (Halliday,Resnick&Walker,p.708,no.42) (Given µ0=4 π x 10-7 H m-1) Solution: a. Given N=2000 turns, l=40x10-2 m, R=16 Ω, V=20 V By applying the equation of magnetic flux density at the centre of the solenoid, thus V µ0 NI and I = R l µ NV B= 0 lR B = 7.9 x10 −3 T B= SF027 35 b. Given l=1.30 m, d=2.60x10-2 m, I=18.0 A, B=23.0x10-3 T By applying the equation of magnetic flux density inside the solenoid, thus µ0 NI B= l Bl N= µ0 I N = 1322 turns Since the shaped of each coil forms the solenoid is circle, then the circumference of one coil is circumference = πd circumference = 8.17 x10 −2 m Therefore the length of the wire forming the solenoid, L is { SF027 L = N (circumference) L = 108 m Example 6 : A closely wound circular coil a diameter of 4.00 cm has 600 turns and carries a current of 0.500 A. Determine the magnitude of the magnetic field a. at the centre of the coil. b. at a point on the axis of the coil 8.00 cm from its centre. (Given µ0=4 π x 10-7 H m-1) (Young&Freedman,p.1098,no.28.28) 36 SF017 Solution: N=600 turns, d=4.00x10-2 m, I=0.500 A a. By using the equation of magnetic flux density at the centre of the coil, thus µ NI d and R = B= 0 2R µ NI B= 0 d B = 9.43 x10 −3 T b. Given x=8.00x10-2 m. 2 By applying the equation of magnetic flux density at a point of distance x on the axis of the circular coil from its centre, thus B= B= ( µ0 NIR 2 2 x +R 2 ) 2 3/ 2 and R= d 2 µ0 NId 2 3/2 d2 8 x 2 + 4 B = 1.35 x10 −4 T SF027 { 37 Example 7 : The segment of wire in figures a and b carry a current of I=5.00 where the radius of the circular arc is R=3.00 A, cm. I R O I O R Fig. b : exercise Fig. a For each figure, find the magnetic flux density at origin (point O) (Given µ0=4 π x 10-7 H m-1) Solution: I=5.00 A, R=3.00x10-2 m Figure a : Sections (1) and (3) are straight line and the angle between dl and the line joining dl to point O is 0°or 180°. Ir 1 Hence B1 and B3 at point O is dl 2 R O SF027 µ0 Idl sin 0 o µ0 Idl sin 180 o B1 = and B3 = 4π ∫ r2 4π ∫ r2 3 B1 = B3 = 0 r dl 38 SF017 Section (2) is a quarter of circular coil hence the B2 at point O is 1 µ NI Use right hand grip and N = 1 B2 = 0 rule 4 2R µ0 I −5 B2 = 2.62 x10 T Direction : into the page B2 = 8R Therefore the total magnetic flux density at point O is BO = B1 + B2 + B3 BO = 2.62 x10 −5 T Direction : into the page Figure b : Ans. :5.24 x 10-5 T into the page. Example 8 : Four long, parallel power wires each carry 100 A current. A cross sectional diagram for this wires is a square, 20.0 cm on each side. For each case in figures a, b and c, { X X X X Fig. a X X X Fig. b :exercise X Fig. c : exercise SF027 39 i. sketch the magnetic field lines pattern on the diagram. ii. calculate the magnetic flux density at the centre of the square. (Given µ0=4 π x 10-7 H m-1) Solution: I1=I2=I3=I4=I=100 Figure (a) : i. X X I2 l X I3 l2 + l2 r1 = r2 = r3 = r4 = r = 2 r = 0.141 m Since r1=r2=r3=r4 and I1=I2=I3=I4 then the magnitude of B1,B2, B3 and B4 at point C are the same and given by µI B1 = B2 = B3 = B4 = B = 0 2πr 40 B = 1.41x10 −4 T X SF027 X A, l=20.0x10-2 m I1 l X ii. r r1 B2 rr2 r B B1 Cr 3 r4 B4 r3 X I4 X SF017 From the diagram, B1 cancel the B3 and B2 cancel the B4, thus r r r r r BC = B1 + B2 + B3 + B4 BC = 0 Figure (b) : Ans. : 0 Figure (c) : Ans. : 4.0 x 10-4 T direction : to the left (180°) 6.6 Magnetic Force on a Moving Charge A stationary electric charge in a magnetic field will not experience any { force. force But if the charge is moving with a velocity, v in a magnetic field, B then it will experience a force. This force known as magnetic force. force The magnitude of the magnetic force can be calculated by using the equation below : F = Bqv sin θ (6.6a) { In vector form, form where SF027 { ( r r r F = q v×B ) (6.6b) F : magnetic force B : magnetic flux density v : velocity of a charge q : magnitude of the charge r r θ : angle