MOVING CHARGES AND MAGNETISM

Transcription

1 4 MOVING CHARGES AND MAGNETISM Moving charges can produce magnetic field. Magnetic field is produced around current carrying conductors also. The SI unit of magnetic induction (magnetic field intensity or magnetic flux density) is tesla(t) or weber /m 2. Magnetic field can exert force on charges. This force is called magnetic force or magnetic Lorentz force Magnetic Lorentz force When a charged particle with charge moves with a velocity in a magnetic field of strength, a magnetic Lorentz force acts on it. = θ = ( ) Here θ is the angle between and.this force acts perpendicular to velocity (or displacement). So the work done by magnetic Lorentz force on a moving charge is zero. (i) When a charged particle moves in a direction perpendicular to magnetic field When a charged particle moves with a velocity v in a magnetic field of strength B, a magnetic Lorentz force acts on it. F = qvb Sinθ F = qvb Here θ = 90 o This force provides centripetal force for the particle and it moves in a circular path. qvb = or r = = r mv, Radius is proportional to momentum. Time period of particle T = = π = π = π Angular frequency of particle, = 2π = 2π = 2π π = Unique Learning Centre, Ulloor, Tvpm. Phone: Page 1

2 (ii) When a charged particle moves at an angle with magnetic field Consider a charged particle moving at an angle with the magnetic field. The velocity of the particle can be resolved into two components -one component parallel to B and one component perpendicular to B. As there is a component of velocity perpendicular to magnetic field, the particle moves in a circular path. No force acts in the horizontal direction as the angle between B and horizontal component of velocity is zero. Therefore the path becomes helical. The distance moved by the particle along the field in one rotation is called pitch. (iii) When a charged particle moves parallel or anti-parallel to magnetic field In this case θ = 0 (for parallel) and θ = 180 (for anti-parallel). = 0 or 180, =0 No force acts on the charged particle. Important When a charged particle enters perpendicular to a uniform magnetic field, the speed and kinetic energy of the particle in the circular path remains the same. Velocity changes and momentum also changes (due to change of direction) An electron is not deflected through a certain region of space. Can we say that there is no magnetic field in that region? No. Electron experiences no force when it moves parallel or anti-parallel to the magnetic field. Care!!! When two different charged particles (electron and proton) having same momentum enter perpendicular to a uniform magnetic field, their paths are equally curved (since radius is proportional to momentum). Note: Electric field can exert force on a charge at rest or in motion. Magnetic field can exert force only if the charge is moving Method of finding the magnetic force on a charge moving perpendicular to a magnetic field The magnetic force on a charge moving perpendicular to a magnetic field can be found by using Fleming s left hand rule. Unique Learning Centre, Ulloor, Tvpm. Phone: Page 2

3 Stretch the fore finger, middle finger and thumb of left hand in mutually perpendicular directions. If fore finger represents magnetic field and middle finger represents the direction of motion of positive charge (current), then the thumb represents the force. It can be found by using right hand palm rule also. Open up the right hand palm. If the thumb represents the direction of motion of positive charge (current), the fingers point in the direction of magnetic field, then palm faces towards force Prove that the work done by magnetic force on a charged particle entering the field is zero. The magnetic force acting on a charged particle entering a magnetic field is = q ( ) is perpendicular to both. As is perpendicular to, F = ma = m. = 0 m. = 0 m. = 0. = = 0. = 0 But. = 2 = 0 = 0 m = 0 or or K = 0 K (Kinetic energy) is a constant. Unique Learning Centre, Ulloor, Tvpm. Phone: Page 3

4 As K is constant, the magnitude of velocity is also constant. By work energy theorem, Work done = Change of kinetic energy Here, Change of kinetic energy = 0 Work done is zero Motion of a charged particle in a combined electric and magnetic field When a charge moves in a combined electric and magnetic field, Force exerted by electric field (Electric Lorentz force) = Eq Force exerted by magnetic field (Magnetic Lorentz force) = q ( v B) Total force(lorentz force) = Eq + q( v B) If the fields are perpendicular and also perpendicular to the velocity of particle as in figure, = Eȷ, = Bk, = vı = qe = qeȷ (along y direction) = q( v B) = q(vı Bk) = qvbı k = qvb( ȷ ) = qvbȷ (along negative y direction) If the magnitude of the force are equal, qvb or vb = Eq = E or v = If a beam containing charged particles are sent to a crossed region of electric and magnetic fields, the particle with velocity = passes through the fields. So the fields act as velocity selector. Velocity selector is otherwise called velocity filter Aurora borealis and aurora australis The electrons and protons of solar wind are trapped near the earth s poles. These particles spiral around the magnetic lines. They bounce back and forth and emit light (green pink colour). It produces a beautiful glow of colours in the sky. This is called aurora. The aurora in the north arctic region is called northern light or aurora borealis. The aurora in the south is called aurora australis. Unique Learning Centre, Ulloor, Tvpm. Phone: Page 4

