Physics 12th Book 1,2

not mean that the average power consumed is zero and Alternating Current that there is no dissipation of electrical energy. As you know, Joule heating is given by i2R and depends on i2 George Westinghouse (which is always positive whether i is positive or negative) (1846 – 1914) A leading and not on i. Thus, there is Joule heating and proponent of the use of dissipation of electrical energy when an alternating current over ac current passes through a resistor. direct current. Thus, The instantaneous power dissipated in the resistor is he came into conflict with Thomas Alva Edison, p = i 2R = im2 R sin2 ω t (7.4) an advocate of direct The average value of p over a cycle is* current. Westinghouse was convinced that the p = <i2R >=< i 2 R sin2 ωt > [7.5(a)] technology of alternating m current was the key to the electrical future. where the bar over a letter (here, p) denotes its average He founded the famous Company named after him value and <......> denotes taking average of the quantity and enlisted the services GEORGE WESTINGHOUSE (1846 – 1914) of Nicola Tesla and inside the bracket. Since, i 2 and R are constants, other inventors in the m development of alternating current motors and p = im2 R < sin2 ωt > [7.5(b)] apparatus for the transmission of high Using the trigonometric identity, sin2 ωt = tension current, pioneering in large scale lighting. 1/2 (1– cos 2ωt ), we have < sin2 ωt > = (1/2) (1– < cos 2ωt >) and since < cos2ωt > = 0**, we have, < sin2 ω t > = 1 2 Thus, p= 1 im2 R [7.5(c)] 2 To express ac power in the same form as dc power (P = I2R), a special value of current is defined and used. It is called, root mean square (rms) or effective current (Fig. 7.3) and is denoted by Irms or I. FIGURE 7.3 The rms current I is related to the peak current im by I = im / 2 = 0.707 im. 1 T * The average value of a function F (t ) over a period T is given by F (t ) = ∫T 0 F (t ) dt ** < cos 2ω t >= 1 T = 1  sin 2ω t T = 1 [sin 2ω T − 0] = 0 235 T T  2ω 0 2ωT ∫ cos 2ω t dt 0 2020-21

Physics It is defined by I = i2 = 1 i 2 = im 2 m 2 = 0.707 im (7.6) In terms of I, the average power, denoted by P is P = p = 1 i 2 R = I 2 R (7.7) 2 m Similarly, we define the rms voltage or effective voltage by V= vm = 0.707 vm (7.8) 2 From Eq. (7.3), we have vm = imR or, vm = im R 22 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 R = V 2 = (220 V)2 = 484Ω P 100 W (b) The peak voltage of the source is EXAMPLE 7.1 v = 2V = 311V m (c) Since, P = I V 236 I = P = 100 W = 0.454A V 220 V 2020-21

Alternating Current 7.3 REPRESENTATION OF AC CURRENT AND VOLTAGE BY ROTATING VECTORS — PHASORS In the previous section, we learnt that the current through a resistor is in phase with the ac voltage. But this is not so in the case of an inductor, a capacitor or a combination of these circuit elements. In order to show phase relationship between voltage and current in an ac circuit, we use the notion of phasors. The analysis of an ac circuit is facilitated by the use of a phasor diagram. A phasor* is a vector which rotates about the origin with angular speed ω, as shown in Fig. 7.4. The vertical components of phasors V and I represent the sinusoidally varying quantities v and i. The magnitudes of phasors V and I represent the amplitudes or the peak values vm and im of these FIGURE 7.4 (a) A phasor diagram for the oscillating quantities. Figure 7.4(a) shows the voltage and current phasors and their circuit in Fig 7.1. (b) Graph of v and relationship at time t1 for the case of an ac source i versus ωt. connected to a resistor i.e., corresponding to the circuit shown in Fig. 7.1. The projection of voltage and current phasors on vertical axis, i.e., vm sinω t and im sinω t, respectively represent the value of voltage and current at that instant. As they rotate with frequency ω, curves in Fig. 7.4(b) are generated. From Fig. 7.4(a) we see that phasors V and I for the case of a resistor are in the same direction. This is so for all times. This means that the phase angle between the voltage and the current is zero. 7.4 AC VOLTAGE APPLIED TO AN INDUCTOR Figure 7.5 shows an ac source connected to an inductor. Usually, inductors have appreciable resistance in their windings, but we shall assume that this inductor has negligible resistance. Thus, the circuit is a purely inductive ac circuit. Let the voltage across the source be v = vm sinω t. Using the Kirchhoff’s loop rule, ∑ ε (t ) = 0 , and since there is no resistor in the circuit, v − L di = 0 (7.10) dt where the second term is the self-induced Faraday FIGURE 7.5 An ac source connected to an inductor. emf in the inductor; and L is the self-inductance of * Though voltage and current in ac circuit are represented by phasors – rotating 237 vectors, they are not vectors themselves. They are scalar quantities. It so happens that the amplitudes and phases of harmonically varying scalars combine mathematically in the same way as do the projections of rotating vectors of corresponding magnitudes and directions. The rotating vectors that represent harmonically varying scalar quantities are introduced only to provide us with a simple way of adding these quantities using a rule that we already know. 2020-21

Physics the inductor. The negative sign follows from Lenz’s law (Chapter 6). Combining Eqs. (7.1) and (7.10), we have di = v = vm sin ω t (7.11) dt L L Equation (7.11) implies that the equation for i(t), the current as a function of time, must be such that its slope di/dt is a sinusoidally varying Interactive animation on Phasor diagrams of ac circuits containing, R, L, C and RLC series circuits: quantity, with the same phase as the source voltage and an amplitude http://www.animations.physics.unsw.edu.au//jw/AC.html given by vm/L. To obtain the current, we integrate di/dt with respect to time: ∫ di dt = vm ∫ sin(ωt )dt dt L and get, i = − vm cos(ωt ) + constant ωL The integration constant has the dimension of current and is time- independent. Since the source has an emf which oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about zero, so that no constant or time-independent component of the current exists. Therefore, the integration constant is zero. Using − cos(ω t ) = sin  ω t − π , we have  2 i = im sin  ωt − π (7.12)  2  where im = vm is the amplitude of the current. The quantity ω L is ωL analogous to the resistance and is called inductive reactance, denoted by XL: (7.13) XL = ω L The amplitude of the current is, then im = vm (7.14) XL The dimension of inductive reactance is the same as that of resistance and its SI unit is ohm (Ω). The inductive reactance limits the current in a purely inductive circuit in the same way as the resistance limits the current in a purely resistive circuit. The inductive reactance is directly proportional to the inductance and to the frequency of the current. A comparison of Eqs. (7.1) and (7.12) for the source voltage and the current in an inductor shows that the current lags the voltage by π/2 or one-quarter (1/4) cycle. Figure 7.6 (a) shows the voltage and the current phasors in the present case at instant t1. The current phasor I is π/2 behind the voltage phasor V. When rotated with frequency ω counter- clockwise, they generate the voltage and current given by Eqs. (7.1) and 238 (7.12), respectively and as shown in Fig. 7.6(b). 2020-21

Alternating Current FIGURE 7.6 (a) A Phasor diagram for the circuit in Fig. 7.5. (b) Graph of v and i versus ωt. We see that the current reaches its maximum value later than the voltage by one-fourth of a period T = π/2 You have seen that an  4 ω  . inductor has reactance that limits current similar to resistance in a dc circuit. Does it also consume power like a resistance? Let us try to find out. The instantaneous power supplied to the inductor is pL = iv = im sin ω t − π  ×vm sin (ωt ) 2  = −imvm cos(ωt ) sin(ωt ) = − imvm sin (2ωt ) 2 So, the average power over a complete cycle is PL = − imvm sin (2ω t ) 2 = − imvm sin (2ω t ) = 0, 2 since the average of sin (2ωt) over a complete cycle is zero. Thus, the average power supplied to an inductor over one complete cycle is zero. Figure 7.7 explains it in detail. Example 7.2 A pure inductor of 25.0 mH is connected to a source of EXAMPLE 7.2 239 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz. Solution The inductive reactance, X L = 2 π ν L = 2 × 3.14 × 50 × 25 × 10–3 Ω = 7.85Ω The rms current in the circuit is I = V = 220 V = 28A X L 7.85 Ω 2020-21

Physics 0-1 Current i through the coil entering at A 1-2 Current in the coil is still positive but is increase from zero to a maximum value. Flux decreasing. The core gets demagnetised and lines are set up i.e., the core gets magnetised. the net flux becomes zero at the end of a half With the polarity shown voltage and current cycle. The voltage v is negative (since di/dt is are both positive. So their product p is positive. negative). The product of voltage and current ENERGY IS ABSORBED FROM THE is negative, and ENERGY IS BEING SOURCE. RETURNED TO SOURCE. One complete cycle of voltage/current. Note that the current lags the voltage. 2-3 Current i becomes negative i.e., it enters 3-4 Current i decreases and reaches its zero at B and comes out of A. Since the direction value at 4 when core is demagnetised and flux of current has changed, the polarity of the is zero. The voltage is positive but the current magnet changes. The current and voltage are is negative. The power is, therefore, negative. both negative. So their product p is positive. ENERGY ABSORBED DURING THE CYCLE ENERGY IS ABSORBED. 2-3 IS RETURNED TO THE SOURCE. 240 FIGURE 7.7 Magnetisation and demagnetisation of an inductor. 2020-21

Alternating Current 7.5 AC VOLTAGE APPLIED TO A CAPACITOR Figure 7.8 shows an ac source ε generating ac voltage v = vm sin ωt connected to a capacitor only, a purely capacitive ac circuit. When a capacitor is connected to a voltage source in a dc circuit, current will flow for the short time required to charge the capacitor. As charge accumulates on the capacitor plates, the voltage across them increases, opposing the current. That is, a capacitor in a dc circuit will limit or oppose the current as it charges. When the capacitor is fully charged, the current in the circuit falls to zero. When the capacitor is connected to an ac source, as in Fig. 7.8, it limits or regulates the current, but FIGURE 7.8 An ac source does not completely prevent the flow of charge. The connected to a capacitor. capacitor is alternately charged and discharged as the current reverses each half cycle. Let q be the charge on the capacitor at any time t. The instantaneous voltage v across the capacitor is v= q (7.15) C From the Kirchhoff’s loop rule, the voltage across the source and the capacitor are equal, vm sin ω t = q C To find the current, we use the relation i = dq dt i = d (vmC sin ω t ) = ω C vm cos(ω t ) dt Using the relation, cos(ω t) = sin  ω t + π , we have  2  i = im sin  ω t + π (7.16)  2  where the amplitude of the oscillating current is im = ωCvm. We can rewrite it as im = vm (1/ω C ) Comparing it to im= vm/R for a purely resistive circuit, we find that (1/ωC) plays the role of resistance. It is called capacitive reactance and is denoted by Xc, (7.17) Xc= 1/ωC so that the amplitude of the current is im = vm (7.18) 241 XC 2020-21