between v and B 41 The direction of the magnetic force can be determined by using the Fleming’s hand rule. Fleming’ z Fleming’s right hand rule : - for negative charge shown in figures 6.6a and 6.6b Fleming’ z Fleming’s left hand rule : - for positive charge r B r F r F r B r v r v Fig. 6.6a Thumb – direction of Force Important Fig. 6.6b First finger – direction of Field { Second finger – direction of Velocity. Velocity. Example 9 : r Determine the direction of the magnetic force, F exerted on a charge in each problems below. r a. b. v r + SF027 r B B − r v 42 SF017 c. X X X X r BX X d. X X rX v X −r v X I − X e. + r v I Solution: a. By using Fleming’s left hand rule, b. By using Fleming’s right hand rule, thus r thus r F (into the r B r page) v v + r F r B − (to the left) d. Using right hand grip rule to c. By using Fleming’s right hand rule, determine the direction of magnetic field forms by the current I on the thus charge position. Then apply the X X X X r Fleming’s right hand rule, thus r F X X −X X X X X X B (to the left) r BX SF027 X r v r F X (to the left) I X X −r v X X 43 e. Using right hand grip rule to determine the direction of magnetic field forms by the current I on the r charge position. Then apply the Fleming’s F (upwards) left hand rule, thus r X XB X Xr X X + I { X X Example 10 : (exercise) Determine the sign of a charge in each problems below. r r r a. b. B F { SF027 v B r v r F r v Ans. : positive charge, positive charge Example 11 : (exercise) Determine the direction of the magnetic force exerted on a positive charge in each figures below when a switch S is closed. 44 SF017 a. b. + Switch, S { r v r v + Switch, S Ans. :a. into the page, b. out of page Example 12 : Calculate the magnitude of the force on a proton travelling 3.0 x107 m s-1 in the uniform magnetic flux density of 1.5 Wb m-2, if : a. the velocity of the proton is perpendicular to the magnetic field. b. the velocity of the proton makes an angle 50° with the magnetic field. (Given the charge of the proton is +1.60 x 10-19 C) Solution: v=3x107 m s-1, B=1.5 T, q=1.60x10-19 C a. Given θ = 90° then by applying the equation of magnetic force on a moving charge, thus F = Bqv sin θ F = 7.2 x10 −12 N SF027 45 b. Given θ = 50° then by applying the equation of magnetic force on a moving charge, thus F = Bqv sin θ F = 5.5 x10 −12 N { { SF027 Example 13 : (exercise) An electron experiences the greatest force as it travels 2.9 x 106 m s-1 in a magnetic field when it is moving north. The force is upward and of magnitude 7.2 x 10-13 N. Find the magnitude and direction of the magnetic field. (Giancolli, pg.705, no.22) (Given the charge of the electron is -1.60 x 10-19 C) Ans : 1.6 T to the east. Example 14 : (exercise) An electron is moving in a magnetic field. At a particular instant, the speed of the electron is 3.0 x 106 m s-1. The magnitude of the magnetic field on the electron is 5.0 x 10-13 N and the angle between the velocity of the electron and the magnetic force is 30°. Calculate the magnitude of the magnetic flux density on the electron in the field. (Given the charge of the electron is -1.60 x 10-19 C) Ans : 1.2 T 46 SF017 6.7 Magnetic Force on a current-carrying conductor { { When a current-carrying conductor is placed in a magnetic field B, thus a magnetic force will acts on that conductor. The magnitude of the magnetic force exerts on the current-carrying conductor is given by (6.7a) F = BIL sin θ In vector form, form where r F { ( ) r r r (6.7b) F = I L× B F : magnetic force B : magnitude of the magnetic flux density I : current L : length of the conductor r θ : angle between direction of I and B The direction of the magnetic force can be determined by using the Fleming’ ’s left hand rule as shown in figure 6.7a. Fleming r B Important Thumb – direction of Force I First finger – direction of Field Second