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ELECTROMAGNETIC INDUCTION Magnetic Flux : φ B r r Magnetic flux φ B through an area ds in a magnetic field B is defined as z r r φ B = B. ds Physically it represents total lines of induction passing through a given area. Regarding magnetic flux it is worth noting that Though lines of force are imaginary, flux is a real scalar physical quantity with dimensions [φ B ] = [ B][ S ] = [φ B ] = i.e. LM MLT N AL −2 LM F OP[ S ] N IL Q OP[ L ] = [ ML T Q 2 2 [as F = BIL sin θ ] −2 A −1 ] As [ ML2 T −2 ] corresponds to energy, SI unit of magnetic flux will be LM N OP Q joule joule × sec coulomb = as ampere = ampere coulomb sec [as (joule/coulomb) = volt] = volt × sec and is called weber = (Wb) or T . m2 (as tesla = Wb / m2 ). And as 1 volt = 108 emu of potential, the CGS unit of flux, maxwell (Mx) is related to weber (Wb) through the relation 1 Wb = 1 V × s = 108 emu of potentical × s = 108 Mx r As magnetic flux associated with elemental area dsr [Fig.] in a field B , r r d φ B = B. ds = B ds cos θ So for a given area (a) it will be maximum when area with r cosθ = maximum = 1, i.e., θ = 00 , i.e., magnetic field B is normal to the ( d φ B ) max = B ds ds ds B B B ds θ P P P d φ B = B ds cos θ (d φ B ) max = B ds ( d φ B ) min = 0 (A) (B) And (b) it will be minimum when cosθ = minimum = 0, i.e., θ = 90 r B is parallel to the area with ( C) 0 , i.e., magnetic field ( dφ B ) min = 0 In case of a body present in a field, either unifom, outward flux is taken to be positive while inward negative. B Negative flux B S N B Positive flux φ = − πR 2 B φ = +πR 2 B n n φ=0 n Total flux = 0 Nonuniform-field Total flux = 0 Uniform-field (A) (B) B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 1 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION z r r r r As φ B = B. ds , so if B = 0 . However the converse may or may not be true, i.e., if φ B = 0, B may or may not be zero. This is because if B ≠ 0 , flux for some elements may be positive while for the other negative giving zero net flux r e.g., for a closed surface enclosing a dipole φ T = 0 but d φ and hence B ≠ 0 . As magnetic lines of force are closed curves (i.e., monopoles do not exist), total magnetic flux linked with a closed surface is always zero, i.e., z r r B. ds = 0 This law is called Gauss's law for magnetism. As φ B = z z r r B. ds = B ds cos θ in case of translatory motion of a straight conductor of length L in a uniform field B, the flux linked with the area generated by the motion of conductor under different situations will be as shown in fig. B A y A B B V L L n L V B φ=0 as S = 0 (A) A V n B φ=0 as θ = 900 (B) φ=BLy Into the page ( C) As magnetic flux linked with a circuit φ B = BS cos θ , so flux linked with a circuit will change only if field B, area S or orientation θ or any combination of these changes. B V N B B n L S V S X Flux changes as B changes Flux changes as θ changes Flux changes as S changes (A) ( C) (B) The flux linked with a loop C will not change with time in situations shown in fig. as here any of B, S and change with time. C C V S X (A) i.e., ω1 V V S N θ does not B C ω2 B S N S X (B) φ = 0 but dφ =0 dt S V ( C) φ = BS cos θ = B ( NπR 2 ) cos 0 = NπR 2 B = constt. Faraday's Experiments : Oersted's discovery in 1820 of the magnetic effect produced by an electric curent revealed a link between electricity and magnetism and forced scientists to look for the inverse effect, i.e., production of electric current by a magnetic field. Faraday through a series of experiments during 1831-32 produced induced current for the first time on 29th August 1831. All his experiments were basically one or the other of the following : (a) Two coils are arranged so that a steady current flows in one and some of its magnetic flux links the other. If the current in the first coil changes, a current is induced in the second see [fig.]. B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 2 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION Current increased Inverse induced emf and current I coil B increases Current decreased Direct induced emf and current I coil B decreases II coil Flux increasing II coil Flux decreasing EMI due to changing field (A) (B) (b) A coil is arranged to link some of the magnetic flux from a source S (which may be either a magnet or a current). If relative motion occurs between coil and source S such that flux linked with the coil changes, a current is induced in it. [Fig.] C C Source Or S I Source Inverse induced emf and current N Direct induced emf and current Or S N S v1 v2 v1 Flux increasing EMI due to motion (A) v2 Flux decreasing (B) (c) If part of a conducting circuit is moving and there by cutting magnetic flux, a current is induced in the circuit see (fig.). Here