3.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

3.10 CELLS, EMF, INTERNAL RESISTANCE We have already mentioned that a simple device to maintain a steady current in an electric circuit is the electrolytic cell. Basically a cell has two electrodes, called the positive (P) and the negative (N), as shown in 93 Reprint 2025-26 Physics Fig. 3.12. They are immersed in an electrolytic solution. Dipped in the solution, the electrodes exchange charges with the electrolyte. The positive electrode has a potential difference V+ (V+ > 0) between itself and the electrolyte solution immediately adjacent to it marked A in the figure. Similarly, the negative electrode develops a negative potential – (V– ) (V– ≥ 0) relative to the electrolyte adjacent to it, marked as B in the figure. When there is no current, the electrolyte has the same potential throughout, so that the potential difference between P and N is V+ – (–V–) = V+ + V– . This difference is called the electromotive force (emf) of the cell and is denoted by e. Thus e = V++V– > 0 (3.36) Note that e is, actually, a potential difference and not a force. The name emf, however, is used because of historical reasons, and was given at a time when the phenomenon was not understood properly. To understand the significance of e, consider a resistor R connected across the cell (Fig. 3.12). A current I flows across R FIGURE 3.12 (a) Sketch of from C to D. As explained before, a steady current is maintained an electrolyte cell with because current flows from N to P through the electrolyte. Clearly, positive terminal P and across the electrolyte the same current flows through the electrolyte negative terminal N. The but from N to P, whereas through R, it flows from P to N. gap between the electrodes The electrolyte through which a current flows has a finite is exaggerated for clarity. A resistance r, called the internal resistance. Consider first the and B are points in the situation when R is infinite so that I = V/R = 0, where V is the electrolyte typically close to potential difference between P and N. Now, P and N. (b) the symbol for V = Potential difference between P and A a cell, + referring to P and – referring to the N + Potential difference between A and B electrode. Electrical + Potential difference between B and N connections to the cell are = e (3.37) made at P and N. Thus, emf e is the potential difference between the positive and negative electrodes in an open circuit, i.e., when no current is flowing through the cell. If however R is finite, I is not zero. In that case the potential difference between P and N is V = V++ V– – I r = e – I r (3.38) Note the negative sign in the expression (I r) for the potential difference between A and B. This is because the current I flows from B to A in the electrolyte. In practical calculations, internal resistances of cells in the circuit may be neglected when the current I is such that e >> I r. The actual values of the internal resistances of cells vary from cell to cell. The internal resistance of dry cells, however, is much higher than the common electrolytic cells. We also observe that since V is the potential difference across R, we have from Ohm’s law V = I R (3.39) 94 Combining Eqs. (3.38) and (3.39), we get Reprint 2025-26 Current Electricity I R = e – I r ε Or, I = (3.40) R + r The maximum current that can be drawn from a cell is for R = 0 and it is Imax = e/r. However, in most cells the maximum allowed current is much lower than this to prevent permanent damage to the cell. 3.113.113.113.113.11 CELLSELLSELLSELLSELLS INININININ SSSERIESSSERIESERIESERIESERIES ANDANDANDANDAND INININININ PPPARALLELPPARALLELARALLELARALLELARALLEL Like resistors, cells can be combined together in an electric circuit. And like resistors, one can, for calculating currents and voltages in a circuit, replace a combination of cells by an equivalent cell. FIGUREFIGUREFIGUREFIGUREFIGURE 3.133.133.133.133.13 Two cells of emf’s e1 and e2 in the series. r1, r2 are their internal resistances. For connections across A and C, the combination can be considered as one cell of emf eeq and an internal resistance req. Consider first two cells in series (Fig. 3.13), where one terminal of the two cells is joined together leaving the other terminal in either cell free. e1, e2 are the emf’s of the two cells and r1, r2 their internal resistances, respectively. Let V (A), V (B), V (C) be the potentials at points A, B and C shown in Fig. 3.13. Then V (A) – V (B) is the potential difference between the positive and negative terminals of the first cell. We have already calculated it in Eq. (3.38) and hence, (3.41) V AB ≡ V ( A ) – V ( B ) = ε1 – I r1 Similarly, V BC ≡ V ( B ) – V ( C ) = ε2 – I r2 (3.42) Hence, the potential difference between the terminals A and C of the combination is V AC ≡ V (A)– V (C) = V ( A ) – V ( B ) + V ( B ) – V ( C ) = (ε1 + ε2 ) – I (r1 + r2 ) (3.43) If we wish to replace the combination by a single cell between A and C of emf eeq and internal resistance req, we would have VAC = eeq– I req (3.44) Comparing the last two equations, we get eeq = e1 + e2 (3.45) and req = r1 + r2 (3.46) In Fig.3.13, we had connected the negative electrode of the first to the 95 positive electrode of the second. If instead we connect the two negatives, Reprint 2025-26 Physics Eq. (3.42) would change to VBC = –e2–Ir2 and we will get eeq = e1 – e2 (e1 > e2) (3.47) The rule for series combination clearly can be extended to any number of cells: (i) The equivalent emf of a series combination of n cells is just the sum of their individual emf’s, and (ii) The equivalent internal resistance of a series combination of n cells is just the sum of their internal resistances. This is so, when the current leaves each cell from the positive electrode. If in the combination, the current leaves any cell from the negative electrode, the emf of the cell enters the expression for eeq with a negative sign, as in Eq. (3.47). Next, consider a parallel combination of the cells (Fig. 3.14). I1 and I2 are the currents leaving the positive electrodes of the cells. At the point B1, I1 and I2 flow in whereas the current I flows out. Since as much charge flows in as out, we have FIGURE 3.14 Two cells in I = I1 + I2 (3.48) parallel. For connections Let V (B1) and V (B2) be the potentials at B1 and B2, respectively. across A and C, the Then, considering the first cell, the potential difference across its combination can be terminals is V (B1) – V (B2). Hence, from Eq. (3.38) replaced by one cell of emf eeq and internal resistances V ≡ V ( B1 ) – V ( B 2 ) = ε1 – I 1r1 (3.49) req whose values are given in Points B1 and B2 are connected exactly similarly to the second Eqs. (3.54) and (3.55). cell. Hence considering the second cell, we also have V ≡ V ( B1 ) – V ( B 2 ) = ε2 – I 2r2 (3.50) Combining the last three equations I = I 1 + I 2 ε 1 – V ε2 – V  ε1 ε2   1 1  = + = + – V + (3.51) r1 r2  r1 r2   r1 r2  Hence, V is given by, ε1r2 + ε2r1 r1r2 V = – I (3.52) r1 + r2 r1 + r2 If we want to replace the combination by a single cell, between B1 and B2, of emf eeq and internal resistance req, we would have V = eeq – I req (3.53) The last two equations should be the same and hence ε1r2 + ε2r1 εeq = (3.54) r1 + r2 r1r2 req = (3.55) r1 + r2 We can put these equations in a simpler way, 96 Reprint 2025-26 Current Electricity 1 1 1 = + (3.56) req r1 r2 εeq ε1 ε2 = + (3.57) req r1 r2 In Fig. (3.14), we had joined the positive terminals together and similarly the two negative ones, so that the currents I1, I2 flow out of positive terminals. If the negative GUSTAVterminal of the second is connected to positive terminal of the first, Eqs. (3.56) and (3.57) would still be valid with e ® 2 –e2 Equations (3.56) and (3.57) can be extended easily. If there are n cells of emf e1, . . . en and of internal ROBERT resistances r1,... rn respectively, connected in parallel, the Gustav Robert Kirchhoff combination is equivalent to a single cell of emf eeq and (1824 – 1887) German internal resistance req, such that physicist, professor at Heidelberg and at 1 1 1 = + ... + Berlin. Mainly known for (3.58) req r1 rn his development of spectroscopy, he also KIRCHHOFF ε eq ε1 εn made many important = + ... + (3.59) contributions to mathe- req r1 rn matical physics, among them, his first and (1824 –3.12 KIRCHHOFF’S RULES second rules for circuits. Electric circuits generally consist of a number of resistors and cells interconnected sometimes in a complicated way. The formulae we have derived earlier for series and parallel combinations of resistors are not always sufficient to determine all the currents and potential differences in the circuit. Two rules, called Kirchhoff’s rules, are very useful for analysis of electric circuits. Given a circuit, we start by labelling currents in each resistor by a symbol, say I, and a directed arrow to indicate that a current I flows along the resistor in the direction indicated. If ultimately I is determined to be positive, the actual current in the resistor is in the direction of the arrow. If I turns out to be negative, the current actually flows in a direction opposite to the arrow. Similarly, for each source (i.e., cell or some other source of electrical power) the positive and negative electrodes are labelled, as well as, a directed arrow with a symbol for the current flowing through the cell. This will tell us the potential difference, V = V (P) – V (N) = e – I r [Eq. (3.38) between the positive terminal P and the negative terminal N; I here is the current flowing from N to P through the cell]. If, while labelling the current I through the cell one goes from P to N, then of course V = e + I r (3.60) Having clarified labelling, we now state the rules and the proof: (a) Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction (Fig. 3.15). 