finger – direction of Current. Current. SF027 47 Fig. 6.7a { Note : z It is clear from eq. (6.7a), { the magnetic force on the conductor has its maximum value when the conductor (and therefore the current) and the magnetic field are perpendicular (at right angles) to each other then θ=90° (shown in figure 6.7b). { the magnetic force on the conductor is zero when the conductor (and therefore the current) is parallel to the magnetic field then θ=0° (shown r in figure 6.7c). r B I θ = 90 o Fmax = BIL sin 90 o Fmax = BIL Fig. 6.7b { SF027 B I θ = 0o F = BIL sin 0 o F =0 Fig. 6.7c One tesla is defined as the magnetic flux density of a field in which a force of 1 newton acts on a 1 metre length of a conductor which carrying a current of 1 ampere and is perpendicular to the field. 48 SF017 { Example 15 : Determine the direction of the magnetic force, exerted on a conductor carrying current, I in each problems below. a. b. X X X X r BX X X X X X X X X X X X r BX X I X X X X X X X X X X (to the right) X X I Solution: For both problems, use Fleming’s left hand rule : b. a. r F X X X X X X X X X X X X X X r BX X (to therleft) BX X I SF027 I r F 49 { Example 16 : A wire of 20 cm long is placed perpendicular to the magnetic field of 0.40 Wb m-2. a. Calculate the magnitude of the force on the wire when a current 12 A is flowing. b. For the same current in (a), determine the magnitude of the force on the wire when its length is extended to 30 cm. c. If the force on the 20 cm wire above is 60 x 10-2 N and the current flows is 12 A, find the magnitude of magnetic field was supplied. Solution: L=20x10-2 a. Given I m, B=0.40 T, θ=90° = 12 A. By applying the equation of magnetic force on a current-carrying conductor, thus F = BIL sin θ F = 0.96 N b. Given I = 12 A and L = 30x10-2 m By applying the equation of magnetic force on a current-carrying conductor, thus F = BIL sin θ F = 1.4 N SF027 50 SF017 c. Given I = 12 A, L = 20x10-2 m , θ=90° and F = 60x10-2 N By applying the equation of magnetic force on a current-carrying conductor, thus F = BIL sin θ B= F IL sin θ B = 0.25 T 6.8 Forces between two current-carrying conductors { r B2 SF027 Fig. 6.8a { Consider two identical straight conductors X and Y carrying currents I1 and I2 with length L are placed parallel to each other as shown in figure 6.8a. { The conductors are in vacuum and X Y their separation is d. I1 { The magnitude of the magnetic flux r I2 r B1 density, B1 at point P on conductor Y F12 due to the current in conductor X is P given by µI r B1 = 0 1 Direction : into the F 21 Q 2πd page/paper { I1 d I2 Conductor Y carries a current I2 and in the magnetic field B1 then conductor Y 51 will experiences a magnetic force, F12. The magnitude of F12 is given by { F12 = B1 I 2 L sin θ and θ = 90 o µ I Use Fleming’ Fleming’s left F12 = 0 1 I 2 L sin 90 o hand rule π 2 d µ I F12 = 0 1 I 2 L Direction : to the left (towards X) 2πd The magnitude of F21 is given by F21 = B2 I 1 L sin θ and θ = 90 o µ I F12 = 0 2 I 1 L sin 90 o 2πd µ I F12 = 0 2 I 1 L Direction : to the right (towards Y) 2πd { Conclusion : r r µII L F12 = F21 = F = 0 1 2 2πd (6.8a) The properties of this force : Attractive force SF027 52 SF017 If the direction of current in conductor Y is change to upside down as shown in figure 6.8b. { X Y I1 r F21 I1 Fig. 6.8b { Q I2 r B1 P r B2 d The magnitude of F12 and F21 can be determined by using eq. 6.8a and its direction by applying Fleming’s left r hand rule. F12 Conclusion : { The properties of this force : Repulsive force I2 Note : The currents are in the same direction – 2 conductors attract each other. z The currents are in opposite direction – 2 conductors repel each other. z SF027 53 { Example 17 : Two very long parallel wires are placed 2.0 cm apart in air. Both wires carry a current of 8.0 A and 10 A respectively. Find a. the magnitude of