Electromagnetic Induction Notes

ELECTROMAGNETIC INDUCTION INTRODUCTION From previous chapters, we know that current produces magnetic field. Is reverse possible i.e. can magnetic field produce electric current ? The answer is ‘yes’. It is found that currents were induced in closed coils when subjected to changing magnetic fields. The phenomenon in which electric current is generated by changing magnetic fields is known as electromagnetic induction. The current so produced is known as induced current. If current is produced in the circuit, this must be due to some emf produced in the circuit. This emf produced in the circuit. This emf produced as a result of change in magnetic field is known as induced emf. The phenomenon of electromagnetic induction (EMI) is the basis of working of power generators, dynamos, transformers etc. Magnetic Flux Magnetic flux though any surface held in a magnetic field is measured as the total number of magnetic field lines crossing the surface. It is a scalar quantity. r Consider a plane area A placed in a magnetic field B as shown in figure. Area vector A makes an angle  with the direction of magnetic field. Now the magnetic flux passing through the area is defined as r r r [for uniform B ]  =B.A or  =BAcos = (B cos ) A  B A where B  B cos is the component of the magnetic field B perpendicular to the face of the area. Direction of area vector is normal to the face of the area. Special Cases (i) if  = 0°, then  =BAcos0   =BA (maximum flux) (ii) if  = 90°, then  = BAcos 90°   = 0 (maximum flux) (iii) if  < 90°, then cos  > 0   > 0 (positive flux) (iv) if  > 90°, then cos  < 0   < 0 (negative flux) If the surface is not plane, we can divide any surface into elements of area dA (as shown in figure). For each element we determine B , the component of normal to the surface at the position of that element, as shown. from above figure, B  B cos , where  is the angle between the direction of B and a line perpendicular to the surface. In general, this component varies from point to point on the surface. We define magnetic flux d B through the area element dA as d   B dA The total magnetic flux through the surface is the sum of the contributions from the individual area elements : r uur    B dA   B cos  dA   B.dA (magnetic flux through a surface) The unit of magnetic flux is tesla-metre2 which is called weber (Wb) in honour of Wilhelm Weber. 1 Wb = 1 Tm2. Clearly, B can be measured in Wb m–2 = 1 T. Sometimes it is referred to as flux density. ELECTROMAGNETIC INDUCTION Ex. A long copper wire carries a current of I ampere. Calculate the magnetic flux per metre of wire for a plane surface S inside the wire as shown in figure. I S R O Sol. Consider an element of width dr at a distance r from the axis of the wire. The field due to the current I in the wire at µ0 I r 2 R 2 and as its direction is perpendicular to the plane surface, flux linked with the element is given by the position of element will be B  I R O n B dr S r µ0 I r (1dr ) cos 0 2 R 2 So, the flux linked with the plane surface is given by d   B ds cos    µ0 I R µI r= 0 2  2 R 0 4 Faraday’s Law of Electromagnetic Induction On the basis of several experimental observations, Michael Faraday came to the following conclusions. 1. Whenever there is a relative motion between a magnet (source of magnetic field) and a closed conducting loop, electric current appears in the loop. It happens because of the change in magnetic flux associated with the loop. 2. Since emf causes current in a circuit, when loop and magnet are brought in relative motion current flows in the loop. This implies that an emf is set up in the loop. This emf is known as induced emf and its magnitude is directly proportional to the rate of change of magnetic flux through the loop with time. Now, we come to the Faraday’s law of electromagnetic induction. In mathematical form induced emf can be given by d the expression    B dt As such Faraday’s law in itself is complete to tell the magnitude and polarity of induced emf but Lenz’s rule is commonly used to determine the polarity of induced emf or direction of induced current. Direction of Induced emf We can find the direction of an induced emf or current   d  B / dt together with some simple sign rules. Here is the procedure : r (1) Define a positive direction for the area vector A . r r (2) From the directions of A and the magnetic field B , determine the sign of the magnetic flux  B and its rate of change d  B / dt . Figure shows several examples. B (increasing) A A B (decreasing) ? ? A A ? ? ? B (increasing) B (decreasing) r A is upward to anticlockwise direction is positive. (ii)  =BAcos (i) dd dd dB dB  ( BA ( BAcos cos) ) AAcos cos   ..... (i) dtdt dtdt dtdt Here  is acute, so cos  is positive. B is increasing, so dB/dt is positive. Then we find that  is negative. So  is clockwise as shown.   For Fig. (b) Here B is decreasing, so dB/dt is negative. Then from eq. (i),  is positive. So  is anticlockwise as shown. For Fig (c) Here  is obtuse, so cos  is negative. B is increasing, so dB/dt is positive. Then from eq. (i)  is positive, so  is anticlockwise as shown. For. Fig. (d) Here  is obtuse, so cos  is negative. B is decreasing, so dB/dt is negative. Then from eq. (i)  is negative, so  is clockwise as shown. Ex. 1. A coil is placed in a constant magnetic field .The magnetic field is parallel to the plane of the coil as shown in figure. Find the emf induced in the coil . B Sol. :  = 0 (always) since area is perpendicular to magnetic field.  emf = 0 Ex. 2. Find the emf induced in the coil shown in figure. The magnetic field is perpendicular to the plane of the coil and is constant. Area = A B Sol. : = BA (always) = const.  emf = 0 ELECTROMAGNETIC INDUCTION Ex. 3. Find the direction of induced current in the coil shown in figure. Magnetic field is perpendicular to the plane of coil and it is increasing with time. B Sol. : Inward flux is increasing with time. To opposite it outward magnetic field should be induced. Hence current will flow anticlockwise. Ex. 4. Figure shows a coil placed in decreasing magnetic field applied perpendicular to the plane of coil .The magnetic field is decreasing at a rate of 10T/s. Find out current in magnitude and direction Sol. :  = B.A emf = A . dB = 2 × 10 = 20 v dt  i = 20/ 5 = 4 amp. From Lenz’s law direction of current will be anticlockwise. Ex. 5. Figure shows a coil placed in a magnetic field decreasing at a rate of 10T/s. There is also a source of emf 30 V in the coil. Find the magnitude and direction of the current in the coil. 30v A=2m2 B R=5 Sol. : Induce emf = 20V equivalent i = 2A clockwise Ex. 6. Figure shows a long current carrying wire and two rectangular loops moving with velocity v. Find the direction of current in each loop. v constant i (i) v (ii) Sol. : In loop (i) no emf will be induced because there is no flux change. In loop (ii) emf will be induced because the coil is moving in a region of decreasing magnetic field inward in direction. Therefore to oppose the flux decrease in inward direction, current will be induced such that its magnetic field will be inwards. For this direction of current should be clockwise. Lenz’s Law Direction of the Induced Current in a Circuit Lenz’s law states that ‘when the magnetic flux through a loop changes, a current is induced in the loop such that the magnetic field due to the induced current opposes the change in the magnetic flux through the loop.’ The above rule can be systematically applied as follows to determine the direction of the induced currents. (i) Identify the loop in which the induced current is to be determined. (ii) Determine the direction of the magnetic field in this loop (i.e in or out of the loop). (iii) The direction of flux is the same as the direction of the magnetic field. Determine if the flux through the loop is increasing or decreasing (because of change in area or change in B). Choose the appropriate current in the loop that will oppose the change in flux. (i) If the flux is into the paper and increasing them the flux due to the induced current should be out of the paper. (ii) If the flux is into the paper and decreasing, the flux due to the induced current should be into the paper. (iii) If the flux is out of the paper and increasing, the flux due to the induced current should be into the paper. (iv) If the flux is out of the paper and decreasing, the flux due to the induced current should be out of the paper. The above description is the physical interpretation of Lenz’s law. We can determine the direction of the induced dB current mathematically by simply applying Lenz’s law ind  with the appropriate conventions. dt The right hand sign convention used is as follows : (i) For counterclockwise current, emf is positive. (ii) For clockwise current, emf is negative. (iii) Magnetic flux out of the paper is positive. (iv) Magnetic flux into the plane of the paper is negative. (v) The rate of change of an increasing positive flux is positive. (vi) The rate of change of an decreasing positive flux is negative. (vii) The rate of change of an increasing negative flux is negative. (viii) The rate of change of an decreasing negative flux is positive. Let us consider some of the cases regarding application of Lenz’s law. Change in B (ix) Suppose north pole of a bar magnet is moved towards a loop as shown in figure. Because of change in magnetic flux associated B (increasing) with the loop current is induced in it. Due to induced current, magnetic field is induced in such a way that it opposes the motion S of the bar magnet. As the north pole is moving towards the loop, hence to oppose the motion of the bar magnet only the north I pole will be induced on that face of the loop which faces the B induced r magnet. The induced current due to the change in B is clockwise, N r as seen from above the loop. The added field Binduced that it causes r is downward, opposing the change in the upward field B . N (x) Consider figure. A rectangular loop ABCD is being pulled out of the magnetic field directed into the plane of the paper. As the loop is dragged out of the field, the flux associated with the loop which is directed into the plane of the paper decreases. The induced current will flow in the loop in the sense to oppose the decrease of this flux. For this to happen, magnetic field due to the induced current in the loop must be directed into the plane of the paper. Hence current in the loop must flow in the clockwise sense. S x x x x x x x x x x x x x x x B x xA x x xC x x x x x x x x B v D ELECTROMAGNETIC INDUCTION Ex. A closed loop with the geometry shown in figure (a) is placed in a uniform magnetic field directed into the plane of the paper. If the magnetic field decreases with time, determine the direction of the induced e.m.f. in this loop. A a x a F Sol. B a a E 2a C xB 2a a D Figure (a) There are two loops that are immersed in the magnetic field, namely, ABEFA and BCDEB. i2 i1 i1 i2- i1 i1 i2 i2 Figure (b) Consider loop ABEFA. magnetic flux is into the plane of the paper. For loop BCDEB too, the magnetic flux is into the plane of the paper. In both loops, the magnetic flux is decreasing with time. Therefore, the induced current in loop ABEFA, i1, will be in a direction so as to induce a flux into the paper. The direction of i1 is clockwise. Like wise in loop BCDEB the current i2 will be in a direction so as to induce a flux into the paper. The direction of i2 is also clockwise. The final induced current in all the arms of the loop are shown Fig. (b). Lenz’s Law (Deciding direction of induced e.m.f.) The induced emf resulting from a changing magnetic flux has a polarity that leads to an induced current whose direction is such that the induced magnetic field opposes the original flux change. We shall discuss later on how it is in accordance with conservation of energy. Ex. Sol. In figure there is a constant magnetic field in a rectangular region of space. This field is directed perpendicularly into the page. Outside this region there is no magnetic field. A copper ring moves through the region from position 1 to position 5. Find the induced current in the ring at it passes through positions. (a) 1(b) 2(c) 3 (d) 4 (e) 5 (a) Since the field is zero outside the rectangular region, no flux passes Position 1 through the region in position 1, there is no change in the flux through the ring, and there is no induced emf or current in the Position 2 ring. (b) In position 2 the flux increases. According to Lenz’s law, the induced current must create an induced magnetic field that op Position 3 poses the increase. To oppose the increase, the induced field must point opposite to the external field and, therefore, must Position 4 point out of the page, for which the induced current must be counter clockwise. Position 5 (c) Here the field is not zero. Hence nonzero flux passes through the ring in position 3 but the flux through the ring is constant, and there is no induced emf or current in the ring. (d) In position 4 the flux decreases. According to Lenz’s law, the induced current must create an induced magnetic field that opposes the decrease. To oppose the decrease, the induced field must point in the direction of external field and, therefore, must point into the page. For which the induced current must be clock-wise. (e) Since the field is zero outside the rectangular region, no flux passes through the ring in position 5, there is no change in the flux through the ring, and there is no induced emf or current in the ring. Note down the following points regarding the Faraday’s law : 1. As we have seen, induced emf is produced only when there is a change in magnetic flux passing through a loop. The flux passing through the loop is given by  = BA cos  Thus, flux can be changed in several ways “ r (i) The magnitude of B can change with time. In the problems if magnetic field is given as a function of time, it implies that the magnetic field is changing. Thus, B = B(t). (ii) The area enclosed by the loop can change with time. This can be done by pulling a loop inside (or outside) a magnetic field. By doing so, the area enclosed by the loop (hatched area) can be changed. x x x x x x x x x x x x x x r (iii) The angle  between B and the normal to the loop can change with time. This can be done by rotating a loop in a magnetic field. x x x x x x x x x x x x (iv) Any combination of the above can occur. 