6-ELECTROMAGNETIC INDUCTION-01-THEORY

ELECTROMAGNETIC INDUCTION When a magnetic field is produced by a current, one calls it magnetic induction, that is a magnetic field is induced by an electric current. When an emf is produced by a change of magnetic flux, it is called electromagnetic induction. Here an electric field is induced by a change in magnetic field. 1. INDUCED EMF: SOME EXPERIMENTAL OBSERVATIONS : The best way to study this is to start from the very simple experiments of Faraday whose conclusions are very profound. (a) (c) S N v=0 S N v=0 i=0 S N Ammeter (b) S N (d) Ammeter v S N (e) v (f) (g) A coil is suspended in an area having uni- form magnetic field. The magnetic field is upwards. There is no relative motion between the two but the magnetic field is varied in time, for example, by decreasing the current flowing in the electromagnet. (h) v B is a constant (i) B is a constant Draw a circle with a thick pen on a piece of thin paper, like this i The current is flowing clockwise. By the right hand thumb rule B is acting downwards, entering into the paper. This is equivalent to a south pole. MAGNETIC FLUX : For a coil in the magnetic field similar to electric flux a quantity is defined known as magnetic flux given as   → ( in uniform magnetic field)   d   → (in non uniform magnetic field). Lenz’s law states that the induced current always opposes any change in the magnetic flux. If one tries to bring a north pole of a magnet near a coil (or a coil tries to come near a north pole), the lines of force entering the coil is opposed by a north pole induced by the current. Therefore the current flows in the anticlockwise direction. If one takes away a north pole, to keep the lines of force, the coil has induced current in the clockwise direction, making a south pole which attracts the lines of force back trying to prevent a change. The magnetic moment of a current carrying loop is → → where → is the area of the loop. M  iA A Induced emf by Faraday’s law   d where  is the magnetic flux. The negative sign shows that dt the induced emf opposes the change. To understand why an induced emf is produced, we have to study (i) Induced emf produced by relative motion and (ii) Induction produced by a varying magnetic field. These two processes are different: 1. In (a) and (f) there is no change of magnetic flux in the coil, but magnetic flux is present. There is also no relative motion of the coil changing the flux (as in (h) and (i)). The observation is that there is no current in the coil because that is no change in the magnetic flux inside the coil in both the cases (a) and (f). 2. In (h) and (i), the magnetic field is a constant but the flux inside the coil is changed by a relative motion of the coil with respect to the field. The observation is that there is an induced current because there is a change in the flux. The change in the flux is caused by the relative motion of the coil. 3. In (g), there is no relative motion but the flux is changed inside the coil by varying the magnetic field. Then also there is an induced emf in the coil. 4. In (b), (c), (d), (e), there is relative motion between the magnet and the coil and because the proxim- ity to the magnet changes, there is also a variation in the magnetic flux produced inside the coil. The current is also observed. i DIRECTION OF THE INDUCED CURRENT: If a current is flowing in the anticlockwise direction, the magnetic lines of force are going upwards as in the north pole of a magnet. Therefore the polarity of the coil looking from above is north pole. The direction of the current induced is such that it always opposes any change in the magnetic flux in the coil. This is Lenz’s Law. 2. MOTIONAL EMF : When the emf induced is due to change in the flux produced by the relative motion between the constant magnetic field and the coil, the induced emf is called motional emf. × × × × × If the arm A B is movable, in the rectangular coil, if A B were to move towards CD, (same as in the earlier experiments) the