the magnetic force in newton, on each metre length of wire. b. the magnetic flux density at point P, midway between the wires if the currents (exercise) i. in the same direction. ii. In opposite direction. (Given µ0=4 π x 10-7 H m-1) Solution: I1=8.0 A, I2=10 A, d=2.0x10-2 m a. Given L = 1.0 m By applying the equation of force for two parallel current-carrying µ0 I 1 I 2 L conductors, thus F= 2πd F = 8.0 x10 −4 N b. Ans. : 0.4 x 10-4 T out of page (downwards), 3.6 x10-4 T into the page (upwards) Hint : Example 4a. SF027 54 SF017 { Example 18 : (exercise) Two long, straight, parallel wires in a vacuum are 0.25 m apart. i. The wires each carry a current of 2.40 A in the same direction. Calculate the force between the wires per metre of their length. Draw a sketch showing clearly the direction of the force on each wire. ii. The current in one of the wires is reduced to 0.64 A. Calculate the current needed in the second wire to maintain the same force between the wires per metre of their length as in (i). (Given µ0=4 π x 10-7 H m-1) Ans. : 4.6 µN, 9.0 A 6.9 Definition of Ampere and Ampere Balance 6.9.1 Definition of Ampere { From the eq. (6.8a), if two long, straight, parallel conductors , 1.0 m apart in vacuum carry equal 1.0 A currents hence the force per unit length that each conductor exerts on the other is F µ0 I 1 I 2 = L 2 πd SF027 F ( 4πx10 −7 )( 1.0 )( 1.0 ) = L 2π( 1.0 ) F = 2.0 x10 −7 N m −1 L 55 The ampere is defined as the constant current that, when it is flowing in each of two infinitely long, straight, parallel conductors conductors which have negligible of cross sectional areas and are 1.0 metre apart in vacuum, would produce a force per unit length between the conductors of 2.0 x 10-7 N m-1. 6.9.2 Ampere (current) Balance { An instrument used to measure a current absolutely, on the basis of the definition of the ampere (due to the forces between two long, straight, parallel conductors). { Figure 6.9a shows a schematic diagram for a current balance where the current can be determined by measuring the force between two conductors carrying the same current. E D d P C Q l Fr F A B { m G r r W = mg SF027 Fig. 6.9a H I A 56 SF017 { { The plane ABCD and light rod PQ are adjusted so that the plane is initially horizontal. Conductors BC and EF are placed parallel to each other, equal in length l and separated by a distance d. { { When a current I flows, conductors BC and EF repels, hence the plane ABCD is unbalanced. The mass m needed to restore the plane ABCD in torque equilibrium where Torque due to the mass = Torque due to the force on BC and l AB = l PQ WlPQ = Fl AB µIIl mg = 0 1 2 2πd µ0 I 2 l mg = 2πd then I= where SF027 and I1 = I 2 = I 2πmgd µ0 l g : gravitional acceleration (9 .81 m s - 2 ) 57 6.10 Motion of a Charged Particle in a Uniform Magnetic Field 6.10.1 A charged particle moves perpendicular to the magnetic field. { Consider a charged particle moving in a uniform magnetic field with its velocity perpendicular to the magnetic field. { As the particle enters the region, it will experiences a magnetic force which the force is perpendicular to the velocity of the particle. Hence the direction of its velocity changes but the magnetic force remains perpendicular to the velocity. { + This magnetic force, FB makes the path of the particle is a circular as shown in figures 6.10a, 6.10b, 6.10c and 6.10d. r v X X X X X r F X B X FB X SF027 Xr X r + X X X Fig. 6.10a v X + + r v r+ FB r v r FB rX v X + r v Fig. 6.10b 58 SF017 − r v X X X X X X r FB −X X X X X r X { r v X r− FB r v v − X X X Fig. 6.10c r v r FB − r FB − r v Fig. 6.10d Since the path is circle therefore the magnetic force FB contributes the centripetal force Fc (net force) in this motion. Thus FB = Fc mv 2 o and θ = 90 Bqv sin θ = r mv where r= m : mass of the charged particle Bq v : magnitude of the velocity r : radius of the circular path q : magnitude of the charged particle SF027 59 { The period of the circular motion, T makes by the particle is given by v = rω T= and 2πr v 2π T mv r= Bq ω= and or T= { Since T= 1 f 2πm Bq thus the frequency of the circular motion makes by the particle is f = Bq 2πm 6.10.2 A charged particle moves not perpendicular to the magnetic field. { If the direction of the initial velocity is not perpendicular to the uniform magnetic field, the velocity component parallel to the field is constant because there is no force parallel to the field. { Therefore the particle moves in a helix path as shown in figure 6.10e. SF027 60 SF017 y r v⊥ r r r v = v⊥ + v// r rwhere vr⊥ : velocity component perpendicular to r the B v// : velocity component parallel to the B r v r v// r B z { { Fig. 6.10e x mv⊥ The radius of the helix, r is given by r = Bq Example 19 : An electron at point A in figure below has a speed v0 of 1.41 x 106 m s-1. v0 A − 10.0 cm B Given e = 1.60 x10-19 C me= 9.11 x 10-31 kg Find a. the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from A to B. b. the time required for the electron to move from A to B. (Young&Freedman,p.1055,no.27.15) SF027 61 Solution: v0=1.41x106 m s-1, d=10.0x10-2 m a. Since the path makes by the electron is semicircular thus the magnitude of the magnetic field is given by d me v0 and r = Be 2 2me v0 B = 1.61x10 −4 T B= ed r= The direction of B : electron (use Fleming’s right hand rule) v0 A − r F B Into the page. b. Since the path is semicircular then the time required for the electron moves from A to B is half of the period and given by SF027 d 2π 1 and r = t = T where v0 = rω; ω = 2 2 T πd t= 2v0 t = 1.11x10 −7 s T= πd v0 62 SF017 6.11 Lorentz Force and Determination of q/m 6.11.1 Lorentz Force { Definition – is defined as the total force acting on a charge q moving with a velocity v in the presence of both an electric field { E and a magnetic field B. Its formula is given by r r r r r r r r F = FE + FB where FE = qE and FB = q v × B r r r r r F = q( E + v × B) where F : Lorentz force ( ) It also known as electromagnetic force. force 6.11.2 Determination of q/m { The value of q/m is constant for any charged particle. { { Consider a positive charged particle with mass m, charge q and speed v enters a region of space where the electric and magnetic fields are perpendicular to the particle’s velocity and to each other as shown in figure 6.11a. SF027 63 + + ++ + + ++ ++Fr++ + + ++ ++ + r B X X XB X X X r r X X X X vX v X + X X X X X X + r v r X X X r X E X −− −− −− − − −− −− −− − −F−− E X Fig. 6.11a { The charged particle will experiences the electric force FE is downwards { magnitude Bqv (fig. 6.11a). If the particle travels in a straight line with constant velocity hence the electric and magnetic r r forces r are equal r in magnitude. Therefore with magnitude qE and the magnetic force FB is upwards with F = FE + FB Bqv = qE E v= B { SF027 and F =0 (6.11a) Only particles with speeds equal to E/B can pass through without being deflected by the fields. Eq. (6.11a) also works for electron or other 64 negative charged particles. SF017 { If no electric field in fig. 6.11a, then the particle will makes a circular path where therefore { FB = FC mv 2 Bqv = r q v = m rB q E = 2 m rB and v= E B (6.11b) In the J.J. Thomson’s experiment , the speed v of electron with mass m is determined by an accelerating potential difference V between two plates where Kinetic energy of the = Electric potential electron energy 1 2 mv = eV 2 2eV v= m (6.11c) Therefore by substituting eq. (6.11c) into eq. (6.11a) then the value of e/m is given by SF027 65 2eV E = m B 2 e E2 = m 2VB 2 { (6.11d) From the experiment the value of e/m is 1.758820174 x 1011 C kg-1. Example 20 : An electron with kinetic energy of 5.0 keV passes perpendicular through a uniform magnetic