it is worth mentioning that Faraday's disc see [fig.] was the first continuous generator. A B R I A B R R B A translating rod A rotating rod (A) (B) Faraday-disc ( C) Faraday's Laws of Electromagnetic Induction : Through experiments (described above) Faraday concluded that : (i) Whenever there is change of flux linked with a circuit or a moving conductor cuts the flux, an emf is induced in it. This phenomenon is called electromagnetic induction and the emf, induced curent. (ii) The induced emf persists only as long as there is change or cutting of flux. (iii) the magnitude of indujced emf is equal to the rate of change of flux, i.e., e = dφ dt Lenz's law : (iv) The direction of induced emf is such as to oppose the change that causes it. (This law is also called Lenz's law), i.e., the direction of induced current is inverse if there is an increase in flux while direct if there is decrease in flux linked with the closed circuit. All the above statements taken together are known as 'Faraday's laws of electromagnetic induction and are expressed analytically as' e=− dφ dt Negative sign shows that if flux increases e is negative, i.e., is in the direction opposite to that of applied emf and vice versa. Induced current and charge Regarding electromagnetic induction it is worthnoting that : In case of electromagnetic induction an emf e = ( d φ / dt ) always exists, either the circuit is closed or open, but the current will exist only if the circuit is closed. If the total resistance of the circuit is R, induced current B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 3 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION I= I= Now as e R .....(i) dq dφ and e = , dt dt So dq 1 d φ = dt R dt or q= i.e., z q 0 dq = 1 R z φ2 φ1 dφ φ 2 − φ1 ∆ φ = . R R .......(ii) i.e., induced charge is independent of the time in which the flux changes or is cut [while induced emf and current depend on time]. r For translatory motion of a rod perpendicular to its length, when it cuts B e = dφ = B υl dt r while in case of uniform rotation of a rod or disc in a field when it cuts B e = dφ 1 = Bl 2 ω = π l 2 fB dt 2 and in case of rotation of a coil in a field perpendicular to its plane e = dφ = NSB ω sin ωt dt A B l θ A B e= e = Bvl (A) 1 Bωl 2 2 (B) B n e = NSB ω sin ωt ( C) Though in general the direction of induced emf or current is determined by Lenz's law, in case of motion of a straight conductor in a magnetic field it can also be determined by the so called 'Fleming's right hand rule', according to which if one stretches the fore finger, the central finger and the thumb of right hand at right in the direction of motion, the central finger will point in the direction of the induced emf or current (fig.). In case of translatory motion of a straight conductor perpendicular to its length in a field B (as discussed in solved) when it cuts the flux, e = B υL . so if a conductor is moving vertically downwards with constant velocity v with its ends pointing east-west, it will cut the horizontal component of earth's field BH . see (Fig.). Thumb B Four finger B Direction of induced emf or current f al in ge r r nt Ce Fleming’s right-hand rule (B) (A) B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 4 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION and hence the flux linked with the area generated by the motion of the conductor, and induced emf will be e= φ H = BH ( ly ) and FG with υ H y = du dt IJ K Fm Fm BV BH BH nE W A vy N BH B W d φH = BH υ y l dt S BV E l BV l l e = BH v y l e=0 e = Bv V l (B) ( C) (A) However in case of vertical motion, if the ends of the conuctor point north-south, both BH and BV will be parallel to the plane of area generated by the motion of the conductor see [fig]; So φ=0 e= and dφ =0 dt And if the wire is moving in a horizontal plane in any direction see [fig.], it will cut flux of Bv (as BH will always be parallel to area) and so φV = BVls e= and d φv = Bv υl dt LMwith υ = ds OP dt Q N While solving problems of motion of a conductor in earth's field remember : tan φ = BV BH , i.e., BV = BH tan φ where φ is the angle of dip (vi) Induced emf for the sake of convenience is divided into two types. If the circuit (or part of it) in which emf is induced remains stationary (i.e, area S and orientation θ remains unchanged) while the field B changes with time the induced emf is called changing field emf see (fig.). In this situation : e= dφ d dB = ( BS cos θ) = ( S cos θ) dt dt dt However, if the field is steady and emf is generated due to cutting of flux by the motion of the circuit or part of it (i.e., either S or θ or both changes), the emf is said to be motional emf see [fig.]