Physics The dimension of capacitive reactance is the same as that of resistance and its SI unit is ohm (Ω). The capacitive reactance limits the amplitude of the current in a purely capacitive circuit in the same way as the resistance limits the current in a purely resistive circuit. But it is inversely proportional to the frequency and the capacitance. FIGURE 7.9 (a) A Phasor diagram for the circuit A comparison of Eq. (7.16) with the in Fig. 7.8. (b) Graph of v and i versus ωt. equation of source voltage, Eq. (7.1) shows that the current is π/2 ahead of voltage. Figure 7.9(a) shows the phasor diagram at an instant t1. Here the current phasor I is π/2 ahead of the voltage phasor V as they rotate counterclockwise. Figure 7.9(b) shows the variation of voltage and current with time. We see that the current reaches its maximum value earlier than the voltage by one-fourth of a period. The instantaneous power supplied to the capacitor is pc = i v = im cos(ωt)vm sin(ωt) = imvm cos(ωt) sin(ωt) = imvm sin(2ωt ) (7.19) 2 So, as in the case of an inductor, the average power PC = imvm sin(2ωt ) = imvm sin(2ωt ) = 0 2 2 since <sin (2ωt)> = 0 over a complete cycle. Figure 7.10 explains it in detail. Thus, we see that in the case of an inductor, the current lags the voltage by π/2 and in the case of a capacitor, the current leads the voltage by π/2. EXAMPLE 7.3 Example 7.3 A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced? Solution When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (1/ωC ) and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before. EXAMPLE 7.4 Example 7.4 A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current? Solution The capacitive reactance is XC = 1 = 1 = 212 Ω 2πνC 2π (50Hz)(15.0 × 10−6 F) 242 The rms current is 2020-21

Alternating Current 0-1 The current i flows as shown and from the 1-2 The current i reverses its direction. The maximum at 0, reaches a zero value at 1. The plate accumulated charge is depleted i.e., the capacitor is A is charged to positive polarity while negative charge discharged during this quarter cycle.The voltage gets q builds up in B reaching a maximum at 1 until the reduced but is still positive. The current is negative. current becomes zero. The voltage vc = q/C is in phase Their product, the power is negative. with q and reaches maximum value at 1. Current THE ENERGY ABSORBED DURING THE 1/4 and voltage are both positive. So p = vci is positive. CYCLE 0-1 IS RETURNED DURING THIS QUARTER. ENERGY IS ABSORBED FROM THE SOURCE DURING THIS QUARTER CYCLE AS THE CAPACITOR IS CHARGED. One complete cycle of voltage/current. Note that the current leads the voltage. 2-3 As i continues to flow from A to B, the capacitor 3-4 The current i reverses its direction at 3 and flows is charged to reversed polarity i.e., the plate B from B to A. The accumulated charge is depleted acquires positive and A acquires negative charge. and the magnitude of the voltage vc is reduced. vc Both the current and the voltage are negative. Their becomes zero at 4 when the capacitor is fully product p is positive. The capacitor ABSORBS discharged. The power is negative.ENERGY ENERGY during this 1/4 cycle. ABSORBED DURING 2-3 IS RETURNED TO THE SOURCE. NET ENERGY ABSORBED IS ZERO. FIGURE 7.10 Charging and discharging of a capacitor. 243 2020-21

PhysicsEXAMPLE 7.4 I = V = 220 V = 1.04 A XC 212 Ω The peak current is im = 2I = (1.41)(1.04 A) = 1.47A This current oscillates between +1.47A and –1.47 A, and is ahead of the voltage by π/2. If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled. Example 7.5 A light bulb and an open coil inductor are connected to an ac source through a key as shown in Fig. 7.11. EXAMPLE 7.5 FIGURE 7.11 The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is inserted. Give your answer with reasons. Solution As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases. 7.6 AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT Figure 7.12 shows a series LCR circuit connected to an ac source ε. As usual, we take the voltage of the source to be v = vm sin ωt. If q is the charge on the capacitor and i the current, at time t, we have, from Kirchhoff’s loop rule: L di + i R + q = v (7.20) dt C We want to determine the instantaneous current i and its phase relationship to the applied alternating voltage v. We shall solve this problem by two methods. First, we use the technique of FIGURE 7.12 A series LCR circuit phasors and in the second method, we solve connected to an ac source. Eq. (7.20) analytically to obtain the time– 244 dependence of i . 2020-21

Alternating Current 7.6.1 Phasor-diagram solution From the circuit shown in Fig. 7.12, we see that the resistor, inductor and capacitor are in series. Therefore, the ac current in each element is the same at any time, having the same amplitude and phase. Let it be i = im sin(ωt+φ ) (7.21) where φ is the phase difference between the voltage across the source and the current in the circuit. On the basis of what we have learnt in the previous sections, we shall construct a phasor diagram for the present case. Let I be the phasor representing the current in the circuit as given by Eq. (7.21). Further, let VL, VR, VC, and V represent the voltage across the inductor, resistor, capacitor and the source, respectively. From previous section, we know that VR is parallel to I, VC is π/2 behind I and VL is π/2 ahead of I. VL, VR, VC and I are shown in Fig. 7.13(a) with apppropriate phase- relations. The length of these phasors or the amplitude of VR, VC and VL are: vRm = im R, vCm = im XC, vLm = im XL (7.22) The voltage Equation (7.20) for the circuit can be written as vL + vR + vC = v (7.23) The phasor relation whose vertical component FIGURE 7.13 (a) Relation between the gives the above equation is phasors VL, VR, VC, and I, (b) Relation VL + VR + VC = V (7.24) between the phasors VL, VR, and (VL + VC) This relation is represented in Fig. 7.13(b). Since for the circuit in Fig. 7.12. VC and VL are always along the same line and in opposite directions, they can be combined into a single phasor (VC + VL) which has a magnitude vCm – vLm. Since V is represented as the hypotenuse of a right-triangle whose sides are VR and (VC + VL), the pythagorean theorem gives: ( )vm2 = vR2m + vCm − vLm 2 Substituting the values of vRm, vCm, and vLm from Eq. (7.22) into the above equation, we have vm2 = (im R)2 + (im X C − im X L )2 = im2 R 2 + (XC − X L )2  or, im = vm [7.25(a)] R2 + ( X C − X L )2 By analogy to the resistance in a circuit, we introduce the impedance Z in an ac circuit: im = vm [7.25(b)] Z where Z = R2 + (XC − X L )2 (7.26) 245 2020-21

Physics Since phasor I is always parallel to phasor VR, the phase angle φ is the angle between VR and V and can be determined from Fig. 7.14: tan φ = vCm − vLm v Rm Using Eq. (7.22), we have tan φ = XC − X L (7.27) R Equations (7.26) and (7.27) are graphically shown in Fig. (7.14). FIGURE 7.14 Impedance This is called Impedance diagram which is a right-triangle with diagram. Z as its hypotenuse. Equation 7.25(a) gives the amplitude of the current and Eq. (7.27) gives the phase angle. With these, Eq. (7.21) is completely specified. If XC > XL, φ is positive and the circuit is predominantly capacitive. Consequently, the current in the circuit leads the source voltage. If XC < XL, φ is negative and the circuit is predominantly inductive. Consequently, the current in the circuit lags the source voltage. Figure 7.15 shows the phasor diagram and variation of v and i with ω t for the case XC > XL. Thus, we have obtained the amplitude and phase of current for an LCR series circuit using the technique of phasors. But this method of analysing ac circuits suffers from certain disadvantages. First, the phasor diagram say nothing about the initial condition. One can take any arbitrary value of t (say, t1, as done throughout this chapter) and draw different phasors which show the relative angle between different phasors. The solution so obtained is called the steady-state solution. This is not a general FIGURE 7.15 (a) Phasor diagram of V and I. solution. Additionally, we do have a (b) Graphs of v and i versus ω t for a series LCR transient solution which exists even for v = 0. The general solution is the sum of the circuit where XC > XL. transient solution and the steady-state solution. After a sufficiently long time, the effects of the transient solution die out and the behaviour of the circuit is described by the steady-state solution. 7.6.2 Analytical solution The voltage equation for the circuit is L di + R i + q = v dt C = vm sin ωt We know that i = dq/dt. Therefore, di/dt = d2q/dt2. Thus, in terms of q, 246 the voltage equation becomes 2020-21

Alternating Current L d2q + R dq + q = vm sin ω t (7.28) dt 2 dt C This is like the equation for a forced, damped oscillator, [see Eq. {14.37(b)} in Class XI Physics Textbook]. Let us assume a solution q = qm sin (ω t + θ ) [7.29(a)] [7.29(b)] so that dq = qm ω cos(ω t + θ) dt and d2q = −qmω2 sin(ω t + θ) [7.29(c)] dt 2 Substituting these values in Eq. (7.28), we get qmω [R cos(ω t + θ) + (XC − X L )sin(ω t + θ)] = vm sin ωt (7.30) where we have used the relation Xc= 1/ωC, XL = ω L. Multiplying and dividing Eq. (7.30) by Z = R2 + ( Xc − X L )2 , we have qm ω Z R cos(ω t + θ) + (XC − XL ) sin(ω t + θ ) = vm sin ω t (7.31)  Z Z Now, let R = cos φ Z and ( XC − X L ) = sin φ Z so that φ = tan−1 XC − X L (7.32) R (7.33) Substituting this in Eq. (7.31) and simplifying, we get: [7.33(a)] qm ω Z cos(ω t + θ − φ) = vm sin ω t Comparing the two sides of this equation, we see that vm = qm ω Z = im Z where im = qm ω and θ −φ = −π or θ = −π +φ [7.33(b)] 2 2 (7.34) Therefore, the current in the circuit is [7.34(a)] i = dq = qm ω cos(ω t + θ) dt = im cos(ωt + θ ) or i = imsin(ωt + φ ) where im = vm = vm Z R 2 + ( X C − X L )2 and φ = tan−1 XC − XL 247 R 2020-21

Physics Thus, the analytical solution for the amplitude and phase of the current in the circuit agrees with that obtained by the technique of phasors. 7.6.3 Resonance An interesting characteristic of the series RLC circuit is the phenomenon of resonance. The phenomenon of resonance is common among systems that have a tendency to oscillate at a particular frequency. This frequency is called the system’s natural frequency. If such a system is driven by an energy source at a frequency that is near the natural frequency, the amplitude of oscillation is found to be large. A familiar example of this phenomenon is a child on a swing. The swing has a natural frequency for swinging back and forth like a pendulum. If the child pulls on the rope at regular intervals and the frequency of the pulls is almost the same as the frequency of swinging, the amplitude of the swinging will be large (Chapter 14, Class XI). For an RLC circuit driven with voltage of amplitude vm and frequency ω, we found that the current amplitude is given by im = vm = vm Z R2 + ( XC − X L )2 with Xc = 1/ωC and XL = ω L . So if ω is varied, then at a particular frequency ( )ω0, Xc = XL, and the impedance is minimum Z = R2 + 02 = R . This frequency is called the resonant frequency: Xc = XL or 1 = ω0 L ω0 C or ω0 = 1 (7.35) LC At resonant frequency, the current amplitude is maximum; im = vm/R. Figure 7.16 shows the variation of im with ω in a RLC series circuit with L = 1.00 mH, C = 1.00 nF for two values of R: (i) R = 100 Ω and (ii) R = 200 Ω. For the source applied vm = 100 V. ω0 for this case is 1 LC = 1.00×106 rad/s. We see that the current amplitude is maximum at the resonant frequency. Since im = vm / R at resonance, the current amplitude for case (i) is twice to that for case (ii). FIGURE 7.16 Variation of im with ω for two Resonant circuits have a variety of cases: (i) R = 100 Ω, (ii) R = 200 Ω, applications, for example, in the tuning mechanism of a radio or a TV set. The antenna of L = 1.00 mH. a radio accepts signals from many broadcasting stations. The signals picked up in the antenna acts as a source in the 248 tuning circuit of the radio, so the circuit can be driven at many frequencies. 2020-21