5 4.07 Cyclotron Cyclotron is a particle accelerator. It was devised by Lawrence and Livingstone. It is used to accelerate positively charged particles like protons, deuterons, alpha particles etc. It consists of two semi-circular metal boxes called dees D 1 and D 2. These are connected to a high frequency alternating source. At the centre, there is a positive ion source. A magnetic field is applied perpendicular to the flat surface of dees. The whole set up is kept inside an evacuated chamber. The ion source emits ions. Suppose D 1 is negative. The ion is attracted towards it. As the ion enters perpendicular to magnetic field, it moves in a circular path. Inside the dees the particle (ion) is not affected by electric field. When it completes a half rotation, the polarity of dees are reversed. When D 2 becomes negative, the ion is attracted towards D 2. When the ion is in the gap between dees, it is accelerated due to the effect of electric field. Its kinetic energy increases. The radius of circular path increases and finally it is taken out with the help of a deflector. The frequency of external source is made equal to the frequency of revolution of the particle. Then only the polarity of dees get reversed when the particle completes a half rotation. This frequency is called cyclotron frequency or magnetic resonance frequency. Time period of particle = But r = T = T = Frequency of particle = π π π = π This will be equal to the frequency of alternating source. Cyclotron frequency = π It is clear that time period and frequency are independent of velocity or radius of orbit. Velocity of the particle = Kinetic energy of the particle = mv = m = Unique Learning Centre, Ulloor, Tvpm. Phone: Page 5

6 Limitations It cannot be used to accelerate uncharged particles. Lighter particles like electrons cannot be accelerated using cyclotron. When the velocity increases, mass of these particles increases. Therefore frequency decreases. Frequency of alternating source cannot be changed in accordance with this decrease of frequency. Note: In a cyclotron, Electric field accelerates the positively charged particles Magnetic field keeps the particle in circular path Note: Mass spectrometer is the device used to separate charged particles Magnetic force on a current carrying conductor kept in a magnetic field Consider a conductor of length and area of cross section containing electrons per unit volume. Volume of the conductor Number of electrons in the conductor If the charge of one electron is q, the total charge = = = When this conductor is connected to a battery and kept in a magnetic field, The force exerted by the magnetic field on each electron of the conductor = q( v B ) Force on all the electrons = ( v B ) But ı =v or Force = (A J B ) ` But = A J = ȷ = v Force = ( B) (Transferred the vector sign from J to ) Force exerted by magnetic field, = ( B) Here is a vector of magnitude equal to the length of conductor and whose direction is the same as that of. Force on a current carrying conductor kept in a magnetic field if current is perpendicular to magnetic field is given by Fleming s left hand rule Biot Savart s law and its mathematical form Magnetic field at a point (db ) due to a small element of a current carrying conductor is directly proportional to the strength of current (), length of element (), sine of the angle (θ) between the element and the line joining element to the point and inversely proportional to the square of distance (r) between the element and the point. Unique Learning Centre, Ulloor, Tvpm. Phone: Page 6

7 db θ db = θ where is the permeability of free space = 4π 10 N/A or Tm/A Current element, () is a vector quantity. It has the same direction as. Magnitude of current element is the product of current and length of the element. Its unit is ampere metre (Am) Vector form of Biot-Savart s law θ We have, db = Π Multiplying numerator and denominator by r db = db = Π Π θ ( ). This is the vector form of Biot Savart s law. It is clear that db is perpendicular to the plane containing dl and rand is given by Right hand scew rule Biot Savart s law in terms of current density J = = = = where is volume. On substituting in db = Π db = Π ( ) ( ) 4.12 Biot Savart s law in terms of charge and velocity dl= dl = = q where is velocity. On substituting in db = Π db = ( ) ( ) Unique Learning Centre, Ulloor, Tvpm. Phone: Page 7