97 Reprint 2025-26 Physics This applies equally well if instead of a junction of several lines, we consider a point in a line. The proof of this rule follows from the fact that when currents are steady, there is no accumulation of charges at any junction or at any point in a line. Thus, the total current flowing in, (which is the rate at which charge flows into the junction), must equal the total current flowing out. (b) Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero FIGURE 3.15 At junction a the current (Fig. 3.15). leaving is I1 + I2 and current entering is I3. This rule is also obvious, since electric potential is The junction rule says I3 = I1 + I2. At point dependent on the location of the point. Thus h current entering is I1. There is only one starting with any point if we come back to the same current leaving h and by junction rule point, the total change must be zero. In a closed that will also be I1. For the loops ‘ahdcba’ loop, we do come back to the starting point and and ‘ahdefga’, the loop rules give –30I1 – hence the rule. 41 I3 + 45 = 0 and –30I1 + 21 I2 – 80 = 0. Example 3.5 A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1 W (Fig. 3.16). Determine the equivalent resistance of the network and the current along each edge of the cube. Z 3.5 FIGURE 3.16 EXAMPLE 98 Reprint 2025-26 Current Electricity Solution The network is not reducible to a simple series and parallel combinations of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network. The paths AA¢, AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further, at the corners A¢, B and D, the incoming current I must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of I, using Kirchhoff’s first rule and the symmetry in the problem. Next take a closed loop, say, ABCC¢EA, and apply Kirchhoff’s second rule: –IR – (1/2)IR – IR + e = 0 where R is the resistance of each edge and e the emf of battery. Thus, 5 e = I R 2 The equivalent resistance Req of the network is ε 5 R eq = = R Similation 3 I 6 for For R = 1 W, Req = (5/6) W and for e = 10 V, the total current (= 3I) in the network is 3I = 10 V/(5/6) W = 12 A, i.e., I = 4 A EXAMPLE The current flowing in each edge can now be read off from the Fig. 3.16. 3.5 application of It should be noted that because of the symmetry of the network, the great power of Kirchhoff’s rules has not been very apparent in Example 3.5. In a general network, there will be no such simplification due to symmetry,and only by application of Kirchhoff’s rules to junctions and closed loops Kirchhoff’s (as many as necessary to solve the unknowns in the network) can we handle the problem. This will be illustrated in Example 3.6. rules: Example 3.6 Determine the current in each branch of the network http://www.phys.hawaii.edu/~teb/optics/java/kirch3/ shown in Fig. 3.17. EXAMPLE FIGURE 3.17 3.6 99 Reprint 2025-26 Physics Solution Each branch of the network is assigned an unknown current to be determined by the application of Kirchhoff’s rules. To reduce the number of unknowns at the outset, the first rule of Kirchhoff is used at every junction to assign the unknown current in each branch. We then have three unknowns I1, I2 and I3 which can be found by applying the second rule of Kirchhoff to three different closed loops. Kirchhoff’s second rule for the closed loop ADCA gives, 10 – 4(I1– I2) + 2(I2 + I3 – I1) – I1 = 0 [3.61(a)] that is, 7I1– 6I2 – 2I3 = 10 For the closed loop ABCA, we get 10 – 4I2– 2 (I2 + I3) – I1 = 0 that is, I1 + 6I2 + 2I3 =10 [3.61(b)] For the closed loop BCDEB, we get 5 – 2 (I2 + I3) – 2 (I2 + I3 – I1) = 0 that is, 2I1 – 4I2 – 4I3 = –5 [3.61(c)] Equations (3.61 a, b, c) are three simultaneous equations in three unknowns. These can be solved by the usual method to give 5 7 I1 = 2.5A, I2 = A, I3 = A 8 18 The currents in the various branches of the network are 5 1 7 AB : A, CA : 2 A, DEB : 1 A 8 2 8 7 1 AD : 18 A, CD : 0 A, BC : 2 2 A It is easily verified that Kirchhoff’s second rule applied to the remaining closed loops does not provide any additional independent equation, that is, the above values of currents satisfy the second rule for every closed loop of the network. For example, the total voltage 3.6 drop over the closed loop BADEB  5   15  5 V +  8 × 4  V −  8 × 4  V EXAMPLE equal to zero, as required by Kirchhoff’s second rule. 3.13 WHEATSTONE BRIDGE As an application of Kirchhoff’s rules consider the circuit shown in Fig. 3.18, which is called the Wheatstone bridge. The bridge has four resistors R1, R2, R3 and R4. Across one pair of diagonally opposite points (A and C in the figure) a source is connected. This (i.e., AC) is called the battery arm. Between the other two vertices, B and D, a galvanometer G (which is a device to detect currents) is connected. This line, shown as BD in the