4.10 Two moving coil meters, M1 and M2 have the following particulars: R1 = 10 Ω, N1 = 30, A1 = 3.6 × 10–3 m2, B1 = 0.25 T R2 = 14 Ω, N2 = 42, A2 = 1.8 × 10–3 m2, B2 = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

4.10 THE MOVING COIL GALVANOMETER Currents and voltages in circuits have been discussed extensively in Chapters 3. But how do we measure them? How do we claim that current in a circuit is 1.5 A or the voltage drop across a resistor is 1.2 V? Figure 4.20 exhibits a very useful instrument for this purpose: the moving coil galvanometer (MCG). It is a device whose principle can be understood on the basis of our discussion in Section 4.9. The galvanometer consists of a coil, with many turns, free to rotate about a fixed axis (Fig. 4.20), 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. (4.20) to be 129 τ = NI AB Reprint 2025-26 Physics 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 =  k  I (4.26) 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 FIGURE 4.20 The moving coil flowing in the circuit. We have come across this usage galvanometer. Its elements are in the Wheatstone’s bridge arrangement. In this usage described in the text. Depending on the neutral position of the pointer (when no current is the requirement, this device can be flowing through the galvanometer) is in the middle of used as a current detector or for the scale and not at the left end as shown in Fig.4.20. measuring the value of the current Depending on the direction of the current, the pointer’s (ammeter) or voltage (voltmeter). 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 rs, called shunt resistance, in parallel with the galvanometer coil; so that most of the current passes through the shunt. The resistance of this arrangement is, ≃ RG rs / (RG + rs) rs if RG >> rs If rs has small value, in relation to the resistance of the rest of the circuit Rc, the effect of introducing the measuring instrument is also small and negligible. This arrangement is schematically shown in Fig. 4.21. FIGURE 4.21 The scale of this ammeter is calibrated and then graduated to read off Conversion of a the current value with ease. We define the current sensitivity of the galvanometer (G) to galvanometer as the deflection per unit current. From Eq. (4.26) this an ammeter by the current sensitivity is, introduction of a NAB φshunt resistance rs of = (4.27) very small value in I k parallel. A convenient way for the manufacturer to increase the sensitivity is to increase the number of turns N. We choose galvanometers having 130 sensitivities of value, required by our experiment. Reprint 2025-26 Moving Charges and Magnetism The galvanometer can also be used as a voltmeter to measure the voltage across a given section of the circuit. For this it must be connected in parallel with that section of the circuit. Further, it must draw a very small current, otherwise the voltage measurement will disturb the original set up by an amount which is very large. Usually we like to keep the disturbance due to the measuring device below one per cent. To ensure this, a large resistance R is connected in series with the galvanometer. This arrangement is schematically depicted in Fig.4.22. Note that the resistance of the voltmeter is now, FIGURE 4.22 RG + R ≃ R : large Conversion of a The scale of the voltmeter is calibrated to read off the voltage value galvanometer (G) to a with ease. We define the voltage sensitivity as the deflection per unit voltmeter by the introduction of avoltage. From Eq. (4.26), resistance R of large φ  NAB  I  NAB  1 value in series. = = (4.28) V  k  V  k  R An interesting point to note is that increasing the current sensitivity may not necessarily increase the voltage sensitivity. Let us take Eq. (4.27) which provides a measure of current sensitivity. If N → 2N, i.e., we double the number of turns, then φ φ → 2 I I Thus, the current sensitivity doubles. However, the resistance of the galvanometer is also likely to double, since it is proportional to the length of the wire. In Eq. (4.28), N →2N, and R →2R, thus the voltage sensitivity, φ φ → V V remains unchanged. So in general, the modification needed for conversion of a galvanometer to an ammeter will be different from what is needed for converting it into a voltmeter. Example 4.12 In the circuit (Fig. 4.23) the current is to be measured. What is the value of the current if the ammeter shown (a) is a galvanometer with a resistance RG = 60.00 Ω; (b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance rs = 0.02 Ω; (c) is an ideal ammeter with zero resistance? EXAMPLE FIGURE 4.23 4.12 131 Reprint 2025-26 Physics Solution (a) Total resistance in the circuit is, RG + 3 = 63 Ω. Hence, I = 3/63 = 0.048 A. (b) Resistance of the galvanometer converted to an ammeter is, R G rs 60 Ω × 0. 