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

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.

6.2 Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.16: (a) A wire of irregular shape turning into a circular shape; 175 Reprint 2025-26 Physics (b) A circular loop being deformed into a narrow straight wire. FIGURE 6.16 6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing? 6.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? 6.5 A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s–1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. 6.6 A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 ´ 10–4 Wb m–2. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential? 6.7 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit. 6.8 A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? Reprint 2025-26 Chapter Seven ALTERNATING CURRENT 7.1 INTRODUCTION We have so far considered direct current (dc) sources and circuits with dc sources. These currents do not change direction with time. But voltages and currents that vary with time are very common. The electric mains supply in our homes and offices is a voltage that varies like a sine function with time. Such a voltage is called alternating voltage (ac voltage) and the current driven by it in a circuit is called the alternating current (ac current)*. Today, most of the electrical devices we use require ac voltage. This is mainly because most of the electrical energy sold by power companies is transmitted and distributed as alternating current. The main reason for preferring use of ac voltage over dc voltage is that ac voltages can be easily and efficiently converted from one voltage to the other by means of transformers. Further, electrical energy can also be transmitted economically over long distances. AC circuits exhibit characteristics which are exploited in many devices of daily use. For example, whenever we tune our radio to a favourite station, we are taking advantage of a special property of ac circuits – one of many that you will study in this chapter. * The phrases ac voltage and ac current are contradictory and redundant, respectively, since they mean, literally, alternating current voltage and alternating current current. Still, the abbreviation ac to designate an electrical quantity displaying simple harmonic time dependance has become so universally accepted that we follow others in its use. Further, voltage – another phrase commonly used means potential difference between two points. Reprint 2025-26 Physics 7.2 AC VOLTAGE APPLIED TO A RESISTOR Figure 7.1 shows a resistor connected to a source ε of ac voltage. The symbol for an ac source in a circuit diagram is . We consider a source which produces sinusoidally varying potential difference across its terminals. Let this potential difference, also called ac voltage, be given by v = vm sin ωt (7.1) where vm is the amplitude of the oscillating potential difference and ω is its angular frequency. Nicola Tesla (1856 – 1943) Serbian-American scientist, inventor and genius. He conceived the idea of the rotating1943) magnetic field, which is the – basis of practically all alternating current machinery, and which(1856 helped usher in the age of FIGURE 7.1 AC voltage applied to a resistor. electric power. He also invented among other To find the value of current through the resistor, we things the induction motor, ε()t = 0 (refer to Section the polyphase system of ac apply Kirchhoff’s loop rule ∑TESLA power, and the high 3.12), to the circuit shown in Fig. 7.1 to get frequency induction coil v m sin ωt = i R (the Tesla coil) used in radio and television sets and v m i = sin ωtNICOLA other electronic equipment. or R The SI unit of magnetic field is named in his honour. Since R is a constant, we can write this equation as i = i m sin ωt (7.2) where the current amplitude im is given by v m i m = (7.3) R Equation (7.3) is Ohm’s law, which for resistors, works equally well for both ac and dc voltages. The voltage across a pure resistor and the current through it, given by Eqs. (7.1) and (7.2) are plotted as a function of time in Fig. 7.2. Note, in particular that both v and i reach zero, minimum and maximum values at the same time. Clearly, the voltage and current are in phase with FIGURE 7.2 In a pure resistor, the voltage and each other. current are in phase. The We see that, like the applied voltage, the current varies minima, zero and maxima sinusoidally and has corresponding positive and negative values occur at the same during each cycle. Thus, the sum of the instantaneous current respective times. values over one complete cycle is zero, and the average current 178 is zero. The fact that the average current is zero, however, does Reprint 2025-26 Alternating Current not mean that the average power consumed is zero and that there is no dissipation of electrical energy. As you know, Joule heating is given by i2R and depends on i2 (which is always positive whether i is positive or negative) and not on i. Thus, there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor. The instantaneous power dissipated in the resistor is p = i 2 R = i m2 R sin 2 ωt (7.4) The average value of p over a cycle is* p = < i 2 R > = < i m2 R sin 2 ωt > [7.5(a)] where the bar over a letter (here, p) denotes its average George Westinghouse value and <......> denotes taking average of the quantity (1846 – 1914) A leading inside the bracket. Since, i2m and R are constants, proponent of the use of p = i m2 R < sin 2 ωt > [7.5(b)] alternating current overUsing the trigonometric identity, sin2 wt = direct current. Thus, GEORGE he came into conflict 1/2 (1– cos 2wt), we have < sin2 wt > = (1/2) (1– < cos 2wt >) with Thomas Alva Edison, and since < cos2wt > = 0**, we have, an advocate of direct 2 1 current. Westinghouse < sin ωt > = was convinced that the 2 technology of alternating Thus, current was the key to 1 2 the electrical future. p = i m R [7.5(c)] He founded the famous 2 Company named after him WESTINGHOUSE To express ac power in the same form as dc power and enlisted the services (P = I2R), a special value of current is defined and used. of Nicola Tesla and It is called, root mean square (rms) or effective current other inventors in the (1846(Fig. 7.3) and is denoted by Irms or I. development of alternating current motors and – apparatus for the transmission of high tension current, pioneering 1914) in large scale lighting. FIGURE 7.3 The rms current I is related to the peak current im by I = mi / 2 = 0.707 im. 1 T F (t ) d t* The average value of a function F (t) over a period T is given by F (t ) = T ∫0 1 T 1  sin 2ωt  T 1 < cos 2ωt > = ∫ cos 2ω t dt = = [ sin 2ω T − 0 ] = 0** T 0 T  2ω  0 2ωT 179 Reprint 2025-26 Physics It is defined by 2 1 2 i m I = i = i m = 2 2 = 0.707 im (7.6) In terms of I, the average power, denoted by P is 1 2 2 P = p = i m R = I R (7.7) 2 Similarly, we define the rms voltage or effective voltage by v m V = = 0.707 vm (7.8) 2 From Eq. (7.3), we have vm = imR v m i m or, = R 2 2 or, V = IR (7.9) Equation (7.9) gives the relation between ac current and ac voltage and is similar to that in the dc case. This shows the advantage of introducing the concept of rms values. In terms of rms values, the equation for power [Eq. (7.7)] and relation between current and voltage in ac circuits are essentially the same as those for the dc case. It is customary to measure and specify rms values for ac quantities. For example, the household line voltage of 220 V is an rms value with a peak voltage of vm = 2 V = (1.414)(220 V) = 311 V In fact, the I or rms current is the equivalent dc current that would produce the same average power loss as the alternating current. Equation (7.7) can also be written as P = V2 / R = I V (since V = I R) Example 7.1 A light bulb is rated at 100W for a 220 V supply. Find (a) the resistance of the bulb; (b) the peak voltage of the source; and (c) the rms current through the bulb. Solution (a) We are given P = 100 W and V = 220 V. The resistance of the bulb is 2 V 2 ( 220 V ) R = = = 484 Ω P 100 W (b) The peak voltage of the source is V 7.1 v m = 2V = 311 (c) Since, P = I V P 100 W I 0.454A EXAMPLE V 220 V Reprint 2025-26 Alternating Current