flux in the coil is changing. Here the area is reduced in the coil so that there is a decrease in the flux in the loop. This is the same as 1(b). The electrons in AB are having a velocity  in the –ve x-direction. The magnetic induction B is in the –z direction. → → → (a) The force acting on a charge due to a magnetic field is   . F qv B The electrons are moving from B to A in the conductor due to Lorentz force. The work done in moving the electrons through the distance 𝑙 is W = q  B . The work done per unit charge, the emf = W  B𝑙v iˆ ˆj kˆ → = - q v o o  (q)( ˆj)(vB)   ˆj qv B F o o B (b) By the right hand palm rule also one finds that the electrons are flowing from B to A i.e. the current is in the clockwise direction, flowing from A to B. (c) According to Faraday’s law    d , where   B𝑙x when the rod moves through a distance dt x. If it moves through x in time t,    B𝑙x  B𝑙v . t The result is identical to the one obtained by Lorentz force method. The – ve sign shows the current is opposing the charge. It is clockwise, looking from the top. As the area is decreased, less lines are entering the area and to have more lines entering the area, it should be south pole. Electrons flow from B to A. The current flows from A to B. 3. INDUCED EMF BY A VARYING MAGNETIC FIELD: → → dB ––→ dE   dB □ E.dl   dt .da where da is the element of area. In the differential form, dx An electric field can be generated by a variable magnetic induction. dt . → In the Lorentz equation of force,  →  →  → . Here it is the first part which produces the in- duced emf. F qE qv B → → d → → is the emf and ( ) is the rate of change of flux. □ E.dl dt B.da The electromagnetic induction, i.e., the production of induced emf due to a variable magnetic field is produced by the first part of the Lorentz force and motional emf (due to a constant B and relative motion between the field and the coil) is due to the second part of the Lorentz force. In (g) for example, there is no relative motion between the coil and the magnetic field but the mag- netic field is varied. The sign : : means out of the paper (‘o’ for out of). If , for example, the field is decreased, less lines will flow out of the loop. To compensate, the north pole is generated which will increase the lines; this means the current is anticlockwise. Even without a coil, one can have an electric field. However the field is detected by a current just as the existence of a magnetic field is detected by a compass needle. i A current flows because of the electric field which applies a force on the electrons to make it move and thus produce current. But E   dV  □ → dV  V. Where is – ve when one measures from a d𝑙 E.d𝑙 d𝑙 higher to a lower value. The independent existence of the electric field is explained by the action of a betatron, for accelerating β particles or electrons, where a variation of a magnetic field causes electron to move in a circle.    d  a2 dB dt dt The resistance of the wire  (r2 )(dB / dt) (r2 ) R  𝑙 ; r2  = specific resistance of the wire i  R  𝑙 . Here 𝑙 = 2 a where a is the radius of the loop. Illustration: A square wire of length l, mass m and resistance R slides without friction down the parallel conducting wires of negligible resistance as shown in figure. The rails are connected to each other at the bottom by a resistanceless rail parallel to the wire so that the wire and rails form a closed rectangular loop. The plane of the rails makes an angle  with horizontal and a uniform vertical field of magnetic induction B exists throughout the region. Show that the wire acquires a steady state velocity of mag- nitude v  mgR sin  . B2𝑙2 cos2  Solution: Force down the plane = mg sin  At any instant if the velocity is v the induced emf = 𝑙 B cos  × v Current in the loop  𝑙Bcosv R Force on the conductor in the horizontal direction  B𝑙 Bcos  v 𝑙 R Component parallel to the incline  B2𝑙2 cos   R B2𝑙2 cos2     If v is constant R v mg sin  v = mRg sin  B2𝑙2 cos2  . 