Alternating Current But to hear one particular radio station, we tune the radio. In tuning, we vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received. When this happens, the amplitude of the current with the frequency of the signal of the particular radio station in the circuit is maximum. It is important to note that resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then do the voltages across L and C cancel each other (both being out of phase) and the current amplitude is vm/R, the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit. Sharpness of resonance The amplitude of the current in the series LCR circuit is given by im = vm  1 2  ω C  R2 + ω L − and is maximum when ω = ω0 = 1/ L C. The maximum value is immax = vm / R . For values of ω other than ω0, the amplitude of the current is less than the maximum value. Suppose we choose a value of ω for which the current amplitude is 1/ 2 times its maximum value. At this value, the power dissipated by the circuit becomes half. From the curve in Fig. (7.16), we see that there are two such values of ω, say, ω1 and ω2, one greater and the other smaller than ω0 and symmetrical about ω0. We may write ω1 = ω0 + ∆ω ω2 = ω0 – ∆ω The difference ω1 – ω2 = 2∆ω is often called the bandwidth of the circuit. The quantity (ω0 / 2∆ω) is regarded as a measure of the sharpness of resonance. The smaller the ∆ω, the sharper or narrower is the resonance. To get an expression for ∆ω, we note that the current amplitude im is 2( )1/ i max for ω1 = ω0 + ∆ω. Therefore, m at ω1, im = vm  1 2  ω 1L ω 1C  R2 + − = i max = vm m 2 R2 249 2020-21

Physics  1 2  ω ω1 C  or R2 + 1L − = R 2  1 2  ω1 ω1 C  or R2 + L − = 2R2 ω 1L − 1 = R ω 1C which may be written as, (ω0 + ∆ω) L − (ω0 1 = R + ∆ω)C ω0L 1 + ∆ω  − 1 ∆ω  = R ω0  ω0C 1 + ω0  Using ω02 = 1 in the second term on the left hand side, we get LC  ∆ω  − ω0L = R ω0L 1 + ω0  1 + ∆ω  ω0  We can approximate 1 + ∆ω  −1 as 1 − ∆ω  since ∆ω ω0  ω0  ω0 <<1. Therefore, ω0L 1 + ∆ω  − ω0L 1 − ∆ω  = R ω0  ω0  or ω0L 2∆ω = R ω0 ∆ω = R [7.36(a)] 2L The sharpness of resonance is given by, ω0 = ω0L [7.36(b)] 2∆ω R The ratio ω0L is also called the quality factor, Q of the circuit. R Q = ω0L [7.36(c)] R 250 From Eqs. [7.36 (b)] and [7.36 (c)], we see that 2∆ω = ω0 . So, larger the Q 2020-21

Alternating Current value of Q, the smaller is the value of 2∆ω or the bandwidth and sharper is the resonance. Using ω02 = 1/ L C , Eq. [7.36(c)] can be equivalently expressed as Q = 1/ω0CR. We see from Fig. 7.15, that if the resonance is less sharp, not only is the maximum current less, the circuit is close to resonance for a larger range ∆ω of frequencies and the tuning of the circuit will not be good. So, less sharp the resonance, less is the selectivity of the circuit or vice versa. From Eq. (7.36), we see that if quality factor is large, i.e., R is low or L is large, the circuit is more selective. Example 7.6 A resistor of 200 Ω and a capacitor of 15.0 µF are connected in series to a 220 V, 50 Hz ac source. (a) Calculate the current in the circuit; (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. Solution Given R = 200 Ω, C = 15.0 µF = 15.0 × 10−6 F V = 220 V, ν = 50 Hz (a) In order to calculate the current, we need the impedance of the circuit. It is Z= R2 + X 2 = R2 + (2π ν C)−2 C = (200 Ω)2 + (2 × 3.14 × 50 × 15.0 × 10−6 F)−2 = (200 Ω)2 + (212.3 Ω)2 EXAMPLE 7.6 251 = 291.67 Ω Therefore, the current in the circuit is I = V = 220 V = 0.755 A Z 291.5 Ω (b) Since the current is the same throughout the circuit, we have VR = I R = (0.755 A)(200 Ω) = 151 V VC = I XC = (0.755 A)(212.3 Ω) = 160.3 V The algebraic sum of the two voltages, VR and VC is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox? As you have learnt in the text, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem: VR+C = VR2 + VC2 = 220 V Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source. 2020-21

Physics 7.7 POWER IN AC CIRCUIT: THE POWER FACTOR We have seen that a voltage v = vm sinωt applied to a series RLC circuit drives a current in the circuit given by i = im sin(ωt + φ) where im = vm and φ = tan−1  XC − XL  Z  R  Therefore, the instantaneous power p supplied by the source is p = v i = (vm sin ω t ) × [im sin(ω t + φ)] = vm im [cos φ − cos(2ω t + φ)] (7.37) 2 The average power over a cycle is given by the average of the two terms in R.H.S. of Eq. (7.37). It is only the second term which is time-dependent. Its average is zero (the positive half of the cosine cancels the negative half). Therefore, P = vmim cos φ = vm im cos φ 2 22 = V I cos φ [7.38(a)] This can also be written as, P = I 2 Z cos φ [7.38(b)] So, the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle φ between them. The quantity cosφ is called the power factor. Let us discuss the following cases: Case (i) Resistive circuit: If the circuit contains only pure R, it is called resistive. In that case φ = 0, cos φ =1. There is maximum power dissipation. Case (ii) Purely inductive or capacitive circuit: If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage and current is π/2. Therefore, cos φ = 0, and no power is dissipated even though a current is flowing in the circuit. This current is sometimes referred to as wattless current. Case (iii) LCR series circuit: In an LCR series circuit, power dissipated is given by Eq. (7.38) where φ = tan–1 (Xc – XL )/ R. So, φ may be non-zero in a RL or RC or RCL circuit. Even in such cases, power is dissipated only in the resistor. Case (iv) Power dissipated at resonance in LCR circuit: At resonance Xc – XL= 0, and φ = 0. Therefore, cosφ = 1 and P = I 2Z = I 2 R. That is, maximum power is dissipated in a circuit (through R) at resonance. 252 EXAMPLE 7.7 Example 7.7 (a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain. (b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain. 2020-21

Alternating Current Solution (a) We know that P = I V cosφ where cosφ is the power factor. To supply a given power at a given voltage, if cosφ is small, we have to increase current accordingly. But this will lead to large power loss (I2R) in transmission. (b)Suppose in a circuit, current I lags the voltage by an angle φ. Then power factor cosφ =R/Z. We can improve the power factor (tending to 1) by making Z tend to R. Let us understand, with the help of a phasor diagram (Fig. 7.17) how this can be achieved. Let us resolve I into two components. Ip along FIGURE 7.17 EXAMPLE 7.7 the applied voltage V and Iq perpendicular to the applied voltage. Iq as you have learnt in Section 7.7, is called the wattless component since corresponding to this component of current, there is no power loss. IP is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit. It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current Iq by an equal leading wattless current I′q. This can be done by connecting a capacitor of appropriate value in parallel so that Iq and I′q cancel each other and P is effectively Ip V. Example 7.8 A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH, and C = 796 µF. Find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor. Solution EXAMPLE 7.8 (a) To find the impedance of the circuit, we first calculate XL and XC. XL = 2 πνL = 2 × 3.14 10–3 Ω Ω × 50 × 25.48 × = 8 XC = 1 253 2πνC 2020-21

Physics = 1 = 4Ω 2 × 3.14 × 50 × 796 × 10−6 Therefore, Z = R2 + (X L − XC )2 = 32 + (8 − 4)2 =5Ω (b) Phase difference, φ = tan–1 XC − X L R = tan −1  4 − 8  = −53.1°  3  Since φ is negative, the current in the circuit lags the voltage across the source. (c) The power dissipated in the circuit is P = I2R EXAMPLE 7.8 Now, I = im = 1  283 = 40A 2 2  5  Therefore, P = (40A)2 × 3 Ω = 4800 W (d) Power factor = cosφ = cos (–53.1°) = 0.6 Example 7.9 Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the impedance, the current, and the power dissipated at the resonant condition. Solution (a) The frequency at which the resonance occurs is ω0 = 1= 1 LC 25.48 × 10−3 × 796 × 10−6 = 222.1rad/s νr = ω0 = 221.1 Hz = 35.4Hz 2π 2 × 3.14 (b) The impedance Z at resonant condition is equal to the resistance: Z = R = 3Ω The rms current at resonance is =V =V =  283 1 = 66.7A Z R  2  3 EXAMPLE 7.9 The power dissipated at resonance is P = I 2 × R = (66.7)2 × 3 = 13.35 kW 254 You can see that in the present case, power dissipated at resonance is more than the power dissipated in Example 7.8. 2020-21

Alternating Current Example 7.10 At an airport, a person is made to walk through the EXAMPLE 7.10 doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work? Solution The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes – resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm. 7.8 LC OSCILLATIONS We know that a capacitor and an inductor can store electrical and magnetic energy, respectively. When a capacitor (initially charged) is connected to an inductor, the charge on the capacitor and the current in the circuit exhibit the phenomenon of electrical oscillations similar to oscillations in mechanical systems (Chapter 14, Class XI). Let a capacitor be charged qm (at t = 0) and connected to an inductor as shown in Fig. 7.18. The moment the circuit is completed, the charge on the capacitor starts decreasing, giving rise to current in the circuit. Let q and i be the charge and current in the circuit at time t. Since di/dt is positive, the induced emf in L will have polarity as shown, i.e., vb < va. According to Kirchhoff’s loop rule, q − L di = 0 (7.39) FIGURE 7.18 At the C dt instant shown, the current i = – (dq/dt ) in the present case (as q decreases, i increases). is increasing; so the polarity of induced emf in Therefore, Eq. (7.39) becomes: the inductor is as shown. d2q + 1 q = 0 (7.40) dt 2 LC This equation has the form d2 x + ω 2 x = 0 for a simple harmonic dt 2 0 oscillator. The charge on the capacitor, therefore, oscillates with a natural frequency ω0 = 1 (7.41) LC and varies sinusoidally with time as q = qm cos(ω0 t + φ) (7.42) where qm is the maximum value of q and φ is a phase constant. Since 255 q = qm at t = 0, we have cos φ =1 or φ = 0. Therefore, in the present case, 2020-21