figure, is called the galvanometer arm. For simplicity, we assume that the cell has no internal resistance. In general there will be currents flowing across all the resistors as well as a current Ig through G. Of special interest, is the case of a balanced bridge where the resistors are such that Ig = 0. We can easily get the balance condition, such that there is no current through G. In this case, the 100 Kirchhoff’s junction rule applied to junctions D and B (see the figure) Reprint 2025-26 Current Electricity immediately gives us the relations I1 = I3 and I2 = I4. Next, we apply Kirchhoff’s loop rule to closed loops ADBA and CBDC. The first loop gives –I1 R1 + 0 + I2 R2 = 0 (Ig = 0) (3.62) and the second loop gives, upon using I3 = I1, I4 = I2 I2 R4 + 0 – I1 R3 = 0 (3.63) From Eq. (3.62), we obtain, I1 R 2 = I 2 R1 whereas from Eq. (3.63), we obtain, I1 R 4 = I 2 R 3 Hence, we obtain the condition R 2 R 4 FIGURE 3.18 = [3.64(a)] R 1 R 3 This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection. The Wheatstone bridge and its balance condition provide a practical method for determination of an unknown resistance. Let us suppose we have an unknown resistance, which we insert in the fourth arm; R4 is thus not known. Keeping known resistances R1 and R2 in the first and second arm of the bridge, we go on varying R3 till the galvanometer shows a null deflection. The bridge then is balanced, and from the balance condition the value of the unknown resistance R4 is given by, R 2 R 4 = R 3 [3.64(b)] R1 A practical device using this principle is called the meter bridge. Example 3.7 The four arms of a Wheatstone bridge (Fig. 3.19) have the following resistances: AB = 100W, BC = 10W, CD = 5W, and DA = 60W. EXAMPLE FIGURE 3.19 3.7 101 Reprint 2025-26 Physics A galvanometer of 15W resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC. Solution Considering the mesh BADB, we have 100I1 + 15Ig – 60I2 = 0 or 20I1 + 3Ig – 12I2= 0 [3.65(a)] Considering the mesh BCDB, we have 10 (I1 – Ig) – 15Ig – 5 (I2 + Ig) = 0 10I1 – 30Ig –5I2 = 0 2I1 – 6Ig – I2 = 0 [3.65(b)] Considering the mesh ADCEA, 60I2 + 5 (I2 + Ig) = 10 65I2 + 5Ig = 10 13I2 + Ig = 2 [3.65(c)] Multiplying Eq. (3.65b) by 10 20I1 – 60Ig – 10I2 = 0 [3.65(d)] From Eqs. (3.65d) and (3.65a) we have 63Ig – 2I2 = 0 I2 = 31.5Ig [3.65(e)] 3.7 Substituting the value of I2 into Eq. [3.65(c)], we get 13 (31.5Ig ) + Ig = 2 410.5 Ig = 2 EXAMPLE Ig = 4.87 mA. SUMMARY 1. Current through a given area of a conductor is the net charge passing per unit time through the area. 2. To maintain a steady current, we must have a closed circuit in which an external agency moves electric charge from lower to higher potential energy. The work done per unit charge by the source in taking the charge from lower to higher potential energy (i.e., from one terminal of the source to the other) is called the electromotive force, or emf, of the source. Note that the emf is not a force; it is the voltage difference between the two terminals of a source in open circuit. 3. Ohm’s law: The electric current I flowing through a substance is proportional to the voltage V across its ends, i.e., V µ I or V = RI, where R is called the resistance of the substance. The unit of resistance is ohm: 1W = 1 V A–1. Reprint 2025-26 Current Electricity 4. The resistance R of a conductor depends on its length l and cross-sectional area A through the relation, ρl R = A where r, called resistivity is a property of the material and depends on temperature and pressure. 5. Electrical resistivity of substances varies over a very wide range. Metals have low resistivity, in the range of 10–8 W m to 10–6 W m. Insulators like glass and rubber have 1022 to 1024 times greater resistivity. Semiconductors like Si and Ge lie roughly in the middle range of resistivity on a logarithmic scale. 6. In most substances, the carriers of current are electrons; in some cases, for example, ionic crystals and electrolytic liquids, positive and negative ions carry the electric current. 7. Current density j gives the amount of charge flowing per second per unit area normal to the flow, j = nq vd where n is the number density (number per unit volume) of charge carriers each of charge q, and vd is the drift velocity of the charge carriers. For electrons q = – e. If j is normal to a cross-sectional area A and is constant over the area, the magnitude of the current I through the area is nevd A. 8. Using E = V/l, I = nevd A, and Ohm’s law, one obtains eE ne 2 = ρ v d m m The proportionality between the force eE on the electrons in a metal due to the external field E and the drift velocity vd (not acceleration) can be understood, if we assume that the electrons suffer collisions with ions in the metal, which deflect them randomly. If such collisions occur on an average at a time interval t, vd = at = eEt/m where a is the acceleration of the electron. This gives m ρ= 2 ne τ 9. In the temperature range in which resistivity increases linearly with temperature, the temperature coefficient of resistivity a is defined as the fractional increase in resistivity per unit increase in temperature. 