4. INDUCTANCE : If there are N coils, the flux associated with all the coils is N times that of a single coil. Taking an ideal solenoid, if a current i is flowing through the coil which has n turns per unit length, by taking an Amperean path ABCD, AB is outside the coil where there is no induction. AD and BC are perpen- dicular to magnetic lines and Amperean path has to be parallel to the lines. Therefore CD is the only effective length. A B → → The magnetic induction is given by Ampere’s law, □ B.dl  u0i where i is the current enclosed. If h is the length of the path CD, B. h = 0in h where n is the number of turns per unit length.  B  0ni Where the current is switched ON, the magnetic field is induced in the coil which increases and due to a change of the magnetic induction due to its own current, an emf is induced which tries to oppose the change of flux. This is a self induced emf. The emf induced di . dt or   L di dt L is called the self inductance of the coil (or just inductance) and the negative sign shows that it opposes the change in flux. The unit of L is henry. 1 henry = 1 volt second / ampere. But  is also = - NdB , where N is the total number of coils. dt The total flux associated is N B = n 𝑙 BA where A is the area of cross-section of the solenoid.  NB  n𝑙(0ni).A d(NB )   n 2𝑙A di  0n 2𝑙.A i dt 0 dt i.e.,   L di . dt Therefore L = 0n 2𝑙A . The inductor coil is a very special device. The magnetic induction inside is B = 0ni independent of the total length or area of cross-section of the coil and independent of the distance from the axis. This means one has uniform magnetic induction inside solenoid just as between a parallel plate capacitor, one has uniform electric field E   0 . In the case of the capacitor  is the charge per unit area and in the case of the solenoid, ni is the current associated with unit length of the coil. Illustration: Calculate the self inductance per unit length of a current loop formed by joining the ends of two long parallel wires of radius r separated by a distance d between their axes, neglecting the end effects and magnetic flux within the wires. Solution: Let A and B are the two long parallel wires as shown which are formed to form a current loop. The two wires would obviously be carrying equal currents in opposite direction. Consider a space length l and thickness dx as shown in figure. The magnetic flux through the area 𝑙 dx due to the currents in the two wires is d  B𝑙dx where B is the intensity of magnetic field due to current carrying wires at a distance x from A. B  0 2i  0 . 2i i 4 x 4 d  x A x P  i  1 1  d  B𝑙dx  0  𝑙dx 2  x d  x  d dx d-x dr  i  1 1  B     0    𝑙dx i r 2  x d  x  0i𝑙 [log x  log (d  x)]dr  0i𝑙 log d  r 2 e e r  e r Magnetic flux linked with unit length of the loop  0il log  e d  r r But  = Li where L is self-inductance. L  0𝑙 log  e d  r r . Illustration: A rod PQ of length L moves with a uniform velocity v perpendicular to its length and parallel to a long straight wire carrying a current i, the end P remain at a distance r from the wire. Calculate the e.m.f. induced across the rod. Solution: Figure shows a rod PQ of length L which moves with a uniform velocity v parallel to a long straight wire carrying a current i. Here a magnetic field is produced by the current carrying wire and rod moves in this field. Consider a small element of length dx of the rod at a distance x and x + dx from the wire. The e.m.f. induced across the element deBv dx …(i) We know that magnetic field B at a distance x from a wire carrying a current i is given by B  0  i 2 x …(ii) From equation (i) and (ii) de  0 i v dx 2 x …(iii) The e.m.f. e induced in the entire length of the rod PQ is given by Q  i e  de  0 v dx P 2 x rL  i   0 v dx r 2 x  rL dx  iv rL  0 .i.v.  0 log x 2  x 2x e r  0iv[log (r  L)  log r] 2  0iv log e e  r  L  . 2 e  r    5. TOROID : A toroid of very large radius r is taken so that the difference between the outer and inner radii can be neglected. Amperian loop By taking an Amperean path ABCD, → →  □ B.dl  B  2π r  μ0ni  2π r where n = N / 2  r. and B  μ0ni Total magnetic flux associated with the toroid = φ  n  2π r  μ0 ni  A tional area of toroid where A = cross-sec-   n.2r . 