Moving Charges and Magnetism PDF

Physics Chapter Four MOVING CHARGES AND MAGNETISM 4.1 INTRODUCTION Both Electricity and Magnetism have been known for more than 2000 years. However, it was only about 200 years ago, in 1820, that it was realised that they were intimately related*....

Physics Chapter Four MOVING CHARGES AND MAGNETISM 4.1 INTRODUCTION Both Electricity and Magnetism have been known for more than 2000 years. However, it was only about 200 years ago, in 1820, that it was realised that they were intimately related*. During a lecture demonstration in the summer of 1820, the Danish physicist Hans Christian Oersted noticed that a current in a straight wire caused a noticeable deflection in a nearby magnetic compass needle. He investigated this phenomenon. He found that the alignment of the needle is tangential to an imaginary circle which has the straight wire as its centre and has its plane perpendicular to the wire. This situation is depicted in Fig.4.1(a). It is noticeable when the current is large and the needle sufficiently close to the wire so that the earth’s magnetic field may be ignored. Reversing the direction of the current reverses the orientation of the needle [Fig. 4.1(b)]. The deflection increases on increasing the current or bringing the needle closer to the wire. Iron filings sprinkled around the wire arrange themselves in concentric circles with the wire as the centre [Fig. 4.1(c)]. Oersted concluded that moving charges or currents produced a magnetic field in the surrounding space. Following this there was intense experimentation. In 1864, the laws obeyed by electricity and magnetism were unified and formulated by 132 * See the box in Chapter 1, Page 3. 2015-16(20/01/2015)  Moving Charges and Magnetism James Maxwell who then realised that light was electromagnetic waves. Radio waves were discovered by Hertz, and produced by J.C.Bose and G. Marconi by the end of the 19th century. A remarkable scientific and technological progress has taken place in the 20th century. This is due to our increased understanding of electromagnetism and the invention of devices for production, amplification, transmission and detection of electromagnetic waves. FIGURE 4.1 The magnetic field due to a straight long current-carrying wire. The wire is perpendicular to the plane of the paper. A ring of compass needles surrounds the wire. The orientation of the needles is shown when (a) the current emer ges out of the plane of the paper, (b) the current moves into the plane of the paper. (c) The arrangement of iron filings around the wire. The darkened ends of the needle represent north poles. The effect of the earth’s magnetic field is neglected. In this chapter, we will see how magnetic field exerts forces on moving charged particles, like electrons, HANS CHRISTIAN OERSTED (1777–1851) protons, and current-carrying wires. We shall also learn how currents produce magnetic fields. We shall see how particles can be accelerated to very high energies in a cyclotron. We shall study how currents and voltages are detected by a galvanometer. In this and subsequent Chapter on magnetism, we adopt the following convention: A current or a field (electric or magnetic) emerging out of the plane of the paper is depicted by a dot (¤). A current or a field going into the plane of the paper is depicted by a cross ( ⊗ )*. Figures. 4.1(a) and 4.1(b) correspond to these two Hans Christian Oersted situations, respectively. (1777–1851) Danish physicist and chemist, 4.2 MAGNETIC F ORCE professor at Copenhagen. He observed that a 4.2.1 Sources and fields compass needle suffers a Before we introduce the concept of a magnetic field B, we deflection when placed near a wire carrying an shall recapitulate what we have learnt in Chapter 1 about electric current. This the electric field E. We have seen that the interaction discovery gave the first between two charges can be considered in two stages. empirical evidence of a The charge Q, the source of the field, produces an electric connection between electric field E, where and magnetic phenomena. * A dot appears like the tip of an arrow pointed at you, a cross is like the feathered tail of an arrow moving away from you. 133 2015-16(20/01/2015)  Physics E = Q r̂ / (4πε0 )r2 (4.1) where r̂ is unit vector along r, and the field E is a vector field. A charge q interacts with this field and experiences a force F given by F = q E = q Q r̂ / (4πε0 ) r 2 (4.2) As pointed out in the Chapter 1, the field E is not just an artefact but has a physical role. It can convey energy and momentum and is not established instantaneously but takes finite time to propagate. The concept of a field was specially stressed by Faraday and was incorporated by Maxwell in his unification of electricity and magnetism. In addition to depending on each point in space, it can also vary with time, i.e., be a HENDRIK ANTOON LORENTZ (1853 – 1928) Hendrik Antoon Lorentz function of time. In our discussions in this chapter, we (1853 – 1928) Dutch will assume that the fields do not change with time. theoretical physicist, The field at a particular point can be due to one or professor at Leiden. He investigated the more charges. If there are more charges the fields add relationship between vectorially. You have already learnt in Chapter 1 that this electricity, magnetism, and is called the principle of superposition. Once the field is mechanics. In order to known, the force on a test charge is given by Eq. (4.2). explain the observed effect Just as static charges produce an electric field, the of magnetic fields on currents or moving charges produce (in addition) a emitters of light (Zeeman magnetic field, denoted by B (r), again a vector field. It effect), he postulated the has several basic properties identical to the electric field. existence of electric charges in the atom, for which he It is defined at each point in space (and can in addition was awarded the Nobel Prize depend on time). Experimentally, it is found to obey the in 1902. He derived a set of principle of superposition: the magnetic field of several transformation equations sources is the vector addition of magnetic field of each (known after him, as individual source. Lorentz transformation equations) by some tangled 4.2.2 Magnetic Field, Lorentz Force mathematical arguments, Let us suppose that there is a point charge q (moving but he was not aware that these equations hinge on a with a velocity v and, located at r at a given time t ) in new concept of space and presence of both the electric field E (r) and the magnetic time. field B (r). The force on an electric charge q due to both of them can be written as F = q [ E (r) + v × B (r)] ≡ F electric +Fmagnetic (4.3) This force was given first by H.A. Lorentz based on the extensive experiments of Ampere and others. It is called the Lorentz force. You have already studied in detail the force due to the electric field. If we look at the interaction with the magnetic field, we find the following features. (i) It depends on q, v and B (charge of the particle, the velocity and the magnetic field). Force on a negative charge is opposite to that on a positive charge. (ii) The magnetic force q [ v × B ] includes a vector product of velocity 134 and magnetic field. The vector product makes the force due to magnetic 2015-16(20/01/2015)  Moving Charges and Magnetism field vanish (become zero) if velocity and magnetic field are parallel or anti-parallel. The force acts in a (sideways) direction perpendicular to both the velocity and the