Physics q = qm cos(ω0t ) (7.43) The current i  = − dq  is given by  dt  i = im sin(ω0t ) (7.44) where im = ω0qm Let us now try to visualise how this oscillation takes place in the circuit. Figure 7.19(a) shows a capacitor with initial charge qm connected to an ideal inductor. The electrical energy stored in the charged capacitor is UE = 1 qm2 . Since, there is no current in the circuit, energy in the inductor 2C is zero. Thus, the total energy of LC circuit is, U = UE = 1 qm2 2C FIGURE 7.19 The oscillations in an LC circuit are analogous to the oscillation of a block at the end of a spring. The figure depicts one-half of a cycle. At t = 0, the switch is closed and the capacitor starts to discharge [Fig. 7.19(b)]. As the current increases, it sets up a magnetic field in the inductor and thereby, some energy gets stored in the inductor in the form of magnetic energy: UB = (1/2) Li2. As the current reaches its maximum value im, (at t = T/4) as in Fig. 7.19(c), all the energy is stored in the magnetic field: UB = (1/2) Li2m. You can easily check that the maximum electrical energy equals the maximum magnetic energy. The capacitor now has no charge and hence no energy. The current now starts charging the capacitor, as in Fig. 7.19(d). This process continues till the capacitor is fully charged (at t = T/2) [Fig. 7.19(e)]. But it is charged with a polarity opposite to its initial state in Fig. 7.19(a). The whole process just described will now repeat itself till the system reverts to its original state. Thus, the energy in the system oscillates between the capacitor 256 and the inductor. 2020-21

Alternating Current The LC oscillation is similar to the mechanical oscillation of a block attached to a spring. The lower part of each figure in Fig. 7.19 depicts the corresponding stage of a mechanical system (a block attached to a spring). As noted earlier, for a block of a mass m oscillating with frequency ω0, the equation is d2 x + ω 2 x = 0 dt 2 0 Here, ω0 = k /m , and k is the spring constant. So, x corresponds to q. In case of a mechanical system F = ma = m (dv/dt) = m (d2x/dt2). For an electrical system, ε = –L (di/dt ) = –L (d2q/dt 2). Comparing these two equations, we see that L is analogous to mass m: L is a measure of resistance to change in current. In case of LC circuit, ω0 = 1/ LC and for mass on a spring, ω0 = k /m . So, 1/C is analogous to k. The constant k (=F/x) tells us the (external) force required to produce a unit displacement whereas 1/C (=V/q ) tells us the potential difference required to store a unit charge. Table 7.1 gives the analogy between mechanical and electrical quantities. TABLE 7.1 ANALOGIES BETWEEN MECHANICAL AND ELECTRICAL QUANTITIES Mechanical system Electrical system Mass m Inductance L Force constant k Reciprocal capacitance 1/C Displacement x Charge q Velocity v = dx/dt Current i = dq/dt Mechanical energy Electromagnetic energy E = 1 kx2 + 1 mv2 U = 1 q2 + 1 L i2 22 2C 2 Note that the above discussion of LC oscillations is not realistic for two 257 reasons: (i) Every inductor has some resistance. The effect of this resistance is to introduce a damping effect on the charge and current in the circuit and the oscillations finally die away. (ii) Even if the resistance were zero, the total energy of the system would not remain constant. It is radiated away from the system in the form of electromagnetic waves (discussed in the next chapter). In fact, radio and TV transmitters depend on this radiation. 2020-21

Physics TWO DIFFERENT PHENOMENA, SAME MATHEMATICAL TREATMENT You may like to compare the treatment of a forced damped oscillator discussed in Section 14.10 of Class XI physics textbook, with that of an LCR circuit when an ac voltage is applied in it. We have already remarked that Eq. [14.37(b)] of Class XI Textbook is exactly similar to Eq. (7.28) here, although they use different symbols and parameters. Let us therefore list the equivalence between different quantities in the two situations: Forced oscillations Driven LCR circuit m d2x + b dx + kx = F cos ωd t L d2q + R dq + q = vm sin ω t dt 2 dt dt 2 dt C Displacement, x Charge on capacitor, q Time, t Time, t Mass, m Self inductance, L Damping constant, b Resistance, R Spring constant, k Inverse capacitance, 1/C Driving frequency, ωd Driving frequency, ω Natural frequency of oscillations, ω Natural frequency of LCR circuit, ω0 Amplitude of forced oscillations, A Maximum charge stored, qm Amplitude of driving force, F0 Amplitude of applied voltage, vm You must note that since x corresponds to q, the amplitude A (maximum displacement) will correspond to the maximum charge stored, q . Equation [14.39 (a)] of Class XI gives m the amplitude of oscillations in terms of other parameters, which we reproduce here for convenience: F0 { }A = m 2(ω 2 − ω 2 )2 + ωd2b 2 1/ 2 d Replace each parameter in the above equation by the corresponding electrical quantity, and see what happens. Eliminate L, C, ω(7,.a3n4d),ωy0,ouusiwngillXLs=eωe Lt,hXaCt= 1/ωC, and ω02 = 1/LC. When you use Eqs. (7.33) and there is a perfect match. You will come across numerous such situations in physics where diverse physical phenomena are represented by the same mathematical equation. If you have dealt with one of them, and you come across another situation, you may simply replace the corresponding quantities and interpret the result in the new context. We suggest that you may try to find more such parallel situations from different areas of physics. One must, of course, be aware of the differences too. 258 2020-21

Alternating Current Example 7.11 Show that in the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the inductor is constant in time. Solution Let q0 be the initial charge on a capacitor. Let the charged capacitor be connected to an inductor of inductance L. As you have studied in Section 7.8, this LC circuit will sustain an oscillation with frquency ω  = 2πν = 1  LC  At an instant t, charge q on the capacitor and the current i are given by: q (t) = q0 cos ωt i (t) = – q0 ω sin ωt Energy stored in the capacitor at time t is UE = 1 C V2 =1 q2 = q02 cos2 (ωt ) 2 2 C 2C Energy stored in the inductor at time t is UM =1 L i2 2 =1 L q02 ω 2 sin2 (ωt ) 2 ( )= q02 sin2 (ωt) 2C ∵ω =1/ LC Sum of energies ( )UE q 02 +UM = 2C cos2 ωt + sin2 ωt = q02 EXAMPLE 7.11 2C This sum is constant in time as qo and C, both are time-independent. Note that it is equal to the initial energy of the capacitor. Why it is so? Think! 7.9 TRANSFORMERS 259 For many purposes, it is necessary to change (or transform) an alternating voltage from one to another of greater or smaller value. This is done with a device called transformer using the principle of mutual induction. A transformer consists of two sets of coils, insulated from each other. They are wound on a soft-iron core, either one on top of the other as in Fig. 7.20(a) or on separate limbs of the core as in Fig. 7.20(b). One of the coils called the primary coil has Np turns. The other coil is called the secondary coil; it has Ns turns. Often the primary coil is the input coil and the secondary coil is the output coil of the transformer. 2020-21

Physics FIGURE 7.20 Two arrangements for winding of primary and secondary coil in a transformer: (a) two coils on top of each other, (b) two coils on separate limbs of the core. When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it. The value of this emf depends on the number of turns in the secondary. We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings. Let φ be the flux in each turn in the core at time t due to current in the primary when a voltage vp is applied to it. Then the induced emf or voltage εs, in the secondary with Ns turns is εs = −Ns dφ (7.45) dt The alternating flux φ also induces an emf, called back emf in the primary. This is εp = −N p dφ (7.46) dt But εp = vp. If this were not so, the primary current would be infinite since the primary has zero resistance (as assumed). If the secondary is an open circuit or the current taken from it is small, then to a good approximation εs = vs where vs is the voltage across the secondary. Therefore, Eqs. (7.45) and (7.46) can be written as vs = −Ns dφ [7.45(a)] dt vp = −N p dφ [7.46(a)] dt (7.47) From Eqs. [7.45 (a)] and [7.46 (a)], we have 260 vs = Ns vp Np 2020-21

Alternating Current Note that the above relation has been obtained using three assumptions: (i) the primary resistance and current are small; (ii) the same flux links both the primary and the secondary as very little flux escapes from the core, and (iii) the secondary current is small. If the transformer is assumed to be 100% efficient (no energy losses), the power input is equal to the power output, and since p = i v, ipvp = isvs (7.48) Although some energy is always lost, this is a good approximation, since a well designed transformer may have an efficiency of more than 95%. Combining Eqs. (7.47) and (7.48), we have ip = vs = Ns (7.49) is vp N p Since i and v both oscillate with the same frequency as the ac source, Eq. (7.49) also gives the ratio of the amplitudes or rms values of corresponding quantities. Now, we can see how a transformer affects the voltage and current. We have: Vs =  Ns  Vp and Is =  Np  Ip (7.50)  Np   N s  That is, if the secondary coil has a greater number of turns than the primary (Ns > Np), the voltage is stepped up (Vs > Vp). This type of arrangement is called a step-up transformer. However, in this arrangement, there is less current in the secondary than in the primary (Np/Ns < 1 and Is < Ip). For example, if the primary coil of a transformer has 100 turns and the secondary has 200 turns, Ns/Np = 2 and Np/Ns=1/2. Thus, a 220V input at 10A will step-up to 440 V output at 5.0 A. If the secondary coil has less turns than the primary (Ns < Np), we have a step-down transformer. In this case, Vs < Vp and Is > Ip. That is, the voltage is stepped down, or reduced, and the current is increased. The equations obtained above apply to ideal transformers (without any energy losses). But in actual transformers, small energy losses do occur due to the following reasons: (i) Flux Leakage: There is always some flux leakage; that is, not all of the flux due to primary passes through the secondary due to poor design of the core or the air gaps in the core. It can be reduced by winding the primary and secondary coils one over the other. (ii) Resistance of the windings: The wire used for the windings has some resistance and so, energy is lost due to heat produced in the wire (I 2R). In high current, low voltage windings, these are minimised by using thick wire. (iii) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core and causes heating. The effect is reduced by using a laminated core. (iv) Hysteresis: The magnetisation of the core is repeatedly reversed by 261 the alternating magnetic field. The resulting expenditure of energy in the core appears as heat and is kept to a minimum by using a magnetic material which has a low hysteresis loss. 2020-21