0ni . A d ()   n2l A di where l  2r dt 0 dt  L   n2lA However when the inner and outer radii are different, the field will not be uniform but the flux will be denser near the inner surfa–c–e→. b  B   B.dA   Bhdr a if the toroid has rectangular cross-section of thickness h. h b  Ni  Nh b B   0 hdr  0 i ln 2r a 2 a L  NB i  N2h b 2 ln a L depends only on geometrical factors as in the case of the capacitance of a parallel plate capacitor. Just as the capacitance of a capacitor can be increased very much by filling the space between the capacitors by a dielectric of high dielectric constant, the inductance can be increased by having a core of material with high magnetic permeability. For example, for a parallel plate capacitor, Cair  0A d with a dielectric of relative permittivity r (which is also called the dielectric constant k), The capacitance C  r0A i.e. C  kC d vac or r C vac Similarly, for a solenoid, with an air core, L = 0n 2𝑙A . For a magnetic core, L = km Lvac where km is the permeability of the material. By having paramag- netic materials or diamagnetic materials, the change in the inductance in only marginal. 6(a) INDUCTANCES IN SERIES : L1 L2 The same current is flowing through L1 and L2 and the induced emf’s of L1 and L2 are added. L di  L di  L di dt 1 dt  L  L1  L2 . 2 dt 6 (b) INDUCTANCES IN PARALLEL L1 when inductances are in parallel,(a) i  i1  i2 and ® i2 (b) e.m.f. is the same. di  di1  di2 dt dt dt  ε  ε  ε  1  1  1 L L1 L2 L L1 L2 7. MUTUAL INDUCTANCE r2 l r1 1 N1 turns 2 N2 turns Let two co-axial coils S1 and S2 have radii r1 and r2. If a current is switched on S2, before the current reaches the maximum, the current is increasing i.e. it varies with time. A magnetic field is set up which varies with the current. This varying magnetic field in turn set up a varying electric field on S1.   M12I2 M12 is the mutual inductance or inductance effect on coil S1 due to coil S2.    d1  M dI2 1 dt 12 dt BS  0 n 2 I2 1 is associated with n1𝑙 coils of S1 due to S2  πr2 μ n I n 𝑙 1 0 2 2 1  μ0n1n 2 πr2 𝑙I2 M12  μ0n1n 2 πr2 𝑙 If I1 is passed through S1, the flux through S2 is  2 = M21 I1 From Faradays laws, 2  M21   r2  n I  n 𝑙 dI1 dt πr2 only is taken because due to the first solenoid, the magnetic induction is only inside the first solenoid. In the earlier case, π r 2 was used to find the flux in that area due to the solenoid S2.  M  2   n n r2𝑙  M 21 0 1 2 1 12 1 (1) Different currents are passing through S1 and S2 but M is independent of I1 or I2. (2) Only the inner area πr 2 is taken ( The overlapping area ). (3) 𝑙 is the overlapping length. If one coil is wound over another having the same length and same area, L1  μ n 2𝑙A, L  μ n 2𝑙A & M   μ0n1n 2𝑙A If the coupling between the two coils is not good, M  K where K, the coupling constant is < 1. Note: Unless M is either given or one is asked to calculate it, when inductances are taken in series or in parallel, one neglects M. Illustration: Space is divided by the line AD into two regions. Region I is field free and the region II has a uniform magnetic field B directed into the plane of paper. ACD is semicircular conducting loop of radius r with centre at O, figure, the plane of the loop being in the plane of the paper. The loop is now made to rotate with a constant angular velocity  about an axis passing through O and perpendicular to the plane of the paper. The effective resistance of the loop is R. (i) Obtain an expression for the magnitude of the induced current in the loop. (ii) Show the direction of the current when the loop is entering into the region II. (iii) Plot a graph between the induced e.m.f. and time of rotation for two periods of rotation. Region I Region II × × × A × × ×  × × × C O × B× × × × × × × × D × × × × × × Solution: (i) When the loop is in region I, the magnetic field linked with the loop is zero. When the loop enters in magnetic field in region II. the magnetic flux linked with it, is given by   BA  e.m.f. induced e   d dt   d(BA) dt Region I Region II × × × E r d× × × × × ×  B dA  B dA dt dt (numerically) × B× × × × × Let d be the angle rotated by the loop in time dt, then dA = Area of the triangle OEA = (1/2) r.r. d e  1 r.r d 1 2 d 1 2 2 dt 2 dt 2  e.m.f.  1 Br2   r × × × × × × × × × Induced current R 2 R dA can also be calculated in the following way: The area corresponding 2 (angle) is r2 . r2  area corresponding to unit angle d  2  d r2 area corresponding to angle d   d 2 r2 1 2 dA   d  r d . 