field of 0.40 x 10-3 T. It is found to follow a circular path. Calculate a. the radius of the circular path. b. the time required for the electron to complete one revolution. (Given 1 eV=1.60x10-19 J, e/m =1.76x1011 C kg-1, me =9.11x10-31 kg) Solution: B=0.40x10-3 T, K=5.0x103(1.60x10-19)=8.0x10-16 a. The speed of the electron is 1 2 mv 2 2K v= m K= SF027 v = 4.2 x107 m s −1 66 J SF017 Since the path made by the electron is circular, thus FB = FC mv 2 o Bev sin θ = and θ = 90 r v r= e B m r = 0.60 m b. The time required for the electron to complete one revolution is 2π v = rω and ω = T 2πr T= T = 9.0 x10 −8 s v { Example 21: (exercise) A proton moving in a circular path perpendicular to a constant magnetic field takes 1.00 µs to complete one revolution. Determine the magnitude of the magnetic field. (Serway & Jewett,pg.921,no.32) (Given mp =1.67x10-27 kg, charge of the proton, q=1.60 x 10-19 C) Ans. : 6.56 x 10-2 T SF027 67 6.12 Torque on a Coil in a Magnetic Field { Consider a rectangular coil (loop) of wire with side lengths a and b that it can turn about axis PQ. The coil is in a magnetic field of flux density B and the plane of the coil makes an angle θ with the direction of the magnetic field. A current I is flowing round the coil as shown in figure r 6.12a. r r F B B I I r r A B P φ θ r F1 r B b a I r F1 r B Q r F I Fig. 6.12a SF027 68 SF017 r B rotation r F1 { r B r F1 r B r A θ φ φ b 2 Q φ b sin φ 2 rotation b sin φ 2 Fig. 6.12b : Side view From the figure 6.12a, the force F1 on the right side of the coil is to the right and B is perpendicular to the current thus the magnitude of the force F1 is given by { { SF027 { { F1 = BIL sin 90 o and L = a F1 = IaB (6.12a) Another force F1 with the same magnitude but opposite direction acts on the opposite side of the coil. (fig. 6.12a) The forces F act on the side length b where the lines of action of both forces lies along the PQ. (fig. 6.12a) 69 The total force on the coil is zero but the net torque is not zero because the forces F1 are not lie along the same line thus the rotation of the coil about an axis PQ is clockwise (fig. 6.12b). The magnitude of the net torque about the axis PQ ( fig. 6.12b) is given by b b τ = − F1 sin φ − F1 sin φ 2 2 b τ = −2 F1 sin φ and F1 = IaB 2 b τ = −2(IaB ) sin φ 2 τ = − IabB sin φ and ab = A(area of coil) τ = IAB sin φ o and φ = 90 − θ (6.12b) or τ = IAB cos θ τ : torque on the coil B : magnetic flux density I : current flows in the coil where SF027 70 SF017 { For a coil of N turns, the torque is given by τ = NIAB sin φ (6.12c) or τ = NIAB cos θ r r φ : angle between vector area A and B r θ : angle between the plane of the coil and B Note : N : number of turns (coils) z The torque is zero when θ=90° or φ=0° but the magnetic flux is maximum as shown in figure 6.12c. r r A τ = NIAB sin 0 o or τ = NIAB cos 90 o B θ = 90 o τ = 0 o but ΦB = BA cos 0 o φ=0 maximum ΦB = BA where { Fig. 6.12c The torque is maximum when θ=0° or φ=90° but the magnetic flux is zero as shown in figure 6.12d. z SF027 71 τ = NIAB sin 90 or τ = NIAB cos 0 maximum τ = NIAB o but ΦB = BA cos 90 ΦB = 0 o z If r r µ = NIA r B o φ = 90 r A o θ = 0o Fig. 6.12d then eq. (6.12c) can be written as τ = µB sin φ Magnitude form In vector form is r r r τ = µ× B where µ is called magnetic moment or electromagnetic moment. moment Magnetic moment is a vector quantity. Its direction can be determined by using right hand grip rule. E.g. Important I { { I SF027 r µ Thumb – direction of magnetic moment Other fingers – direction of current in the 72 coil SF017 z In a radial field, field the plane of the coil is always parallel to the magnetic field for any orientation of the coil about the vertical axis as shown in figure 6.12e. Fig. 6.12e : Plan view of moving coil meter N radial field fixed soft iron cylinder S θ = 0o or φ = 90 o coil Hence the torque