magnetic field. Its direction is given by the screw rule or right hand rule for vector (or cross) product as illustrated in Fig. 4.2. (iii) The magnetic force is zero if charge is not moving (as then |v|= 0). Only a moving charge feels the magnetic force. The expression for the magnetic force helps us to define the unit of the magnetic field, if one takes q, F and v, all to be unity in the force FIGURE 4.2 The direction of the magnetic equation F = q [ v × B] =q v B sin θ n̂ , where θ force acting on a charged particle. (a) The is the angle between v and B [see Fig. 4.2 (a)]. force on a positively charged particle with The magnitude of magnetic field B is 1 SI unit, velocity v and making an angle θ with the when the force acting on a unit charge (1 C), magnetic field B is given by the right-hand moving perpendicular to B with a speed 1m/s, rule. (b) A moving charged particle q is is one newton. deflected in an opposite sense to –q in the Dimensionally, we have [B] = [F/qv] and the unit presence of magnetic field. of B are Newton second / (coulomb metre). This unit is called tesla ( T) named after Nikola Tesla (1856 – 1943). Tesla is a rather large unit. A smaller unit (non-SI) called gauss (=10–4 tesla) is also often used. The earth’s magnetic field is about 3.6 × 10–5 T. Table 4.1 lists magnetic fields over a wide range in the universe. T ABLE 4.1 ORDER OF MAGNITUDES OF MAGNETIC FIELDS IN A VARIETY OF PHYSICAL SITUATIONS Physical situation Magnitude of B (in tesla) Surface of a neutron star 10 8 Typical large field in a laboratory 1 Near a small bar magnet 10 –2 On the earth’s surface 10 –5 Human nerve fibre 10 –10 Interstellar space 10 –12 4.2.3 Magnetic force on a current-carrying conductor We can extend the analysis for force due to magnetic field on a single moving charge to a straight rod carrying current. Consider a rod of a uniform cross-sectional area A and length l. We shall assume one kind of mobile carriers as in a conductor (here electrons). Let the number density of these mobile charge carriers in it be n. Then the total number of mobile charge carriers in it is nlA. For a steady current I in this conducting rod, we may assume that each mobile carrier has an average 135 2015-16(20/01/2015)  Physics drift velocity v d (see Chapter 3). In the presence of an external magnetic field B, the force on these carriers is: ×B F = (nlA)q vd ×× where q is the value of the charge on a carrier. Now nqvd is the current density j and |(nq vd )|A is the current I (see Chapter 3 for the discussion of current and current density). Thus, F = [(nqvd )l A] × B = [ jAl ] × B = Il × B (4.4) where l is a vector of magnitude l, the length of the rod, and with a direction identical to the current I. Note that the current I is not a vector. In the last step leading to Eq. (4.4), we have transferred the vector sign from j to I. Equation (4.4) holds for a straight rod. In this equation, B is the external magnetic field. It is not the field produced by the current-carrying rod. If the wire has an arbitrary shape we can calculate the Lorentz force on it by considering it as a collection of linear strips dlj and summing F = ∑ Idl j × B j This summation can be converted to an integral in most cases. O N PERMITTIVITY AND PERMEABILITY In the universal law of gravitation, we say that any two point masses exert a force on each other which is proportional to the product of the masses m1, m2 and inversely proportional to the square of the distance r between them. We write it as F = Gm 1m2 /r 2 where G is the universal constant of gravitation. Similarly in Coulomb’s law of electrostatics we write the force between two point charges q1 , q2, separated by a distance r as F = kq 1q2/r 2 where k is a constant of proportionality. In SI units, k is taken as 1/4πε where ε is the permittivity of the medium. Also in magnetism, we get another constant, which in SI units, is taken as µ/4π where µ is the permeability of the medium. Although G, ε and µ arise as proportionality constants, there is a difference between gravitational force and electromagnetic force. While the gravitational force does not depend on the intervening medium, the electromagnetic force depends on the medium between the two charges or magnets. Hence while G is a universal constant, ε and µ depend on the medium. They have different values for different media. The product εµ turns out to be related to the speed v of electromagnetic radiation in the medium through εµ =1/ v 2. Electric permittivity ε is a physical quantity that describes how an electric field affects and is affected by a medium. It is determined by the ability of a material to polarise in response to an applied field, and thereby to cancel, partially, the field inside the material. Similarly, magnetic permeability µ is the ability of a substance to acquire magnetisation in magnetic fields. It is a measure of the extent to which magnetic field can penetrate matter. EXAMPLE 4.1 Example 4.1 A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B (Fig. 4.3). What is the magnitude of the magnetic field? 136 2015-16(20/01/2015)  Moving Charges and Magnetism FIGURE 4.3 Solution From Eq. (4.4), we find that there is an upward force F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity: m g = I lB mg B= Il E XAMPLE 4.1 0.2 × 9.8 http://www.phys.hawaii.edu/~teb/optics/java/partmagn/index.html Interactive demonstration: Charged particles moving in a magnetic field. = = 0.65 T 2 × 1.5 Note that it would have been sufficient to specify m/l, the mass per unit length of the wire. The earth’s magnetic field is approximately 4 × 10–5 T and we have ignored it. Example 4.2 If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig. 4.4), which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge). FIGURE 4.4 E XAMPLE 4.2 Solution The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so v × B is along the z-axis (screw rule or right-hand thumb rule). So, (a) for electron it will be along –z axis. (b) for a positive charge (proton) the force is along +z axis. 4.3 MOTION IN A MAGNETIC F IELD We will now consider, in greater detail, the motion of a charge moving in a magnetic field. We have learnt in Mechanics (see Class XI book, Chapter 6) that a force on a particle does work if the force has a component along (or opposed to) the direction of motion of the particle. In the case of motion 137 2015-16(20/01/2015)  Physics of a charge in a magnetic field, the magnetic force is perpendicular to the velocity of the particle. So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed). [Notice that this is unlike the force due to an electric field, q E, which can have a component parallel (or antiparallel) to motion and thus can transfer energy in addition to momentum.] We shall consider motion of a charged particle in a uniform magnetic field. First consider the case of v perpendicular to B. The perpendicular force, q v × B, acts as a centripetal force and produces a circular motion perpendicular to the magnetic field. The particle will describe a circle if v and B are perpendicular to each other (Fig. 4.5). If velocity has a component along B, this component FIGURE 4.5 Circular motion remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to B is as before a circular one, thereby producing a helical motion (Fig. 4.6). You have already learnt in earlier classes (See Class XI, Chapter 4) that if r is the radius of the circular path of a particle, then a force of m v2 / r, acts perpendicular to the path towards the centre of the circle, and is called