10. Ohm’s law is obeyed by many substances, but it is not a fundamental law of nature. It fails if (a) V depends on I non-linearly. (b) the relation between V and I depends on the sign of V for the same absolute value of V. (c) The relation between V and I is non-unique. An example of (a) is when r increases with I (even if temperature is kept fixed). A rectifier combines features (a) and (b). GaAs shows the feature (c). 11. When a source of emf e is connected to an external resistance R, the voltage Vext across R is given by ε R Vext = IR = R + r where r is the internal resistance of the source. 103 Reprint 2025-26 Physics 12. Kirchhoff’s Rules – (a) Junction Rule: At any junction of circuit elements, the sum of currents entering the junction must equal the sum of currents leaving it. (b) Loop Rule: The algebraic sum of changes in potential around any closed loop must be zero. 13. The Wheatstone bridge is an arrangement of four resistances – R1, R2, R3, R4 as shown in the text. The null-point condition is given by R1 R 3 = R 2 R 4 using which the value of one resistance can be determined, knowing the other three resistances. Physical Quantity Symbol Dimensions Unit Remark Electric current I [A] A SI base unit Charge Q, q [T A] C Voltage, Electric V [M L2 T –3 A –1] V Work/charge potential difference Electromotive force e [M L2 T –3 A –1] V Work/charge Resistance R [M L2 T –3 A –2] W R = V/I Resistivity r [M L3 T –3 A –2] W m R = rl/A Electrical s [M –1 L–3 T 3 A 2] S s = 1/r conductivity –3 –1 Electric force Electric field E [M L T A–1] V m charge –1 e E τ Drift speed vd [L T –1] m s vd = m Relaxation time t [T] s Current density j [L–2 A] A m–2 current/area Mobility m [M L3 T –4 A –1] m 2 V –1s –1 v d / E POINTS TO PONDER 1. Current is a scalar although we represent current with an arrow. Currents do not obey the law of vector addition. That current is a scalar also follows from it’s definition. The current I through an area of cross-section is given by the scalar product of two vectors: I = j . DS 104 where j and DS are vectors. Reprint 2025-26 Current Electricity 2. Refer to V-I curves of a resistor and a diode as drawn in the text. A resistor obeys Ohm’s law while a diode does not. The assertion that V = IR is a statement of Ohm’s law is not true. This equation defines resistance and it may be applied to all conducting devices whether they obey Ohm’s law or not. The Ohm’s law asserts that the plot of I versus V is linear i.e., R is independent of V. Equation E = ρ j leads to another statement of Ohm’s law, i.e., a conducting material obeys Ohm’s law when the resistivity of the material does not depend on the magnitude and direction of applied electric field. 3. Homogeneous conductors like silver or semiconductors like pure germanium or germanium containing impurities obey Ohm’s law within some range of electric field values. If the field becomes too strong, there are departures from Ohm’s law in all cases. 4. Motion of conduction electrons in electric field E is the sum of (i) motion due to random collisions and (ii) that due to E. The motion due to random collisions averages to zero and does not contribute to vd (Chapter 10, Textbook of Class XI). vd , thus is only due to applied electric field on the electron. 5. The relation j = ρ v should be applied to each type of charge carriers separately. In a conducting wire, the total current and charge density arises from both positive and negative charges: j = ρ+ v+ + ρ– v– ρ = ρ+ + ρ– Now in a neutral wire carrying electric current, ρ+ = – ρ– Further, v+ ~ 0 which gives ρ = 0 j = ρ– v Thus, the relation j = ρ v does not apply to the total current charge density. 6. Kirchhoff’s junction rule is based on conservation of charge and the outgoing currents add up and are equal to incoming current at a junction. Bending or reorienting the wire does not change the validity of Kirchhoff’s junction rule. EXERCISES