2 2 (ii) According to Lenz’s law, the direction of current induced is to oppose the change in magnetic flux. So that magnetic field induced must be upward. In this way, the direction of current must be anticlockwise. (iii) The graph is shown in figure. When the loop enters the magnetic field, the magnetic flux liked increases and e.m.f. e  (1/ 2)Br2  in one direction. When the loop comes out of the field, the flux decreases and e.m.f. is induced in opposite sense. 1 Br 2 2  1 Br2 2 8. ENERGY, ENERGY DENSITY: (a) The work done by a pure inductance (that is the energy stored on the inductance) dW  i  L di i  Li di dt dt dt dW  L i di   W  1 Li2 2 i W   L i di 0 This energy that is stored in the inductance is completely magnetic energy. (b) Inside a capacitor, it is electric field energy that is stored: Let at some stage, the charge on each of capacitor plates be q. The potential difference be- q tween the plates is C . If an additional charge dq is added to the plate, the work done dW is Vdq qmax q dW  W   dq 0  W  1 q2  1 2 C 2  1 q2    1 CV2 2  1 q V. 2 Here q is qmax A capacitor stores electric energy. This is something like a spring. Comparing their values in series and in parallel, 1  1  1 in series and C = C + C in parallel. C C1 C2 1 2 ( 1  1  1 for springs in series and K = K  K when they are in parallel. K is the spring K K1 K 2 1 2 constant). Energy density = Energy per unit volume, 1 Li2 uB  2 A𝑙 But L = μ0n 2𝑙A B inside a solenoid = 0ni i  B ; μ0 n 1  n2𝑙A B2  uB  0 2 A𝑙 1 B2  uB  2   2n2 0 The electric field energy stored between the capacitor plates  Energy per unit volume, U  1 CV2 . 2 1 CV2 1  A V2 1  V 2 1 u   0  ε0    ε0E2 E 2 Ad 2 d Ad 2  d  2 The values of uB and uE are valid in general for any magnetic and electric field.  1 B2 1 Magnetic field energy per unit volume 2  . Electric field energy per unit volume  2 0 E . The characteristic of the conductor to oppose the magnetic flux by self induced emf is shared by the capacitor also. C As the capacitor continues to be charged by the battery, the potential difference between the parallel plates increases and it opposes the current. It acts like a variable voltage battery connected opposite to the cell. The opposition to the flow of charge increases in the capacitor till it reaches its maximum charge. Then it stops the current totally. In the case of the inductor also the induced emf opposes the current due to the battery. However the opposition is maximum initially and finally opposition is negligible. Illustration: A long coaxial cable consists of two concentric cylinders of radii a and b. The central con- ductor of the cable carries a steady current i and the outer conductor provides the return path of the current. Calculate (i) the energy stored in the magnetic field of length l of such a cable, (ii) the self inductance of this length l of the cable. Solution: See figure. The magnetic field B in the space between the two conductors is given by B  0i , …(i) 2r where i = current in either of two conductors r = distance of the point from the axis. Because or □ B.dl  0  B.2r 0i dr current enclosed by the path (Ampere’s law) a r O The energy density in the space between the conductors b B2 1   i 2  i2 joule u   2 2  2r   0 82 r2 metre3 0 0 Consider a volume element dV in the form of a cylindrical shell of radii r and (r + dr) as shown in the figure. Energy dW = u dV =  0i2l  dr   i2 0  2r l dr 82 r2 4  r    Total magnetic energy can be obtained by integrating this expression between the limits r = a to r = b. Hence  i2l b  dr  W  dW  0    i2l  0 4 4 log  b    a  r  If L be the self inductance of length 𝑙 of the cable, then the energy in magnetic field will be (1/ 2)Li2 . Hence 1 Li2  i2l  0 log  b  2 4 L  0l log      b  . 