. In this situation : e= dφ d = B ( S cos θ) dt dt In practice the emf induced apart from the changing field type or motional can also be combination of the two. (vii) As by definition of emf r r r r dW F . dl e= = = E . dl q q z z z And as by Faraday's law of electromagnetic inmduction e = −( d φ / dt ) So z r r dφ e = E . dl = − ≠0 dt B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 5 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION r i.e., in case of electromagnetic induction, line-integral of induced field E round a closed path is not zero, i.e., induced electric field is non-conservative. The electric field produced by stationary charges is called electrostatic field and for it z r r E . dl = 0 i.e., it is conserva- tive and so it gives path independent potential difference. (viii) If the induced emf is due to changing magnetic field, z r r r dφ dB r E . dl = − =− . ds dt dt z which in advanced physics is written as r r dB curl E = − dt and is one of the four famous maxwell field equations and expresses the fact that a changing magnetic field (in absence of any charge) produces an electric field in free space in the form of closed lines of force as shown in fig. So if a charged particle is free to move in a circular path enclosing a changing magnetic field, the speed and hence kinetic energy of the charged particle will increase. This actually happens in betaron. This is also why a changing magnetic field can accelerate a charged particle atr rest though a steady magnetic field can never. B B Increasing Decreasing E E (B) (A) Note : Here it is worthy to note that : (a) In case of non-conservative electric field : (1) The work done round of a closed path is not zero (2) Lines of force are closed curves and can exist in free space even in absence of any charge (b) Inverse of the fact that a changing magnetic field produces an electric field is also found to be true, i.e., a changing electric field produces magnetic field and this fact is expressed mathematically by the relation z where I d = ε 0 r r d φE B. dl = µ 0 ε 0 = µ0 Id dt dφ e and is calloed displacement current. dt (ix) In case of motional emf, the motion of the conductor in the field exerts a force on the free charge (i.e., valence electron) in the conductor so that one end of the conductor becomes positive while the other negative resulting in a PD accross its ends due to which a non-conservative electric field is set up in the conductor. In steady state the magnetic force on the free charge is blanced by the electric force due to induced field. For example, in case of motion of a conductor r r r AB in a magnetic field B , as shown in Fig., magnetic force valence electrons, Fm = − q ( v × B ) = q υ B towards end B; so end B will become negative while A positive. If V is the PD between ends A and B which are separated by a distance l, E = (V / l ) and as due to this induced electric field electrons in the conductor will experience a force FE = qE r opposite to the field E , i.e., towards A, so in equilibrium q (V / l ) = q υ B , i.e., V = Bυl which is in agreement with Faraday's law e= dφ d = ( Bls) = B υ l dt dt FG with υ = dsIJ H dt K B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 6 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION Note : here it is worthy to note that the rod AB is acting as a source of emf and inside a source of emf direction of current is from lower potential to higher potential; so the point A of the rod is at higher potential than B through the current in the rod AB is from B to A. A A A Fe l I F V Fm Fm I B B (B) B (A) Fm ( C) (x) Phenomenon of electromagnetic induction in accordance with Lenz's law represents conservation of energy. e.g., if a conducting rod of length l is moved in a magnetic field B and the current I is induced in the Fig. the magnetic force opposing the motion of the rod will be FM = Bil . So rate of doing work in maintaining the motion of the rod by the pulling force F : dW = P = F υ = BIl υ = VI dt [as V = Bυl ] which is exactly equal to (V 2 / R ) [as I = (V / R ) ], i.e., the rate at which heat is produced in the circuit. So the phenomenon of electromagnetic induction in accordance with Lenz's law represents conversion of mechanical energy first to electrical and then to thermal. DISCUSSION : (i) A copper ring is held horizontally and a bar magnet is dropped through the ring with its length along the axis of the ring. Will the acceleration of the falling magnet be equal to, greater than or lesser than that due to gravity ? Answer : As shown in Fig. when the magnet approaches the coil, in accordance with Lenz's law a current will be induced in the coil which will oppose, i.e., repel the approaching magnet, so a= mg − F F = g− < g m m When the magnet has passed through the coil, the coil through electromagnetic induction will oppose the change, i.e., will attract the magnet and hence