3.3 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10–4 °C–1. 105 Reprint 2025-26 Physics 3.4 A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10–7 m2, and its resistance is measured to be 5.0 W. What is the resistivity of the material at the temperature of the experiment? 3.5 A silver wire has a resistance of 2.1 W at 27.5 °C, and a resistance of 2.7 W at 100 °C. Determine the temperature coefficient of resistivity of silver. 3.6 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10–4 °C–1. 3.7 Determine the current in each branch of the network shown in Fig. 3.20: FIGURE 3.20 3.8 A storage battery of emf 8.0 V and internal resistance 0.5 W is being charged by a 120 V dc supply using a series resistor of 15.5 W. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit? 3.9 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m–3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10–6 m2 and it is carrying a current of 3.0 A. Reprint 2025-26 Chapter Four MOVING CHARGES AND MAGNETISM 4.1 INTRODUCTION Both Electricity and Magnetism have been known for more than 2000 years. However, it was only about 200 years ago, in 1820, that it was realised that they were intimately related. During a lecture demonstration in the summer of 1820, Danish physicist Hans Christian Oersted noticed that a current in a straight wire caused a noticeable deflection in a nearby magnetic compass needle. He investigated this phenomenon. He found that the alignment of the needle is tangential to an imaginary circle which has the straight wire as its centre and has its plane perpendicular to the wire. This situation is depicted in Fig.4.1(a). It is noticeable when the current is large and the needle sufficiently close to the wire so that the earth’s magnetic field may be ignored. Reversing the direction of the current reverses the orientation of the needle [Fig. 4.1(b)]. The deflection increases on increasing the current or bringing the needle closer to the wire. Iron filings sprinkled around the wire arrange themselves in concentric circles with the wire as the centre [Fig. 4.1(c)]. Oersted concluded that moving charges or currents produced a magnetic field in the surrounding space. Following this, there was intense experimentation. In 1864, the laws obeyed by electricity and magnetism were unified and formulated by Reprint 2025-26 Physics James Maxwell who then realised that light was electromagnetic waves. Radio waves were discovered by Hertz, and produced by J.C.Bose and G. Marconi by the end of the 19th century. A remarkable scientific and technological progress took place in the 20th century. This was due to our increased understanding of electromagnetism and the invention of devices for production, amplification, transmission and detection of electromagnetic waves. FIGURE 4.1 The magnetic field due to a straight long current-carrying wire. The wire is perpendicular to the plane of the paper. A ring of compass needles surrounds the wire. The orientation of the needles is shown when (a) the current emerges out of the plane of the paper, (b) the current moves into the plane of the paper. (c) The arrangement of iron filings around the wire. The darkened ends of the needle represent north poles. The effect of the earth’s magnetic field is neglected. In this chapter, we will see how magnetic field exerts forces on moving charged particles, like electrons, protons, and current-carrying wires. We shall also learn how currents produce magnetic fields. We shall see how particles can be accelerated to very high energies in a cyclotron. We shall study how currents and voltages are detected by a galvanometer.(1777–1851) In this and subsequent Chapter on magnetism, we adopt the following convention: A current or a field (electric or magnetic) emerging out of the plane of the paper is depicted by a dot (¤). A current or a field going into the plane of the paper is depicted by a cross ()*. Hans Christian Oersted Figures. 4.1(a) and 4.1(b) correspond to these twoOERSTED (1777–1851) Danish situations, respectively. physicist and chemist, professor at Copenhagen. 4.2 MAGNETIC FORCE He observed that a compass needle suffers a 4.2.1 Sources and fields deflection when placed Before we introduce the concept of a magnetic field B, weCHRISTIAN near a wire carrying an electric current. This shall recapitulate what we have learnt in Chapter 1 about discovery gave the first the electric field E. We have seen that the interaction empirical evidence of a between two charges can be considered in two stages.HANS connection between electric The charge Q, the source of the field, produces an electric and magnetic phenomena. field E, where * A dot appears like the tip of an arrow pointed at you, a cross is like the feathered 108 tail of an arrow moving away from you. Reprint 2025-26 Moving Charges and Magnetism E = Q ˆr / (4pe0)r2 (4.1) where ˆr is unit vector along r, and the field E is a vector field. A charge q interacts with this field and experiences a force F given by F = q E = q Q ˆr / (4pe0) r 2 (4.2) As pointed out in the Chapter 1, the field E is not just an artefact but has a physical role. It can convey energy and momentum and is not established instantaneously but takes finite time to propagate. The concept of a field was specially stressed by Faraday and was incorporated by Maxwell in his unification of electricity and magnetism. In addition to depending on each point in space, it can also vary with time, i.e., be a function of time. In our Hendrik Antoon Lorentz discussions in this chapter, we will assume that the fields (1853 – 