on the coil in a radial field is always constant and maximum given by τ = NIAB sin 90 oor τ = NIAB cos 0 o τ = NIAB maximum Radial field is used in moving coil galvanometer. SF027 73 6.13 Moving-Coil Galvanometer and Direct Current (DC) Motor 6.13.1 Moving-Coil Galvanometer { A galvanometer consists of a coil of wire suspended in the magnetic field of a permanent magnet. The coil is rectangular and consists of many turns of fine wire as shown in figure 6.13a. { When the current I flows through the coil, the magnetic field exerts a torque on the coil as given by τ = NIAB { This torque is opposed by a spring which exerts a torque, τs given by where τ s = kθ k : torsional constant θ : rotation angle of the coil in radian Fig. 6.13a SF027 { The coil and pointer will rotate only to the point where the spring torque balances the torque due to 74 magnetic field, thus SF017 τ = τs NIAB = kθ kθ I= NAB 6.13.2 Direct-Current (DC) Motor { A motor is an instrument that converts electrical energy to mechanical energy. { A motor works on the same principle as a galvanometer, except that there is no spring so the coil can rotate continuously in one direction. { When a current flows in the coil, a torque is produced, which causes the coil PQRS to rotate as shown in figure 6.13b. N R I S Fig. 6.13b SF027 { { { { SF027 I B1 Q I C1, C2 : Commutators B1, B2 : Brushes P C1 C2 B2 S I PQRS : Rectangular coil 75 The commutators also rotate with the coil PQRS but the brushes are stationary with the circuit. When the coil rotates half revolution (180°), each commutator changes its connection to the other brush where C1 touching B2 and C2 touching B1. This arrangement will cause the direction of the current through the coil to be reversed after every half revolution and ensures that the direction of the torque is always the same. Therefore the coil can turn continuously. The torque on the coil PQRS of the motor is given by τ = NIAB sin φ or τ = NIAB cos θ Example 22: A 20 turns rectangular coil with sides 6.0 cm x 4.0 cm is placed vertically in a uniform horizontal magnetic field of magnitude 1.0 T. If the current flows in the coil is 5.0 A, determine the torque acting on the coil when the plane of the coil is a. perpendicular to the field, b. parallel to the field, c. at 60° to the field. 76 SF017 Solution: N=20 turns, A=24x10-4 m2,B=1.0 T, I=5.0 A r B o From the figure, θ =90° and φ =0° , θ = 90r thus the torque on the coil is A τ = NIAB cos θ or τ = NIAB sin φ τ =0 r r A B From the figure, θ =0° and φ =90° , φ = 90 o thus the torque on the coil is τ = NIAB cos θ or τ = NIAB sin φ τ = 0.24 N m a. b. r B c. θ = 60 o φ = 30 r A o From the figure, θ =60° and φ =30° , thus the torque on the coil is τ = NIAB cos θ or τ = NIAB sin φ τ = 0.12 N m SF027 77 6.14 Hall Effect Definition – is defined as the production of a potential difference within a conductor or semiconductor through which a current flowing when there is a strong transverse magnetic field. 6.14.1 Explanation of Hall Effect { { Consider a flat conductor (such as copper) carrying a current I in the direction of +x-axis and is placed in a uniform magnetic field B (-z axis) perpendicular to the plane of the conductor as shown in figure 6.14a. +y VH SF027 r B h P − − − − r− − r− − − −− − d vd FB I I +x r −r E F + + + + + + + E+ + ++ + Q r d : width of the conductor B h : thickness of the conductor +z Fig. 6.14a 78 SF017 { In metal the charge carrier is electron. The electrons drift with a drift velocity vd in the opposite direction of the current I (shown in figure 6.14a). { { { The magnetic force FB acts on the electron in the direction upwards (Fleming’s right hand rule) and cause the electron deflected to the upper surface (P). As time passes, more and more electrons will accumulate on the upper surface (P) and left behind positive charges at the lower