the acceleration of the magnet. a= mg − F F = g− < g m m F a N g S a g F S a N (A) g (B) a g ( C) Note : If the ring in the above problem has a slot, an emf (which is changing field type) will be induced in the ring but no current will flow; so the coil can no more oppose the approach of the magnet and the magnet will fall through the ring with an acceleration a = g . (ii). Two identical co-axial circular loops carry equal currents circulating in the same direction. What will happen to the current in each loop if the loop's approach each other B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 7 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION Answer : As the field at an axial point due to a current carrying coil is given by µ 0 2 πNIR 2 B= 4 π ( R 2 + x 2 ) 3/ 2 so when the coils approach each other the flux linked with each coil will increase. So in accordance with Len's law a current will be induced in each coil which will try to decrease the flux, i.e., the induced current in each coils will be opposite to initial current. So the current in each coil will decrease as the coil approach each other y I I C2 C1 I’ I’ C2 C1 (A) y < x S θ I '< I (B) Note : If the coil moves away from each other, the flux linked with each coil will decrease and so the induced current will try to increre the flux and hence current in each coil will increase. (iii) Three identical coils A, B and C are placed with their plances parallel to one onther.Coils A and C carry equal currents as shown in Fig. Coils B and C are fixed in position and coil A is moved towards B with uniform motion. Is there any induced current in B ? If no, give reason; if yes, mark the direction induced current in the diagram. A B C A (A) B C (B) Answer : Due to motion of coil A towards B, the fludx linked with B due to A will increase while that due to C will remain constant. So in accordance with Lenz's law a current will be induced in coil B which will oppose the change, i.e., approach of A towards it. So a current will be induced in B which will be in opposite direction to that in A (as opposite currents repel each other) as shown in Fig. (iv) In what way should the conductor AB be moved in a magnetic field such that the A current flows as shown in Fig. Which of the points A or B is at higher potential ? Answer : In accordance with Fleming's right hand rule, for the given direction of current I B and field B, the direction of motion of conductor AB is into the page. S N r I Now as due to motion of conductor AB, the free electrons in it due to field B will experience B a force along AB towards A and hence end A will experience a force along i.e., at a lower potential than end B. So end B will be at higher potential. Note : Here the conductor AB (through the phenomenon of electromagnetic induction) is acting as a source of emf and inside a source of emf current flows from lower to flows higher potential so the point B will be at higher potential and when the circuit is completed, the current outside the rod will be from B to A. (v) (a) A curent from A to B is increasing in magnetude. What is the direction of induced current, if any in the loop shown in Fig. (b) If inslead of current it is an electron, what will happen ? Answer : (a) When current in the wire AB increases, the flux linked with the loop (which is out of the page)will increase, and hence the induced current in the loop will be inverse, i.e., clockwise and will try to decrease the flux linked with it, i.e., will repel the conductor AB as shown in Fig. (b) If instead of current the moving charge is an electron moving from left to right, the flux linked with the loop (which is into the page) will first increase and then decrease as the electron passes by. So the induced current I i in the loop will be first anticlockwise and will change direction (i.e., will become clockwise) as the electron [i.e., crosses point C]. B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 8 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION Bout Ii Ii B A I I = Constant φ = Constant I i = Zero B A B A C I ACW CW C I = Decreasing I = Increasing φ = Decreasing Ii = Direct, ACW φ = Increasing I i = Inverse i.e. CW (A) ( C) (B) (vi) A bar magnet is pulled rapidly through a conducting coil along its axis with uniform velocity, with its south pole entering the coil first. Sketch qualitatively (a) the induced current and (b) the joule heating, as a function of time. Answer : (a) When the magnet is moved towards the coil, the flux through the coil will increase nonlinearly from zero to a maximum and then will decrease as B ∝ (1 / x ) 3 as shown in Fig. Now as induced emf e=− dφ = − slope of (φ / t ) curve dt So emf [and hence current (e/R)] will vary with time as shown in Fig. φ P A C A B t C A t B t C S (A) ( C) (B) (b) as Joule heating, i.e.,heat produced per sec., P = ei = FG as i = e IJ H RK e2 R it will vary with time as shown in Fig. (vii). A rectangular flat loop of wire with dimensions l and b has N turns and a total resistance R. The loop moves with constant velocity from P to Q through a region of constant magnetic field B as shown in Fig. Discuss the variation of (a) the flux linked with the loop (b) the emf induced in the loop and (c) the external force acting on the loop as a function of position of the loop in the field. Answer : (a) As flux linked with the loop : P ( x = 0) x = 2b Q z r r φ = B. ds = BS A l [as B = constt. and So the positions in the field in a tabular form is given below : S. No. 1. 2. 3. 4. 5. θ = D0 ] Position of side Flux linked with AB from P the loop φ x<0 0< x <b b < x < 2b 2b < x < 3b x > 3b B b 0 (as B = 0) C b 2b The emf induced in the loop e = − dφ 1 dt 0 BNlx −BυlN BNlb 0 BNl ( 3b − x ) BυlN 0 (as B = 0) 0 B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 9 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION From this table it is clear that as the loop enters the field, the flux linked with it frirst increases linearly and then after becoming constant (when the whole loop is in the field) be gins to decrease linearly (when the loop comes out of the field). This is shown graphically in Fig. (b) As in case of electromagnetic induction for constant field, the motional emf e= dφ d ds = B ( S cos θ) = B dt dt dt [ as θ = 0 here] So the inverse (i.e., CW) when it is leaving the field. when the whole loop is in the field and is moving through it cutting the flux, the induced emf will be zero (as the flux linked with it remains constant). All this is shown in the above table and Fig. BNlb Bv 2 l 2 BvlN N φ F R b 0 0 0 2b 3b 4b 2b 3b 4b b 2b 3b 4b (A) (B) ( C) (c) As the net force on a current loop in a uniform field is always zero, the loop will experience a force : FM = BIl = Bl FG e IJ = B l υ N H RK R 2 2 [as e = BυlN ] opposing its motion only when it is entering or leaving the field. So in maintaining uniform motion of the loop an external force B 2l 2 υ F = FM = N R parallel to the motion of the loop is applied when it is entering or leaving the field as shown in Fig. Note : In this problem the induced current and emf in the loop is zero even through the moving conductor is cutting the flux. This is because the total flux linked with the loop (NBlb) remains unchanged when the loop A B moves through the field. Actually this situation is electrically equivalent to two identical sources of emf connected to each other as shown in Fig. giving net zero current and emf in the loop. Types of Electromagnetic Induction and Inductance Electromagnetic induction has been divided into the following two types : D C (A) Self-Induction Whenever the electric current passing througha coil or circuit changes, the magnetic flux linked with it will also change. As a result of this, in accordance with Faraday's laws of electromagnetic induction, an emf is induced in the coil or thge circuit which opposes the change that causes it. This phenomenon is called 'self induction' and the emf induced, back emf. Circuit is made off or I decreasing Circuit is made on or I increasing I e= L dI dt I e= L dI dt I I (B) (A) So if I is the current flowing through a circuit, φ ∝ I i.e., φ = LI So coefficient of self-induction of a coil or circuit is numerically equal to the flux linked with it when unit current is passing through it.The configuration having inductance is called inductor and is represented by the symbol (B) Mutual Induction Whenever the current passing through a coil or circuit changes, the magnetic flux linked with a neighbouring coil or B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 10 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION circuit will also change. Hence and emf will be induced in the neighbouring coil or circuit. This phenomenon is called 'mutual induction'. The coil or circuit in which the current changes is called 'primary' while the other in which emf is set up is called 'secondary'. IP IS P IP S IP S P ~ Variable current Load P R S M (B) (A) In case of mutual inductance for two coils situated close to each other, flux linked with the secondary due to current in the primary. or φS ∝ I P φ S = MI P where M is a constant of proportionality and is called coefficient of mutual, induction or simply mutual inductance and as from M = φS if IP = 1 i.e., coefficient of mutual inductance of two coils or circuitsx is numerically equal to the flux linked with one circuit or coil when unit current flows through the other. Regarding self and mutual induction it is worth-noting that : (1) In case of self or mutual induction as φ = LI or φ S = MI P and according to