1928) Dutch do not change with time. theoretical physicist, The field at a particular point can be due to one or professor at Leiden. He investigated themore charges. If there are more charges the fields add HENDRIK relationship between vectorially. You have already learnt in Chapter 1 that this electricity, magnetism, and is called the principle of superposition. Once the field is mechanics. In order to known, the force on a test charge is given by Eq. (4.2). explain the observed effect Just as static charges produce an electric field, the of magnetic fields on ANTOONcurrents or moving charges produce (in addition) a emitters of light (Zeeman magnetic field, denoted by B (r), again a vector field. It effect), he postulated the existence of electric chargeshas several basic properties identical to the electric field. in the atom, for which he It is defined at each point in space (and can in addition was awarded the Nobel Prize depend on time). Experimentally, it is found to obey the in 1902. He derived a set of LORENTZprinciple of superposition: the magnetic field of several transformation equations sources is the vector addition of magnetic field of each (known after him, as individual source. Lorentz transformation equations) by some tangled (1853 4.2.2 Magnetic Field, Lorentz Force mathematical arguments, – but he was not aware thatLet us suppose that there is a point charge q (moving these equations hinge on a with a velocity v and, located at r at a given time t) in new concept of space andpresence of both the electric field E (r) and the magnetic 1928) time. field B (r). The force on an electric charge q due to both of them can be written as F = q [ E (r) + v × B (r)] º Felectric +Fmagnetic (4.3) This force was given first by H.A. Lorentz based on the extensive experiments of Ampere and others. It is called the Lorentz force. You have already studied in detail the force due to the electric field. If we look at the interaction with the magnetic field, we find the following features. (i) It depends on q, v and B (charge of the particle, the velocity and the magnetic field). Force on a negative charge is opposite to that on a positive charge. (ii) The magnetic force q [ v × B ] includes a vector product of velocity 109 and magnetic field. The vector product makes the force due to magnetic Reprint 2025-26 Physics field vanish (become zero) if velocity and magnetic field are parallel or anti-parallel. The force acts in a (sideways) direction perpendicular to both the velocity and the magnetic field. Its direction is given by the screw rule or right hand rule for vector (or cross) product as illustrated in Fig. 4.2. (iii) The magnetic force is zero if charge is not moving (as then |v|= 0). Only a moving charge feels the magnetic force. The expression for the magnetic force helps us to define the unit of the magnetic field, if one FIGURE 4.2 The direction of the magnetic takes q, F and v, all to be unity in the force force acting on a charged particle. (a) The equation F = q [ v × B] =q v B sin q ˆn , where q is force on a positively charged particle with the angle between v and B [see Fig. 4.2 (a)]. The velocity v and making an angle q with the magnitude of magnetic field B is 1 SI unit, when magnetic field B is given by the right-hand the force acting on a unit charge (1 C), moving rule. (b) A moving charged particle q is perpendicular to B with a speed 1m/s, is one deflected in an opposite sense to –q in the newton. presence of magnetic field. Dimensionally, we have [B] = [F/qv] and the unit of B are Newton second / (coulomb metre). This unit is called tesla (T) named after Nikola Tesla (1856 – 1943). Tesla is a rather large unit. A smaller unit (non-SI) called gauss (=10–4 tesla) is also often used. The earth’s magnetic field is about 3.6 × 10–5 T. 4.2.3 Magnetic force on a current-carrying conductor We can extend the analysis for force due to magnetic field on a single moving charge to a straight rod carrying current. Consider a rod of a uniform cross-sectional area A and length l. We shall assume one kind of mobile carriers as in a conductor (here electrons). Let the number density of these mobile charge carriers in it be n. Then the total number of mobile charge carriers in it is nlA. For a steady current I in this conducting rod, we may assume that each mobile carrier has an average drift velocity vd (see Chapter 3). In the presence of an external magnetic field B, the force on these carriers is: F = (nlA)q vd ´ B where q is the value of the charge on a carrier. Now nq vd is the current density j and |(nq vd)|A is the current I (see Chapter 3 for the discussion of current and current density). Thus, F = [(nq vd )lA] × B = [ jAl ] ´ B = Il ´ B (4.4) where l is a vector of magnitude l, the length of the rod, and with a direction identical to the current I. Note that the current I is not a