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12. Blectromagnetic Induction 12th Science 12th - Physics Syllabus 1, Introduction 2. Faraday’s Laws of Electromagnetic Induction 3. Lenz’s Law 7 3-1 Applications of Lenz’s law 3-2 Motion of a Magnet Toward a Loop 3.3 Energy Conservation in Lenz’s Law 3.4 Lenz’s Law and Faraday’s Law 4. Flux of the Field 5. Motional Electromotive Force 6, Induced emf in a Stationary Coil ina Changing Magnetic Field 7. Generators 8. Back emf and back torque 9. Induction and Energy Transfer 10. Eddy Currents 11. SelfInductance 12. Energy Stored in a Magnetic Field 13. Energy Density of a Magnetic Field 14. Mutual Inductance (M) 15. Transformer ‘| ‘Theory Notes Q. Explain Faraday’s exper- | Faraday’s experiment tment of electromagnetic a) Magnet and coil experiment induction. Consider a coil connected with a placed near it. galvanometer and a bar magnet coil -> bar magnet coil — bar magnet 1) When magnet is moved towards the coil, current is produced in the coil. This current is called induced current. ; 2) When magnet is moved away from the coil, current is produced in the opposite direction. 3) If the coil is moved towards or away from the magnet, then also induced current is produced in the coil. 4) If the polarity of the moving magnet is reversed, the direction of the induced current in the coil is also reversed. 5) The magnitude of the induced current depends upon the relative speed of the coil and the magnet. It also depends upon the number of turns of the coil. b) Coil and coil experiment Consider two coils with their planes facing each other. One coil is connected in series with a battery, rheostat and key. This is primary coil. Other coil is connected to a galvanometer. This is secondary coil. primary coil secondary coil II- ge 1) When the key is closed in primary coil, momentary defiection is seen in secondary coil, 2) When the key is opened in primary coil, momentary deflection in opposite direction is observed. 3) If the current in the primary coil is increasing, induced current is produced in the secondary coil. 4) If the current in the primary coil is decreasing, induced current is produced in the opposite direction in the secondary coil. 5) If there is relative motion between two coils, induced current is produced in the secondary coil. Conclusion Whenever there is a change in magnetic flux associated with the coil, an emf is induced in the coil. This emf is called induced emf and the resulting current is called the induced current. This phenomenon is called as electromagnetic induction. Q. Lenz's law is incorporated| Lenz’s law and Faraday’s coe into Faraday’s law. Explain | According to the Faraday’s 1° do C= rr A of the loop perpendicular to the plane Consider tht are a Fo pra! = 0) tthe magnet Bel B. dBA) uke b= FAS ee dt . — a\B\ “e=-lA at Since both !4 land ag! are positive, the -ve sign one shows that the quantity on RHS is -ve. Hence quantity on must be negative in order to be equal to the RHS. db , ‘ Thus the negative sign in the equation ¢ ~~ "qj incorporates Lenz's law into Faraday’s law. ————— Q. Explain flux of the field. | Flux of the field = i B Consider a small element of area da situated in a magnetic field B. | The magnetic flux through the area da is given by- do = B.da=\B \.\da \cos@ Where @ is the angle between the direction of magnetic field B and the direction assigned to the area da. _ Now consider a finite area S of which the element da is a part. The flux passing through S is calculated as- $= {Bde 5 If the magnetic field at every point changes with time, then the flux will also change with time. b= b= [Bad Q. Derive the equation for Translational motion of a conductor motional electromotive force Consider a rectangular conductive wire frame ABCD of area (Ix) for the translational motion placed in uniform magnetic field B which is perpendicular to the of a conductor. plane of the paper and directed inside. Let the wire BC of length | is Q. Prove that induced emfis | moved out with velocity v through distance dx in time dt. e= Bly. ®,8 8 ® @ ® ele © 8 |S @® l Ql] @ (3) ®@ ees eve E @ ® Q ® @ @ A = ip @ ® ® ® ® @ Q. Derive the equation for motional electromotive force for a rotating bar. A charge g which is carried along by the moving wire BC, experi- ences Lorentz force given = q(vxB) P= gvBsin 0 Where @/ is the angle between B and, This force is perpendicular to both ¥ and B and hence is parallel to wire BC. It is constant along the