2   9. CHARGING AND DISCHARGING OF CAPACITORS AND SWITCHING ON AND OFF THE CURRENT IN INDUCTOR CIRCUITS. It is interesting to study the charging and discharging of capacitors and the behavior of inductors when the current is switched ON and OFF. A i  B  C D C Applying Kirchhoff’s laws to the loop ABCD,  R i  q  ε  0 C  R dq  q    0 dt C  dq RC  q  ε C  0 dt solving this equation as dy  Py  Q, where dx P  1 RC and Q  εC RC one gets q  q 1 et /RC  . This is similar to the artificial radioactivity equation or filling a leaking tank. As more and more charge is put, current flows back, due to the capacitor itself. RC is the time constant, similar to the mean life of radioactivity. 1 RC  λ, the activity constant. Compare N = N0 1  eλt , in induced radioactivity when study- ing the production of isotopes. q o Time q reaches q asymptotically at t =  As dq  E et / RC , i  i et / RC max (compare A = A0 eλt ) dt R 0 i imax o Switching on the L – R circuit Time i R -  L + Kirchhoff’s equation gives  i R  L di  ε  0 dt di But the induced emf is decreasing as dt   R  L di  ε  0 dt  L di  iR  ε dt is decreasing  di  i   ε . 1  dε L / R    R L / R  ε is i . L/R is the time constant.   1 is “mean life” in the language of radioactivity. R max L L / R  i  imax (1 et / ) Current imax Time The current build-up in the L-R circuit The induced emf L di dt Time as a function of time The discharge of a capacitor and an inductor in C – R and L – R circuits [Precaution: First after fully charging, the C – R and L – R circuit should be made and then the battery connection is broken]. Illustration: Show that if the flux of induction through a coil changes from 1 to 2 , then the charge q that flows through the circuit of total resistance R is given by q  2  1 . R Solution: Let  be the instantaneous flux. Then d is the instantaneous rate of change of flux and dt that is also the instantaneous emf and obviously d / R dt current is the rate of flow of charge, that is, i  dq dt is the instantaneous current. Since q  idt t  d  or q   / R dt t0   where  is the time during which change take place. But at t = 0,   1 and t  ,   2 .  q  1 2 1 d  2  1 R 10. DECAY OF THE CURRENT IN L – R CIRCUIT : A B A B D C Before the battery is cut off. D C After the battery is cut off. When the battery is cut off, the self inductor will try to oppose the change. The Kirchhoff’s equation for the loop ABCD gives,  L di  i dt  di   R i R  0   dt    di  λ i dt where λ  1 L / R the negative sign shows that i is decreasing. L/R is the mean life. This is called time constant for L- R circuit. Illustration: A conductor of length l and mass m can slide without fric- tion along two vertical conductors connected at the top through a capacitor. Perpendicular to the plane of the fig- ure, a uniform magnetic field of induction B is set up. Find the voltage across the capacitor in terms of the distance x through which it falls. Solution:  , the voltage across the capacitor = induced emf across the slider = Blv. As the capacitor does not permit the flow of current, there will be no magnetic force on the rod to prevent its free fall. C × × × × × ×  v2  2gx    Bl or v  . OSCILLATIONS : 11. L – C CIRCUIT: L C L is equivalent to inertia or mass in mechanics and is C equivalent to a spring where 1/C = k. The charge q is equivalent to displacement and dq  i is equivalent to the velocity. dt P.E. + K.E. = const.  1 kx2  1 m2  E   2 2  1 q2  1 2 C 2 Differentiating,   Li2  E 1 2q dq  1 L 2 i di  0 2 C dt 2 dt  q dq  C dt dq d2q L   0 dt dt2  d2q    1  dt 2  .q LC   This is of the form, acceleration = - ω2 × displacement. This is S.H.M., where ω2 is 1/LC As   2f , f  where f is the frequency of the electrical oscillations.