Faraday's law of electromagnetic induction e = −(dφ / dt ) , e = −L so dI dt eS = − M or dI P dt i.e., coefficient of self induction is numerically equal to emf induced in a coil when the rate of change of current in it is unity. while coefficient of mutual induction is numerically equal to emf induced in one cdoil when the rate of change of current in the other is unity. (2) As e = L FG dI IJ or H dt K FG dI IJ , the dimensions of inductance, i.e., [L] or [M] will be H dt K LF dI I O L ML T × T O = [ ML T A ] [ e] × MG J P = M MNH dt K PQ N AT A PQ e =M −1 2 −2 2 Note : As magnetic energy stored in a coil U = −2 −2 1 2 LI , so 2 [ L] = [U / I 2 ] = [ ML2 T −2 A−2 ] And so SI unit of L or M will be : kg m2 s −2 = J V ×s Wb Tm2 = = ohm × s = = A2 A A A and is called henry (H). (3) An inductance is said to be ideal if it has no resistance. In partice due to finite resistance of conductors an inductance always has a resistance, i.e., one cannot have inductance without having resistance. However, converse may or may not be true, B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 11 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION R ≠ 0 and L ≠ 0 R ≠ 0 but L = 0 (A) (B) i.e., one can have a resistance with or without having inductance. A resistance without inductance is called 'noninductive resistance' and is shown in Fig. (4) The mutual inductance M of two coils or circuits having self-inductance L1 and L2 is given by M = k L1 L2 where k is a constant called 'coefficient of coupling'. If the coils are wound over each other the coupling is said to be 'right' otherwise 'losse'. For tight coupling k = 1 and so M = L2 L2 while for loose coupling 0 < k < 1 and hence M < L1 L2 . Furthermore form expression (6) it is also clear that if L = 0 , M will be zero, i.e., a system cannot have mutual inductance without having self-inductances. However, converse may or may not be true, i.e., if mutual inductance of a system is zero it may or may not have self-inductances as M = 0 can beatisfied either by setting k = 0 or L = 0 . (5) Self-inductance in some cases of interest : (a) Self-inductance of a coil : As for a coil of radius R having N turns, field at the centre of the coil, B= µ 0 2 π NI 4π R so flux linked with the coil due to its own current, φ = BS = FG H IJ K µ 0 2 π NI µ ( π R 2 N ) = 0 ( 2 π 2 N 2 R) I 4π 4π R But as by definition φ = LI , so LC = µ0 1 (2 π 2 N 2 R) = µ 0 πN 2 R 4π 2 (b) Self inductance of a solenoid : In case of a solenoid having n turns per unit B= FG IJ H K µ0 N (4 πnI ) = µ 0nI = µ 0 I 4π l FG as n = N IJ H l K and so if the cross-sectional area of solenoid is S F NI φ = B( NS ) = µ G J I × ( NS ) = µ HlK 0 0 N2 SI l But as by definition φ = LI , so LS = µ 0 FG N IJ S = µ n Sl HlK 2 2 0 (6) Energy stored in a coil When current in a coil is changing, due to opposition by the coil through its self inductance L, work done in time dt, dW = P dt = eI dt = LI dI FG as e = L dI IJ H dt K B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 12 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION So work done in estabilising a current I in the coil z I W = LI dI = 0 1 2 LI 2 This work is stored as 'magnetic potential energy' U M which is not localised but is distributed in the field associated with the current-carrying coil, i.e., UM =W = 1 2 LI 2 From it is clear that L = 2W if I = 1 . So coefficient of self induction of a coil is also numerically equal to twice the work required to establish a unit current in the coil. Note : As for a solenoid L = µ 0n 2 SI , magnetic energy per unit volume, 1 2 LI 1 U uM = = 2 = µ 0n 2 I 2 volume lS 2 and as for a solenoid B = µ 0 nI , FG IJ H K 1 B uM = µ 0 × 2 µ0 2 = 1 B2 2 µ0 This expression is general and is called 'energy density' of magnetic field and is a magnetic analogue of uE = 1 ε0 E 2 . 2 (7) Coils in Series and Parallel (a) Coils in Series : If two coils of inductances L1 and L2 are connected in series with coefficient of coupling k = 0 , then as in series, the potential divides, i.e., e = e1 + e2 or LS dI dI dI = L1 1 + 2 dt dt dt LMas e = − L dI OP dt Q N However in series as current remains same I = I1 = I 2 dI dI1 dI 2 = = dt dt dt i.e., LS = L1 + L2 So I1 L1 L2 e1 LS I2 I e2 (B) (A) (b) Coils in Parallel : As in case of parallel combination of coils, the current divides, i.e., I = I1 + I 2 i.e., or dI dI1 dI 2 = + dt dt dt LMas e = − L dI , i. e., dI = − e OP dt dt LQ N e e e = 1 + 2 LP L1 L2 B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 13 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION I I1 I2 L1 L2 I LP I (A) (B) However, in parallel, as potential remains same, i.e., e = e1 = e2 , So 1 1 1 = + LP L1 L2 i.e., L1 L2 ( L1 + L2 ) LP = From expressions it is eviudent that 'grouping of inductance's is governed by