the centripetal force. If the velocity v is perpendicular to the magnetic field B, the magnetic force is perpendicular to both v and B and acts like a centripetal force. It has a magnitude q v B. Equating the two expressions for centripetal force, m v 2/r = q v B, which gives r = m v / qB (4.5) FIGURE 4.6 Helical motion for the radius of the circle described by the charged particle. The larger the momentum, the larger is the radius and bigger the circle described. If ω is the angular frequency, then v = ω r. So, ω = 2π ν = q B/ m [4.6(a)] which is independent of the velocity or energy. Here ν is the frequency of rotation. The independence of ν from energy has important application in the design of a cyclotron (see Section 4.4.2). The time taken for one revolution is T= 2π/ω ≡ 1/ν. If there is a component of the velocity parallel to the magnetic field (denoted by v ||), it will make the particle move along the field and the path of the particle would be a helical one (Fig. 4.6). The distance moved along the magnetic field in one rotation is called pitch p. Using Eq. [4.6 (a)], we have p = v ||T = 2πm v || / q B [4.6(b)] The radius of the circular component of motion is called the radius of 138 the helix. 2015-16(20/01/2015)  Moving Charges and Magnetism Example 4.3 What is the radius of the path of an electron (mass 9 × 10-31 kg and charge 1.6 × 10–19 C) moving at a speed of 3 ×107 m/s in a magnetic field of 6 × 10 –4 T perpendicular to it? What is its –19 frequency? Calculate its energy in keV. ( 1 eV = 1.6 × 10 J). Solution Using Eq. (4.5) we find EXAMPLE 4.3 r = m v / (qB ) = 9 ×10–31 kg × 3 × 107 m s–1 / ( 1.6 × 10–19 C × 6 × 10–4 T ) = 26 × 10–2 m = 26 cm ν = v / (2 πr) = 2×106 s–1 = 2×106 Hz =2 MHz. 2 –31 14 2 2 –17 E = (½ ) mv = (½ ) 9 × 10 kg × 9 × 10 m /s = 40.5 ×10 J ≈ 4×10 J = 2.5 keV. –16 HELICAL MOTION OF CHARGED PARTICLES AND AURORA BOREALIS In polar regions like Alaska and Northern Canada, a splendid display of colours is seen in the sky. The appearance of dancing green pink lights is fascinating, and equally puzzling. An explanation of this natural phenomenon is now found in physics, in terms of what we have studied here. Consider a charged particle of mass m and charge q, entering a region of magnetic field B with an initial velocity v. Let this velocity have a component vp parallel to the magnetic field and a component vn normal to it. There is no force on a charged particle in the direction of the field. Hence the particle continues to travel with the velocity vp parallel to the field. The normal component vn of the particle results in a Lorentz force (vn × B) which is perpendicular to both vn and B. As seen in Section 4.3.1 the particle thus has a tendency to perform a circular motion in a plane perpendicular to the magnetic field. When this is coupled with the velocity parallel to the field, the resulting trajectory will be a helix along the magnetic field line, as shown in Figure (a) here. Even if the field line bends, the helically moving particle is trapped and guided to move around the field line. Since the Lorentz force is normal to the velocity of each point, the field does no work on the particle and the magnitude of velocity remains the same. During a solar flare, a large number of electrons and protons are ejected from the sun. Some of them get trapped in the earth’s magnetic field and move in helical paths along the field lines. The field lines come closer to each other near the magnetic poles; see figure (b). Hence the density of charges increases near the poles. These particles collide with atoms and molecules of the atmosphere. Excited oxygen atoms emit green light and excited nitrogen atoms emits pink light. This phenomenon is called Aurora Borealis in physics. 139 2015-16(20/01/2015)  Physics 4.4 MOTION IN COMBINED ELECTRIC AND MAGNETIC FIELDS 4.4.1 Velocity selector You know that a charge q moving with velocity v in presence of both electric and magnetic fields experiences a force given by Eq. (4.3), that is, F = q (E + v × B) = FE + FB We shall consider the simple case in which electric and magnetic fields are perpendicular to each other and also perpendicular to the velocity of the particle, as shown in Fig. 4.7. We have, ( ) FE = qE = qE ˆj, FB = qv × B , = q v iˆ × Bk ˆ = –qB ˆj Therefore, F = q ( E – vB ) ˆj. Thus, electric and magnetic forces are in opposite directions as shown in the figure. Suppose, we adjust the value of E and B such that magnitudes of the two forces are equal. Then, total force on FIGURE 4.7 the charge is zero and the charge will move in the fields undeflected. This happens when, E qE = qvB or v = (4.7) B http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=33.0 This condition can be used to select charged particles of a particular velocity out of a beam containing charges moving with different speeds (irrespective of their charge and mass). The crossed E and B fields, therefore, serve as a velocity selector. Only particles with speed E/B pass undeflected through the region of crossed fields. This method was employed by J. J. Thomson in 1897 to measure the charge to mass ratio (e/m) of an electron. The principle is also employed in Mass Spectrometer – a device that separates charged particles, usually ions, according to their charge to mass ratio. 4.4.2 Cyclotron Interactive demonstration: The cyclotron is a machine to accelerate charged particles or ions to high energies. It was invented by E.O. Lawrence and M.S. Livingston in 1934 to investigate nuclear structure. The cyclotron uses both electric and magnetic fields in combination to increase the energy of charged particles. As the fields are perpendicular to each other they are called crossed fields. Cyclotron uses the fact that the frequency of revolution of the Cyclotron charged particle in a magnetic field is independent of its energy. The particles move most of the time inside two semicircular disc-like metal containers, D1 and D2, which are called dees as they look like the letter D. Figure 4.8 shows a schematic view of the cyclotron. Inside the metal boxes the particle is shielded and is not acted on by the electric field. The magnetic field, however, acts on the particle and makes it go round in a circular path inside a dee. Every time the particle moves from one dee to another it is acted upon by the electric field. The sign of the electric field is changed alternately in tune with the circular motion of the particle. This ensures that the particle is always accelerated by the electric field. 