2. When the magnetic field passing through a loop is changed, an induced emf and hence an induced current is produced in the circuit. If R is the resistance of the circuit, then induced current is given by e 1   d B  i   R R  dt  Current starts flowing in the circuit, i.e., flow of charge takes place. Charge flown in the circuit in time dt will be given by 1 dq  i.dt  ( d B ) R Thus, for a time interval t we can write e    B 1  B  1 ,i   and q  it  (B ) t R  dt  R From these equations we can see that e and i are inversely proportional to t and q is independent of t. It depends on the magnitude of change in flux, not the time taken by it. Ex. Sol. x x x x x x x A circular loop of radius a having n turns is kept in a horizontal plane. A uniform x x x x x x x magnetic field B exists in a vertical direction shown in figure. Find the emf induced x x x x x x x in the loop if the loop is rotated with a uniform angular velocity  about x x x x x x x (a) an exist passing through the centre and perpendicular to the plane of the loop. x x x x x x x (b) the diameter. (a) The emf induces when there is change of flux. As in this case(figure) there is no change of flux, hence no emf will be induced in the coil. (b) If the loop is rotated about the diameter there will be change of flux with time. In this case e.m.f. will be induced in the coil. The area of the loop is A – a2. If the normal of the loop makes an angle  = 0 with the magnetic field at t = 0, this angle will become  = t at time t. The flux of the magnetic field at this time is ELECTROMAGNETIC INDUCTION   nB a 2 cos   nB a 2 cos  t d   na 2 B sin  t dt e is coming out to be positive, so direction of induced emf will be as shown in figure. Because here this sense is positive if we look at direction of area vector. The induced e.m.f. is    Ex. Figure (a) shows two circular rings of radii a and b (a > b) joined together with wires of negligible resistance. Figure (b) shows the pattern obtained by folding the small loop in the plane of the large loop. x x x x x x x x x x x x x x x x x x x x x x x x a (a) x x x x x x x x xa x x x x x x x x x x b a x x b x (b) x x (c) The pattern shown in figure (c) is obtained by twisting the small loop of Fig. (a) through 180°. dB  k , perpendicular to the plane All the three arrangements are placed in a uniform time varying magnetic field dt of the loops. If the resistance per unit length of the wire is , then determine the induced current in each case. Sol. (a) B   ( a 2  b 2 ) B  d B dB   (a 2  b 2 )    (a 2  b 2 )k dt dt Induced current, I   R   (a 2  b 2 )k k (a 2  b 2 )  [2 ( a  b)] 2 (a  b) (b) B   a 2 B   b 2 B   B ( a 2  b 2 )  d B  dB   (a 2  b 2 )   k (a 2  b2 ) dt dt Induced current, I   R   k ( a 2  b 2 ) k ( a  b)  2 (a  b) 2 x x x x x x x x x x x x xa x x x x x x x x x x x x x x x b a b x x x x x x x x x x x x x x x x x x a b x x (c) Here induced emf in both loops will oppose each other, hence effective flux is given by  B   a 2 B   b 2 B   B(a 2  b 2 )   dB   k (a 2  b 2 ) dt Induced current, I  Ex. Sol.  R  k (a  b) 2 Space is divided by the line AD into two refions. Region I is field free and region II has a uniform magnetic field B directed into the plane of the paper. BCE is a semicircular conducting loop of radius r with centre at O, 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 the 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 Region II. (iii) Plot a graph between the induced emf and the time of rotation for the two periods of the rotation. (i) When the loop is rotated about an axis passing centre O and perpendicular to the plane of the paper, the angle between magnetic field vector B and area A is always 0°. When the loop is region I, the magnetic flux linked with loop = BA cos 0 = 0 the (since B = 0 in region I). (a) (b) When the loop enters the magnetic field in region II, the magnetic flux linked with it is given by  = BA where A 1 2 r  . Therefore, emf induced 2 e d d dA  Br 2 d   ( BA)   B  dt dt dt 2 dt  Br 2  2 A resistance of the loop is R, the current induced is given by  e i e 1 Br 2  R 2 R This is the required expression for current induced in the loop. (ii) According to Lenz’s law, the direction of current induced is to oppose the change in magnetic flux. So, when entering into region II the field produced by the current induced must be upward. For this, the current in the loop must be anticlockwise as shown in Fig. (a). ELECTROMAGNETIC INDUCTION 1 2 Br  is 2 induced in one direction. When the loop comes out of the field, the flux decreases and emf is induced in opposite sense. The graph for representing the emf induced versus time for two periods (T = 2/) is shown in Fig. (b). Here we have taken anticlockwise direction as positive. (iii) When the loop enters the magnetic field the magnetic flux linked with it increases and the emf i  MOTIONAL ELECTROMOTIVE FORCE When a conductor moves in a magnetic field, then charges inside conductor experience magnetic force given by r r r r F  qv  B , where v is the velocity of conductor (here we assume that velocity of charges inside conductor is same as the velocity of conductor and random motion of charges inside the conductor is neglected). Due to this force charges start moving in a particular direction in the conductor or we say that an emf is induced in the conductor. This emf is known as motional emf. r Consider a conductor of arbitrary shape as shown in figure moving in some magnetic field B . r The different parts of the conductor may have different velocity. Consider an element dl have velocity vr . Force due to magnetic field on a charge in conductor. r r r F  qv  B r work done by this force on charge q in passing through dl r r r r r dW  F .dl  q(v  B).dl r dW r r r So emf induced within dl : d    (v  B).dl q b r r r Integrate this to find net emf :    (v  B).dl a This emf is directed from a to b. Other Approach : Due to the magnetic force, charges star moving in the conductor an electric field is set up in the conductor. In r equilibrium force of electric field balances the force of magnetic field. Let electric field induced is E , then r r r r r r  qv  B  qE  0 E  v  B Now potential difference developed across the ends of the conductor : b r b r r r Vb  Va    E.dl   (v  B ).dl a a If circuit is incomplete, then this terminal potential difference will be equal to emf induced. Hence emf induced : b r r r    (v  B).dl a b r r r Note : We can also write    B.(dl  v ) a r r r r r r As dl  v is the area swept per unit time by length dl and hence B.(dl  v ) is the flux of induction through the area. Therefore, the motional e.m.f. is equal to the flux of induction cut by the conductor per unit time. * We can write the expression for induced emf in various forms : b r r r b r r r b r r r    (v  B ).dl   ( B  dl ).v   (dl  v ).B a a a r r r if any two out of v , B and dl become parallel or antiparallel,  will become zero. Explanation of emf induced in rod on the basis of magnetic force: If a rod is moving with velocity v in a magnetic field B ,as shown , the free electrons in a rod will experience a magnetic force in downward direction and hence free electrons will accumulate at the lower end and there will be a deficiency of free electrons and hence a surplus of positive charge at the upper end. These charges at the ends will produce an electric field in downward direction which will exert an upward force on electron. If the rod has been moving for quite some time enough charges will accumulate at the ends so that the two forces qE and qvB will balance each other. Thus E = v B. VP – VQ= V B  The moving rod is equivalent to the following diagram, electrically. Figure shows a closed coil ABCA moving in a uniform magnetic field