02 Ω = ≃ 0.02Ω R G + rs ( 60 + 0 .02 )Ω 4.12 Total resistance in the circuit is, 0.02 Ω+ 3 Ω= 3.02 Ω. Hence, I = 3/3.02 = 0.99 A. (c) For the ideal ammeter with zero resistance, EXAMPLE I = 3/3 = 1.00 A SUMMARY 1. The total force on a charge q moving with velocity v in the presence of magnetic and electric fields B and E, respectively is called the Lorentz force. It is given by the expression: F = q (v × B + E) The magnetic force q (v × B) is normal to v and work done by it is zero. 2. A straight conductor of length l and carrying a steady current I experiences a force F in a uniform external magnetic field B, F = I l × B where|l| = l and the direction of l is given by the direction of the current. 3. In a uniform magnetic field B, a charge q executes a circular orbit in a plane normal to B. Its frequency of uniform circular motion is called the cyclotron frequency and is given by: q B νc = 2 π m This frequency is independent of the particle’s speed and radius. This fact is exploited in a machine, the cyclotron, which is used to accelerate charged particles. 4. The Biot-Savart law asserts that the magnetic field dB due to an element dl carrying a steady current I at a point P at a distance r from the current element is: r d l µ × 0 I dB = 3 r 4 π To obtain the total field at P, we must integrate this vector expression over the entire length of the conductor. 5. The magnitude of the magnetic field due to a circular coil of radius R carrying a current I at an axial distance x from the centre is Reprint 2025-26 Moving Charges and Magnetism 2 IR µ 0 B = 2 2 2( x + R )3/2 At the centre this reduces to µ0 I B = 2 R 6. Ampere’s Circuital Law: Let an open surface S be bounded by a loop B.d l = µ0 I where I refers to C. Then the Ampere’s law states that ∫CÑ the current passing through S. The sign of I is determined from the right-hand rule. We have discussed a simplified form of this law. If B is directed along the tangent to every point on the perimeter L of a closed curve and is constant in magnitude along perimeter then, BL = µ0 Ie where Ie is the net current enclosed by the closed circuit. 7. The magnitude of the magnetic field at a distance R from a long, straight wire carrying a current I is given by: µ0 I B = 2 π R The field lines are circles concentric with the wire. 8. The magnitude of the field B inside a long solenoid carrying a current I is B = µ0nI where n is the number of turns per unit length. 9. Parallel currents attract and anti-parallel currents repel. 10. A planar loop carrying a current I, having N closely wound turns, and an area A possesses a magnetic moment m where, m = N I A and the direction of m is given by the right-hand thumb rule : curl the palm of your right hand along the loop with the fingers pointing in the direction of the current. The thumb sticking out gives the direction of m (and A) When this loop is placed in a uniform magnetic field B, the force F on it is: F = 0 And the torque on it is, τ = m × B In a moving coil galvanometer, this torque is balanced by a counter- torque due to a spring, yielding kφ = NI AB where φ is the equilibrium deflection and k the torsion constant of the spring. 11. A moving coil galvanometer can be converted into a ammeter by introducing a shunt resistance rs, of small value in parallel. It can be converted into a voltmeter by introducing a resistance of a large value in series. 133 Reprint 2025-26 Physics Physical Quantity Symbol Nature Dimensions Units Remarks Permeability of free µ0 Scalar [MLT –2A–2] T m A–1 4π × 10–7 T m A–1 space Magnetic Field B Vector [M T –2A–1] T (telsa) Magnetic Moment m Vector [L2A] A m2 or J/T Torsion Constant k Scalar [M L2T –2] N m rad–1 Appears in MCG POINTS TO PONDER 1. Electrostatic field lines originate at a positive charge and terminate at a negative charge or fade at infinity. Magnetic field lines always form closed loops. 2. The discussion in this Chapter holds only for steady currents which do not vary with time. When currents vary with time Newton’s third law is valid only if momentum carried by the electromagnetic field is taken into account. 3. Recall the expression for the Lorentz force, F = q (v × B + E) This velocity dependent force has occupied the attention of some of the greatest scientific thinkers. If one switches to a frame with instantaneous velocity v, the magnetic part of the force vanishes. The motion of the charged particle is then explained by arguing that there exists an appropriate electric field in the new frame. We shall not discuss the details of this mechanism. However, we stress that the resolution of this paradox implies that electricity and magnetism are linked phenomena (electromagnetism) and that the Lorentz force expression does not imply a universal preferred frame of reference in nature. 4. Ampere’s Circuital law is not independent of the Biot-Savart law. It can be derived from the Biot-Savart law. Its relationship to the Biot-Savart law is similar to the relationship between Gauss’s law and Coulomb’s law. EXERCISES 4.1 A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil? 4.2 A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire? 4.3 A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B 134 at a point 2.5 m east of the wire. Reprint 2025-26 Moving Charges and Magnetism

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