140 Each time the acceleration increases the energy of the particle. As energy 2015-16(20/01/2015)  Moving Charges and Magnetism increases, the radius of the circular path increases. So the path is a spiral one. The whole assembly is evacuated to minimise collisions between the ions and the air molecules. A high frequency alternating voltage is applied to the dees. In the sketch shown in Fig. 4.8, positive ions or positively charged particles (e.g., protons) are released at the centre P. They move in a semi-circular path in one of the dees and arrive in the gap between the dees in a time interval T/2; where T, the period of revolution, is given by Eq. (4.6), 1 2 πm T = = νc qB qB or ν c = (4.8) 2πm This frequency is called the cyclotron frequency for obvious reasons and is denoted by νc. The frequency νa of the applied voltage is adjusted so that the polarity of the dees is reversed in the same time that it takes the ions to complete one half of the revolution. The requirement νa = ν c is called the resonance condition. The phase of the supply is adjusted so that when the positive ions arrive at the edge of D1, D2 is at a lower potential and the ions are accelerated across the gap. Inside the dees the particles travel in a region free of the electric field. The increase in their kinetic energy is qV each time they cross from one dee to another (V refers to the voltage across the dees at that time). From Eq. (4.5), it is clear that the radius of their path goes on increasing each time their kinetic energy increases. The ions are repeatedly accelerated across the dees until they have the required energy to have a radius approximately that of the dees. They are then deflected by a magnetic field and leave the system via an exit slit. From Eq. (4.5) we have, qBR v = (4.9) m where R is the radius of the trajectory at exit, and equals the radius of a dee. Hence, the kinetic energy of the ions is, FIGURE 4.8 A schematic sketch of the cyclotron. There is a source of charged 1 q2 B 2 R 2 particles or ions at P which move in a mv 2 = (4.10) 2 2m circular fashion in the dees, D1 and D 2, on account of a uniform perpendicular The operation of the cyclotron is based on the magnetic field B. An alternating voltage fact that the time for one revolution of an ion is source accelerates these ions to high independent of its speed or radius of its orbit. speeds. The ions are eventually ‘extracted’ The cyclotron is used to bombard nuclei with at the exit port. energetic particles, so accelerated by it, and study 141 2015-16(20/01/2015)  Physics the resulting nuclear reactions. It is also used to implant ions into solids and modify their properties or even synthesise new materials. It is used in hospitals to produce radioactive substances which can be used in diagnosis and treatment. Example 4.4 A cyclotron’s oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating protons? If the radius of its ‘dees’ is 60 cm, what is the kinetic energy (in MeV) of the proton beam produced by the accelerator. –19 –27 –13 (e =1.60 × 10 C, mp = 1.67 × 10 kg, 1 MeV = 1.6 × 10 J). Solution The oscillator frequency should be same as proton’s cyclotron frequency. Using Eqs. (4.5) and [4.6(a)] we have E XAMPLE 4.4 B = 2π m ν/q =6.3 ×1.67 × 10 –27 × 107 / (1.6 × 10–19) = 0.66 T Final velocity of protons is v = r × 2π ν = 0.6 m × 6.3 ×107 = 3.78 × 107 m/s. 2 –27 14 –13 E = ½ mv = 1.67 ×10 × 14.3 × 10 / (2 × 1.6 × 10 ) = 7 MeV. ACCELERATORS IN I NDIA India has been an early entrant in the area of accelerator- based research. The vision of Dr. Meghnath Saha created a 37" Cyclotron in the Saha Institute of Nuclear Physics in Kolkata in 1953. This was soon followed by a series of Cockroft-Walton type of accelerators established in Tata Institute of Fundamental Research (TIFR), Mumbai, Aligarh Muslim University (AMU), Aligarh, Bose Institute, Kolkata and Andhra University, Waltair. The sixties saw the commissioning of a number of Van de Graaff accelerators: a 5.5 MV terminal machine in Bhabha Atomic Research Centre (BARC), Mumbai (1963); a 2 MV terminal machine in Indian Institute of Technology (IIT), Kanpur; a 400 kV terminal machine in Banaras Hindu University (BHU), Varanasi; and Punjabi University, Patiala. One 66 cm Cyclotron donated by the Rochester University of USA was commissioned in Panjab University, Chandigarh. A small electron accelerator was also established in University of Pune, Pune. In a major initiative taken in the seventies and eighties, a Variable Energy Cyclotron was built indigenously in Variable Energy Cyclotron Centre (VECC), Kolkata; 2 MV Tandem Van de Graaff accelerator was developed and built in BARC and a 14 MV Tandem Pelletron accelerator was installed in TIFR. This was soon followed by a 15 MV Tandem Pelletron established by University Grants Commission (UGC), as an inter-university facility in Inter-University Accelerator Centre (IUAC), New Delhi; a 3 MV Tandem Pelletron in Institute of Physics, Bhubaneshwar; and two 1.7 MV Tandetrons in Atomic Minerals Directorate for Exploration and Research, Hyderabad and Indira Gandhi Centre for Atomic Research, Kalpakkam. Both TIFR and IUAC are augmenting their facilities with the addition of superconducting LINAC modules to accelerate the ions to higher energies. Besides these ion accelerators, the Department of Atomic Energy (DAE) has developed many electron accelerators. A 2 GeV Synchrotron Radiation Source is being built in Raja Ramanna Centre for Advanced Technologies, Indore. The Department of Atomic Energy is considering Accelerator Driven Systems (ADS) for power production and fissile material breeding as future options. 142 2015-16(20/01/2015)  Moving Charges and Magnetism 4.5 MAGNETIC FIELD DUE TO A CURRENT ELEMENT, B IOT-SAVART LAW All magnetic fields that we know are due to currents (or moving charges) and due to intrinsic magnetic moments of particles. Here, we shall study the relation between current and the magnetic field it produces. It is given by the Biot-Savart’s law. Figure 4.9 shows a finite conductor XY carrying current I. Consider an infinitesimal element dl of the conductor. The magnetic field dB due to this element is to be determined at a point P which is at a distance r from it. Let θ be the angle between dl and the displacement vector r. According to Biot-Savart’s law, the magnitude of the magnetic field dB is proportional to the current I, the element length |dl|, and inversely proportional to the square of the distance r. Its direction* is perpendicular to the plane containing dl and r. Thus, in vector notation, I dl × r dB ∝ r3 FIGURE 4.9 Illustration of the Biot-Savart law. The µ0 I d l × r = [4.11(a)] current element I dl 4π r3 produces a field d B at a where µ0/4π is a constant of proportionality. The above distance r. The ⊗ sign expression holds when the medium is vacuum. indicates that the field is perpendicular The magnitude of this field is, to the plane of this µ 0 I dl sin θ page and directed dB = [4.11(b)] 4π r2 into it. where we have used the property of cross-product. Equation [4.11 (a)] constitutes our basic equation for the magnetic field. The proportionality constant in SI units has the exact value, µ0 = 10 −7 Tm/A [4.11(c)] 4π We call µ0 the permeability of free space (or vacuum). The Biot-Savart law for the magnetic field has certain similarities as well as differences with the Coulomb’s law for the electrostatic field. Some of these are: (i) Both are long range, since both depend inversely on the square of distance from the source to the point of interest. The principle of superposition applies to both fields. [In this connection, note that the magnetic field is linear in the source I dl just as the electrostatic field is linear in its source: the electric charge.] (ii) The electrostatic field is produced by a scalar source, namely, the electric charge. The magnetic field is produced by a vector source I dl. * The sense of dl × r is also given by the Right Hand Screw rule : Look at the plane containing vectors dl and r. Imagine moving from the first vector towards second vector. If the movement is anticlockwise, the resultant is towards you. If it is clockwise, the resultant is away from you. 143 2015-16(20/01/2015)  Physics (iii) The electrostatic field is along the displacement vector joining the source and the field point. The magnetic field is perpendicular to the plane containing the displacement vector r and the current element I dl. (iv) There is an angle dependence in the Biot-Savart law which is not present in the electrostatic case. In