B with a velocity v. The flux passing through the coil is a constant and therefore the induced emf is zero. Now consider rod AB , which is a part of the coil. Emf induced in the rod =B L v Suppose the emf induced in part ACB is E , as shown. A E C VBL Since the emf in the coil is zero, Emf (in ACB) + Emf (in BA) = 0 or -E + vBL = 0 or E = vBL Thus emf induced in any path joining A and B is same, provided the magnetic field is uniform. Also the equivalent emf between A and B is BLv (here the two emf’s are in parallel) ELECTROMAGNETIC INDUCTION Ex. Figure shows an irregular shaped wire AB moving with velocity v, as shown. Find the emf induced in the wire. Sol. The same emf will be induced in the straight imaginary wire joining A and B , which is Bv sin  Ex. A rod of length l is kept parallel to a long wire carrying constant current i. It is moving away from the wire with a velocity v. Find the emf induced in the wire when its distance from the long wire is x. Sol. E=BlV=  0i l v 2x OR Emf is equal to the rate with which magnetic field lines are cut . In dt time the area swept by the rod is l v dt. The magnetic field lines cut in dt time = B l vdt = 0i l vdt . 2x t  0i l v  The rate with which magnetic field lines are cut = 2x Ex. i l Const B V x A rectangular loop ,as shown in the figure ,moves away from an infinitely long wire carrying a current i. Find the emf induced in the rectangular loop. v L i b x Sol. E = B1 LV – B2Lv =  0iLbv  0i  0i Lv  Lv = 2xx  b 2 x 2  x  b  B1LV B2LV Aliter: Consider a small segment of width dy at a distance y from the wire. Let flux through the segment be d   0i d = 2y L dy    0 iL 2 xb  x dy y dy y L Constant current i =  0 iL n x  b   nx  2 x V Now  Ex. d   0 iL  1 dx 1 dx   0 iL   b     0 ibLv    v =      dt 2   x  b dt x dt  2   x x  b  2  x x  b   0 ibLv induced emf = 2x x  b  A rod of length l is placed perpendicular to a long wire carrying current i. The rod is moved parallel to the wire with a velocity v. Find the emf induced in the rod , if its nearest end is at a distance ‘a’ from the wire. v a Sol. Consider a segment of rod of length dx , at a distance x from the wire. Emf induced in the segment  0i dE = dx.v 2x a  E=  a Ex. – B +  0 ivdx  0iv    a  n   = 2 2 x  a  A rectangular loop is moving parallel to a long wire carrying current i with a velocity v. Find the emf induced in the loop , if its nearest end is at a distance ‘a’ from the wire. Draw equivalent electrical diagram. v Const i a Sol. b emf = 0 ;   a  b  0 iv n   2 a    0 iv  a  b  n   2 a   Induced emf due to rotation Rotation of the Rod Consider a conducting rod of length  rotating in a uniform magnetic field. Emf induced in a small segment of length dr, of the rod = v B dr = rB dr    emf induced in the rod = B rdr  0 1 B2 2 ELECTROMAGNETIC INDUCTION or Ex. Sol. equivalent of this rod is as following d  dt d flux through the area swept by the rod in time dt   dt dt 1 2 B  dt 1 2 = 2 = B 2 dt A rod PQ of length 2l is rotating about one end P in a uniform magnetic field B which is perpendicular to the plane of rotation of the rod. Point M is the mid point of the rod. Find the induced emf between M & Q if that between P & Q = 100V .   B  2 EMQ + 2 Ex. EPM = Bw2 = 100 2 B2  2 EMQ = 3 3 B2 = ×100 V = 75 V 8 4 2 = ×B A rod of length  and resistance r rotates about one end as shown in figure. Its other end touches a conducting ring a of negligible resistance. A resistance R is connected between centre and periphery. Draw the electrical equivalence and find the current in the resistance R. There is a uniform magnetic field B directed as shown. R r Sol.   i R 1 B l 2 2 current i = Rr Ex. Q M P 2  EMQ + EPM = EPQ B=uniform Solve the above question if the length of rod is 2L and resistance 2r and it is rotating about its centre. Both ends of the rod now touch the conducting ring r r O Sol. R E r A  C,E,A C r E R E 1 BL2 2 i  r r R R 2 2  O  Ex. A rod of length l is rotating with an angular speed  about its one end which is at a distance ‘a’ from an infinitely long wire carrying current i. Find the emf induced in the rod at the instant shown in the figure. i  a Sol. Consider a small segment of rod of length dx , at a distance x from one end of the rod. Emf induced in the segment xw a i w dx x   0i dE = (xdx 2( x  a) Ex. E=   0i  2(x  a ) (xdx = 0  0 i     a   .   a.n 2   a  conducting rails (closed by a resistance R) and it is projected with an initial velocity u .Find its velocity as a function of time. Sol. t=0 A rod of mass m and resistance r is placed on fixed, resistanceless, smooth  m,r ×B u R Let at an instant the velocity of the rod be v .The emf induced in the rod will be vBl. The electrically equivalent circuit is shown in the following diagram. t=0  Current in the circuit i = At time t m,r ×B u R Bv Rr Magnetic force acting on the rod is F = i B, opposite to the motion of the rod. iB = – m i= Bv Rr dV dt ...(1) ...(2) ELECTROMAGNETIC INDUCTION Now solving these two equation dV B 2 2 V =–m. dt Rr let 2 2 B  =k (R  r )m v  u Ex. Sol.  dV V =   K.dt – K . dt =  v ln   = – Kt  u 0 –Kt u V dV V  t V = ue B 2 2 dV – . dt = (R  r )m V t V = ue–Kt In the above question find the force required to move the rod with constant velocity v, and also find the power delivered by the external agent . The force needed to keep the velocity constant Fext = iB = B 2 2 v Rr Power due to external force = B2 2 v 2 2 = = i2(R+r) Rr Rr Note that the power delivered by the external agent is converted into joule heating in the circuit. That means magnetic field helps in converting the mechanical energy into joule heating. Ex. In the above question if a constant force F is applied on the rod. Find the velocity of the rod as a function of time assuming it started with zero initial velocity. Sol. let m dv = F – i B dt m dv B 2 2 v =F– dt Rr B 2 2 K= Rr ....(1) v   F  kV  Kt  =– n   F  m V= Ex.  0  dV F  Kv = t dt m i= Bv Rr – 0 ....(2) 1 n(F – KV )0V  t K m V  F – KV = F e kt / m Vmax F (1 – e kt / m ) K t A rod PQ of mass m and resistance r is moving on two fixed, resistanceless, smooth conducting rails (closed on both sides by resistances R1 and R2) .Find the current in the rod at the instant its velocity is v. P R1 × B r V Q R2 Sol. i= BV R1R 2 r R1  R 2 this circuit is equivalent to the following diagram. R1 Blv R2 r EMF Induced due to rotation of a coil Ex. A ring rotates with angular velocity  about an axis perpendicular to the plane of V the ring passing through the center of the ring. A constant magnetic field B exists parallel to the axis. Find the emf induced in the ring w B× Sol. Flux passing through the ring  = B.A is a constant here, therefore emf induced in the coil is zero. Every point of this ring is at the same potential, by symmetry. Ex. A ring rotates with angular velocity  about an axis in the plane of the ring and which passes through the center of the ring. A constant magnetic field B exists perpendicular to the plane of the ring. Find the emf induced in the ring as a function of time. Sol. At any time t ,  = BA cos  = BA cos t Now induced emf in the loop e= d = BA  sin t dt If there are N turns emf = BAN sin t BA N is the amplitude of the emf e = em sin t i= e em = sin wt = im sin wt R R em R The rotating coil thus produces a sinusoidally varying current or alternating current. This is also the principle used in generator. im = Ex. Sol. Figure shows a wire frame PQSTXYZ placed in a time varying magnetic field given as B=t, where is a positive constant. Resistance per unit length of the wire is Find the current induced in the wire and draw its electrical equivalent diagram. Y b Z b>a ×B X S T ×B Induced emf in part PQST =  a (in anticlockwise direction, from Lenz’s Law) P a Similarly Induced emf in part TXYZ =  b2 (in anticlockwise direction, from Lenz’s Law) Total resistance of the part PQST =4a. Total resistance of the part TXYZ = 4b. The equivalent circuit is as shown in the following diagram. 