4.9 A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

4.13 (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.) 135 Reprint 2025-26 Physics Chapter Five MAGNETISM AND MATTER 5.1 INTRODUCTION Magnetic phenomena are universal in nature. Vast, distant galaxies, the tiny invisible atoms, humans and beasts all are permeated through and through with a host of magnetic fields from a variety of sources. The earth’s magnetism predates human evolution. The word magnet is derived from the name of an island in Greece called magnesia where magnetic ore deposits were found, as early as 600 BC. In the previous chapter we have learned that moving charges or electric currents produce magnetic fields. This discovery, which was made in the early part of the nineteenth century is credited to Oersted, Ampere, Biot and Savart, among others. In the present chapter, we take a look at magnetism as a subject in its own right. Some of the commonly known ideas regarding magnetism are: (i) The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south to the north. (ii) When a bar magnet is freely suspended, it points in the north-south direction. The tip which points to the geographic north is called the north pole and the tip which points to the geographic south is called 136 the south pole of the magnet. Reprint 2025-26 Magnetism and Matter (iii) There is a repulsive force when north poles ( or south poles ) of two magnets are brought close together. Conversely, there is an attractive force between the north pole of one magnet and the south pole of the other. (iv) We cannot isolate the north, or south pole of a magnet. If a bar magnet is broken into two halves, we get two similar bar magnets with somewhat weaker properties. Unlike electric charges, isolated magnetic north and south poles known as magnetic monopoles do not exist. (v) It is possible to make magnets out of iron and its alloys. We begin with a description of a bar magnet and its behaviour in an external magnetic field. We describe Gauss’s law of magnetism. We next describe how materials can be classified on the basis of their magnetic properties. We describe para-, dia-, and ferromagnetism. 5.25.25.25.25.2 TTTHETTHEHEHEHE BBBARBBARARARAR MMMAGNETMMAGNETAGNETAGNETAGNET We begin our study by examining iron filings sprinkled on a sheet of glass placed over a short bar magnet. The arrangement of iron filings is shown in Fig. 5.1. The pattern of iron filings suggests that the magnet has two poles similar to the positive and negative charge of an electric dipole. As mentioned in the introductory section, one pole is designated the North pole and the other, the South pole. When suspended freely, these poles point approximately towards the geographic north and south poles, respectively. A similar pattern of iron filings is observed around a current carrying solenoid. FIGUREFIGUREFIGUREFIGUREFIGURE 5.15.15.15.15.1 The arrangement 5.2.15.2.15.2.15.2.15.2.1 TheTheTheTheThe magneticmagneticmagneticmagneticmagnetic fieldfieldfieldfieldfield lineslineslineslineslines of iron filings surrounding a bar magnet. The pattern mimics The pattern of iron filings permits us to plot magnetic field lines. The pattern the magnetic field lines*. This is shown both suggests that the bar magnet is for the bar-magnet and the current- a magnetic dipole. carrying solenoid in Fig. 5.2. For comparison refer to the Chapter 1, Figure 1.14(d). Electric field lines of an electric dipole are also displayed in Fig. 5.2(c). The magnetic field lines are a visual and intuitive realisation of the magnetic field. Their properties are: (i) The magnetic field lines of a magnet (or a solenoid) form continuous closed loops. This is unlike the electric dipole where these field lines begin from a positive charge and end on the negative charge or escape to infinity. * In some textbooks the magnetic field lines are called magnetic lines of force. This nomenclature is avoided since it can be confusing. Unlike electrostatics the field lines in magnetism do not indicate the direction of the force on a 137 (moving) charge. Reprint 2025-26 Physics FIGURE 5.2 The field lines of (a) a bar magnet, (b) a current-carrying finite solenoid and (c) electric dipole. At large distances, the field lines are very similar. The curves labelled i and ii are closed Gaussian surfaces. (ii) The tangent to the field line at a given point represents the direction of the net magnetic field B at that point. (iii) The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field B. In Fig. 5.2(a), B is larger around region ii than in region i . (iv) The magnetic field lines do not intersect, for if they did, the direction of the magnetic field would not be unique at the point of intersection. One can plot the magnetic field lines in a variety of ways. One way is to place a small magnetic compass needle at various positions and note its orientation. This gives us an idea of the magnetic field direction at various points in space. 