Fig. 4.9, the magnetic field at any point in the direction of dl (the dashed line) is zero. Along this line, θ = 0, sin θ = 0 and from Eq. [4.11(a)], |dB| = 0. There is an interesting relation between ε 0, the permittivity of free space; µ , the permeability of free space; and c, the speed of light in 0 vacuum:  µ0   1  ε 0 µ0 = ( 4 πε 0 )    4 π = 9  10  9 × 10  −7 = ( 1 )8 2 = (3 ×10 ) 1 c2 We will discuss this connection further in Chapter 8 on the electromagnetic waves. Since the speed of light in vacuum is constant, the product µ0ε 0 is fixed in magnitude. Choosing the value of either ε 0 or µ0, fixes the value of the other. In SI units, µ0 is fixed to be equal to 4π × 10–7 in magnitude. Example 4.5 An element ∆ l = ∆x ˆi is placed at the origin and carries a large current I = 10 A (Fig. 4.10). What is the magnetic field on the y-axis at a distance of 0.5 m. ∆x = 1 cm. FIGURE 4.10 Solution µ0 I dl sin θ |dB | = [using Eq. (4.11)] 4π r2 Tm dl = ∆x = 10 −2 m , I = 10 A, r = 0.5 m = y, µ0 / 4π = 10− 7 A θ = 90° ; sin θ = 1 10−7 × 10 × 10 −2 dB = = 4 × 10–8 T 25 × 10−2 The direction of the field is in the +z-direction. This is so since, EXAMPLE 4.5 ( ) dl × r = ∆x iˆ × y ˆj = y ∆ x iˆ × ˆj = y ∆x k ˆ We r emind you of the following cyclic property of cr oss-products, ˆi × ˆj = k ˆ ; ˆj × k ˆ = ˆi ; k ˆ × ˆi = ˆj 144 Note that the field is small in magnitude. 2015-16(20/01/2015)  Moving Charges and Magnetism In the next section, we shall use the Biot-Savart law to calculate the magnetic field due to a circular loop. 4.6 MAGNETIC FIELD ON THE AXIS OF A CIRCULAR CURRENT LOOP In this section, we shall evaluate the magnetic field due to a circular coil along its axis. The evaluation entails summing up the effect of infinitesimal current elements (I dl) mentioned in the previous section. We assume that the current I is steady and that the evaluation is carried out in free space (i.e., vacuum). Figure 4.11 depicts a circular loop carrying a steady current I. The loop is placed in the y-z plane with its centre at the origin O and has a radius R. The x-axis is the axis of the loop. We wish to calculate the magnetic field at the point P on this axis. Let x be the distance of P from the centre O of the loop. Consider a conducting element dl of the loop. This is shown in Fig. 4.11. The magnitude dB of the magnetic field due to dl is given by the Biot-Savart law [Eq. 4.11(a)], µ0 I dl × r dB = (4.12) 4π r3 FIGURE 4.11 Magnetic field on the Now r 2 = x 2 + R 2. Further, any element of the loop axis of a current carrying circular will be perpendicular to the displacement vector from loop of radius R. Shown are the the element to the axial point. For example, the element magnetic field dB (due to a line dl in Fig. 4.11 is in the y-z plane whereas the element dl ) and its displacement vector r from dl to the axial point P is in components along and the x-y plane. Hence |dl × r|=r dl. Thus, perpendicular to the axis. µ0 I dl dB = ( 4π x 2 + R 2 ) (4.13) The direction of dB is shown in Fig. 4.11. It is perpendicular to the plane formed by dl and r. It has an x-component dB x and a component perpendicular to x-axis, dB ⊥. When the components perpendicular to the x-axis are summed over, they cancel out and we obtain a null result. For example, the dB ⊥ component due to dl is cancelled by the contribution due to the diametrically opposite dl element, shown in Fig. 4.11. Thus, only the x-component survives. The net contribution along x-direction can be obtained by integrating dBx = dB cos θ over the loop. For Fig. 4.11, R cos θ = (4.14) ( x + R 2 )1/ 2 2 From Eqs. (4.13) and (4.14), µ0 Idl R dBx = (x ) 3 /2 4π 2 + R2 145 2015-16(20/01/2015)  Physics The summation of elements dl over the loop yields 2πR, the circumference of the loop. Thus, the magnetic field at P due to entire circular loop is µ0I R 2 B = B x ˆi = ˆ 3 /2 i ( 2 x 2 + R2 ) (4.15) As a special case of the above result, we may obtain the field at the centre of the loop. Here x = 0, and we obtain, µ0 I ˆ B0 = i (4.16) 2R The magnetic field lines due to a circular wire form closed loops and are shown in Fig. 4.12. The direction of the magnetic field is given by (another) right-hand thumb rule stated below: Curl the palm of your right hand around the circular wire with the fingers pointing in the direction of the current. The right-hand thumb gives the direction of the magnetic field. FIGURE 4.12 The magnetic field lines for a current loop. The direction of the field is given by the right-hand thumb rule described in the text. The upper side of the loop may be thought of as the north pole and the lower side as the south pole of a magnet. Example 4.6 A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2.0 cm as shown in Fig. 4.13(a). Consider the magnetic field B at the centre of the arc. (a) What is the magnetic field due to the straight segments? (b) In what way the contribution to B from the semicircle differs from that of a circular loop and in what way does it resemble? (c) Would your answer be dif ferent if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Fig. 4.13(b)? EXAMPLE 4.6 146 FIGURE 4.13 2015-16(20/01/2015)  Moving Charges and Magnetism Solution (a) dl and r for each element of the straight segments are parallel. Therefore, dl × r = 0. Straight segments do not contribute to |B|. (b) For all segments of the semicircular arc, dl × r are all parallel to each other (into the plane of the paper). All such contributions EXAMPLE 4.6 add up in magnitude. Hence direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a –4 circular loop. Thus B is 1.9 × 10 T normal to the plane of the paper going into it. (c) Same magnitude of B but opposite in direction to that in (b). Example 4.7 Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil? Solution Since the coil is tightly wound, we may take each circular EXAMPLE 4.7 element to have the same radius R = 10 cm = 0.1 m. The number of turns N = 100. The magnitude of the magnetic field is, µ 0 NI 4 π × 10 –7 × 102 × 1 B= = −4 = 2 π × 10 −4 = 6.28 × 10 T 2R 2 × 10–1 4.7 AMPERE’S C IRCUITAL LAW There is an alternative and appealing way in which the Biot-Savart law may be expressed. Ampere’s circuital law considers an open surface with a boundary (Fig. 4.14). The surface has current passing through it. We consider the boundary to be made up of a number of small line elements. Consider one such element of length dl. We take the value of the tangential component of the magnetic field, B t, at this element and multiply it by the length of that element dl [Note: Bt dl=B.d l]. All such products are added together. We consider the limit as the lengths of elements get smaller and their number gets larger. The sum then tends to an integral. Ampere’s law states that this integral is equal to µ0 times the total current FIGURE 4.14 passing through the surface, i.e., ∫ Bidl = µ I 0 [4.17(a)] where I is the total current through the surface. The integral is taken over the closed loop coinciding with the boundary C of the surface. The relation above involves a sign-convention, given by the right-hand rule. Let the fingers of the right-hand be curled in the sense the boundary is traversed in the loop integral “B.d l. Then the direction of the thumb gives the sense in which the current I is regarded as positive. For several applications, a much simplified version of Eq. [4.17(a)] proves sufficient. We shall assume that, in such cases, it is possible to choose the loop (called an amperian loop) such that at each point of the loop, either 147 