2 a Q ELECTROMAGNETIC INDUCTION l 4b writing KVL along the current flow b2 –a2 – 4ai –4bi = 0 i=  (b – a) 4 i 2 b i i 2 a i i l 4a EMF Induced in a rotating disc : Consider a disc of radius r rotating in a magnetic field B. Consider an element dx at a distance x form the centre. This element is moving with speed v = x.  Induced emf across dx = B(dx) v = Bdxx = Bxdx  emf between the centre and the edge of disc. r  = Bxdx  0 B r 2 2 A Conductor of Arbitary Shape Consider a conductor ab of arbitary shape translating in a uniform magnetic field B. Then induced emf in this conductor will be same as in a straight conductor connected between a and b, i.e.   BvL sin  Proof : Consider an arbitrary shaped conductor ab Emf induced in the conductor is given by : r r r r    (v  B ).dl   vBnˆ.dl  vB  dl cos   vBh  vBL sin  A Rotating Conductor of Arbitary Shape : Consider an arbitary shaped conductor ab which is rotating about end a with angular velocity  as shown. We want to find induced emf between ends a and b. Induced emf developed between ends a and b is given by : r r r    (v  B ).dl  r r r r    [(  r )  B ].dl r   [{ kˆ  ( xiˆ  yjˆ)}  (  Bkˆ)]. dl r   [(  xjˆ   yiˆ)  (  Bkˆ )]. dl r r   [(  xBiˆ   Byjˆ ).dl  B  ( xiˆ  yjˆ).dl   L r r  B  r . dl  B  r dl cos   B  rdr 0 1 B L2 2 Motional EMF when the magnetic Field is Non-Uniform In some of the cases motion of a conductor may be in a non-uniform magnetic field. Take the following steps while calculating motional emf. Step 1 : Determine the magnetic field at all points on the rod. Step 2 : Consider a small element at some distance from one end of the rod. Step 3 : Assuming B to be uniform over this element, calculate the potential difference across this element using the procedures outlined earlier. Step 4 : Integrate over the entire rod to calculate the total induced emf. Let us learn to calculate the induced emf through the examples given below. Ex. A copper rod of length 0,19 m is moving with uniform velocity 10 m s–1 parallel to a long straight wire carrying a current of 0.5 A. The rod is perpendicular to the wire with its ends at distances 0.01 and 0.2 m from it. Calculate the emf induced in the rod. Sol. As shown in figure, consider an element of length dy at a distance y from the wire, then at this position of the element, the magnetic field due to the current-carrying wire PQ will be B  So, the emf induced in the element d   Bv dy = due to its motion in the field of the wire. µ0 2 I into the plane of the paper.. 4 y µ0 2 I v dy and hence the emf induced across the ends of the rod 4 y ELECTROMAGNETIC INDUCTION I b  = d  a µ µ0 b dy 2 Iv  , i.e.   0 2 Iv log e    4 4 y a a b 0.01 m y Substituting the given data with b = (a + ),   107  2  5  10 log e A v dy 0.20  105  log e 20  30 µV 0.01 0.19 m B Ex. An infinite wire carries a current I. An ‘S’ shaped conducting rod of two semicircles each of radius r is placed at an angle q to the wire. The centre of the conductors at a distance d from the wire. If the rod translates parallel to the wire with a velocity v as shown in figure, calculate the emf induced across the ends OB of the rod. Sol. Join the end point O and B and replace the two semicircles by a straight rod length 4r (in figure). Induced emf in straight rod OB will be same as in the actual conductor. Now this straight rod can be replaced by a combination of two rods OA and AB. Induced emf in OB will be same as induced emf in OA+ induced emf in AB will be zero because its velocity will be parallel to its length. Hence induced emf in OA will be the net induced emf in actual conductor. v 4r O  B 4r sin  4r cos  A We now have a rod of effective length 4r cosq translating in a non-uniform magnetic field. Consider a small eleµI ment of the wire of length dx located at a distance x from the wire. The magnetic field at this element is B  0 2 x µ0 I v dx . The potential difference across this element is dV  2 x The potential difference across the ends of the rod can be calculated by intergating over the whole end. Therefore, V   dV  or V d  2 r cos   d  2 r cos  µ0 I v dx 2 x µ0 Iv  d  2r cos   In   2  d  2r cos   From the right hand rule we see that electrons will accumulate at end B. Therefore, end O is at a high potential than B. Induced Electric Field r We have seen that the electric potential difference between two points A and B in an electric field E can be written as B r r V  VB  VA    E.ds A When the electric field is conservative, as is the case of electrostatics, the line integral of it is path-independent, r r which implies Ñ  E.ds  0 Faraday’s law show that as magnetic flux changes with time, an induced current begins to flow. What cause the charges to move ? It is the induced emf which is the work done per unit charge. However, since magnetic field can do no work, as we have shown, the work done on the mobile charges must be electric, and the electric field in this situation cannot be conservative because the line integral of a conservative field must vanish. Therefore, we r r r conclude that there is a non-conservative electric field Enc associated with an induced emf   Ñ  Enc .ds Combining with Faraday’s law then yields : r dB r .ds   dt The above expression implies that a changing magnetic flux will induce a non-conservative electric field which can vary with time. It is important to distinguish between the induced, non-conservative electric field and the conservative electric field which arises from electric charges. As an example, let’s consider a uniform magnetic field which points into the page and is confined to a circular region r with radius R, as shown in Figure. Suppose the magnetic of B increase with time, i.e., dB/dt>0. let’s find the induced electric field everywhere due to the changing magnetic field. Ñ E nc Since the magnetic field is confined to a circular region, from symmetry arguments we choose the integration path to be a circle of radius r. The magnir tude of the induced field Enc at all points on a circle is the same. According to r Lenz’s law, the direction Enc of must be such that it would drive the induced Bin R current to produce a magnetic field opposing the change in magnetic flux. With r the area vector A pointing out of the page, the magnetic flux is negative or inward. With dB > dt > 0, the inward magnetic flux is increasing. Therefore, to counteract this change the induced current must flow counterclockwise to r produce more outward flux. The direction of Enc is shown in Figure. r Let’s proceed to find the magnitude of Enc . In the region r > R the rate of change of magnetic flux is : dB d r r d  dB  2  ( B. A)  (  BA)     r dt dt dt  dt  Using equation, we have r d  B  dB  2 r .ds  Enc (2 r )     r dt  dt  which implies Ñ E nc Enc  r dB 2 dt r B into page R C ELECTROMAGNETIC INDUCTION Similarly, for r > R the induced electric field may