5.2.2 Bar magnet as an FIGURE 5.3 Calculation of (a) The axial field of a equivalent solenoid finite solenoid in order to demonstrate its In the previous chapter, we have similarity to that of a bar magnet. (b) A magnetic explained how a current loop acts as a needle in a uniform magnetic field B. The arrangement may be used to determine either B magnetic dipole (Section 4.9). We or the magnetic moment m of the needle. mentioned Ampere’s hypothesis that all magnetic phenomena can be explained in 138 terms of circulating currents. Reprint 2025-26 Magnetism and Matter The resemblance of magnetic field lines for a bar magnet and a solenoid suggest that a bar magnet may be thought of as a large number of circulating currents in analogy with a solenoid. Cutting a bar magnet in half is like cutting a solenoid. We get two smaller solenoids with weaker magnetic properties. The field lines remain continuous, emerging from one face of the solenoid and entering into the other face. One can test this analogy by moving a small compass needle in the neighbourhood of a bar magnet and a current-carrying finite solenoid and noting that the deflections of the needle are similar in both cases. To make this analogy more firm we may calculate the axial field of a finite solenoid depicted in Fig. 5.3 (a). We can demonstrate that at large distances this axial field resembles that of a bar magnet. The magnitude of the field at point P due to the solenoid is µ0 2 m B = 3 (5.1) 4π r This is also the far axial magnetic field of a bar magnet which one may obtain experimentally. Thus, a bar magnet and a solenoid produce similar magnetic fields. The magnetic moment of a bar magnet is thus equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field. 5.2.3 The dipole in a uniform magnetic field Let’s place a small compass needle of known magnetic moment m allowing it to oscillate in the magnetic field. This arrangement is shown in Fig. 5.3(b). The torque on the needle is [see Eq. (4.23)], τ = m × B (5.2) In magnitude τ = mB sinθ Here τ is restoring torque and θ is the angle between m and B. An expression for magnetic potential energy can be obtained on lines similar to electrostatic potential energy. The magnetic potential energy Um is given by (θ)d θ U m = ∫τ sin θ d θ = −mB cos θ = ∫mB = –m.B (5.3) We have emphasised in Chapter 2 that the zero of potential energy can be fixed at one’s convenience. Taking the constant of integration to be zero means fixing the zero of potential energy at θ = 90°, i.e., when the needle is perpendicular to the field. Equation (5.3) shows that potential energy is minimum (= –mB) at θ = 0° (most stable position) and maximum (= +mB) at θ = 180° (most unstable position). Example 5.1 (a) What happens if a bar magnet is cut into two pieces: (i) transverse to its length, (ii) along its length? (b) A magnetised needle in a uniform magnetic field experiences a EXAMPLE torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why? 5.1 139 Reprint 2025-26 Physics (c) Must every magnetic configuration have a north pole and a south pole? What about the field due to a toroid? (d) Two identical looking iron bars A and B are given, one of which is definitely known to be magnetised. (We do not know which one.) How would one ascertain whether or not both are magnetised? If only one is magnetised, how does one ascertain which one? [Use nothing else but the bars A and B.] Solution (a) In either case, one gets two magnets, each with a north and south pole. (b) No force if the field is uniform. The iron nail experiences a non- uniform field due to the bar magnet. There is induced magnetic moment in the nail, therefore, it experiences both force and torque. The net force is attractive because the induced south pole (say) in the nail is closer to the north pole of magnet than induced north pole. (c) Not necessarily. True only if the source of the field has a net non-zero magnetic moment. This is not so for a toroid or even for a straight infinite conductor. (d) Try to bring different ends of the bars closer. A repulsive force in some situation establishes that both are magnetised. If it is always attractive, then one of them is not magnetised. In a bar magnet the intensity of the magnetic field is the strongest at the two ends (poles) and weakest at the central region. This fact may be used to determine whether A or B is the magnet. In this case, to see which one of the two bars is a magnet, pick up one, (say, A) and lower one of its ends; first on one of the ends of the 5.1 other (say, B), and then on the middle of B. If you notice that in the middle of B, A experiences no force, then B is magnetised. If you do not notice any change from the end to the middle of B, EXAMPLE then A is magnetised. 