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

8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to 213 a 230 V ac supply with a (angular) frequency of 300 rad s–1. Reprint 2025-26 Physics (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. FIGURE 8.6 8.3 What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500m? 8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? 8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? 8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? 8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? 8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine, B0,w, k, and l. (b) Find expressions for E and B. 8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hn (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation? 8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s–1.] Reprint 2025-26

4.5 MAGNETIC FIELD ON THE AXIS OF A CIRCULAR CURRENT LOOP In this section, we shall evaluate the magnetic field due to a circular coil along its axis. The evaluation entails summing up the effect of infinitesimal current elements (I dl) mentioned in the previous section. We assume that the current I is steady and that the evaluation is carried out in free space (i.e., vacuum). Fig. 4.9 depicts a circular loop carrying a steady current I. The loop is placed in the plane with its centre at the origin O and has a radius R. The x-axis is the axis of the loop. We wish to calculate the magnetic field at the point P on this axis. Let x be the distance of P from the centre O of the loop. Consider a conducting element dl of the loop. This is FIGURE 4.9 Magnetic field on the shown in Fig. 4.9. The magnitude dB of the magnetic axis of a current carrying circular field due to dl is given by the Biot-Savart law [Eq. 4.7(a)], loop of radius R. Shown are the µ0 I d l × r magnetic field dB (due to a line dB = 3 (4.8) element dl ) and its 4π r components along and Now r2 = x2 + R2 . Further, any element of the loop perpendicular to the axis. will be perpendicular to the displacement vector from the element to the axial point. For example, the element dl in Fig. 4.9 is in the plane, whereas, the displacement vector r from dl to the axial point P is in the plane. Hence |dl × r|=r dl. Thus, µ0 Idl d B = 4 π x 2 + R 2 (4.9) 115 ( ) Reprint 2025-26 Physics The direction of dB is shown in Fig. 4.9. It is perpendicular to the plane formed by dl and r. It has an x-component dBx and a component perpendicular to x-axis, dB⊥. When the components perpendicular to the x-axis are summed over, they cancel out and we obtain a null result. For example, the dB⊥ component due to dl is cancelled by the contribution due to the diametrically opposite dl element, shown in Fig. 4.9. Thus, only the x-component survives. The net contribution along x-direction can be obtained by integrating dBx = dB cos θ over the loop. For Fig. 4.9, R cosθ= 2 2 1/2 (4.10) ( x + R ) From Eqs. (4.9) and (4.10), µ0 Idl R d B x = 4 π x 2 + R 2 3/2 ( ) The summation of elements dl over the loop yields 2πR, the circumference of the loop. Thus, the magnetic field at P due to entire circular loop is 2 µ 0 I R ˆ i B = B x ˆi = 3/2 (4.11) 2 2 2 x + R ( ) As a special case of the above result, we may obtain the field at the centre of the loop. Here x = 0, and we obtain, B 0 = µ0 I ˆi (4.12) 2 R The magnetic field lines due to a circular wire form closed loops and are shown in Fig. 4.10. The direction of the magnetic field is given by (another) right-hand thumb rule stated below: Curl the palm of your right hand around the circular wire with the fingers pointing in the direction of the current. The right-hand thumb gives the direction of the magnetic field. FIGURE 4.10 The magnetic field lines for a current loop. The direction of the field is given by the right-hand thumb rule described in the text. The upper side of the loop may be thought of as the north pole and the lower 116 side as the south pole of a magnet. Reprint 2025-26 Moving Charges and Magnetism Example 4.5 A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2.0 cm as shown in Fig. 4.11(a). Consider the magnetic field B at the centre of the arc. (a) What is the magnetic field due to the straight segments? (b) In what way the contribution to B from the semicircle differs from that of a circular loop and in what way does it resemble? (c) Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Fig. 4.11(b)? FIGURE 4.11 Solution (a) dl and r for each element of the straight segments are parallel. Therefore, dl × r = 0. Straight segments do not contribute to |B|. (b) For all segments of the semicircular arc, dl × r are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop. Thus B is 1.9 × 10–4 T normal to the plane of the EXAMPLE paper going into it. (c) Same magnitude of B but opposite in direction to that in (b). 4.5 Example 4.6 Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil? Solution Since the coil is tightly wound, we may take each circular element to have the same radius R = 10 cm = 0.1 m. The number of turns N = 100. The magnitude of the magnetic field is, EXAMPLE –7 NI µ 4 π × 10 × 10 2 × 1 0 B = = –1 = 2 π × 10−4 = 6.28 × 10 −4 T 4.6 R 2 10 2 ×