be obtained as Enc (2 r )   d  B  dB  2   r dt  dt  or r R 2 dB Enc  2r dt The pattern of Enc as a function of r is shown in figure. A Plot of Enc as a function of r is shown in figure. Enc R r Eddy Currents We have seen that when a conducting loop moves through a magnetic field, current is induced as the result of changing magnetic flux. If a solid conductor were used instead of a loop, as shown in Figure. current can also be induced along any closed loop in the conductor. The induced current are called an early current. The induced eddy currents also generate a magnetic force that opposes the motion, making it more difficult to move the conductor across the magnetic field (Figure). Since the conductor has non-vanashing resistance R, joule heating causes a loss of power by an amount P = 2 / R. Therefore by increasing the value of R, power loss can be reduced. One way to increase R is to laminate the conducting slab, or construct the slab by using gluing together thin strips that are insulated from one another (see Fig.a). Another way is to make cuts in the slab, thereby disrupting the conducting path (Figure - b). Figure (a) Figure (b) There are important applications of eddy currents. For example, the currents can be used to suppress unwanted mechanical oscillations. Another application is the magnetic braking system in high-speed transit cars. Self-Inductance Consider again a coil consisting of N turns and carrying current I in the counterclockwise direction, as shown in Figure. If the current is already, then the magnetic flux through the loop will remain constant. However, suppose the current I changes with time, then according to faraday’s law, an induced emf will arise to oppose the change. dI dI  0 . The property of  0 , and counterclockwise if dt dt the loop in which its own magnetic field opposes any change in current is called :self-inductance”, and the emf generated is called the self-induced emf or back emf, which we denote as L. All current-carrying loops exhibit this property. In particular, an inductor is a circuit element (symbol ______ ) which has a large self-inductance. According to lenz law the induced emf will be clockwise if Mathematically, the self-induced emf can be written as dB d r r L  N   N  B.dA dt dt and is related to the self-inductance L by dI dt The above two expression can be combined to yield  L  L N B I Physically, the inductance L is a measure of an inductor’s “opposition” to the change of current ; larger the value of L, lower the rate of change of current. L ELECTROMAGNETIC INDUCTION Ex. Compute the self-inductance of a solenoid with N turns, length l , and the radius R with a current I flowing through each turn, as shown in Figure. Ignore edge effects Sol. Ignoring edge effects and applying Ampere’s law, the magnetic field inside a solenoid is given by Eq. : r µ NI B= 0 kˆ = µ0 nI kˆ l where n = N /  is the number of turns per unit length. The magnetic flux through each turn is  B  BA  µ0 nI .( R 2 )  µ0 nI  R 2 N B ˆ k = µ0 n 2 R 2 l l We see that L depends only on the geometrical factors (n, R and  ) and is independent of the current I. Thus, the self-inductance is Note Ex. L= A long coaxial cable as consisting of two thin concentric cylindrical conducting shells of radii a and b and length l. The conducting shells carry the same current I in opposite directions. Imagine that the inner conductor carries current to a device and that the outer one acts as a return path carrying the current back to the source. Calculate the self-inductance L of this cable. I dr b a I r B ? Sol. Imagine a thin radial slice of the coaxial cable, such as the shaded rectangle in Figure. The magnetic field is perpendicular to the rectangle of length l and width b-a, the cross section of interest. Divide this rectangle into strips of width dr. We see that the area of each strip is Idr and that the flux through each strip is B dA = BIdr Hence, we find the total flux through the entire cross section by integrating  B   B dA   b a µ0 I µ Il l dr  0 2 r 2  b a dr µ0 I l  b   In   r 2 a Using this result, we find that the self-inductance of the cable is  µ l b L  B  0 In   I 2  a  Self Inductance of Solenoid r A >> r i Let the volume of the solenoid be V , the number of turns per unit length be n. Let a current I be flowing in the solenoid. Magnetic field in the solenoid is given as B=0ni.The magnetic flux through one turn of solenoid = 0 n i A. The total magnetic flux through the solenoid = N  = N0ni A = 0n2i A   L = 0 n2  A = 0 n2 V  = µ0 n i r2 (n)  = µ0 n2 r2. i Inductance per unit volume = µ0n2. L= Self inductance is the physical property of the loop due to which it opposes the change in current that means it tries to keep the current constant. Current can not change suddenly in the inductor. Inductor It is represent by electrical equivalence of loop L R L  L i i If current i through the inductor is increasing the induced emf will oppose the increase in current and hence will be opposite to the current. If current i through the inductor is decreasing the induced emf will oppose the decrease in current and hence will be in the direction of the current. i (increasing) L A i > + di  VB dt L di dt L i Induced emf Over all result VA  L i(decreasing) – induced emf B ELECTROMAGNETIC INDUCTION Note : If there is a resistance in the inductor (resistance of the coil of inductor) then : A Ex. A B is a part of circuit. Find the potential difference vA – vB if (i) (ii) (iii) Sol. current i = 2A and is constant current i = 2A and is increasing at the rate of 1 amp/sec. current i = 2A and is decreasing at the rate 1 amp/sec. – + A 1H di di L =1 dt dt writing KVL from A to B di VA – 1 – 5 – 2i = VB . dt (i) (ii) (iii)  B L, R + 2i – 5volt + B – di =0 dt VA – 5 – 4 = VB  VA – VB = 9 volt di Put i = 2 , = 1 ; V A – 1 – 5 – 4 = VB or VA – VB = 10 V0 dt di Put i = 2 , = – 1 ; VA + 1 – 5 – 2 × 2 = VB or VA = 8 volt. dt Put i = 2, Energy Stored in an Inductor If current in an inductor at an instant is i and is increasing at the rate di/dt,the induced emf will oppose the current . Its behaviour is shown in the figure. working as a load L i,increasing di/dt di Power consumed by the inductor = i L dt di Energy consumed in dt time = i L dt dt i L di/dt   total energy consumed as the current increases from 0 to I = 0 1 2 1 Li  U = L 2 2 2 This energy is stored in the magnetic field with energy density = Note : dU B 2 B2   dV 2 2 0  r Total energy U =  B2 dV 2 0 r 1  iLdi = 2 L 2 Ex. Sol. A circuit contains an ideal cell and an inductor with a switch. Initially the switch is open .It is closed at t=0. Find the current as a function of time. i di  = L dt  0 t = Li Ex.   dt =  t i= L i  Ldi 0 i1 B i3 C i2 L t=0 e di Find the currents i1 ,i2 ,i3 and 3 at t=0 and at t = . dt Initially all currents are zero. Sol. R A In the following circuit, the switch is closed at t = 0. R R F E D At t = 0 i3 is zero ,since current cannot suddenly change due to the inductor. applying KVL in the part ABEF we get i1= i2=  i1= i2 (from KCL)  . 