8 4.13 Comparison between Biot Savart s law and Couloumb s law Both are long ranged. Both are inverse square laws. Principle of super position can be applied to both. Electro static field is produced by a scalar source, charge. Magnetic field is produced by a vector source, current element. Electrostatic field is along the line joining the source charge and the point. Magnetic field is perpendicular to the plane containing the displacement vector and the current element. Biot Savert s law is angle dependent. Coulomb s law is angle independent Magnetic field on the axis of a circular loop of current Consider a circular loop of radius R carrying a current. Take a small element on it. The magnetic field at a point P which is at a distance r from it, db = Π ( ) Here the angle between dl and r is 90 0 r = Sin90 = db = db = Π Π From figure, r 2 = x 2 + R 2 db = Π ( ) The direction of db is perpendicular to the plane containing and. r db can be resolved into two components db x (along the axis) and db y (perpendicular to the axis). The y components, due to diametrically opposite elements cancel each other. The net magnetic field is along the axis due to one element, db x = db Cosθ The angle between and is (90-θ). Unique Learning Centre, Ulloor, Tvpm. Phone: Page 8

9 From figure, Sin (90-θ) = or Cosθ = Cosθ = ( ) db x = db Cosθ = Π ( ) ( ) Number of elements in the ring = Magnetic field due to all the elements, π B = Π ( ) ( ) π B = ( ) ( along the axis) Vectorially B = ( ) ı where ı is a unit vector along the axis of the circular loop. Field at the centre of the loop Here x = 0 B = B = ( ) ı ı 4.15 Magnetic field lines due to a circular loop of current Direction of magnetic field due to a circular loop of current If the direction of circular current is represented by the curly fingers of right hand then the stretched thumb gives the direction of magnetic field Magnetic field at the centre of concentric circular currents Unique Learning Centre, Ulloor, Tvpm. Phone: Page 9

10 (i) B = + (ii) B = 4.17 Clock rule (Used to find the direction of magnetic field around a current carrying loop) Imagine facing one end of a current carrying loop. If the current at that end is clockwise, that end is south. If it is anti- clockwise, that end is north. Variation of magnetic field with distance along the axis of a circular loop of current Magnetic field due to a long straight conductor carrying current Consider a long conductor carrying a current. Consider a small element of length. P is a point at a distance from it. Magnetic field at P due to is, = Draw AC BP (A) From ACB, Sinθ = AC = θ (1) From the sector PAC, dφ = AC = φ (2) From (1) and (2) θ = φ (3) Unique Learning Centre, Ulloor, Tvpm. Phone: Page 10

11 Equation (A) can be modified as db db = = From POA, cos φ = OR Putting in (4) db = I = (4) Magnetic field due to the entire wire, B = B = π sin sin B = π sin + sin B = π sin + sin B B B = π cos d = π sin = π sin sin ( ) If the conductor is infinitely long, = = 90 B = π B = B π = π The direction of this field is obtained from right hand thumb rule. Here the field at P is into the plane of paper. Vectorially, B = π n where n is a unit vector perpendicular to the plane of paper into it Direction of magnetic field around a current carrying conductor Direction of magnetic field around a current carrying conductor can be obtained from right hand thumb rule or Maxwell s right handed screw rule. Right hand thumb rule Imagine holding the current carrying conductor in right hand with thumb pointing in the direction of current. Then the curly fingers represent the direction of magnetic field. Unique Learning Centre, Ulloor, Tvpm. Phone: Page 11

12 Maxwell s right handed screw rule Imagine the rotation of a right handed screw along the current carrying conductor. If the direction of advancement of its tip represents the direction of current, then the direction of motion of thumb gives the direction of magnetic field Ampere s swimming rule: (Used to find the direction of deflection of magnetic needle kept under a current carrying conductor) Imagine a person swimming along the current carrying conductor in the direction of current facing a magnetic needle under it. Then the north pole of the needle will be deflected towards his left hand side. For a current carrying conductor, the magnetic field at a distance d is, B = B 4.21 Variation of magnetic field with distance for a current carrying conductor for a point in the interior and exterior of the conductor Magnetic field at a point outside at distance from the centre of a conductor of radius is. Magnetic field at a point inside at distance from the centre of a conductor of radius is 4.22 Force between two straight parallel conductors carrying current Consider two conductors of length l 1 and l 2 carrying current i 1 and i 2, in the same direction. Let the distance between them be d. F 21 F 12 d Consider the first conductor. Take a point at a distance d from it (at right side) l 1 i 1 l 2 i 1 Unique Learning Centre, Ulloor, Tvpm. Phone: Page 12