2015-16(20/01/2015)  Physics (i) B is tangential to the loop and is a non-zero constant B, or (ii) B is normal to the loop, or (iii) B vanishes. Now, let L be the length (part) of the loop for which B is tangential. Let Ie be the current enclosed by the loop. Then, Eq. (4.17) reduces to, BL =µ0I e [4.17(b)] When there is a system with a symmetry such as for a straight infinite current-carrying wire in Fig. 4.15, the Ampere’s law enables an easy evaluation of the magnetic field, much the same way Gauss’ law helps in determination of the electric field. This is exhibited in the Andre Ampere (1775 – Example 4.9 below. The boundary of the loop chosen is 1836) Andre Marie Ampere a circle and magnetic field is tangential to the was a French physicist, circumference of the circle. The law gives, for the left hand mathematician and side of Eq. [4.17 (b)], B. 2πr. We find that the magnetic chemist who founded the field at a distance r outside the wire is tangential and science of electrodynamics. given by Ampere was a child prodigy who mastered advanced B × 2πr = µ0 I, mathematics by the age of B = µ0 I/ (2πr) (4.18) 12. Ampere grasped the significance of Oersted’s The above result for the infinite wire is interesting discovery. He carried out a from several points of view. large series of experiments (i) It implies that the field at every point on a circle of to explore the relationship radius r, (with the wire along the axis), is same in between current electricity magnitude. In other words, the magnetic field and magnetism. These possesses what is called a cylindrical symmetry. The investigations culminated field that normally can depend on three coordinates in 1827 with the ANDRE AMPERE (1775 –1836) publication of the depends only on one: r. Whenever there is symmetry, ‘Mathematical Theory of the solutions simplify. Electrodynamic Pheno- (ii) The field direction at any point on this circle is mena Deduced Solely from tangential to it. Thus, the lines of constant magnitude Experiments’. He hypo- of magnetic field form concentric circles. Notice now, thesised that all magnetic in Fig. 4.1(c), the iron filings form concentric circles. phenomena are due to These lines called magnetic field lines form closed circulating electric loops. This is unlike the electrostatic field lines which currents. Ampere was originate from positive charges and end at negative humble and absent- minded. He once forgot an charges. The expression for the magnetic field of a invitation to dine with the straight wire provides a theoretical justification to Emperor Napoleon. He died Oersted’s experiments. of pneumonia at the age of (iii) Another interesting point to note is that even though 61. His gravestone bears the wire is infinite, the field due to it at a nonzero the epitaph: Tandem Felix distance is not infinite. It tends to blow up only when (Happy at last). we come very close to the wire. The field is directly proportional to the current and inversely proportional to the distance from the (infinitely long) current 148 source. 2015-16(20/01/2015)  Moving Charges and Magnetism (iv) There exists a simple rule to determine the direction of the magnetic field due to a long wire. This rule, called the right-hand rule*, is: Grasp the wire in your right hand with your extended thumb pointing in the direction of the current. Your fingers will curl around in the direction of the magnetic field. Ampere’s circuital law is not new in content from Biot-Savart law. Both relate the magnetic field and the current, and both express the same physical consequences of a steady electrical current. Ampere’s law is to Biot-Savart law, what Gauss’s law is to Coulomb’s law. Both, Ampere’s and Gauss’s law relate a physical quantity on the periphery or boundary (magnetic or electric field) to another physical quantity, namely, the source, in the interior (current or charge). We also note that Ampere’s circuital law holds for steady currents which do not fluctuate with time. The following example will help us understand what is meant by the term enclosed current. Example 4.8 Figure 4.15 shows a long straight wire of a circular cross-section (radius a) carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region r < a and r > a. FIGURE 4.15 Solution (a) Consider the case r > a. The Amperian loop, labelled 2, is a circle concentric with the cross-section. For this loop, L = 2π r Ie = Current enclosed by the loop = I The result is the familiar expression for a long straight wire B (2π r) = µ 0I µ0 I B= [4.19(a)] 2 πr 1 EXAMPLE 4.8 B∝ (r > a) r (b) Consider the case r < a. The Amperian loop is a circle labelled 1. For this loop, taking the radius of the circle to be r, L =2πr * Note that there are two distinct right-hand rules: One which gives the direction of B on the axis of current-loop and the other which gives direction of B for a straight conducting wire. Fingers and thumb play different roles in the two. 149 2015-16(20/01/2015)  Physics Now the current enclosed I e is not I, but is less than this value. Since the current distribution is uniform, the current enclosed is,  πr 2  Ir 2 Ie = I  2  = 2  πa  a I r2 Using Ampere’s law, B (2 π r ) = µ 0 a2  µI  B =  0 2 r [4.19(b)]  2 πa  B∝r (r < a) FIGURE 4.16 Figure (4.16) shows a plot of the magnitude of B with distance r from the centre of the wire. The direction of the field is tangential to EXAMPLE 4.8 the respective circular loop (1 or 2) and given by the right-hand rule described earlier in this section. This example possesses the required symmetry so that Ampere’s law can be applied readily. It should be noted that while Ampere’s circuital law holds for any loop, it may not always facilitate an evaluation of the magnetic field in every case. For example, for the case of the circular loop discussed in Section 4.6, it cannot be applied to extract the simple expression B = µ0I/2R [Eq. (4.16)] for the field at the centre of the loop. However, there exists a large number of situations of high symmetry where the law can be conveniently applied. We shall use it in the next section to calculate the magnetic field produced by two commonly used and very useful magnetic systems: the solenoid and the toroid. 4.8 THE SOLENOID AND THE TOROID The solenoid and the toroid are two pieces of equipment which generate magnetic fields. The television uses the solenoid to generate magnetic fields needed. The synchrotron uses a combination of both to generate the high magnetic fields required. In both, solenoid and toroid, we come across a situation of high symmetry where Ampere’s law can be 150 conveniently applied. 2015-16(20/01/2015)  Moving Charges and Magnetism 4.8.1 The solenoid We shall discuss a long solenoid. By long solenoid we mean that the solenoid’s length is large compared to its radius. It consists of a long wire wound in the form of a helix where the neighbouring turns are closely spaced. So each turn can be regarded as a circular loop. The net magnetic field is the vector sum of the fields due to all the turns. Enamelled wires are used for winding so that turns are insulated from each other. FIGURE 4.17 (a) The magnetic field due to a section of the solenoid which has been stretched out for clarity. Only the exterior semi-circular part is shown. Notice how the circular loops between neighbouring turns tend to cancel. (b) The magnetic field of a finite solenoid. Figure 4.17 displays the magnetic field lines for a finite solenoid. We show a section of this solenoid in an enlarged manner in Fig. 4.17(a). Figure 4.17(b) shows the entire finite solenoid with its magnetic field. In Fig. 4.17(a), it is clear from the circular loops that the field between two neighbouring turns vanishes. In Fig. 4.17(b), we see that the field at the interior mid-point P is uniform, strong and along the axis of the solenoid. The field at the exterior mid-point Q is weak and moreover is along the axis of the solenoid with no perpendicular or normal component. As the solenoid is made longer it appears like a long cylindrical metal sheet. Figure 4.18 represents this idealised picture. The field outside the solenoid approaches zero. We shall assume that the field outside is zero. The field inside becomes everywhere parallel to the axis. FIGURE 4.18 The magnetic field of a very long solenoid. We consider a rectangular Amperian loop abcd to determine the field. 