2R At t=  i3 will become constant and hence potential difference across the inductor will be zero. It is just like a simple wire and the circuit can be solved assuming it to be like shown in the following diagram. Before S2 is closed and S1 is opened current in the left part of the circuit = Now when S2 closed S1 opened, current through the inductor can not change suddenly , current  will continue to move in the inductor.. R Applying KVL in loop 1. L di  + (2R ) + 4 = 0 dt R  di 6 =– dt L R R E s1  . R C i2 F In the circuit shown in the figure,S1 remains closed for a long time and S2 remains open. Now S2 is closed and S1 is opened. Find out the di/dt just after that moment. Sol. t=0 e  2 i2= i3= ,i = . 3R 1 3R Ex. i1 B i3 R A s2 R D 2R i 4e L e 2R R /R Ldi dt i=/R R 1 4e -4e ELECTROMAGNETIC INDUCTION R-L DC Circuit Current Growth (i) EMF equation E  IR  L (ii) Current at any instant I L R dI dt + – E switch When key is closed the current in circuit increases exponentially with respect to time. The current in circuit at any t   instant ‘t’ given by I  I0 1  e     t = 0 (just after the closing of key)  I = 0 t =(some time after closing of key)  (iii) I  I0 Just after the closing of the key inductance behaves like open circuit and current in circuit is zero. Open circuit, t = 0, I = 0 (iv) + Inductor provide infinite resistence Some time after closing of the key inductance behaves like simple R connecting wire (short circuit) and current in circuit is constant. + – E Short circuit, t , I I0 , Inductor provide zero resistence I0  R (Final, steady, maximum or peak value of current) or ultimate current Note : Peak value of current in circuit does not depends on self inductance of coil. (v) Time constant of circuit ()  (vi) L It is a time in which current increases up to 63% or 0.63 times of peak current value. R sec. Half life (T) It is a time in which current increases upto 50% or 0.50 times of peak current value. I = I0 (1 – e–t/), t = T, I = 1 I0 I0  = I0 (1 – e–T/) e–T/ =  eT/ = 2 2 2 2 T L logee = loge2  T  0.6 9 3   T  0 .6 9 3 R sec.  (vii) Rate of growth of current at any instant :–  dI  E  t /  )  dt   L (e   – t=0  E  dI    dt    max L t =    dI   dt   0   Note : Maximum or initial value of rate of growth of current does not depends upon resistance of coil. ( ) Current Decay dI 0 dt (i) Emf equation IR  L (ii) Current at any instant Once current acquires its final max steady value, if suddenly switch is put off then current start decreasing exponentially wrt to time. At switch put off condition t = 0, I = I0, source emf E is cut off from circuit I  I0 (e  t /  ) (iii) Just after opening of key t=0 Some time after opening of key t  I = I0 = E R   I  0 Time constant () It is a time in which current decreases up to 37% or 0.37 times of peak current value. (iv) Half life (T) It is a time in which current decreases upto 50% or 0.50 times of peak current value. (v) Rate of decay of current at any instant  dI   E   t /    dt    L  e     t = 0 E  dI  = t   dt  L   max. Graph for R–L circuit Current Growth (a) (b) Current decay (a) (b)   dI    dt   0   ELECTROMAGNETIC INDUCTION Ex. At t = 0 switch is closed (shown in figure) after a long time suddenly the inductance of the inductor is made  times lesser ( L ) then its initial value, find out instant current just after the operation.  R L S Sol. Using above result (note 4)  R In the following circuit the switch is closed at t = 0. Initially there is no current in inductor. Find out current the inductor coil as a function of time. L1 i1 = L2 i2 Ex. Sol. :  i2 = At any time t  + i1R – (i – i1) R = 0  + 2i1R – iR = 0 iR   i1 = 2R di Now,  + i1R + iR + L. =0 dt R – t=0 + (i–i1)R – – iR + – R L i–i 1 + – di  3iR + = –L . dt 2 2     3iR   dt = –L . di  2   – di  dt =    3iR 2L  –     3iR  1 t  = ln  3R    2L  3Rt    i=+ 1  e 2L 3R    0 dt 2L = i di     3iR 0     3iR  3Rt  = – ln     2L     Figure shows a circuit consisting of a ideal cell, an inductor L and a resistor R,connected in series. Let the switch S be closed at t = 0. Suppose at t = 0 current in the inductor is i0 then find out equation of current as a function of time Let an instant t current in the circuit is i which is increasing at the rate di/dt. R L C i B S di dt i i1  – Sol. R +  iR    di  + iR + L .  +  =0 dt  2  t Ex. i1R A R i0 L C B S A Writing KVL along the circuit , we have  L i di    iR dt   Equivalent self inductance i > A i > A + L1 Parallel Combination t dt L 0    iR  Rt  =–  ln     i R L 0     (  i0R )e Rt / L R – + – di    iR i= L di dt VA  VB L di / dt VA – L1 from (1) and (2)  i0   – iR = ( – i0R) e Rt / L Series combination di –iR=0 dt –L B ..(1) + L2 – B di di – L2 = VB dt dt ....(2) L = L1 + L2 ( neglecting mutual inductance) A i + + i1 i2 > > > di1 di 2 From figure VA – VB = L1 = L2 .....(3) dt dt also i = i1 + i2 di di1 di 2   or dt dt dt VA  VB VA  VB VA  VB or = + L1 L2 L 1 1 1   ( neglecting mutual inductance) L L1 L 2 L1 B – L2 – Mutual Induction Whenever the current passing through primary coil or circuit change then magnetic flux neighbouring secondary coil or circuit will also change. Acc. to Lenz for opposition of flux change, so an emf induced in the neighbouring coil or circuit. This phenomenon called as 'Mutual induction'. In case of mutual inductance for two coils situated close to each other, flux linked with the secondary due to current in primary. Due to Air gap always Case - I 2<1 and 2 = B1A2 (=0°). When current through primary is constant Total flux of secondary is directly proportional to current flow through the primary coil N2 2 I1  N2 2 = MI1, M  ( ) N 2 2 N B A  2 1 2  T s where M : is coefficient of mutual induction. I1 I1 Ip ELECTROMAGNETIC INDUCTION Case - II When current through primary changes with respect to time If dI1 dB1 d 1 d 2    Static EMI dt dt dt dt N22 = MI1 –N2 d  d 2 dI1  = –M ,  N 2 dt  dt dt  called total mutual induced emf of secondary coil em. • The units and dimension of M are same as ‘L’. • The mutual inductance does not depends upon current through the primary and it is constant for circuit system. 'M' depends on : (a) Number of turns (N1, N2). (b) Cofficient of self inductance (L1, L2). (c) Area of cross section. (d) Magnetic permeabibility of medium (r). (e) Distance between two coils (As d  = M  ). (f) Orientation between two coils. (g) Coupling factor 'K' between two coils. Different Coefficient of Mutual Inductance (a) (a) In terms of their number of turns (b) In terms of their coefficient of self inductances (c) In terms of their nos of turns (N1 , N2) Two co–axial solenoids : (MS S ) 1 2 Coefficient of mutual inductance between two solenoids M s1s2 = (b) N 2 B1 A N   0 N1 I1  = 2  A  I1 I1  l   N N A  Ms s   0 1 2  l   1 2 Two plane concentric coils : (MC C ) 1 2 N1 N2 r2 O M c1c2 = r1 I (r1>>r2) N 2 B1 A 2 I1 where B1 = 0 N1 I1 , A2 = r22 2r1  N N r 2  N I  M c1c2 = N2  0 1 1  (r22) Mc c  0 1 2 2 2 r1 I1  2r1  1 2 Two concentric loop r2 M  2 (r1 >> r2) r1 Two concentric square loops r2 O r1 b2 M a a tiny b A square and a circular loop r2 M a tiny r a In terms of L1 and L2 : For two magnetically coupled coils :- M  K L1 L 2 here 'K' is coupling factor between two coils and its range 0  K1 • For ideal coupling Kmax = 1  Mmax  L1 L 2 (where M is geometrical mean of L1 and L2) • For real coupling (0 < K < 1) M  K L1 L 2 • Value of coupling factor 'K' decided from fashion of coupling. Different Fashion of Coupling P S K=1 M = L1L2 (max) ideal coupling (coaxial fashion) plane are parallel: 0° fashion air gap P S 0 NP Step up transformer : Converts low voltage high current in to High voltage low current Step down transformer : Converts High voltage low current into low voltage high current. Power transmission is carried out always