length / of the wire and is zero elsewhere. When the charge q moves a distance / along the wire, the work done will be- = F.1= qvBsin @.l work charge woo = yBsing! q emp = For maximum induced emf, sin@ = | * Cm = BV...) Also, The flux ¢ through the frame ABCD is Blix. Therefore the magnitude of the induced emf will be- = 4 _ dippnya ; lel=—y = £(Bix) = Brat = = Blv Both the equations (1) and (2) prove that, (2) e=Blv Motional emf in a rotating bar Consider a bar rotating in uniform magnetic field B whichis perpen- dicular to the plane of the paper and directed inside. Consider a small segment dr of the bar at a distance r from the pivot moving with velocity v. 8 c) 8 @ ® 8 pivots, 2 8 8 8 The induced emf de in the small segment dr of the rotating conductor will be- de = Bvdr The total induced emf in rotating bar will be- = ofa = i] Bydr =/ Borrdr cutie v= wr} es obo rdr .e= Bot = S Bol Q. Give theory of AC gener- ator and derive an expres- sion for an alternating emf. Q. Obtain an expression for the emf induced in a coil rotating with a uniform angular velocity in a uniform magnetic field. Show graphi- cally the variation of the emf with time t. (Mar. 15, Mark 4) Q. Obtain an expression for the induced emf ina coil rotating with uniform angular velocity in a uniform magnetic field. Plot a graph of the variation of the induced emf against phase (8 = wt) over one cycle. (Oct. 13, Mark 4) frequency(v) = __l__. Jreq time period(T ) - P= bsin z= 5 = Osin Ft Differentiating w.r.t.t dO 2a ap 8ocos i( _ dO _ 2nby a OTC 5 2at T At mean position where @ = 0, emf is maximum (eo) and a Therefore equation (1) will become a=(E) (3) ) Thus for the given magnet-coil system, the peak induced emf (e0) is directly proportional to angular amplitude (6) and inversely proportional to time period (7). Measurement of induced emf , In order to measure the induced emf, a capacitor (C) and diode (D) are connected across the coil. R K diode mA During half a cycle of motion of the magnet two peak values i.e. one negative and one positive are obtained. Diode conducts only during the ‘positive’ pulse and will be cut off during the ‘negative’ pulse. Thus during each swing, the capacitor is charged to some extent and within few oscillations, it will be charged up to its peak value é. At peak value ammeter shows no further increase. The voltage developed across the capacitor is measured with the help of voltmeter (V). AC Generator (emf induced in a rotating coil) AC generator is a machine that converts mechanical energy into electrical energy in the form of AC current. It consists of a conducting coil of area A and having N number of turns. The coil is rotating with uniform angular velocity w in a uniform magnetic field B. The axis of rotation is in the plane of the coil and perpendicular to the uniform magnetic field. Q. What is back emf? in time f. Let the coil rotates through an he 0 in “b= O i through the coil is given as - The magnetic 1s o= NBA cos @ = NBA cos wt Due to the rotation of the coil, magnetic flux linked with the coil is i in it. changes and emf is induced in i -—db = 4 (NBAcoswt) ..e = NBAwsinwt af =+1 For sinwr =+ o=+NBAw =4& 2.@ = @sinwt -.@ = esin 2nft tth 7 i ncy of revolution of the coil, This tne expreseon for emf induced in the coil. This induced emf varies with sinwr. Thus it is called as sinusoidal or alternating emf, As the value of sinwt varies between +1 and —1, the polarity of the emf changes with time. Hence the direction of the induced current changes periodically and it is called as alternating current (AC), The frequency of AC is equal to the number of times per second, the current changes from positive to negative and back again. ; The variation of the induced emf with the angle of rotation @ or time t is given below- i 0 TIA T/2 3/T4 T ST/4 37/2 @ 0 a2 m 3/2 2a 3x/2 3x e 0 & 0 =e 0 eo ) te, 2n 30 —_ or a si ~ gy Thusinduced emfhas maximum value é,when@ = 14,32 52 etc. i.e. when plane of coil is parallel to the directior? of the thagnetic field. The induced emf has minimum value 0, when @ = 17, 37, 5, wu... etc. ie. when plane of the coil is perpendicular to the direction of the magnetic field. Back emf ; In a motor, the coil rotates in magnetic field due to the current in the coil. As the coil rotates, induced emf is produced in the coil. According to the Lenz’s law the induced emf opposes the change which causes it. Hence the induced emf opposes the current in the coil. This induced emf produced due to the generator action in the motor is called back emf. Q. What are eddy currents? Q. Explain the phenomenon of self-inductance. 