Physics The large scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is stepped-up (so that current is reduced and consequently, the I 2R loss is cut down). It is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes. SUMMARY 1. An alternating voltage v = v sin ωt applied to a resistor R drives a m current i = im sinωt in the resistor, im = vm . The current is in phase with R the applied voltage. 2. For an alternating current i = im sin ωt passing through a resistor R, the average power loss P (averaged over a cycle) due to joule heating is ( 1/2 )i 2mR. To express it in the same form as the dc power (P = I 2R), a special value of current is used. It is called root mean square (rms) current and is donoted by I: I = im = 0.707 im 2 Similarly, the rms voltage is defined by V = vm = 0.707 vm 2 We have P = IV = I 2R 3. An ac voltage v = vm sin ωt applied to a pure inductor L, drives a current in the inductor i = im sin (ωt – π/2), where im = vm/XL. XL = ωL is called inductive reactance. The current in the inductor lags the voltage by π/2. The average power supplied to an inductor over one complete cycle is zero. 4. An ac voltage v = vm sinωt applied to a capacitor drives a current in the capacitor: i = im sin (ωt + π/2). Here, im = vm , XC = 1 is called capacitive reactance. XC ωC The current through the capacitor is π/2 ahead of the applied voltage. As in the case of inductor, the average power supplied to a capacitor over one complete cycle is zero. 5. For a series RLC circuit driven by voltage v = vm sin ωt, the current is given by i = im sin (ωt + φ ) im = vm R2 + XC − XL 2 ( )where and φ = tan−1 XC − X L R 262 Z = ( )R2 + XC − X L 2 is called the impedance of the circuit. 2020-21

Alternating Current The average power loss over a complete cycle is given by P = V I cosφ The term cosφ is called the power factor. 6. In a purely inductive or capacitive circuit, cosφ = 0 and no power is dissipated even though a current is flowing in the circuit. In such cases, current is referred to as a wattless current. 7. The phase relationship between current and voltage in an ac circuit can be shown conveniently by representing voltage and current by rotating vectors called phasors. A phasor is a vector which rotates about the origin with angular speed ω. The magnitude of a phasor represents the amplitude or peak value of the quantity (voltage or current) represented by the phasor. The analysis of an ac circuit is facilitated by the use of a phasor diagram. 8. An interesting characteristic of a series RLC circuit is the phenomenon of resonance. The circuit exhibits resonance, i.e., the amplitude of the current is maximum at the resonant frequency, ω0 = 1 . The quality factor Q defined by LC Q = ω0L = 1 is an indicator of the sharpness of the resonance, R ω0CR the higher value of Q indicating sharper peak in the current. 9. A circuit containing an inductor L and a capacitor C (initially charged) with no ac source and no resistors exhibits free oscillations. The charge q of the capacitor satisfies the equation of simple harmonic motion: d2q + 1 q = 0 dt 2 LC and therefore, the frequency ω of free oscillation is ω0 = 1 . The LC energy in the system oscillates between the capacitor and the inductor but their sum or the total energy is constant in time. 10. A transformer consists of an iron core on which are bound a primary coil of Np turns and a secondary coil of Ns turns. If the primary coil is connected to an ac source, the primary and secondary voltages are related by Vs =  N s  Vp  N p  and the currents are related by Is =  Np  Ip  Ns  If the secondary coil has a greater number of turns than the primary, the voltage is stepped-up (Vs > Vp). This type of arrangement is called a step- up transformer. If the secondary coil has turns less than the primary, we have a step-down transformer. 263 2020-21

Physics Physical quantity Symbol Dimensions Unit Remarks rms voltage V [M L2 T–3 A–1] V V= vm , vm is the 2 rms current amplitude of the ac voltage. Reactance: Inductive I [ A] A I= im , im is the amplitude of Capacitive 2 Impedance the ac current. Resonant XL [M L2 T –3 A–2] XL = ω L frequency XC [M L2 T –3 A–2] XC = 1/ ω C Z [M L2 T –3 A–2] Quality factor Depends on elements present in the circuit. Power factor ωr or ω0 [T –1] Hz ω0 = 1 for a LC series RLC circuit Q Dimensionless Q = ω0 L = 1 for a series Dimensionless R ω0 C R RLC circuit. = cosφ, φ is the phase difference between voltage applied and current in the circuit. 264 POINTS TO PONDER 1. When a value is given for ac voltage or current, it is ordinarily the rms value. The voltage across the terminals of an outlet in your room is normally 240 V. This refers to the rms value of the voltage. The amplitude of this voltage is vm = 2V = 2(240) = 340 V 2. The power rating of an element used in ac circuits refers to its average power rating. 3. The power consumed in an ac circuit is never negative. 4. Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current? It cannot be derived from the mutual attraction of two parallel wires carrying ac currents, as the dc ampere is derived. An ac current changes direction 2020-21

Alternating Current with the source frequency and the attractive force would average to zero. Thus, the ac ampere must be defined in terms of some property that is independent of the direction of the current. Joule heating is such a property, and there is one ampere of rms value of alternating current in a circuit if the current produces the same average heating effect as one ampere of dc current would produce under the same conditions. 5. In an ac circuit, while adding voltages across different elements, one should take care of their phases properly. For example, if VR and VC are voltages across R and C, respectively in an RC circuit, then the total voltage across RC combination is VRC = VR2 + VC2 and not VR + VC since VC is π/2 out of phase of VR. 6. Though in a phasor diagram, voltage and current are represented by vectors, these quantities are not really vectors themselves. They are scalar quantities. It so happens that the amplitudes and phases of harmonically varying scalars combine mathematically in the same way as do the projections of rotating vectors of corresponding magnitudes and directions. The ‘rotating vectors’ that represent harmonically varying scalar quantities are introduced only to provide us with a simple way of adding these quantities using a rule that we already know as the law of vector addition. 7. There are no power losses associated with pure capacitances and pure inductances in an ac circuit. The only element that dissipates energy in an ac circuit is the resistive element. 8. In a RLC circuit, resonance phenomenon occur when XL = XC or ω0 = 1 . For resonance to occur, the presence of both L and C LC elements in the circuit is a must. With only one of these (L or C ) elements, there is no possibility of voltage cancellation and hence, no resonance is possible. 9. The power factor in a RLC circuit is a measure of how close the circuit is to expending the maximum power. 10. In generators and motors, the roles of input and output are reversed. In a motor, electric energy is the input and mechanical energy is the output. In a generator, mechanical energy is the input and electric energy is the output. Both devices simply transfor m energy from one form to another. 11. A transformer (step-up) changes a low-voltage into a high-voltage. This does not violate the law of conservation of energy. The current is reduced by the same proportion. 12. The choice of whether the description of an oscillatory motion is by means of sines or cosines or by their linear combinations is unimportant, since changing the zero-time position transforms the one to the other. 265 2020-21

Physics EXERCISES 7.1 A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. 7.2 7.3 (a) What is the rms value of current in the circuit? 7.4 (b) What is the net power consumed over a full cycle? 7.5 7.6 (a) The peak voltage of an ac supply is 300 V. What is the rms voltage? 7.7 (b) The rms value of current in an ac circuit is 10 A. What is the 7.8 7.9 peak current? A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine 7.10 the rms value of the current in the circuit. 7.11 A 60 µF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit. In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer. Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0H, C = 32 µF and R = 10 Ω. What is the Q-value of this circuit? A charged 30 µF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit? Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially ? What is the total energy at later time? A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 µF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle? A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 µH, what must be the range of its variable capacitor? [Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.] Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80µF, R = 40 Ω. 266 FIGURE 7.21 (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency. 2020-21

Alternating Current ADDITIONAL EXERCISES 7.12 An LC circuit contains a 20 mH inductor and a 50 µF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. 7.13 Let the instant the circuit is closed be t = 0. 7.14 7.15 (a) What is the total energy stored initially? Is it conserved during 7.16 LC oscillations? 7.17 7.18 (b) What is the natural frequency of the circuit? (c) At what time is the energy stored 7.19 (i) completely electrical (i.e., stored in the capacitor)? (ii) completely magnetic (i.e., stored in the inductor)? 267 (d) At what times is the total energy shared equally between the inductor and the capacitor? (e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat? A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply. (a) What is the maximum current in the coil? (b) What is the time lag between the voltage maximum and the current maximum? Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state? A 100 µF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply. (a) What is the maximum current in the circuit? (b) What is the time lag between the current maximum and the voltage maximum? Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state. Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency. A circuit containing a 80 mH inductor and a 60 µF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible. (a) Obtain the current amplitude and rms values. (b) Obtain the rms values of potential drops across each element. (c) What is the average power transferred to the inductor? (d) What is the average power transferred to the capacitor? (e) What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.] Suppose the circuit in Exercise 7.18 has a resistance of 15 Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed. 2020-21

Physics 7.20 A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply. 7.21 7.22 (a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value. 7.23 7.24 (b) What is the source frequency for which average power absorbed 7.25 by the circuit is maximum. Obtain the value of this maximum 7.26 power. 268 (c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies? (d) What is the Q-factor of the given circuit? Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 µF, and R = 7.4 Ω. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way. Answer the following questions: (a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage? (b) A capacitor is used in the primary circuit of an induction coil. (c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L. (d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line. (e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil? A power transmission line feeds input power at 2300 V to a step- down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V? At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m3s–1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms–2 ). A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town. (a) Estimate the line power loss in the form of heat. (b) How much power must the plant supply, assuming there is negligible power loss due to leakage? (c) Characterise the step up transformer at the plant. Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred? 2020-21

Chapter Eight ELECTROMAGNETIC WAVES 8.1 INTRODUCTION In Chapter 4, we learnt that an electric current produces magnetic field and that two current-carrying wires exert a magnetic force on each other. Further, in Chapter 6, we have seen that a magnetic field changing with time gives rise to an electric field. Is the converse also true? Does an electric field changing with time give rise to a magnetic field? James Clerk Maxwell (1831-1879), argued that this was indeed the case – not only an electric current but also a time-varying electric field generates magnetic field. While applying the Ampere’s circuital law to find magnetic field at a point outside a capacitor connected to a time-varying current, Maxwell noticed an inconsistency in the Ampere’s circuital law. He suggested the existence of an additional current, called by him, the displacement current to remove this inconsistency. Maxwell formulated a set of equations involving electric and magnetic fields, and their sources, the charge and current densities. These equations are known as Maxwell’s equations. Together with the Lorentz force formula (Chapter 4), they mathematically express all the basic laws of electromagnetism. The most important prediction to emerge from Maxwell’s equations is the existence of electromagnetic waves, which are (coupled) time- varying electric and magnetic fields that propagate in space. The speed of the waves, according to these equations, turned out to be very close to 2020-21