5.2.4 The electrostatic analog Comparison of Eqs. (5.1), (5.2) and (5.3) with the corresponding equations for electric dipole (Chapter 1), suggests that magnetic field at large distances due to a bar magnet of magnetic moment m can be obtained from the equation for electric field due to an electric dipole of dipole moment p, by making the following replacements: 1 µ0 E → B , p → m , → 4 πε0 4 π In particular, we can write down the equatorial field (BE) of a bar magnet at a distance r, for r >> l, where l is the size of the magnet: µ0 m B E = − 4 π r 3 (5.4) Likewise, the axial field (BA) of a bar magnet for r >> l is: µ0 2 m B A =140 4 π r 3 (5.5) Reprint 2025-26 Magnetism and Matter Equation (5.5) is just Eq. (5.1) in the vector form. Table 5.1 summarises the analogy between electric and magnetic dipoles. TABLE 5.1 THE DIPOLE ANALOGY Electrostatics Magnetism 1/ε0 µ0 Dipole moment p m Equatorial Field for a short dipole –p/4πε0r 3 – µ0 m / 4π r 3 Axial Field for a short dipole 2p/4πε0r 3 µ0 2m / 4π r 3 External Field: torque p × E m × B External Field: Energy –p.E –m.B Example 5.2 Figure 5.4 shows a small magnetised needle P placed at a point O. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q. (a) In which configuration the system is not in equilibrium? (b) In which configuration is the system in (i) stable, and (ii) unstable equilibrium? (c) Which configuration corresponds to the lowest potential energy among all the configurations shown? FIGURE 5.4 Solution Potential energy of the configuration arises due to the potential energy of one dipole (say, Q) in the magnetic field due to other (P). Use the result that the field due to P is given by the expression [Eqs. (5.4) and (5.5)]: µ0 m P B P = − 3 (on the normal bisector) 4π r µ0 2 m P B P = 3 (on the axis) 4π r where mP is the magnetic moment of the dipole P. EXAMPLE Equilibrium is stable when mQ is parallel to BP, and unstable when it is anti-parallel to BP. 5.2 141 Reprint 2025-26 Physics For instance for the configuration Q3 for which Q is along the perpendicular bisector of the dipole P, the magnetic moment of Q is parallel to the magnetic field at the position 3. Hence Q3 is stable. 5.2 Thus, (a) PQ1 and PQ2 (b) (i) PQ3, PQ6 (stable); (ii) PQ5, PQ4 (unstable) EXAMPLE (c) PQ6 5.3 MAGNETISM AND GAUSS’S LAW In Chapter 1, we studied Gauss’s law for electrostatics. In Fig 5.2(c), we see that for a closed surface represented by i , the number of lines leaving the surface is equal to the number of lines entering it. This is consistent with the fact that no net charge is enclosed by the surface. However, in the same figure, for the closed surface ii , there is a net outward flux, since it does include a net (positive) charge.1855) The situation is radically different for magnetic fields – which are continuous and form closed loops. Examine the Gaussian surfaces represented by i or ii in Fig 5.2(a) or Karl Friedrich Gauss Fig. 5.2(b). Both cases visually demonstrate that the(1777 (1777 – 1855) He was a number of magnetic field lines leaving the surface is child prodigy and was gifted balanced by the number of lines entering it. The net in mathematics, physics, magnetic flux is zero for both the surfaces. This is true engineering, astronomy for any closed surface.GAUSS and even land surveying. The properties of numbers fascinated him, and in his work he anticipated major mathematical development of later times. Along with Wilhelm Welser, he built theFRIEDRICH first electric telegraph in 1833. His mathematical theory of curved surface laid the foundation for theKARL later work of Riemann. FIGURE 5.5 Consider a small vector area element ∆S of a closed surface S as in Fig. 5.5. The magnetic flux through ÄS is defined as ∆φB = B.∆S, where B is the field at ∆S. We divide S into many small area elements and calculate the individual flux through each. Then, the net flux φB is, B.∆S = 0 (5.6) φB = ∑ ∆ φB = ∑ ’ all ’ ’ all ’ where ‘all’ stands for ‘all area elements ∆S′. Compare this with the Gauss’s law of electrostatics. The flux through a closed surface in that case is given by E.