2 v sess el) i he work done in pullin, i 4) and (6), it is clear that t ; e te ot areagh the a ones field appears as heat energy in the loop. —— ts a 2 ty thecirenating currents generated ina psp Satay a Plats, . ing magnetic field, are called as e urrents, 2) egies ‘also broduced when metal block is moving in uniform magnetic field. : 3) Actording to Lenz’s law, eddy currents oppose the change in the magnetic flux. ents . bles contaiee lange number of free electrons. The force acting on electrons due to varying magnetic field acts as centripetal force, Therefore electrons move in circular path inside the conductor. Thus eddy currents are generated. Drawback of eddy currents Eddy currents produce heat in conductor and cause power loss. Minimization of eddy currents Eddy currents can be reduced by discontinuity in the structure of the conductor. Self-Inductance (L) Production of induced emf in a coil due to the changing current in the same coil is called as self-inductance. According to Lenz’s law, the direction of this induced emf is opposite to the direction of the current passing through conductor. Therefore it is also called as back emf. The magnetic flux () linked with the coil is directly proportional to the current (7) in the coil. hel “G@=L1 Where L is aconstant of proportionality and is termed as self-induc- tance (or coefficient of self induction) of the coil. I Hence the self-inductance of a magnetic flux linked with the According to Faraday’s law, th A circuit is defined as the ratio of circuit to the current flowing in it. € induced emf is given as- = db edt e-4 1) -—, 4 We ag LD = La =-(s) dijdt Hence the self-inductance of a circuit is also defined as the ratio of | induced emf in the circuit to the rate of change of current in it. Unit of L | Magnitude of self-inductance is given as- Q. Define Henry (for self-inductance). Q. Show that the self-induc- tance of a circuit is numer- ically equal to twice the work done in establishing the magnetic flux associ- ated with unit current in the circuit. Q. Give the analogy between mass and self-inductance. le| 0 adil , unit of € yolt o ie y= __ — —©" _ = Henry want OF I. unit of Tunit of { ampli sec y SI unit of self-inductance is Henry. Dimensions are [L’M'T’?] Henry (H) (for self-inductance) lvolt_ I henry(H) = ‘Tampere/I second The self-inductance of a coil is said to be 1 henry, 1 volt is induced in a coil due to the rate of change © ampere per second in the same coil. if the back emf of f current of 1 Ans. Self-inductance produced in a circuit is given as- = (2 — (aia) When the current increases in the cir opposite to it. Consequently, the work will ha to establish the magnetic flux associated wi in the circuit. Work done in time df is- cuit, an induced emf acts 11 have to be done in order ith a steady current Jo dW = e.ldt di wep ee pal 2 dW = (IE \ Dd aco OF eal = ab =u at =-L)dl Total work done can be obtained as- he w= faw= [-ual Magnitude of work done will be- w=4Lh If 4, = 1, then ai KS 2 7 L=t+2Ww Thus self-inductance of a circuit is numerically equal to twice the work done in establishing the magnetic flux associated with unit current in the circuit. Ans. Self-inductance of a circuit is numerically equal to twice the work done in establishing the magnetic flux associated with unit current in the circuit. L=+2Ww Wath For /, current, W= 5h This work done represents the energy of the circuit. B=418.....(1) Q. Give the analogy between energy stored in a magnetic field and energy stored in electric field. Q. Derive the equation for the energy density of a magnetic field. dW =~ e.dq = Ldha.. 4: a = } The total work done can be obtained as- Ww= [aw = [iat “Wa dLP This work done is stored in the form of energy in magnetic field. 