Physics James Clerk Maxwell the speed of light( 3 ×108 m/s), obtained from optical measurements. This led to the remarkable conclusion (1831 – 1879) Born in that light is an electromagnetic wave. Maxwell’s work thus unified the domain of electricity, magnetism and Edinburgh, Scotland, light. Hertz, in 1885, experimentally demonstrated the existence of electromagnetic waves. Its technological use was among the greatest by Marconi and others led in due course to the physicists of the revolution in communication that we are witnessing today. In this chapter, we first discuss the need for displacement current and its consequences. Then we present a descriptive account of electromagnetic waves. The broad spectrum of electromagnetic waves, stretching from γ rays (wavelength ~10–12 m) to long radio waves (wavelength ~106 m) is described. How the electromagnetic waves are sent and received for communication is discussed in Chapter 15. nineteenth century. He 8.2 DISPLACEMENT CURRENT derived the thermal velocity distribution of We have seen in Chapter 4 that an electrical current produces a magnetic field around it. Maxwell showed molecules in a gas and that for logical consistency, a changing electric field must also produce a magnetic field. This effect is of great was among the first to importance because it explains the existence of radio waves, gamma rays and visible light, as well as all other obtain reliable forms of electromagnetic waves. estimates of molecular To see how a changing electric field gives rise to a magnetic field, let us consider the process of parameters from charging of a capacitor and apply Ampere’s circuital law given by (Chapter 4) measurable quantities JAMES CLERK MAXWELL (1831–1879) like viscosity, etc. Maxwell’s greatest acheivement was the unification of the laws of electricity and magnetism (discovered by Coulomb, Oersted, “B.dl = µ0 i (t ) (8.1) Ampere and Faraday) into a consistent set of to find magnetic field at a point outside the capacitor. Figure 8.1(a) shows a parallel plate capacitor C which equations now called is a part of circuit through which a time-dependent current i (t ) flows . Let us find the magnetic field at a Maxwell’s equations. point such as P, in a region outside the parallel plate capacitor. For this, we consider a plane circular loop of From these he arrived at radius r whose plane is perpendicular to the direction of the current-carrying wire, and which is centred the most important symmetrically with respect to the wire [Fig. 8.1(a)]. From symmetry, the magnetic field is directed along the conclusion that light is circumference of the circular loop and is the same in magnitude at all points on the loop so that if B is the an electromagnetic magnitude of the field, the left side of Eq. (8.1) is B (2π r). So we have wave. Interestingly, Maxwell did not agree with the idea (strongly suggested by the Faraday’s laws of electrolysis) that electricity was particulate in nature. 270 B (2πr) = µ0i (t ) (8 .2) 2020-21

Electromagnetic Waves Now, consider a different surface, which has the same boundary. This is a pot like surface [Fig. 8.1(b)] which nowhere touches the current, but has its bottom between the capacitor plates; its mouth is the circular loop mentioned above. Another such surface is shaped like a tiffin box (without the lid) [Fig. 8.1(c)]. On applying Ampere’s circuital law to such surfaces with the same perimeter, we find that the left hand side of Eq. (8.1) has not changed but the right hand Fsiidg.e8i.s1z(be)roanadn(dc)n. oStoµw0ie, since no current passes through the surface of have a contradiction; calculated one way, there is a magnetic field at a point P; calculated another way, the magnetic field at P is zero. Since the contradiction arises from our use of Ampere’s circuital law, this law must be missing something. The missing term must be such that one gets the same magnetic field at point P, no matter what surface is used. We can actually guess the missing term by looking carefully at Fig. 8.1(c). Is there anything passing through the surface S between the plates of the capacitor? Yes, of course, the electric field! If the plates of the capacitor have an area A, and a total charge Q, the magnitude of the electric field E between the plates is (Q/A)/ε0 (see Eq. 2.41). The field is perpendicular to the surface S of Fig. 8.1(c). It has the same magnitude over the area A of the capacitor plates, and vanishes outside it. So what is the electric flux ΦE through the surface S ? Using Gauss’s law, it is ΦE = E A = 1 Q A = Q (8.3) ε0 A ε0 Now if the charge Q on the capacitor plates changes with time, there is a current i = (dQ/dt), so that using Eq. (8.3), we have dΦE = d  Q  = 1 dQ FIGURE 8.1 A dt dt  ε0  ε0 dt parallel plate capacitor C, as part of This implies that for consistency, a circuit through which a time ε0  dΦE  =i (8.4) dependent current dt i (t) flows, (a) a loop of This is the missing term in Ampere’s circuital law. If we generalise radius r, to determine magnetic field at a this law by adding to the total current carried by conductors through point P on the loop; the surface, another term which is ε0 times the rate of change of electric (b) a pot-shaped flux through the same surface, the total has the same value of current i surface passing for all surfaces. If this is done, there is no contradiction in the value of B through the interior between the capacitor obtained anywhere using the generalised Ampere’s law. B at the point P plates with the loop shown in (a) as its is non-zero no matter which surface is used for calculating it. B at a rim; (c) a tiffin- point P outside the plates [Fig. 8.1(a)] is the same as at a point M just shaped surface with inside, as it should be. The current carried by conductors due to flow of the circular loop as its rim and a flat charges is called conduction current. The current, given by Eq. (8.4), is a circular bottom S new term, and is due to changing electric field (or electric displacement, between the capacitor an old term still used sometimes). It is, therefore, called displacement plates. The arrows show uniform electric current or Maxwell’s displacement current. Figure 8.2 shows the electric field between the and magnetic fields inside the parallel plate capacitor discussed above. capacitor plates. The generalisation made by Maxwell then is the following. The source 271 of a magnetic field is not just the conduction electric current due to flowing 2020-21

Physics charges, but also the time rate of change of electric field. More precisely, the total current i is the sum of the conduction current ddet)n).oSteodwbeyhica, avend the displacement current denoted by id (= ε0 (dΦE/ i =ic + id =ic + ε0 dΦE (8.5) dt In explicit terms, this means that outside the capacitor plates, wncuoercrhoeannvtd,eui.coetn.i,oliynd =ccou0n.rrOdeunncttt,hiioe.neo.t,chiuce=rrrh0ean,natdni,cdin=thsiie,dreeatnihsdeocnnaolpyaddciisistppollraa, cctheemmereeennistt currTehnetg, esnoetrhaalitseidd=(ai.nd correct) Ampere’s circuital law has the same form as Eq. (8.1), with one difference: “the total current passing through any surface of which the closed loop is the perimeter” is the sum of the conduction current and the displacement current. The generalised law is ∫ Bidl = µ0 ic + µ0 ε0 dΦE (8.6) dt and is known as Ampere-Maxwell law. In all respects, the displacement current has the same physical effects as the conduction current. In some cases, for example, steady electric fields in a conducting wire, the displacement current may FIGURE 8.2 (a) The be zero since the electric field E does not change with time. In other electric and magnetic cases, for example, the charging capacitor above, both conduction fields E and B between and displacement currents may be present in different regions of the capacitor plates, at space. In most of the cases, they both may be present in the same the point M. (b) A cross region of space, as there exist no perfectly conducting or perfectly sectional view of Fig. (a). insulating medium. Most interestingly, there may be large regions of space where there is no conduction current, but there is only a displacement current due to time-varying electric fields. In such a region, we expect a magnetic field, though there is no (conduction) current source nearby! The prediction of such a displacement current can be verified experimentally. For example, a magnetic field (say at point M) between the plates of the capacitor in Fig. 8.2(a) can be measured and is seen to be the same as that just outside (at P). The displacement current has (literally) far reaching consequences. One thing we immediately notice is that the laws of electricity and magnetism are now more symmetrical*. Faraday’s law of induction states that there is an induced emf equal to the rate of change of magnetic flux. Now, since the emf between two points 1 and 2 is the work done per unit charge in taking it from 1 to 2, the existence of an emf implies the existence of an electric field. So, we can rephrase Faraday’s law of electromagnetic induction by saying that a magnetic field, changing with time, gives rise to an electric field. Then, the fact that an electric field changing with time gives rise to a magnetic field, is the symmetrical counterpart, and is 272 * They are still not perfectly symmetrical; there are no known sources of magnetic field (magnetic monopoles) analogous to electric charges which are sources of electric field. 2020-21

Electromagnetic Waves a consequence of the displacement current being a source of a magnetic field. Thus, time- dependent electric and magnetic fields give rise to each other! Faraday’s law of electromagnetic induction and Ampere-Maxwell law give a quantitative expression of this statement, with the current being the total current, as in Eq. (8.5). One very important consequence of this symmetry is the existence of electromagnetic waves, which we discuss qualitatively in the next section. MAXWELL’S EQUATIONS IN VACUUM ∫1. E dA =Q /ε0 (Gauss’s Law for electricity) (Gauss’s Law for magnetism) 2. ∫B dA =0 (Faraday’s Law) (Ampere – Maxwell Law) ∫3. E dl = –dΦB dt ∫4. B dl = µ0 ic + µ0 ε0 dΦE dt Example 8.1 A parallel plate capacitor with circular plates of radius 1 m has a capacitance of 1 nF. At t = 0, it is connected for charging in series with a resistor R = 1 M Ω across a 2V battery (Fig. 8.3). Calculate the magnetic field at a point P, halfway between the centre and the periphery of the plates, after t = 10–3 s. (The charge on the capacitor at time t is q (t) = CV [1 – exp (–t/τ )], where the time constant τ is equal to CR.) FIGURE 8.3 Solution The time constant of the CR circuit is τ = CR = 10–3 s. Then, EXAMPLE 8.1 we have q(t) = CV [1 – exp (–t/τ)] = 2 × 10–9 [1– exp (–t/10–3)] The electric field in between the plates at time t is E = q (t ) = q ; A = π (1)2 m2 = area of the plates. πε0 ε0A Consider now a circular loop of radius (1/2) m parallel to the plates 273 passing through P. The magnetic field B at all points on the loop is 2020-21

Physics along the loop and of the same value. The flux ΦE through this loop is ΦE = E × area of the loop = E × π ×  12 2 = πE = q 4 4ε0 The displacement current id = ε0 dΦE = 1 dq = 0.5 × 10–6 exp (–1) dt 4 dt EXAMPLE 8.1 at t = 10–3s. Now, applying Ampere-Maxwell law to the loop, we get B × 2π ×  12 = µ0 (ic + id ) = µ0 (0 + id ) = 0.5×10–6 µ0exp(–1) or, B = 0.74 × 10–13 T 274 8.3 ELECTROMAGNETIC WAVES 8.3.1 Sources of electromagnetic waves How are electromagnetic waves produced? Neither stationary charges nor charges in uniform motion (steady currents) can be sources of electromagnetic waves. The former produces only electrostatic fields, while the latter produces magnetic fields that, however, do not vary with time. It is an important result of Maxwell’s theory that accelerated charges radiate electromagnetic waves. The proof of this basic result is beyond the scope of this book, but we can accept it on the basis of rough, qualitative reasoning. Consider a charge oscillating with some frequency. (An oscillating charge is an example of accelerating charge.) This produces an oscillating electric field in space, which produces an oscillating magnetic field, which in turn, is a source of oscillating electric field, and so on. The oscillating electric and magnetic fields thus regenerate each other, so to speak, as the wave propagates through the space. The frequency of the electromagnetic wave naturally equals the frequency of oscillation of the charge. The energy associated with the propagating wave comes at the expense of the energy of the source – the accelerated charge. From the preceding discussion, it might appear easy to test the prediction that light is an electromagnetic wave. We might think that all we needed to do was to set up an ac circuit in which the current oscillate at the frequency of visible light, say, yellow light. But, alas, that is not possible. The frequency of yellow light is about 6 × 1014 Hz, while the frequency that we get even with modern electronic circuits is hardly about 1011 Hz. This is why the experimental demonstration of electromagnetic wave had to come in the low frequency region (the radio wave region), as in the Hertz’s experiment (1887). Hertz’s successful experimental test of Maxwell’s theory created a sensation and sparked off other important works in this field. Two important achievements in this connection deserve mention. Seven years after Hertz, Jagdish Chandra Bose, working at Calcutta (now Kolkata), 2020-21