surface (Q). This results in an electric field E acting in the direction upwards and the electrons will experience electric force FE in the direction downwards. { { The electric force FE will gradually increase as more electrons accumulate at the upper surface. An equilibrium will be reached when the magnitude of the electric force FE becomes equal to the magnitude of the magnetic force FB and then { the electrons will drift along the –x axis without any more deflection. At this instant, the upper surface (P) will be negative potential and the lower surface (Q) will be positive potential. Hence a potential difference will exist and known as Hall potential difference (voltage) VH. SF027 79 { If the flat conductor is replaced with the flat semiconductor and the charge carrier is positive charge, explain the phenomenon of hall effect in this flat based on figure 6.14b. (exercise) +y r B h P VH I d + +z r B Q d : width of the conductor Fig. 6.14b 6.14.2 Equation of Hall Voltage , VH { { h : thickness of the conductor According to the figures 6.14a and 6.14b, the Hall potential difference (voltage) is VH = Ed When the equilibrium is reached between the electric and magnetic force then FB = FE where FE = qE and FB = Bqvd sin 90 o Bqvd = qE SF027 I +x and E= VH d 80 SF017 V Bqvd = q H d VH = Bvd d { (6.14a) From the definition of the drift velocity vd, I where e = q nAe I vd = n(dh )q vd = { and Therefore eq. 6.14a can be written as I d VH = B ndhq 1 BI VH = nq h where SF027 A = dh (6.14b) 1 : Hall coefficient nq h : thickness of the flat conductor/semiconductor n : number of charge carriers per unit volume 81 (charge carrier density) 6.14.3 Uses of the Hall Effect { The Hall effect can be used to z determine the sign of the charge carriers in a conductor / semiconductor. z determine the density of the charge carriers in a conductor/ semiconductor. z measure the magnetic flux density B that is known as Hall probe. { Example 23: A current of 3.00 A flows through a piece of metal of width 0.800 cm and thickness 625 µm. The metal is placed in a magnetic field of flux density 5.00 T perpendicular to the plane of the metal. If the Hall voltage across the width of the metal is 24.0 µV, determine a. the drift velocity of the electron in the metal. b. the density of the conduction electron in the metal. (Given charge of the electron, e= 1.60 x 10-19 C) Solution: I=3.00 A, d=0.800x10-2 m, h=625x10-6 m, B=5.00 T, VH=24.0x10-6 V. a. When the equilibrium is reached between the electric and magnetic force then FB = FE SF027 Bevd = eE and E= VH d 82 SF017 VH Bd vd = 6.00 x10 −4 m s −1 vd = b. By applying the equation of the Hall voltage, thus 1 BI VH = ne h BI n= VH eh n = 6.30 x10 28 electron m −3 { SF027 83 { { SF027 Example 24: (exercise : unit 6.12) A rectangular coil of 10 cm x 4.0 cm in a galvanometer has 50 turns and a magnetic flux density of 5.0 x 10-2 T. The resistance of the coil is 40 Ω and a potential difference of 12 V is applied across the galvanometer, calculate the maximum torque on the coil. Ans. : 3.0 x 10-3 N m. Example 25: (exercise) An electron moving at a steady speed of 0.50 x 106 m s-1 passes between two flat, parallel metal plates 2.0 cm apart with a p.d. of 100 V between them. The electron is kept travelling in a straight line perpendicular to the electric field between the plates by applying a magnetic field perpendicular to the electron’s path and to the electric field. Calculate : a. the intensity of the electric field. b. the magnetic flux density needed. Ans. : 0.50 x 104 V m-1, 0.010 T Example 26: (exercise) A moving coil meter has a 50 turns coil measuring 1.0 cm by 2.0 cm. It is held in a radial magnetic field of flux density 0.15 T and its suspension has a torsional constant of 3.0 x 10-6 N m rad-1. Find the current is required to give a deflection of 0.5 rad. Ans. :1.0 mA 84