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 magneticequation F = q [ v × B] =q v B sin θnˆ, where θ force acting on a charged particle. (a) Theis the angle between v and B [see Fig. 4.2 (a)]. force on a positively charged particle withThe 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. TABLE 4.1 ORDEROFMAGNITUDESOFMAGNETICFIELDSINAVARIETYOFPHYSICALSITUATIONS Physical situation Magnitude of B (in tesla) Surface of a neutron star 108 Typical large field in a laboratory 1 10–2Near a small bar magnet 10–5On the earth’s surface 10–10Human nerve fibre 10–12Interstellar space 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 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, 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, qv × 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 componentFIGURE 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 v 2 / 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 = 2πν = 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 = 2πmv || / q B [4.6(b)] The radius of the circular component of motion is called the radius of the helix. 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 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)= 26 × 10–2 m = 26 cm ν = v / (2 πr) = 2×106 s–1 = 2×106 Hz =2 MHz. 2 –311422–17E = (½ )mv = (½ ) 9 × 10 kg × 9 × 10 m/s = 40.5 ×10 J ≈ 4×10–16 J = 2.5 keV. HELICALMOTIONOFCHARGEDPARTICLESANDAURORABOREALIS 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 v p parallel to the magnetic field and a component v n 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 v p parallel to the field. The normal component v n of the particle results in a Lorentz force (v n×B) which is perpendicular to both v n 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. EXAMPLE 4.3 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. 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 = 2 π r I = Current enclosed by the loop = I eThe result is the familiar expression for a long straight wire B (2π r) = µ0I µ0 I B = [4.19(a)]2 π r 1 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 = 2 π r * 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 149the two. EXAMPLE 4.8 Moving Charges and Magnetism µ 0NI B = (4.21)2πr 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 = µ0n 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. MAGNETICCONFINEMENT 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. Physics EXAMPLE 4.9 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 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 = µn I0 = 4π × 10–7 × 103 × 5= 6.28 × 10–3 T 4.9 FORCEBETWEEN 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 aFIGURE 4.20 Two long straight and b separated by a distance d and carrying (parallel)parallel conductors carrying steady currents I and I, respectively. The conductor ‘a’currents Ia and Ib and separated by a abdistance d. B is the magnetic field set produces, the same magnetic field B at all points alonga aup 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, B =µ0Ia a 2π d The conductor ‘b’ carrying a current I b will experience a sideways force due to the field B . The direction of this force is towards the a conductor ‘a’ (Verify this). We label this force as F ba, the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by Eq. (4.4), 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 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 f = F/l = I B sinθ (a) When the current is flowing from east to west, θ = 90° Hence, f = I B= 1 × 3 × 10–5 = 3 × 10–5 N m–1 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, F = F = I b B12But 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, aa τ= F1 sinθ+ F2 sinθ 22 FIGURE 4.22 (a) The area vector of the loop ABCD makes an arbitrary angle θ with = I ab B sin θthe magnetic field. (b) Top view of = I AB sin θ (4.27)the loop. The forces F1 and F2 acting As θ ‚ 0, the perpendicular distance betweenon the arms AB and CD 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 vector product of the magnetic moment of the coil and the magnetic field. We define the magnetic moment of the current loop as, m = I A (4.28) where the direction of the area vector A is given by the right-hand thumb rule and is directed into the plane of the paper in Fig. 4.21. Then as the angle between m and B is θ , Eqs. (4.26) and (4.27) can be expressed by one expression τ= m × B (4.29) This is analogous to the electrostatic case (Electric dipole of dipole moment pe in an electric field E). τ= pe × E As is clear from Eq. (4.28), the dimensions of the magnetic moment are [A][L2] and its unit is Am2. From Eq. (4.29), we see that the torque τ vanishes when m is either parallel or antiparallel to the magnetic field B. This indicates a state of equilibrium as there is no torque on the coil (this also applies to any object with a magnetic moment m). When m and B are parallel the Physics (ii) is subject to torque like a magnetic needle. This led Ampere to suggest that all magnetism is due to circulating currents. This seems to be partly true and no magnetic monopoles have been seen so far. However, elementary particles such as an electron or a proton also carry an intrinsic magnetic moment, not accounted by circulating currents. 