formulae similar to that of grouping of resistances, i.e., RP = RS = R1 + R2 and R1 R2 ( R1 + R2 ) Note : In case oof series grouping of two inductors, if mutual inductance is also taken into account with k = 1 , then L = L1 + L2 ± 2 M as shown in fig. L1 L2 M L1 M L2 L = L1 + L2 + 2 M L = L1 + L2 − 2 M Current in the two coils is in same direction Current in the two coils is in opposite direction (A) (B) (8) Growth and Decay of Current in L-R circuit (a) Growth of Current : When a circuit containing an inductance L and a resistance R in series is connected to a battery of emf E as shown in Fig., the current in the ciruit will grow with time. So if I is the curent in the circuit at any time t, from Kirchhoff's loop rule we have : I E b a E R E 0.693 I 0 L R I0 = dI dt I = 0 at t = 0 t (A) t (B) ( C) Growth of current in L-R circuit E−L or dI − RI = 0 dt dI 1 = dt ( E − RI ) L dI + RI = E dt i.e., L i.e., − R dI R = − dt ( E − RI ) L The above expression on integration with initial condition I = 0 at t = 0 yields : I = I 0 (1 − e − t / τ ) with I0 = E R and τ= L R This is the required result and from this it is clear that : (i) The current in the circuit grows exponentially with time from 0 to the maximum value I 0 ( = E / R ) as shown in fig. This in turn implies that just after closing the switch as I = 0 , inductors act as open circuit, i.e., broken wires Fig., and long after the switch has been closed as I = I 0 , the inductors act as a short circuit, i.e., a simple connecting wire as shown in Fig. B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 14 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION R I =0 E =0 E I= I =0 e=0 E R E R e=0 L S Initially S S Just after closing the switch S Long after closing the switch S (A) ( C) dI I 0 − t / τ = e dt τ (ii) As I = I 0 (1 − e − t / τ ) , i.e., (B) LMas I N dI E R − t / τ = × e dt R L 0 E L and τ = R R = OP Q dI E = L × e − t / τ = Ee − t / τ dt L From this expression it is clear that as t ≥ 0, e ≤ E , i.e., induced emf in a circuit can never be greater than applied e =L So emf. Also as thjis expression is independent of current, we come to the conclusion that an inductor can be have induced emf even in the absence of current through it, e.g., at t = 0, e = E but I = I 0 (1 − e −0/ τ ) = 0 . (iii) The constant τ = ( L / R ) has the units and dimensions of time as τ= L H Ω×s = = =s R Ω Ω and hence called 'inductive time constant' of the circuit and if in eqn. t = τ I = I 0 (1 − e −1 ) = I 0 [1 − (2.718) −1 ] = I 0 [1 − 0.3678] = 0.632 I 0 So inductive time constant of a L − R circuit is numerically equal to the time taken by the current to grow from 0 to 0.632 times of its maximum value in the circuit. Greater the time constant more slowly will the current rise or fall in the circuit. (b) Decay of current : When the current in the L − R circuit has reached its maximum value I 0 ( = E / R ) , if the battery is removed from the circuit, the current in the circuit cannot abruptly drop to zero due to opposition by inductance and decay to zero over time. E I0 = I = I θ at t = 0 b a E R I R L t Decay of current in L-R circuit (B) (A) The differential equation governing the decay will be obtained by putting E = 0 in Eqn., i.e., for this case L= dI + RI = 0 dt Proceeding in a similar fashion as in case (a), on integration with initial condition I = I 0 at t = 0 the above equation yields : I = I 0e − t / τ I0 = with E R and τ= L R B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 15 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION This is the required result and from this it is clear that current in a L − R circuit decays exponentially with time as shown in Fig. (9) Oscillatory discharge of L-C circuit When a charged capacitor C having an initial charge q0 is discharged through an inductance L, Kirchhoff's loop rule for the circuit yields : q0 C O T/2 T q0ω L t I (B) (A) VC + V L = 0 i.e., q dI −L =0 C dt However, as here charge is decreasing with tim, i.e., I = −( dq / dt ) , so the above equation reduces to FG H IJ K q d 2q −L − 2 =0 C dt i.e. 1 d 2q + q=0 2 dt LC This equation is of the same form as equation for simple harmonic motion : ( d 2 x / dt 2 ) + ω 2 x = 0 . So the charge oscillates in the circuit with a natural angular frequency ω= 1 LC and in general its variation with time will be given by : q = q0 sin(ω t + φ ) However, as here at t = 0, q = q 0 so sin φ = 1 i.e., φ = ( π / 2 ) and hence FG H q = q0 sin ωt + IJ K π = q0 cos ωt 2 The current I in the circuit at any time t will therefore be dq d = (q0 cos ω t ) = − q0ω sin ω t dt dt π I = I 0 cos ω t + with I 0 = q 0ω 2 I= or FG H IJ K From Eqn. it is clear that in cale of oscillatory discharge of a capacitor through an inductor current leads the charge in phase by ( π / 2 ) radian. this