Electromagnetic Waves succeeded in producing and observing electromagnetic waves of much shorter wavelength (25 mm to 5 mm). His experiment, like that of Hertz’s, was confined to the laboratory. At around the same time, Guglielmo Marconi in Italy followed Hertz’s work and succeeded in transmitting electromagnetic waves over distances of many kilometres. Marconi’s experiment marks the beginning of the field of communication using electromagnetic waves. 8.3.2 Nature of electromagnetic waves It can be shown from Maxwell’s equations that electric Heinrich Rudolf Hertz HEINRICH RUDOLF HERTZ (1857–1894) and magnetic fields in an electromagnetic wave are perpendicular to each other, and to the direction of (1857 – 1894) German propagation. It appears reasonable, say from our discussion of the displacement current. Consider physicist who was the Fig. 8.2. The electric field inside the plates of the capacitor is directed perpendicular to the plates. The magnetic first to broadcast and field this gives rise to via the displacement current is along the perimeter of a circle parallel to the capacitor receive radio waves. He plates. So B and E are perpendicular in this case. This is a general feature. produced electro- In Fig. 8.4, we show a typical example of a plane magnetic waves, sent electromagnetic wave propagating along the z direction (the fields are shown as a function of the z coordinate, them through space, and at a given time t). The electric field Ex is along the x-axis, and varies sinusoidally with z, at a given time. The measured their wave- magnetic field By is along the y-axis, and again varies sinusoidally with z. The electric and magnetic fields Ex length and speed. He and By are perpendicular to each other, and to the direction z of propagation. We can write Ex and By as showed that the nature follows: of their vibration, reflection and refraction was the same as that of light and heat waves, establishing their identity for the first time. He also pioneered research on discharge of electricity through gases, Ex= E0 sin (kz–ωt ) [8.7(a)] and discovered the photoelectric effect. HerBe yk=isB0resliante(kdzt–oωtth) e wave length λ of the [8.7(b)] wave by the usual equation k = 2π (8.8) λ and ω is the angular frequency. k is the magnitude of the wave vector (or propagation vector) k and its direction describes the direction of propagation of the FIGURE 8.4 A linearly polarised electromagnetic wave, propagating in the z-direction with the oscillating electric field E wave. The speed of propagation along the x-direction and the oscillating magnetic field B along of the wave is (ω/k ). Using the y-direction. Eqs. [8.7(a) and (ebq)]ufaotrioEnx as,ndonBey and Maxwell’s 275 finds that 2020-21

Physics ω = ck, where, c = 1/ µ0ε0 [8.9(a)] The relation ω = ck is the standard one for waves (see for example, Section 15.4 of class XI Physics textbook). This relation is often written in terms of frequency, ν (=ω/2π) and wavelength, λ (=2π/k) as 2πν = c  2π  or λ νλ = c [8.9(b)] It is also seen from Maxwell’s equations that the magnitude of the Simulate propagation of electromagnetic waves electric and the magnetic fields in an electromagnetic wave are related as (i) http://www.amanogawa.com/waves.html (ii) http://www.phys.hawaii.edu/~teb/java/ntnujava/emWave/emWave.html B0 = (E0/c) (8.10) We here make remarks on some features of electromagnetic waves. They are self-sustaining oscillations of electric and magnetic fields in free space, or vacuum. They differ from all the other waves we have studied so far, in respect that no material medium is involved in the vibrations of the electric and magnetic fields. Sound waves in air are longitudinal waves of compression and rarefaction. Transverse elastic (sound) waves can also propagate in a solid, which is rigid and that resists shear. Scientists in the nineteenth century were so much used to this mechanical picture that they thought that there must be some medium pervading all space and all matter, which responds to electric and magnetic fields just as any elastic medium does. They called this medium ether. They were so convinced of the reality of this medium, that there is even a novel called The Poison Belt by Sir Arthur Conan Doyle (the creator of the famous detective Sherlock Holmes) where the solar system is supposed to pass through a poisonous region of ether! We now accept that no such physical medium is needed. The famous experiment of Michelson and Morley in 1887 demolished conclusively the hypothesis of ether. Electric and magnetic fields, oscillating in space and time, can sustain each other in vacuum. But what if a material medium is actually there? We know that light, an electromagnetic wave, does propagate through glass, for example. We have seen earlier that the total electric and magnetic fields inside a medium are described in terms of a permittivity ε and a magnetic permeability µ (these describe the factors by which the total fields differ from the external fields). These replace ε0 and µ0 in the description to electric and magnetic fields in Maxwell’s equations with the result that in a material medium of permittivity ε and magnetic permeability µ, the velocity of light becomes, v= 1 (8.11) µε Thus, the velocity of light depends on electric and magnetic properties of the medium. We shall see in the next chapter that the refractive index of one medium with respect to the other is equal to the ratio of velocities of light in the two media. The velocity of electromagnetic waves in free space or vacuum is an important fundamental constant. It has been shown by experiments on 276 electromagnetic waves of different wavelengths that this velocity is the 2020-21

Electromagnetic Waves same (independent of wavelength) to within a few metres per second, out of a value of 3×108 m/s. The constancy of the velocity of em waves in vacuum is so strongly supported by experiments and the actual value is so well known now that this is used to define a standard of length. Namely, the metre is now defined as the distance travelled by light in vacuum in a time (1/c) seconds = (2.99792458 × 108)–1 seconds. This has come about for the following reason. The basic unit of time can be defined very accurately in terms of some atomic frequency, i.e., frequency of light emitted by an atom in a particular process. The basic unit of length is harder to define as accurately in a direct way. Earlier measurement of c using earlier units of length (metre rods, etc.) converged to a value of about 2.9979246 × 108 m/s. Since c is such a strongly fixed number, unit of length can be defined in terms of c and the unit of time! Hertz not only showed the existence of electromagnetic waves, but also demonstrated that the waves, which had wavelength ten million times that of the light waves, could be diffracted, refracted and polarised. Thus, he conclusively established the wave nature of the radiation. Further, he produced stationary electromagnetic waves and determined their wavelength by measuring the distance between two successive nodes. Since the frequency of the wave was known (being equal to the frequency of the oscillator), he obtained the speed of the wave using the formula v = νλ and found that the waves travelled with the same speed as the speed of light. The fact that electromagnetic waves are polarised can be easily seen in the response of a portable AM radio to a broadcasting station. If an AM radio has a telescopic antenna, it responds to the electric part of the signal. When the antenna is turned horizontal, the signal will be greatly diminished. Some portable radios have horizontal antenna (usually inside the case of radio), which are sensitive to the magnetic component of the electromagnetic wave. Such a radio must remain horizontal in order to receive the signal. In such cases, response also depends on the orientation of the radio with respect to the station. Do electromagnetic waves carry energy and momentum like other waves? Yes, they do. We have seen in chapter 2 that in a region of free Celheacptrtiecrfi6e,ldasEs,otchiearteediswainthenaemrgaygdneentiscitfyie(lεd0EB2/is2)a. space with Similarly, as seen in magnetic aennedrmgyadgennestiictyfi(eBl2d/s2, µt0h)e. rAesiselaecntroonm-zaegrnoeetnicerwgayvdeecnosnittyaiansssboocitahteedlecwtirtihc it. Now consider a plane perpendicular to the direction of propagation of the electromagnetic wave (Fig. 8.4). If there are, on this plane, electric charges, they will be set and sustained in motion by the electric and magnetic fields of the electromagnetic wave. The charges thus acquire energy and momentum from the waves. This just illustrates the fact that an electromagnetic wave (like other waves) carries energy and momentum. Since it carries momentum, an electromagnetic wave also exerts pressure, called radiation pressure. If the total energy transferred to a surface in time t is U, it can be shown that the magnitude of the total momentum delivered to this surface (for complete absorption) is, p=U (8.12) 277 c 2020-21

Physics When the sun shines on your hand, you feel the energy being absorbed from the electromagnetic waves (your hands get warm). Electromagnetic waves also transfer momentum to your hand but because c is very large, the amount of momentum transferred is extremely small and you do not feel the pressure. In 1903, the American scientists Nicols and Hull succeeded in measuring radiation pressure of visible light and verified Eq. (8.12). It was found to be of the order of 7 × 10–6 N/m2. Thus, on a surface of area 10 cm2, the force due to radiation is only about 7 × 10–9 N. The great technological importance of electromagnetic waves stems from their capability to carry energy from one place to another. The radio and TV signals from broadcasting stations carry energy. Light carries energy from the sun to the earth, thus making life possible on the earth. Example 8.2 A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, E = 6.3 ˆj V/m. What is B at this point? Solution Using Eq. (8.10), the magnitude of B is B=E c = 6.3 V/m = 2.1 × 10 –8 T 3 ×108 m/s EXAMPLE 8.2 To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction bpTeehruaplseo,nndgicxBu-ld=airr2et.co1tibo×no1t.h0S–ix8n-ckˆaen,Td(+yˆj-)a×xe(+s.kˆU) s=inˆig,vBecitsoralaolgnegbrtah,eEz-×dBireschtoiuonld. Example 8.3 The magnetic field in a plane electromagnetic wave is given by By = (2 × 10–7) T sin (0.5×103x+1.5×1011t). (a) What is the wavelength and frequency of the wave? (b) Write an expression for the electric field. Solution (a) Comparing the given equation with By = B0 sin 2π  x + t   λ T  We get, λ = 2π m = 1.26 cm, 0.5 × 103 278 EXAMPLE 8.3 ( )and 1 = ν = 1.5 ×1011 /2π = 23.9 GHz T (b) E0 = B0c = 2×10–7 T × 3 × 108 m/s = 6 × 101 V/m The electric field component is perpendicular to the direction of propagation and the direction of magnetic field. Therefore, the electric field component along the z-axis is obtained as Ez = 60 sin (0.5 × 103x + 1.5 × 1011 t) V/m 2020-21

Electromagnetic Waves Example 8.4 Light with an energy flux of 18 W/cm2 falls on a non- reflecting surface at normal incidence. If the surface has an area of 20 cm2, find the average force exerted on the surface during a 30 minute time span. Solution The total energy falling on the surface is U = (18 W/cm2) × (20 cm2) × (30 × 60 s) = 6.48 × 105 J Therefore, the total momentum delivered (for complete absorption) is p = U = 6.48 × 105 J = 2.16 × 10–3 kg m/s c 3 × 108m/s EXAMPLE 8.4 The average force exerted on the surface is F= p = 2.16 × 10−3 = 1.2 × 10−6 N t 0.18 × 104 How will your result be modified if the surface is a perfect reflector? Example 8.5 Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m. Assume that the efficiency of the bulb is 2.5% and it is a point source. Solution The bulb, as a point source, radiates light in all directions uniformly. At a distance of 3 m, the surface area of the surrounding sphere is A = 4 πr 2 = 4π (3)2 = 113 m2 The intensity I at this distance is I = Power = 100 W × 2.5 % Area 113 m2 = 0.022 W/m2 Half of this intensity is provided by the electric field and half by the magnetic field. ( )1 1 2 2I = ε 0 Er2msc ( )=1 2 0.022 W/m2 0.022 8.85 × 10−12 3 × 108 ( )( )Erms = V/m = 2.9 V/m EXAMPLE 8.5 279 The value of E found above is the root mean square value of the electric field. Since the electric field in a light beam is sinusoidal, the peak electric field, E0 is E0 = 2Erms = 2 × 2.9 V/m = 4.07 V/m Thus, you see that the electric field strength of the light that you use for reading is fairly large. Compare it with electric field strength of TV or FM waves, which is of the order of a few microvolts per metre. 2020-21