13 Magnetic field at the point B 1 = µ π The second conductor is kept there. (into the plane of paper by right hand thumb rule) Force acting on it = θ Here i = i 2, l = l 2, B = B 1, θ = 90 0 Force = i 2 l 2 µ Sin 90 π Force on unit length, F 21 = µ π (towards left by Fleming s left hand rule) Similarly magnetic field at a point d from second conductor (at left side) is B 2 = µ (outwards from the plane of paper by right hand thumb rule) π The first conductor is kept here. Force on it = sinθ Here i = i 1, l = l 1, B = B 2, θ = 90 0 Force = i 1 l 1 µ π sin 900 Force on unit length, F 12 = µ π (towards right by Fleming s left hand rule) F 12 = F 21 Thus if the currents are in the same direction the wires attract. If the currents are in the opposite direction they repel Definition of one ampere When = = = 1A and d = 1m Force on unit length = 2 10 One ampere is that current, which when passed in each of the two parallel infinitely long conductors of negligible cross section placed in vacuum at a distance of 1 m from each other produces between them a force of 2 10 newton per unit length Magnetic field at a point, half way between two wires carrying current of same magnitude If currents are in the same direction, B = = 0 If currents are in the opposite direction B = + = Magnetic field around parallel current carrying wires Unique Learning Centre, Ulloor, Tvpm. Phone: Page 13

14 4.26 Torque acting on a current loop which is kept in a magnetic field Consider a rectangular loop carrying current. It is kept in a magnetic field. The sides are a and b respectively. The angle between perpendicular to the plane of loop and magnetic field is θ. The force acting on arm PQ = θ Here =, =, =, θ = 90 0 Force, = (into the plane of paper and perpendicular by Fleming s left hand rule) The force acting on RS, = θ Here =, =, =,θ = 90 0 Force = ( outwards from the plane of paper and perpendicular by Fleming s left hand rule) The force acting on arms QR and SP = θ = θ These act along the axis of the coil in the opposite directions.hence they cancel each other. The forces on PQ and RS constitute a couple. Moment of couple=torque() = one of the forces X perpendicular distance between their lines of action. = X Sin θ But X = (area of the loop) = θ If there are N turns, = θ = θ where m is magnetic moment. Work done in rotating a coil through an angle from magnetic field direction W = mb(1-cos θ) 4.27 Ampere s circuital law and its application to find the magnetic field due to a current carrying straight wire, solenoid and toroid Ampere s circuital law Line integral of magnetic field over a closed path in free space is equal to times the total current enclosed by the path.. = Magnetic field due to an infinitely long straight conductor carrying current Unique Learning Centre, Ulloor, Tvpm. Phone: Page 14

15 Consider a long conductor carrying current. P is a point at a distance r from it. Including P we consider a closed path. It is a circle of radius r. This is the Amperian loop. Its plane is perpendicular to the conductor. At P, the angle between and a small element is zero.. = = B = B 2 = Solenoid B = An insulated wire wound in the form of a helix is called solenoid. For a solenoid, turns must be closely wound and diameter of coil is smaller than length. Consider an infinitely long solenoid carrying current. Let be the length. To find the magnetic field on the axis, we take a rectangular Amperian loop.. = On the arm ab, the angle between and a small element is zero (as the magnetic field is along the axis).. = On the arm bc the angle between and a small element is = Cos = On the arm cd the value of B=0. = 0 On the arm da the angle between and a small element is = 90 0 = 0. =. = = By Ampere s circuital law. = = = where N is the total number of turns = = But the number of turns in unit length, = Or = Unique Learning Centre, Ulloor, Tvpm. Phone: Page 15

16 Magnetic field due to a toroid Toroid is an endless solenoid in the form of a ring. Consider a toroid of radius r carrying current. Magnetic field has constant magnitude everywhere inside the toroid. But the magnetic field is zero in the open space interior(point P) and exterior point(point Q). On the axis, the magnetic field is along the axis of toroid. We consider three circular Amperian loops. (i)for the open space inside the toroid The amperian loop is taken as 1 and let its radius be. If is the magnetic field, By Ampere s circuital law,. = = As = 0 = 0 Or = 0 (ii)for the exterior of the toroid The Amperian loop is taken as 3 and let its radius be. If is the magnetic field, By Ampere s circuital law,. = = Here the current enclosed is zero as the current entering is cancelled by the current coming out. As = 0 = 0 Or = 0 (iii)for the point inside the toroid The Amperian loop is taken as 2 and let its radius be = r. If is the magnetic field, By Ampere s circuital law,. = = B = where N is the total number of turns B 2 = B = = B = 4.28 Finding polarity of a current carrying solenoid It can be found by using right hand grip rule or clock rule. Here thumb gives the direction of magnetic field along the axis of solenoid. Unique Learning Centre, Ulloor, Tvpm. Phone: Page 16