2Up= LP Ans, The total electrostatic energy U: the plates of a capacitor with capacitance C andc potential V is- stored in the medium between harge q held at Ure EV) The energy Uy stored in a magnetic field due to a circuit carrying current J and having inductance L is- Up = LLP oo (2) From equations (1) and (2), it is observed that wee field is analogous to the energy § eld. the energy stored in tored in the electric Energy density of a magnetic field The energy U, stored in a magnetic current / and having inductance L is- Up =F LP ol) Consider a long solenoid having length / near the middle, cross-sec~ tional area A and carrying current i through it. It has uniform magnetic field B inside the solenoid while outside it is almost zero. The energy stored, per unit volume, in the magnetic field is called energy density of the magnetic field. energy stored in magnetic field (Us) field due to a circuit carrying . energy density(ue) = volume (AD) Ue = 4. sl) - peti talt cue GLP ay = (Fg 8) Inductance per unit volume of a long solenoid is given as- J = pig Aen fA) e number of turns per unit length the medium (for air //o = 47% X 10° Tm/A) From equations (1) and (2) L } Lr Where ni is th Jig is permeability constant of From equation (3) and (4) Un = fo AZZ ee ery Ug = yor LD netic field inside the solenoid is given as B= poln _ BB 2pbo The mag) tts This is the equation for energy density for the magnetic field B. 12. Electromagnetic Induction Q. Explain the phenomenon of mutual-inductance. Q. Define Henry (for mutual-inductance). Mutual-Inductance (M) Production of induced emf in a coil due to the changing current in the neighboring coil is called as mutual-inductance. : - Consider two coils placed close to each other. The coil in which current is provided by an external source is referred as primary coil while the other coil is referred as secondary coil. secondary coil primary coil When the current in the primary coil (/,) changes, magnetic flux linked with the secondary coil (¢,) changes. obs Ip 2 Gc=M Ip Where M is a constant of proportionality and i al-inductance (or coefficient of mutual induction; coil with respect to primary coil. Mae 1 Hence the mutual-inductance of a circ magnetic flux linked with the circuit to (nearby) circuit. According to Faraday’s law, given as- __ dbs dt s termed as mutu- ) of the secondary uit is defined as the ratio of the current flowing in other the induced emf in secondary coil is + wa(— M ( di, dt ) ctance of a circuit is also defined as the ratio Hence the mutual-indu' cuit to the rate of change of current in other of induced emf in the cir (nearby) circuit. Unit of M Magnitude of mutual-inductance is given as- __lel M= [dl,/at | aint ___unitofeé___voll_ _ unit of M unit of I,/unit of t amp! sec Henry SI unit of mutual-inductance is Henry. Dimensions are [eM T?r?] — Henry (H) (for mutual-inductance) ____l volt 1 henry) = Tampere/isecond tance between two coils is said to be 1 henry ifan secondary coil due to rate of change nd in the primary coil. The mutual indu emf of 1 volt is induced in the of current of 1 ampere per seco Q. What is step up and step down transformer? Theory . . When an alternating emf is applied to the primary coil, changing | magnetic flux is produced around it. This changing magnetic flux is linked with the secondary coil through soft iron core. Thus alter- nating emf is induced in the secondary coil. . The magnetic flux linked with each coil is proportional to their number of turns. br bs b= Os =F Bre oe ol Where, ¢» =flux linked with primary coil, ip @s = flux linked with secondary coil. Differentiating above equation w.r.t. time, dbs _— Ns dbr (yy dt Ne dpe According to Faraday’s laws of E.M.1. de ep =~ tbe = emf induced in primary coil __ dbs _ . : ‘ es=—~ 7, = emf induced in secondary coil Therefore equation (2) becomes 6 @s = kep....... (3) Where, k= = is called transformation ratio. e For a step up transformer k is greater than 1 and for a step down transformer & is less than 1. For an idea! transformer, input power = output power cls = Cplp fale (4) e it Where, i, = current in primary coil, i, = current in secondary coil. In step - up transformer, és > ep .', is < ip ie. voltage increases by deceasing the current. In step-down transformer, @s < ep." is > ip ie. current increases by decreasing the voltage. In other words a transformer simply transforms the voltages and currents obeying the law of conservation of energy. The transformers are of two types- 1) Step-up transformer 1) It converts low voltage at high current into high voltage at low current. 2) In this transformer the number of turns in secondary coil (Ns) is greater than the number of turns in primary coil (N,). (i.e. Ns > Ne). 2) Step-down transformer 1) It converts high voltage at low current into a low voltage at high current. 2) In this transformer the number of turns in secondary coil (Ns) is less than the number of turns in primary coil (N».) (ie. Ny < No) | Important Formulae ee 1) Faraday Lenz’s law - 42 dA e dt’ — B=, dre =- 4 2) The emt induced in a straight conductor e= Biv 3) Back emf (self induction) __ di Mon A ern: a, a = nli 4) Mutual induction —_ MonnmA ud M=——+—.,.