Physics Now, let us calculate the strength of the magnetic field. It is EXAMPLE 8.5 Brms = Erms = 2.9 V m−1 = 9.6 × 10–9 T c 3 × 108 m s−1 Again, since the field in the light beam is sinusoidal, the peak magnetic field is B0 = 2 Brms = 1.4 × 10–8 T. Note that although the energy in the magnetic field is equal to the energy in the electric field, the magnetic field strength is evidently very weak. Electromagnetic spectrum 8.4 ELECTROMAGNETIC SPECTRUM http://www.fnal.gov/pub/inquiring/more/light http://imagine.gsfc.nasa.gov/docs/science/ At the time Maxwell predicted the existence of electromagnetic waves, the only familiar electromagnetic waves were the visible light waves. The existence of ultraviolet and infrared waves was barely established. By the end of the nineteenth century, X-rays and gamma rays had also been discovered. We now know that, electromagnetic waves include visible light waves, X-rays, gamma rays, radio waves, microwaves, ultraviolet and infrared waves. The classification of em waves according to frequency is the electromagnetic spectrum (Fig. 8.5). There is no sharp division between one kind of wave and the next. The classification is based roughly on how the waves are produced and/or detected. 280 FIGURE 8.5 The electromagnetic spectrum, with common names for various part of it. The various regions do not have sharply defined boundaries. 2020-21

Electromagnetic Waves We briefly describe these different types of electromagnetic waves, in order of decreasing wavelengths. 8.4.1 Radio waves Radio waves are produced by the accelerated motion of charges in conducting wires. They are used in radio and television communication systems. They are generally in the frequency range from 500 kHz to about 1000 MHz. The AM (amplitude modulated) band is from 530 kHz to 1710 kHz. Higher frequencies upto 54 MHz are used for short wave bands. TV waves range from 54 MHz to 890 MHz. The FM (frequency modulated) radio band extends from 88 MHz to 108 MHz. Cellular phones use radio waves to transmit voice communication in the ultrahigh frequency (UHF) band. How these waves are transmitted and received is described in Chapter 15. 8.4.2 Microwaves Microwaves (short-wavelength radio waves), with frequencies in the gigahertz (GHz) range, are produced by special vacuum tubes (called klystrons, magnetrons and Gunn diodes). Due to their short wavelengths, they are suitable for the radar systems used in aircraft navigation. Radar also provides the basis for the speed guns used to time fast balls, tennis- serves, and automobiles. Microwave ovens are an interesting domestic application of these waves. In such ovens, the frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the waves is transferred efficiently to the kinetic energy of the molecules. This raises the temperature of any food containing water. MICROWAVE OVEN The spectrum of electromagnetic radiation contains a part known as microwaves. These waves have frequency and energy smaller than visible light and wavelength larger than it. What is the principle of a microwave oven and how does it work? Our objective is to cook food or warm it up. All food items such as fruit, vegetables, meat, cereals, etc., contain water as a constituent. Now, what does it mean when we say that a certain object has become warmer? When the temperature of a body rises, the energy of the random motion of atoms and molecules increases and the molecules travel or vibrate or rotate with higher energies. The frequency of rotation of water molecules is about 2.45 gigahertz (GHz). If water receives microwaves of this frequency, its molecules absorb this radiation, which is equivalent to heating up water. These molecules share this energy with neighbouring food molecules, heating up the food. One should use porcelain vessels and not metal containers in a microwave oven because of the danger of getting a shock from accumulated electric charges. Metals may also melt from heating. The porcelain container remains unaffected and cool, because its large molecules vibrate and rotate with much smaller frequencies, and thus cannot absorb microwaves. Hence, they do not get heated up. Thus, the basic principle of a microwave oven is to generate microwave radiation of appropriate frequency in the working space of the oven where we keep food. This way energy is not wasted in heating up the vessel. In the conventional heating method, the vessel on the burner gets heated first, and then the food inside gets heated because of transfer of energy from the vessel. In the microwave oven, on the other hand, energy is directly delivered to water molecules which is shared by the entire food. 281 2020-21

Physics 8.4.3 Infrared waves Infrared waves are produced by hot bodies and molecules. This band lies adjacent to the low-frequency or long-wave length end of the visible spectrum. Infrared waves are sometimes referred to as heat waves. This is because water molecules present in most materials readily absorb infrared waves (many other molecules, for example, CO2, NH3, also absorb infrared waves). After absorption, their thermal motion increases, that is, they heat up and heat their surroundings. Infrared lamps are used in physical therapy. Infrared radiation also plays an important role in maintaining the earth’s warmth or average temperature through the greenhouse effect. Incoming visible light (which passes relatively easily through the atmosphere) is absorbed by the earth’s surface and re- radiated as infrared (longer wavelength) radiations. This radiation is trapped by greenhouse gases such as carbon dioxide and water vapour. Infrared detectors are used in Earth satellites, both for military purposes and to observe growth of crops. Electronic devices (for example semiconductor light emitting diodes) also emit infrared and are widely used in the remote switches of household electronic systems such as TV sets, video recorders and hi-fi systems. 8.4.4 Visible rays It is the most familiar form of electromagnetic waves. It is the part of the spectrum that is detected by the human eye. It runs from about 4 × 1014 Hz to about 7 × 1014 Hz or a wavelength range of about 700 – 400 nm. Visible light emitted or reflected from objects around us provides us information about the world. Our eyes are sensitive to this range of wavelengths. Different animals are sensitive to different range of wavelengths. For example, snakes can detect infrared waves, and the ‘visible’ range of many insects extends well into the utraviolet. 8.4.5 Ultraviolet rays It covers wavelengths ranging from about 4 × 10–7 m (400 nm) down to 6 × 10–10m (0.6 nm). Ultraviolet (UV) radiation is produced by special lamps and very hot bodies. The sun is an important source of ultraviolet light. But fortunately, most of it is absorbed in the ozone layer in the atmosphere at an altitude of about 40 – 50 km. UV light in large quantities has harmful effects on humans. Exposure to UV radiation induces the production of more melanin, causing tanning of the skin. UV radiation is absorbed by ordinary glass. Hence, one cannot get tanned or sunburn through glass windows. Welders wear special glass goggles or face masks with glass windows to protect their eyes from large amount of UV produced by welding arcs. Due to its shorter wavelengths, UV radiations can be focussed into very narrow beams for high precision applications such as LASIK (Laser- assisted in situ keratomileusis) eye surgery. UV lamps are used to kill germs in water purifiers. Ozone layer in the atmosphere plays a protective role, and hence its 282 depletion by chlorofluorocarbons (CFCs) gas (such as freon) is a matter of international concern. 2020-21

Electromagnetic Waves 8.4.6 X-rays Beyond the UV region of the electromagnetic spectrum lies the X-ray region. We are familiar with X-rays because of its medical applications. It covers wavelengths from about 10–8 m (10 nm) down to 10–13 m (10–4 nm). One common way to generate X-rays is to bombard a metal target by high energy electrons. X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer. Because X-rays damage or destroy living tissues and organisms, care must be taken to avoid unnecessary or over exposure. 8.4.7 Gamma rays They lie in the upper frequency range of the electromagnetic spectrum and have wavelengths of from about 10–10m to less than 10–14m. This high frequency radiation is produced in nuclear reactions and also emitted by radioactive nuclei. They are used in medicine to destroy cancer cells. Table 8.1 summarises different types of electromagnetic waves, their production and detections. As mentioned earlier, the demarcation between different regions is not sharp and there are overlaps. TABLE 8.1 DIFFERENT TYPES OF ELECTROMAGNETIC WAVES Type Wavelength range Production Detection Radio Receiver’s aerials Microwave > 0.1 m Rapid acceleration and Infra-red decelerations of electrons Point contact diodes Light in aerials Thermopiles Ultraviolet 0.1m to 1 mm Klystron valve or Bolometer, Infrared X-rays magnetron valve photographic film Gamma rays The eye 1mm to 700 nm Vibration of atoms Photocells and molecules Photographic film 700 nm to 400 nm Electrons in atoms emit Photocells 400 nm to 1nm light when they move from Photographic film 1nm to 10–3 nm one energy level to a lower energy level Photographic film Geiger tubes Inner shell electrons in Ionisation chamber atoms moving from one -do- energy level to a lower level 283 X-ray tubes or inner shell electrons <10–3 nm Radioactive decay of the nucleus 2020-21

Physics SUMMARY 1. Maxwell found an inconsistency in the Ampere’s law and suggested the existence of an additional current, called displacement current, to remove this inconsistency. This displacement current is due to time-varying electric field and is given by id = ε0 dΦΕ dt and acts as a source of magnetic field in exactly the same way as conduction current. 2. An accelerating charge produces electromagnetic waves. An electric charge oscillating harmonically with frequency ν, produces electromagnetic waves of the same frequency ν. An electric dipole is a basic source of electromagnetic waves. 3. Electromagnetic waves with wavelength of the order of a few metres were first produced and detected in the laboratory by Hertz in 1887. He thus verified a basic prediction of Maxwell’s equations. 4. Electric and magnetic fields oscillate sinusoidally in space and time in an electromagnetic wave. The oscillating electric and magnetic fields, E and B are perpendicular to each other, and to the direction of propagation of the electromagnetic wave. For a wave of frequency ν, wavelength λ, propagating along z-direction, we have E = Ex (t) = E0 sin (kz – ω t ) = E0 sin 2π  z − νt   = E0 sin 2π  z − t   λ  λ T  B = By(t) = B0 sin (kz – ω t) = B0 sin 2π  z − νt   = B0 sin 2π  z − t   λ  λ T  They are related by E0/B0 = c. 5. The speed c of electromagnetic wave in vacuum is related to µ0 and ε0 (the free space permeability and permittivity constants) as follows: c = 1/ µ0 ε0 . The value of c equals the speed of light obtained from optical measurements. Light is an electromagnetic wave; c is, therefore, also the speed of light. Electromagnetic waves other than light also have the same velocity c in free space. The speed of light, or of electromagnetic waves in a material medium is given by v = 1/ µ ε 284 where µ is the permeability of the medium and ε its permittivity. 6. Electromagnetic waves carry energy as they travel through space and this energy is shared equally by the electric and magnetic fields. Electromagnetic waves transport momentum as well. When these waves strike a surface, a pressure is exerted on the surface. If total energy transferred to a surface in time t is U, total momentum delivered to this surface is p = U/c. 7. The spectrum of electromagnetic waves stretches, in principle, over an infinite range of wavelengths. Different regions are known by different 2020-21