17 Variation of B with the length of solenoid Magnetic field at each end of a solenoid = Magnetic field inside a solenoid is uniform. But field outside is non uniform. When a current is passed through a solenoid, as the currents in the neighbouring turns are parallel, the turns attract each other. Hence the solenoid contracts Limitations of Ampere s circuital law It is not a universal law. It deals with steady currents only. For the Amperian loop, at every point of this loop B either disappears or is perpendicular to the loop or is uniform and tangential to the loop A loop of flexible wire having irregular shape and carrying current is placed in a magnetic field. What will be the expected shape of the loop? Circle with plane perpendicular to the field direction. Circle has maximum area and hence maximum number of field lines pass through it. Tension developed in the wire = Magnetic dipole moment of a current loop Magnetic moment of the loop, = If there are N turns m = Unit : Am 2 = 4.31 Magnetic moment of a revolving electron Consider an electron revolving round a nucleus in an orbit of radius r. Current produced = But T = = = Unique Learning Centre, Ulloor, Tvpm. Phone: Page 17

18 The magnetic moment associated with this circulating current is = = = Multiplying and dividing by = But = (angular momentum of electron) = = (1) (in to the plane of paper) is called gyro magnetic ratio. But angular momentum = = = Putting in (1) This is the expression for magnetic moment associated with an electron in an orbit. If n = 1 = On putting values = 9.27 x Am 2. This value is called Bohr magneton. The electron possesses orbital magnetic moment and spin magnetic moment Moving coil galvanometer Moving coil galvanometer consists of a fine insulated coil which is wound over an aluminum frame. It is free to move within the pole pieces of a cylindrical magnet. At the centre of the coil there is a soft iron cylinder. Soft iron helps in concentrating magnetic field lines and to keep the field radial so that the angle between the perpendicular to the plane of coil and magnetic field is always A pointer is attached to the coil. The coil is connected to a phosphor bronze spring. When current is passed through the coil, it experiences torque and it turns through an angle. = θ = 90 = It stops at a position. Then the restoring torque or anti torque in the spring balances the torque. Restoring torque = Kφ where K is the torsinal constant or spring constant K φ = Or φ = Deflection for unit current, φ is called current sensitivity.its unit is radian/ampere. To increase current sensitivity we have to increase the value of A, B,N and decrease the value of K. Voltage sensitivity = φ = φ Voltage Sensitivity = φ = where R is the resistance of the coil. Unique Learning Centre, Ulloor, Tvpm. Phone: Page 18

19 Figure of merit of galvanometer It is defined as the current which produces a deflection of one scale division in the galvanometer. G = φ = Properties of phosphor bronze It is a good conductor of electricity. It does not get oxidized when current is passed. It is perfectly elastic. It is non- magnetic. Its torsion constant (restoring torque per unit twist) is small Conversion of galvanometer into ammeter Galvanometer can be converted into ammeter by connecting a low resistance called shunt(low resistance) parallel to it. Let be the main current, i g be the current through galvanometer. The current through shunt = PD across galvanometer = PD across shunt i g R G = ( ) r s r s = 4.34 Use of shunt resistor It protects galvanometer from excess current It is used to covert galvanometer to ammeter It is used to increase the range of ammeter. Very Important points For an ideal ammeter, resistance = 0 As ammeter has small resistance, it is connected in series in a circuit. Due to its small resistance, the current is not affected. If ammeter is connected parallel, it draws large current due to low resistance and may get damaged. Resistance of ammeter is smaller than that of galvanometer. Ammeter of low range has more resistance than that of more range. Range of ammeter can be increased but cannot be decreased. If the range of ammeter is to be increased from to, the value of shunt, r s = Unique Learning Centre, Ulloor, Tvpm. Phone: Page 19

20 4.35 Conversion of galvanometer into voltmeter Galvanometer can be converted to voltmeter by connecting a high resistance in series with it. Let i g be the current. The same current passes through galvanometer and high resistance. V = i g R G + i g R V i g R G = i g R Or R = Very Important points Resistance of an ideal voltmeter is infinity. Resistance of voltmeter is high. When it is connected parallel, it draws only small current. Hence the P.D across the element is unaffected. Voltmeter of low range has low resistance. If the voltage range of voltmeter is to be increased n times then the value of high resistance, = ( 1) If the range of voltmeter is to be increased from to, the value of high resistance,r = ( ) The range of voltmeter can be increased or decreased. If an ammeter of range and resistance is to be converted to a voltmeter of range V then the resistance R to be connected R = - If a voltmeter of range and resistance is to be converted to an ammeter of range then the shunt S to be connected S = ( ) 4.36 Hall effect The creation of an electric field ( ) in a metal slab perpendicular to both magnetic field ( ) and current density ( ) when a magnetic field ( ) is applied perpendicular to current density ( ) is called Hall effect. Here or = where is Hall constant. Unique Learning Centre, Ulloor, Tvpm. Phone: Page 20