at "High voltage low current" so that voltage drop and power losses are minimum in transmission line. voltage drop = IL RL , IL = line current RL = total line resistance, IL = power to be transmission line voltage power losses = I2L R L High voltage coil having more number of turns and always made of thin wire and high current coil having less number of turns and always made of thick wires. Ideal Transformer : ( = 100%) (a) No flux leakage s = p   dp  ds = dt dt es = ep = e induced emf per turn of each coil is also same. total induced emf for secondary Es = Nse total induced emf for primary Ep = Npe Ns Es = =n Np Ep (b) where n : turn ratio , p : transformation ratio No load condition VP = EP and (c) or p ES = VS VS N S  VP N P from (i) and (ii) VS N S  =n VP N P or p No power loss Pout = Pin and VSIS = VPIP from equation (iii) and (iv) VS IP  VP IS valid only for ideal transformer VS IP N S   n VP IS N P or p Note : Generally transformers deals in ideal condition i.e. Pin = Pout, if other information are not given. Real transformer ( 100%) Some power is always lost due to flux leakage, hysteresis, eddy currents, and heating of coils. hence Pout < Pin always. efficiency of transformer    Pout V I  S . S  10 0 Pin VP IP Applications The most important application of a transformer is in long distance transmission of electric power from generating station to consumers hundreds of kilometers away through transmission lines at reduced loss of power. Transmission lines having resistance R and carrying current  have loss of power = 2R. This loss is reduced by reducing the current by stepping up the voltage at generating station. This high voltage is transmitted through high-tension transmission lines supported on robust pylons (iron girder pillars). The voltage is stepped down at consumption station. A typical arrangement is shown below : ELECTROMAGNETIC INDUCTION Raja Park Sub-Station Kota Jaipur Jaipur Area Station Step up transformer House 200 V to 11 kV Kota thermal Power Station Ex. Step up transformer 11kV to 135 kV Step down Step down transformer transformer 135 kV 11 kV to 11kV to 220kV A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric power plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5  per km. The town gets from the line through a 4000 – 220 V step down transformer at a sub-station in the town. (a) Estimate the line power loss in the form of heat. (b) How much power must be plant supply, assuming there is a negligible power loss due to leakage? (c) Characterise the step up transformer at the plant. Sol. The diagram shows the network : For sub-station, P = 800 kW = 800 × 103 watts V = 220 V S = P 8 00  10 3 40 = = × 103 A. V 2 20 11 Primary current (p) in sub-station transformer will be given by 2 20  40  10 3 = 200 A 1 1  4 000 Hence transmission line current = 200 A 4000 × P = 220 × S, P = (a) transmission line resistance = 2 × 15 × 0.5 = 15  transmission line power loss = 2R = 200 × 200 × 15 = 6 × 105 watt = 600 kW. (b) power to be supplied by plant = power required at substation + loss of power of transmission = 800 + 600 = 1400 kW. 1400 kW 1 40 0  1 0 00 Power = = = 7000 V 200 A Current 200 Step-up transformer at power plant has characteristics 440 - 7000 V. (c) Voltage in secondary at power plant has characteristics = Ex. A power transmission line feeds input power at 2300 V to a step down transformer having 4000 turns in its primary. What should be the number of turns in the secondary to get output power at 230 V? Sol. Ep = 2300 V ; NP = 4000, ES = 230 V Ex. The output voltage of an ideal transformer, connected to a 240 V a.c. mains is 24 V. When this transformer is used to light a bulb with rating 24V, 24W calculate the current in the primary coil of the circuit. Sol. EP = 240 V, ES = 24 V, E 2 30 ES NS  4 00  N S  N P  S  4 00 0   EP 2 30 0 EP N P ES IS = 24 W Current in primary coil IP = E S IS 24   0 .1 A EP 24 0 Losses of Transformer (a) (b) Copper or joule heating losses Where : There losses occurs in both coils of shell part Reason : Due to heating effect of current (H = I2Rt) Remmady : To minimise these losses, high current coil always made up with thick wire and for removal of produced heat circulation of mineral oil should be used. Flux leakage losses Where : There losses occurs in between both the coil of shell part. Cause Due to air gap between both the coils. : Remmady : (c) (i) To minimise there losses both coils are tightly wound over a common soft iron core (high magnetic permeability) so a closed path of magnetic field lines formed itself within the core and tries to makes coupling factor K  1 Iron losses Where : There losses occurs in core part. Types (i) Hysteresis losses (ii) Eddy currents losses : Hysteresis losses Cause : Remmady : Transformer core always present in the effect of alternating magnetic field (B = B0sint) so it will magnetised & demagnetised with very high frequeny (f = 50 Hz).During its demagnetization a part of magnetic energy left inside core part in form of residual magnetic field. Finally this residual energy waste as heat. To minimise these losses material of transformer core should be such that it can be easily magnetised & demagnetised. For this purpose soft ferromagnetic material should be used. For Ex. soft iron (low retentivity and low coercivity) Eddy Currents (or Focalt's currents) (F1<<> diameter) where m = magnetic permeability of the core material n = number of turns in the solenoid per unit length i Load For two coils in parallel L  = current in the solenoid Secondary winding I I axis of solenoid Core Self inductance of a solenoid L = 0 n2AI A = area of cross section of solenoid. 13. Super Conduction Loop in Magnetic Field : R = 0;  = 0. Therefore  total = constant. Thus in a superconducting loop flux never changes. (or it opposes 100 %) 1 2 LI 2 14. Energy Stored in an Inductor : W  15. Energy of interaction of two loops U = I12  I 2 1  MI1I 2 where M is mutual inductance 16. Growth of a Current in an L - R Circuit : L E I  1  e Rt L  . [If initial current = 0] R L = time constant of the circuit. R I0  (a) E . R L behaves as open circuit at t = 0 [If i = 0] I R t=0 ELECTROMAGNETIC INDUCTION (b) L behaves as short circuit at t =  always. L Curve (1)  Large R Curve (2)  17. I L Small R (2) (1) Decay of Current : Initial current through the inductor = I0 ; Current at any instant i = I0e–Rt/L I I0 18. An emf is induced in a closed loop where magnetic flux is varied. The induced electric field is not conservative field ur r because for induced electric field, the line integral Ñ  E.d l around a closed path is non- zero. 19. Acceleration of a magnet falling through a long solenoid decrease because the induced current produced in a circuit always flows in such direction that it opposes the change or the cause that produces it. 20. The mutual inductance of two coils is doubled if the self inductance of the primary and secondary coil is doubled because mutual inductance M  L1L2 . 21. The possibility of an electric bulb fusing is higher at the time of switching ON and OFF because inductive effects produce a surge at the time of switch-off and switch-on. 22. Motional emf : If a conductor is moved in a magnetic field then motional emf will be E = B l eff v Here v  l eff & v  B & B  l eff l eff  effective length between the end points of conductor which is perpendicular to the velocity..