4.10.3 The magnetic dipole moment of a revolving electron In Chapter 12 we shall read about the Bohr model of the hydrogen atom. You may perhaps have heard of this model which was proposed by the Danish physicist Niels Bohr in 1911 and was a stepping stone to a new kind of mechanics, namely, quantum mechanics. In the Bohr model, the electron (a negatively charged particle) revolves around a positively charged nucleus much as a planet revolves around the sun. The force in the former case is electrostatic (Coulomb force) while it is gravitational for the planet-Sun case. We show this Bohr picture of the electron in Fig. 4.23. The electron of charge (–e) (e = + 1.6 × 10–19 C) performs uniform circular motion around a stationary heavy nucleus of charge +Ze. This constitutes a current I, where, e I = (4.32)FIGURE 4.23 In the Bohr model T of hydrogen-like atoms, the and T is the time period of revolution. Let r be the orbital negatively charged electron is radius of the electron, and v the orbital speed. Then,revolving with uniform speed around a centrally placed positively charged (+Ze) T = 2πr v (4.33) nucleus. The uniform circular Substituting in Eq. (4.32), we have I = ev/2πr.motion of the electron There will be a magnetic moment, usually denoted by µ,constitutes a current. The lassociated with this circulating current. From Eq. (4.28) itsdirection of the magnetic 2moment is into the plane of the magnitude is, µl = Iπr = evr/2. paper and is indicated The direction of this magnetic moment is into the plane separately by ⊗. of the paper in Fig. 4.23. [This follows from the right-hand rule discussed earlier and the fact that the negatively charged electron is moving anti-clockwise, leading to a clockwise current.] Multiplying and dividing the right-hand side of the above expression by the electron mass me, we have, e µ=(m vr )l 2me e e = l[4.34(a)]2m e Here, l is the magnitude of the angular momentum of the electron about the central nucleus (“orbital” angular momentum). Vectorially, e µ =− ll [4.34(b)]2m e The negative sign indicates that the angular momentum of the electron 162 is opposite in direction to the magnetic moment. Instead of electron with Physics coil galvanometer (MCG). It is a device whose principle can be understood on the basis of our discussion in Section 4.10. The galvanometer consists of a coil, with many turns, free to rotate about a fixed axis (Fig. 4.24), in a uniform radial magnetic field. There is a cylindrical soft iron core which not only makes the field radial but also increases the strength of the magnetic field. When a current flows through the coil, a torque acts on it. This torque is given by Eq. (4.26) to be τ = NI AB where the symbols have their usual meaning. Since the field is radial by design, we have taken sin θ = 1 in the above expression for the torque. The magnetic torque NIAB tends to rotate the coil. A spring Sp provides a counter torque kφ that balances the magnetic torque NIAB; resulting in a steady angular deflection φ. In equilibrium kφ = NI AB where k is the torsional constant of the spring; i.e. the restoring torque per unit twist. The deflection φ is indicated on the scale by a pointer attached to the spring. We have ⎛ NAB ⎞φ=⎜ ⎟ I (4.38)⎝ k ⎠ The quantity in brackets is a constant for a given galvanometer. The galvanometer can be used in a number of ways. It can be used as a detector to check if a current is flowing in the circuit. We have come across this usage in the Wheatstone’s bridge arrangement. In this usage the neutral position of the pointer (when no current is flowing through the galvanometer) is in the middle of the scale and not at the left end as shown in Fig.4.24. Depending on the direction of the current, the pointer deflection is either to the right or the left. The galvanometer cannot as such be used as an ammeter to measure the value of the current in a given circuit. This is for two reasons: (i) Galvanometer is a very sensitive device, it gives a full-scale deflection for a current of the order of µA. (ii) For measuring currents, the galvanometer has to be connected in series, and as it has a large resistance, this will change the value of the current in the circuit. To overcome these difficulties, one attaches a small resistance r ,sFIGURE 4.24 The moving coil called shunt resistance, in parallel with galvanometer. Its elements are the galvanometer coil; so that most of the current described in the text. Depending on passes through the shunt. The resistance of this the requirement, this device can be arrangement is,used as a current detector or for R r/ (R + r) ≃ r if R >> rmeasuring the value of the current Gs GssGs (ammeter) or voltage (voltmeter). If rs has small value, in relation to the resistance of the rest of the circuit R , the effect of introducing the 164 cmeasuring instrument is also small and negligible. This