∆S =142 ∑ q ε0 Reprint 2025-26 Magnetism and Matter where q is the electric charge enclosed by the surface. The difference between the Gauss’s law of magnetism and that for electrostatics is a reflection of the fact that isolated magnetic poles (also called monopoles) are not known to exist. There are no sources or sinks of B; the simplest magnetic element is a dipole or a current loop. All magnetic phenomena can be explained in terms of an arrangement of dipoles and/or current loops. Thus, Gauss’s law for magnetism is: The net magnetic flux through any closed surface is zero. Example 5.3 Many of the diagrams given in Fig. 5.6 show magnetic field lines (thick lines in the figure) wrongly. Point out what is wrong with them. Some of them may describe electrostatic field lines correctly. Point out which ones. EXAMPLE FIGURE 5.6 5.3 143 Reprint 2025-26 Physics Solution (a) Wrong. Magnetic field lines can never emanate from a point, as shown in figure. Over any closed surface, the net flux of B must always be zero, i.e., pictorially as many field lines should seem to enter the surface as the number of lines leaving it. The field lines shown, in fact, represent electric field of a long positively charged wire. The correct magnetic field lines are circling the straight conductor, as described in Chapter 4. (b) Wrong. Magnetic field lines (like electric field lines) can never cross each other, because otherwise the direction of field at the point of intersection is ambiguous. There is further error in the figure. Magnetostatic field lines can never form closed loops around empty space. A closed loop of static magnetic field line must enclose a region across which a current is passing. By contrast, electrostatic field lines can never form closed loops, neither in empty space, nor when the loop encloses charges. (c) Right. Magnetic lines are completely confined within a toroid. Nothing wrong here in field lines forming closed loops, since each loop encloses a region across which a current passes. Note, for clarity of figure, only a few field lines within the toroid have been shown. Actually, the entire region enclosed by the windings contains magnetic field. (d) Wrong. Field lines due to a solenoid at its ends and outside cannot be so completely straight and confined; such a thing violates Ampere’s law. The lines should curve out at both ends, and meet eventually to form closed loops. (e) Right. These are field lines outside and inside a bar magnet. Note carefully the direction of field lines inside. Not all field lines emanate out of a north pole (or converge into a south pole). Around both the N-pole, and the S-pole, the net flux of the field is zero. (f ) Wrong. These field lines cannot possibly represent a magnetic field. Look at the upper region. All the field lines seem to emanate out of the shaded plate. The net flux through a surface surrounding the shaded plate is not zero. This is impossible for a magnetic field. The given field lines, in fact, show the electrostatic field lines around a positively charged upper plate and a negatively charged 5.3 carefullylower plate.grasped.The difference between Fig. [5.6(e) and (f)] should be (g) Wrong. Magnetic field lines between two pole pieces cannot be precisely straight at the ends. Some fringing of lines is inevitable. Otherwise, Ampere’s law is violated. This is also true for electric EXAMPLE field lines. Example 5.4 (a) Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). Do the magnetic field lines also represent the lines of force on a moving charged particle at every point? (b) If magnetic monopoles existed, how would the Gauss’s law of 5.4 magnetism be modified? (c) Does a bar magnet exert a torque on itself due to its own field? Does one element of a current-carrying wire exert a force on another EXAMPLE element of the same wire? Reprint 2025-26 Magnetism and Matter (d) Magnetic field arises due to charges in motion. Can a system have magnetic moments even though its net charge is zero? Solution (a) No. The magnetic force is always normal to B (remember magnetic force = qv × B). It is misleading to call magnetic field lines as lines of force. (b) Gauss’s law of magnetism states that the flux of B through any closed surface is always zero ∫s B .∆s = 0 . If monopoles existed, the right hand side would be equal to the monopole (magnetic charge) qm enclosed by S. [Analogous to Gauss’s law of electrostatics, B.∆s = µ0qm where qm is the ∫S (monopole) magnetic charge enclosed by S.] (c) No. There is no force or torque on an element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire. (For the special case of a straight wire, this force is zero.) (d) Yes. The average of the charge in the system may be zero. Yet, the mean of the magnetic moments due to various current loops may not be zero. We will come across such examples in connection EXAMPLE with paramagnetic material where atoms have net dipole moment 5.4 through their net charge is zero.