= nMip 5) Flux link with coil ¢ =nAB cos@ 6) EMF induced in rotating coil e = esinwt,i = insinwt & = 27fnAB, 16) Transformation ratio Ds _ & -&_h Ke o e ft 17) Efficiency of transformer _ output power _ el, — “input power eply 18) Displacement current [eg ae deinttinn From Text Book (Unsolved) = a na Faraday disc dy yt y disc dynamo, a metal disc of radius R rotates with an ang an axis perpendicular to the plane of the d ular velocity @ about placed in a magnetic field B acti and passing through its centre. The othe eee ian B acting perpendicular to the plane of the disc. Determine the een the rim and the axis of the disc Ans: + (BoR’) A horizontal wire 2 . 2) 0 m long extending from east to west is falling with a velocity of 10 m/s normal to the Earth’ wire? arth's magnetic field of0.5"10-"7, What is the value of induced emf in the Ans: 10 mV I disc i -_ 3} A ete cond made to spin at 20 revolutions per second about an axis passing through ; vaenetic field nome to its plane. The disc has a radius of 30 cm and spins in a uniform oat per second by 3 T, which is parallel to the axis of rotation. Calculate (a) The area swept i € radius of the dis r y dius of the disc, (c) The induced emf in the disc, c, (b) The flux cut per second by a ra ‘Ans: (a) 5.65 m?, (b) 1.130 Wb, (°) 1.130V 4) A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 10 Ain 0.2 s, what is the change of flux linkage with the other coil? Ans: 75 Wb 5) Along solenoid consisting of 1.5*10° turns/m has an area of cross-section of 25cm’. Acoil C, consisting of 150 turns (/N.) is wound tightly around the centre of the solenoid. Calculate for a current of 3.0 A in the solenoid (a) the magnetic flux density at the centre of the solenoid, (b) the flux linkage in the coil C, (c) the average emf induced in coil C if the current in the solenoid is reversed in direction in a time of 0.5 s.(ulo = 4x 107 T.mlA) ‘Ans: (a) 5.66%10° T, (b) 2.12x10° Wb, (c) 8.48x10° V 6) A search coil having 2000 turns with area 1.5 emis placed in a magnetic field of 0.60T. The coil is moved rapidly out of the field in a time of 0.2 second. Calculate the induced emf across the search coil. Ans: 0.9 V 7) An aircraft of wing span of 50 m flies horizontally in earth’s magnetic field of 6x10°Tata speed of 400 m/s. Calculate the emf generated between the tips of the wings of the aircraft. Ans: 1.2 V 8) A stiff semi-circular wire of radius R is rotated in a uniform magnetic field B about an axis passing through its ends. If the frequency of rotation of wire be f, calculate the amplitude of alternating emf induced in the wire. Ans: @ = 7 BR°f 9) Calculate the value of induced emf between the ends of an axle of a railway carriage 1.75 m long travelling on level ground with a uniform velocity of 50 km per hour. The vertical component of Earth’s magnetic field (Bv) is given to beSx10°7. Ans: 0.122 mV 10) The value of mutual inductance of two coils is 10 mH. If the current in one of the coil changes from 5A to 1A in 0.2 s, calculate the value of emf induced in the other coil. Also calculate the value of induced charge passing through the coil if its resistance is 5 ohm. Ans: 8 mC eee 11) An emf of 96.0 mV is induced in the windings of a coil when the current in a nearby coil is increasing at the rate of 1.20 A/s. What is the mutual inductance (M) of the two coils? Ans: 80 mH 12) A long solenoid of length /, crosssectional area A and having N, turns (primary coil), has a small coil of N, turns (secondary coil) wound about its centre. Determine the Mutual induc- tance (M) of the two coils. Ans: M = p, N, N,A 13) The primary and secondary coil of a transformer each have an inductance of 200x10°H. The mutual inductance (M) between the windings is 4x10°H. What percentage of the flux from one coil reaches the other? Ans: 2% 14) A toroidal ring, made from a bar of length (1)1 mand diameter (d) 1 cm, is pent into a circle. It is wound tightly with 100 turns per cm. If the permeability of bar is equal to that of free space (fo), calculate the magnetic field inside the bar (B) when the current (i) circulating through the turns is 100 A. Also determine the self-inductance (L) of the coil. Ans: 1.256T, 10mH 15) A uniform magnetic field B(1 ), pointing upward fills a circular region of radius, s in hori- zontal plane. If B is changing with time, find the induced electric field. 7 GB Ans: ~7 5°77 20