151 2015-16(20/01/2015)  Physics Consider a rectangular Amperian loop abcd. Along cd the field is zero as argued above. Along transverse sections bc and ad, the field component is zero. Thus, these two sections make no contribution. Let the field along ab be B. Thus, the relevant length of the Amperian loop is, L = h. Let n be the number of turns per unit length, then the total number of turns is nh. The enclosed current is, Ie = I (n h), where I is the current in the solenoid. From Ampere’s circuital law [Eq. 4.17 (b)] BL = µ0I e, B h = µ0I (n h) B = µ0 n I (4.20) The direction of the field is given by the right-hand rule. The solenoid is commonly used to obtain a uniform magnetic field. We shall see in the next chapter that a large field is possible by inserting a soft iron core inside the solenoid. 4.8.2 The toroid The toroid is a hollow circular ring on which a large number of turns of a wire are closely wound. It can be viewed as a solenoid which has been bent into a circular shape to close on itself. It is shown in Fig. 4.19(a) carrying a current I. We shall see that the magnetic field in the open space inside (point P) and exterior to the toroid (point Q) is zero. The field B inside the toroid is constant in magnitude for the ideal toroid of closely wound turns. Figure 4.19(b) shows a sectional view of the toroid. The direction of the magnetic field inside is clockwise as per the right-hand thumb rule for circular loops. Three circular Amperian loops 1, 2 and 3 are shown by dashed lines. By symmetry, the magnetic field should be tangential to each of them and constant in magnitude for a given loop. The circular areas bounded by loops 2 and 3 both cut the toroid: so that each turn of current carrying wire is cut once by the loop 2 and twice by the loop 3. Let the magnetic field along loop 1 be B 1 in magnitude. Then in Ampere’s circuital law [Eq. 4.17(a)], L = 2π r1. However, the loop encloses no current, so Ie = 0. Thus, B 1 (2 π r1) = µ0(0), B1 = 0 FIGURE 4.19 (a) A toroid carrying Thus, the magnetic field at any point P in the open space a current I. (b) A sectional view of inside the toroid is zero. the toroid. The magnetic field can be obtained at an arbitrary We shall now show that magnetic field at Q is likewise distance r from the centre O of zero. Let the magnetic field along loop 3 be B 3. Once again the toroid by Ampere’s circuital from Ampere’s law L = 2 π r3. However, from the sectional law. The dashed lines labelled cut, we see that the current coming out of the plane of the 1, 2 and 3 are three circular paper is cancelled exactly by the current going into it. Thus, Amperian loops. I e= 0, and B3 = 0. Let the magnetic field inside the solenoid be B. We shall now consider the magnetic field at S. Once again we employ Ampere’s law in the form of Eq. [4.17 (a)]. We find, L = 2π r. The current enclosed Ie is (for N turns of toroidal coil) N I. 152 B (2πr) = µ0NI 2015-16(20/01/2015)  Moving Charges and Magnetism µ 0NI B= (4.21) 2π r We shall now compare the two results: for a toroid and solenoid. We re-express Eq. (4.21) to make the comparison easier with the solenoid result given in Eq. (4.20). Let r be the average radius of the toroid and n be the number of turns per unit length. Then N = 2πr n = (average) perimeter of the toroid × number of turns per unit length and thus, B = µ0 n I, (4.22) i.e., the result for the solenoid! In an ideal toroid the coils are circular. In reality the turns of the toroidal coil form a helix and there is always a small magnetic field external to the toroid. M AGNETIC CONFINEMENT We have seen in Section 4.3 (see also the box on helical motion of charged particles earlier in this chapter) that orbits of charged particles are helical. If the magnetic field is non-uniform, but does not change much during one circular orbit, then the radius of the helix will decrease as it enters stronger magnetic field and the radius will increase when it enters weaker magnetic fields. We consider two solenoids at a distance from each other, enclosed in an evacuated container (see figure below where we have not shown the container). Charged particles moving in the region between the two solenoids will start with a small radius. The radius will increase as field decreases and the radius will decrease again as field due to the second solenoid takes over. The solenoids act as a mirror or reflector. [See the direction of F as the particle approaches coil 2 in the figure. It has a horizontal component against the forward motion.] This makes the particles turn back when they approach the solenoid. Such an arrangement will act like magnetic bottle or magnetic container. The particles will never touch the sides of the container. Such magnetic bottles are of great use in confining the high energy plasma in fusion experiments. The plasma will destroy any other form of material container because of it’s high temperature. Another useful container is a toroid. Toroids are expected to play a key role in the tokamak, an equipment for plasma confinement in fusion power reactors. There is an international collaboration called the International Thermonuclear Experimental Reactor (ITER), being set up in France, for achieving controlled fusion, of which India is a collaborating nation. For details of ITER collaboration and the project, you may visit http://www.iter.org. 153 2015-16(20/01/2015)  Physics Example 4.9 A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid? Solution The number of turns per unit length is, 500 n = = 1000 turns/m 0.5 EXAMPLE 4.9 The length l = 0.5 m and radius r = 0.01 m. Thus, l/a = 50 i.e., l >> a. Hence, we can use the long solenoid formula, namely, Eq. (4.20) B = µ0n I = 4π × 10–7 × 103 × 5 = 6.28 × 10–3 T 4.9 FORCE BETWEEN TWO PARALLEL CURRENTS , THE AMPERE We have learnt that there exists a magnetic field due to a conductor carrying a current which obeys the Biot-Savart law. Further, we have learnt that an external magnetic field will exert a force on a current-carrying conductor. This follows from the Lorentz force formula. Thus, it is logical to expect that two current-carrying conductors placed near each other will exert (magnetic) forces on each other. In the period 1820-25, Ampere studied the nature of this magnetic force and its dependence on the magnitude of the current, on the shape and size of the conductors as well as the distances between the conductors. In this section, we shall take the simple example of two parallel current- carrying conductors, which will perhaps help us to appreciate Ampere’s painstaking work. Figure 4.20 shows two long parallel conductors a FIGURE 4.20 Two long straight parallel conductors carrying steady and b separated by a distance d and carrying (parallel) currents Ia and Ib and separated by a currents I a and Ib , respectively. The conductor ‘a’ distance d. Ba is the magnetic field set produces, the same magnetic field Ba at all points along up by conductor ‘a’ at conductor ‘b’. the conductor ‘b’. The right-hand rule tells us that the direction of this field is downwards (when the conductors are placed horizontally). Its magnitude is given by Eq. [4.19(a)] or from Ampere’s circuital law, µ0I a Ba = 2πd The conductor ‘b’ carrying a current I b will experience a sideways force due to the field B a. The direction of this force is towards the conductor ‘a’ (Verify this). We label this force as Fba, the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by 154 Eq. (4.4), 2015-16(20/01/2015)  Moving Charges and Magnetism F ba = Ib L B a µ0 I a I b = L (4.23) 2π d It is of course possible to compute the force on ‘a’ due to ‘b’. From considerations similar to above we can find the force Fab, on a segment of length L of ‘a’ due to the current in ‘b’. It is equal in magnitude to Fba, and directed towards ‘b’. Thus, Fba = –F ab (4.24) Note that this is consistent with Newton’s third Law. Thus, at least for parallel conductors and steady currents, we have shown that the Biot-Savart law and the Lorentz force yield results in accordance with Newton’s third Law*. We have seen from above that currents flowing in the same direction attract each other. One can show that oppositely directed currents repel each other. Thus, Parallel currents attract, and antiparallel currents repel. This rule is the opposite of what we find in electrostatics. Like (same sign) charges repel each other, but like (parallel) currents attract each other. Let f ba represent the magnitude of the force Fba per unit length. Then, from Eq. (4.23), µ0 I a I b f ba = (4.25) 2πd The above expression is used to define the ampere (A), which is one of the seven SI base units. The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2 × 10–7 newtons per metre of length. This definition of the ampere was adopted in 1946. It is a theoretical definition. In practice one must eliminate the effect of the earth’s magnetic field and substitute very long wires by multiturn coils of appropriate geometries. An instrument called the current balance is used to measure this mechanical force. The SI unit of charge, namely, the coulomb, can now be defined in terms of the ampere. When a steady current of 1A is set up in a conductor, the quantity of charge that flows through its cross-section in 1s is one coulomb (1C). * It turns out that when we have time-dependent currents and/or charges in motion, Newton’s third law may not hold for forces between charges and/or conductors. An essential consequence of the Newton’s third law in mechanics is conservation of momentum of an isolated system. This, however, holds even for the case of time-dependent situations with electromagnetic fields, provided the momentum carried by fields is also taken into account. 155 2015-16(20/01/2015)  Physics ROGET ’S SPIRAL FOR ATTRACTION BETWEEN PARALLEL CURRENTS Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor µ. Hence it is difficult to demonstrate attraction or repulsion between currents. Thus for 5 A current in each wire at a separation of 1cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5 – 1s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring, and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply, and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit, and the cycle continues with tick, tick, tick,.... In the beginning, you may require some small adjustments to get a good effect. Keep your face away from mercury vapours as they are poisonous. Do not inhale mercury vapours for long. Example 4.10 The horizontal component of the earth’s magnetic field at a certain place is 3.0 ×10–5 T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west; (b) south to north? Solution F = I l × B F = IlB sinθ The force per unit length is EXAMPLE 4.10 f = F/l = I B sinθ (a) When the current is flowing from east to west, θ = 90° Hence, f=IB 156 = 1 × 3 × 10–5 = 3 × 10 –5 N m–1 2015-16(20/01/2015)  Moving Charges and Magnetism This is larger than the value 2×10–7 Nm–1 quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth’s magnetic field and other stray fields while standardising the ampere. EXAMPLE 4.10 The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors. (b) When the current is flowing from south to north, o θ=0 f=0 Hence there is no force on the conductor. 4.10 TORQUE ON CURRENT LOOP, MAGNETIC DIPOLE 4.10.1 Torque on a rectangular current loop in a uniform magnetic field We now show that a rectangular loop carrying a steady current I and placed in a uniform magnetic field experiences a torque. It does not experience a net force. This behaviour is analogous to that of electric dipole in a uniform electric field (Section 1.10). We first consider the simple case when the rectangular loop is placed such that the uniform magnetic field B is in the plane of the loop. This is illustrated in Fig. 4.21(a). The field exerts no force on the two arms AD and BC of the loop. It is perpendicular to the arm AB of the loop and exerts a force F1 on it which is directed into the plane of the loop. Its magnitude is, F1 = I b B Similarly it exerts a force F2 on the arm CD and F 2 is directed out of the plane of the paper. F 2 = I b B = F1 Thus, the net force on the loop is zero. There is a torque on the loop due to the pair of forces F1 and F2. Figure 4.21(b) shows a view of the loop from the AD end. It shows that the torque on the loop tends to rotate it anti-clockwise. This torque is (in magnitude), a a τ = F1 + F2 2 2 a a FIGURE 4.21 (a) A rectangular = IbB + IbB = I (ab ) B current-carrying coil in uniform 2 2 magnetic field. The magnetic moment =IAB (4.26) m points downwards. The torque τ is where A = ab is the area of the rectangle. along the axis and tends to rotate the coil anticlockwise. (b) The couple We next consider the case when the plane of the loop, acting on the coil. is not along the magnetic field, but makes an angle with it. We take the angle between the field and the normal to 157 2015-16(20/01/2015)  Physics the coil to be angle θ (The previous case corresponds to θ = π/2). Figure 4.22 illustrates this general case. The forces on the arms BC and DA are equal, opposite, and act along the axis of the coil, which connects the centres of mass of BC and DA. Being collinear along the axis they cancel each other, resulting in no net force or torque. The forces on arms AB and CD are F1 and F2. They too are equal and opposite, with magnitude, F1 = F2 = I b B But they are not collinear! This results in a couple as before. The torque is, however, less than the earlier case when plane of loop was along the magnetic field. This is because the perpendicular distance between the forces of the couple has decreased. Figure 4.22(b) is a view of the arrangement from the AD end and it illustrates these two forces constituting a couple. The magnitude of the torque on the loop is, a a τ = F1 sin θ + F2 sin θ 2 2 FIGURE 4.22 (a) The area vector of the loop ABCD makes an arbitrary angle θ with = I ab B sin θ the magnetic field. (b) Top view of = I A B sin θ (4.27) the loop. The forces F1 and F2 acting on the arms AB and CD As θ à 0, the perpendicular distance between are indicated. the forces of the couple also approaches zero. This makes the forces collinear and the net force and torque zero. The torques in Eqs. (4.26) and (4.27) can be expressed as vec