3.3 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10–4 °C–1. 105 Reprint 2025-26 Physics 3.4 A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10–7 m2, and its resistance is measured to be 5.0 W. What is the resistivity of the material at the temperature of the experiment? 3.5 A silver wire has a resistance of 2.1 W at 27.5 °C, and a resistance of 2.7 W at 100 °C. Determine the temperature coefficient of resistivity of silver. 3.6 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10–4 °C–1. 3.7 Determine the current in each branch of the network shown in Fig. 3.20: FIGURE 3.20 3.8 A storage battery of emf 8.0 V and internal resistance 0.5 W is being charged by a 120 V dc supply using a series resistor of 15.5 W. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit? 3.9 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m–3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10–6 m2 and it is carrying a current of 3.0 A. Reprint 2025-26 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, 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 Reprint 2025-26 Physics 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 took place in the 20th century. This was 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 emerges 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, 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.(1777–1851) 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 ()*. Hans Christian Oersted Figures. 4.1(a) and 4.1(b) correspond to these twoOERSTED (1777–1851) Danish situations, respectively. physicist and chemist, professor at Copenhagen. 4.2 MAGNETIC FORCE He observed that a compass needle suffers a 4.2.1 Sources and fields deflection when placed Before we introduce the concept of a magnetic field B, weCHRISTIAN near a wire carrying an electric current. This shall recapitulate what we have learnt in Chapter 1 about discovery gave the first the electric field E. We have seen that the interaction empirical evidence of a between two charges can be considered in two stages.HANS connection between electric The charge Q, the source of the field, produces an electric and magnetic phenomena. field E, where * A dot appears like the tip of an arrow pointed at you, a cross is like the feathered 108 tail of an arrow moving away from you. Reprint 2025-26 Moving Charges and Magnetism E = Q ˆr / (4pe0)r2 (4.1) where ˆr 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 ˆr / (4pe0) 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 function of time. In our Hendrik Antoon Lorentz discussions in this chapter, we will assume that the fields (1853 – 1928) Dutch do not change with time. theoretical physicist, The field at a particular point can be due to one or professor at Leiden. He investigated themore charges. If there are more charges the fields add HENDRIK 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 ANTOONcurrents 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 existence of electric chargeshas several basic properties identical to the electric field. 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 LORENTZprinciple 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 (1853 4.2.2 Magnetic Field, Lorentz Force mathematical arguments, – but he was not aware thatLet us suppose that there is a point charge q (moving these equations hinge on a with a velocity v and, located at r at a given time t) in new concept of space andpresence of both the electric field E (r) and the magnetic 1928) 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)] º Felectric +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 109 and magnetic field. The vector product makes the force due to magnetic Reprint 2025-26 Physics 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 FIGURE 4.2 The direction of the magnetic takes q, F and v, all to be unity in the force force acting on a charged particle. (a) The equation F = q [ v × B] =q v B sin q ˆn , where q is force on a positively charged particle with the angle between v and B [see Fig. 4.2 (a)]. The velocity v and making an angle q with the magnitude of magnetic field B is 1 SI unit, when magnetic field B is given by the right-hand the force acting on a unit charge (1 C), moving rule. (b) A moving charged particle q is perpendicular to B with a speed 1m/s, is one deflected in an opposite sense to –q in the newton. presence of magnetic field. Dimensionally, we have [B] = [F/qv] and the unit 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. 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 drift velocity vd (see Chapter 3). In the presence of an external magnetic field B, the force on these carriers is: F = (nlA)q vd ´ B where q is the value of the charge on a carrier. Now nq vd 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 = [(nq vd )lA] × 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 l. 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. Reprint 2025-26 Moving Charges and Magnetism 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? 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 m g B = I l Interactive Charged 0.2 × 9.8 = = 0.65 T 2 × 1.5 Note that it would have been sufficient to specify m/l, the mass per EXAMPLE particles unit length of the wire. The earth’s 4 × 10–5 T and we have ignored it. magnetic field is approximately 4.1 moving demonstration: in a 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 magnetic charge), (b) a proton (positive charge). field. http://www.phys.hawaii.edu/~teb/optics/java/partmagn/index.html FIGURE 4.4 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 EXAMPLE axis. (b) for a positive charge (proton) the force is along +z axis. 4.2 111 Reprint 2025-26 Physics 4.3 MOTION IN A MAGNETIC FIELD 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 5) 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 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, qE, 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 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 3) 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 FIGURE 4.5 Circular motion 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) 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 w is the angular frequency, then v = w r. So, w = 2p n = q B/ m [4.6(a)] which is independent of the velocity or energy . Here n is the frequency of rotation. The independence of n from energy has important application in the design of a cyclotron. The time taken for one revolution is T= 2p/ w º 1/n. If there is a component of the velocity FIGURE 4.6 Helical motion parallel to the magnetic field (denoted by v||), 112 it will make the particle move along the field and the path of the Reprint 2025-26 Moving Charges and Magnetism 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 = 2pm v|| / q B [4.6(b)] The radius of the circular component of motion is called the radius of the helix. 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 frequency? Calculate its energy in keV. ( 1 eV = 1.6 × 10–19 J). Solution Using Eq. (4.5) we find r = m v / (qB) = 9 ×10–31 kg × 3 × 107 m s–1 / ( 1.6 × 10–19 C × 6 × 10–4 T) = 28 × 10–2 m = 28 cm n = v / (2 pr) = 17×106 s–1 = 17×106 Hz =17 MHz. 2 EXAMPLE E = (½ )mv = (½ ) 9 × 10–31 kg × 9 × 1014 m2/s2 = 40.5 ×10–17 J ≈ 4×10–16 J = 2.5 keV. 4.3