in turn implies that when charge is maximum (or minimum) current is zero and vice-versa. Note : here it is worthy to note that if a charged capacitor is discharged through an inductor L and a resistor R in series, the discharge will be oscillatory only if FG IJ H K 1 R > LC 2L 2 and the motion of charge in the circuit will be damped oscillatory with angular frequency ω= FG IJ H K 1 R − LC 2L 2 B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 16 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION Question VIII. What will happen to the inductance of a solenoid (a) when the number of turns and the length are doubled keeping the area of cross-section same, (b) when the air inside the solenoid is replaced by iron of relative permeability µ r ? Answer : In case of a solenoid as B = µ 0 nI , φ = B ( nlS ) = µ 0 n 2 lSI and hence L= FG as n = N IJ H l K φ N2 = µ 0 n 2 ls = µ 0 S I l So (a) when N and l are doubled, (2 N )2 N2 L' = µ 0 S = 2µ 0 S = 2L 2l l i.e., inductance of the solenoid will be doublled, (b) When air is replaced by iron, µ 0 will change to µ , so that L' = µ n 2 lS L' µ = = µ r i.e., L µ0 and hence L' = µ r L So inductance will become µ r times of its initial value. Some Applications of Elkectromagnetic Induction (A) Eddy Currents When a changing magnetic flux is applied to a bulk piece of conducting material circulating currents called eddy currents are induced in the magerial. Because the resistance of the bulk conductor is usually low, eddy currents often have large magnitudes and heat up the conductor. F F (B) (A) When a metal plate of non-magnetic material such as copper of aluminium is allowed to swing through a strong magnetic field, then in entering or leaving the field the eddy currents are set up in the plate which opposes the motion as shown in Fig. These currents dissipate energy within the conductor as heat. The source of the dissipated energy is the kinetic energy of the moving conducting plate. So the kinetic energy of the moving conducting plate is reduced and the plate slows down resulting in electromagnetic damping. This electromagnetic damping is used to damp the oscillations of a galvanometer coil or chemical blance and in braking electric trains. The heat produced by eddy currents in metals in used in cooking and smelting (Induction furnace). Eddy currents are often undesirable. To reduce their effects slots are sometimes cut into moving metallic parts of machinery Fig. These slots intercept the conducting paths and decreases the magnitudes of the induced currents. Most transformers and motors have parts constructed from alternate layers of conducting and nonconducting substances such as lacquer, shellac or metallic oxide. Such laminations serve to interrupt the conducting paths and reduce the conversion of electrical energy to thermal energy due to I 2 R heating caused by eddy currents. Here it is worthy to not that eddy current losses can be minimised but can never be made zero (like radiation losses). B) Back EMF of Motors An electric motor converts electrical energy into mechanical energy, i.e., (rotational kinetic energy) and is based on the fact that a current-carrying coil in a uniform magnetic field experiences a torque. As the coil rotates in the magnetic field, the flux linked with the rotating coil will change, and hence an emf called 'back emf' is developed in the coil. The back emf B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 17 PDF created with pdfFactory Pro trial version www.pdffactory.com ELECTROMAGNETIC INDUCTION e is proportional to the angular speed of the coil motor, armature current I= E −e R ω and opposes the impressed voltage E. So if R is the resistance of e ∝ω with When the motor is first turned on, the coil is at rest and so there in no back emf. The 'start up' current can be quite large because it is limited only by the resistance of the coil. As the rotation rate increases the back emf increases and hence the current reduces. If there is no load attached to the motor, the angular speed increases until the input energy just balances the frictional and resistive losses. At this stage as back emf is maximum, current is quite small. When a load is applied, the angular speed decreases which in turn reduces the back emf. As a result of this the current increases. The additional power supplied by the external source of emf is converted into mechanical power of the motor. If the load is too large the back emf is reduced still further. The larger current that results may cause the motor to "burn out". B-38, 2nd Floor, Shivalik, Malviya Nagar, New Delhi-110017, email - info@topperselearning.com, Tel.: 011-32713055 18 PDF created with pdfFactory Pro trial version www.pdffactory.com

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