vector. In the last step leading to Eq. (4.4), we have transferred the vector sign from j to l. Equation (4.4) holds for a straight rod. In this equation, B is the external magnetic field. It is not the field produced by the current-carrying rod. If the wire has an arbitrary shape we can calculate the Lorentz force on it by considering it as a collection of linear strips dlj and summing F  Idl j × B j This summation can be converted to an integral in most cases. Reprint 2025-26 Moving Charges and Magnetism Example 4.1 A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B (Fig. 4.3). What is the magnitude of the magnetic field? FIGURE 4.3 Solution From Eq. (4.4), we find that there is an upward force F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity: m g = I lB m g B = I l Interactive Charged 0.2 × 9.8 = = 0.65 T 2 × 1.5 Note that it would have been sufficient to specify m/l, the mass per EXAMPLE particles unit length of the wire. The earth’s 4 × 10–5 T and we have ignored it. magnetic field is approximately 4.1 moving demonstration: in a Example 4.2 If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig. 4.4), which way would the Lorentz force be for (a) an electron (negative magnetic charge), (b) a proton (positive charge). field. http://www.phys.hawaii.edu/~teb/optics/java/partmagn/index.html FIGURE 4.4 Solution The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so v × B is along the z-axis (screw rule or right-hand thumb rule). So, (a) for electron it will be along –z EXAMPLE axis. (b) for a positive charge (proton) the force is along +z axis. 4.2 111 Reprint 2025-26 Physics 4.3 MOTION IN A MAGNETIC FIELD We will now consider, in greater detail, the motion of a charge moving in a magnetic field. We have learnt in Mechanics (see Class XI book, Chapter 5) that a force on a particle does work if the force has a component along (or opposed to) the direction of motion of the particle. In the case of motion of a charge in a magnetic field, the magnetic force is perpendicular to the velocity of the particle. So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed). [Notice that this is unlike the force due to an electric field, qE, which can have a component parallel (or antiparallel) to motion and thus can transfer energy in addition to momentum.] We shall consider motion of a charged particle in a uniform magnetic field. First consider the case of v perpendicular to B. The perpendicular force, q v × B, acts as a centripetal force and produces a circular motion perpendicular to the magnetic field. The particle will describe a circle if v and B are perpendicular to each other (Fig. 4.5). If velocity has a component along B, this component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to B is as before a circular one, thereby producing a helical motion (Fig. 4.6). You have already learnt in earlier classes (See Class XI, Chapter 3) that if r is the radius of the circular path of a particle, then a force of m v2 / r, acts perpendicular to the path towards the centre of the circle, and is called the centripetal force. If the FIGURE 4.5 Circular motion velocity v is perpendicular to the magnetic field B, the magnetic force is perpendicular to both v and B and acts like a centripetal force. It has a magnitude q v B. Equating the two expressions for centripetal force, m v 2/r = q v B, which gives r = m v / qB (4.5) for the radius of the circle described by the charged particle. The larger the momentum, the larger is the radius and bigger the circle described. If w is the angular frequency, then v = w r. So, w = 2p n = q B/ m [4.6(a)] which is independent of the velocity or energy . Here n is the frequency of rotation. The independence of n from energy has important application in the design of a cyclotron. The time taken for one revolution is T= 2p/ w º 1/n. If there is a component of the velocity FIGURE 4.6 Helical motion parallel to the magnetic field (denoted by v||), 112 it will make the particle move along the field and the path of the Reprint 2025-26 Moving Charges and Magnetism particle would be a helical one (Fig. 4.6). The distance moved along the magnetic field in one rotation is called pitch p. Using Eq. [4.6 (a)], we have p = v||T = 2pm v|| / q B [4.6(b)] The radius of the circular component of motion is called the radius of the helix. Example 4.3 What is the radius of the path of an electron (mass 9 × 10-31 kg and charge 1.6 × 10–19 C) moving at a speed of 3 ×107 m/s in a magnetic field of 6 × 10–4 T perpendicular to it? What is its frequency? Calculate its energy in keV. ( 1 eV = 1.6 × 10–19 J). Solution Using Eq. (4.5) we find r = m v / (qB) = 9 ×10–31 kg × 3 × 107 m s–1 / ( 1.6 × 10–19 C × 6 × 10–4 T) = 28 × 10–2 m = 28 cm n = v / (2 pr) = 17×106 s–1 = 17×106 Hz =17 MHz. 2 EXAMPLE E = (½ )mv = (½ ) 9 × 10–31 kg × 9 × 1014 m2/s2 = 40.5 ×10–17 J ≈ 4×10–16 J = 2.5 keV. 4.3