P1-Allens Made Physics Theory {PART-1}

JEE-Physics  When  path  difference  between  the  secondary  wavelets  coming  from pilnancied ewnatv e    lens  2  2 A  and  B  is    n or  2n   or    even  multiple  of    then  minima  occurs For  minima  a sin n  2 n     where  n  =  1,  2,  3  ... long  diffraction   2  narrow slit pattern  When  path  difference  between  the  secondary  wavelets  coming  from A and B is  (2n+ 1)   or  odd  multiple  of    then  maxima  occurs 22 For  maxima  where  n  =  1,  2,  3  ... asin  n =  (2n  +  1)  2 n  =  2     second  maxima n  =  1       first  maxima   and  In  alternate  order  minima  and  maxima  occurs  on  both  sides  of  central  maxima. For  nth  minima P' xn If  distance  of  nth  minima  from  central  maxima  =  x O n n n P distance  of  slit  from  screen  =  D  ,  width  of  slit  =  a D P\" 2n sin n  n Path  difference   =  a  sinn  =  2  a In  POP'  tan n  xn If n is small sin n  tan n  n D xn  nD n  xn  n First  minima  occurs  both  sides  on  central  maxima. a D a For  first  minima  x  D and  x  a Da  Linear  width  of  central  maxima w  =  2x  wx =  2 D  x a  Angular  width  of  central  maxima w= 2  2  a SPECIAL  CASE Lens  L2  is  shifted  very  near  to  slit  AB.  In  this  case  distance  between  slit  and  screen  will  be  nearly  equal to  the focal  length  of  lense  L     (i.e.  D     f  )  n =  xn  n  nf NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Wave Optics\\Eng\\02_Diffraction of Light & Polarisation.p65 2 f a x =  n  a A L2 P B D~f 2f 2x 2 w =    and  angular  width  of  central  maxima  w   =   x  a B f a Fringe  width    :  Distance  between  two  consecutive  maxima  (bright  fringe)  or  minima  (dark  fringe)  is  known  as fringe  width.  Fringe  width  of  central  maxima  is  doubled  then  the  width  of  other  maximas  i.e., D nD D   =  x   –  x  =  (n  +  1)    –  =  n  +  n a a a 1   50 E

JEE-Physics Intensity  curve  of  Fraunhofer's  diffraction I0 Intensity  of  maxima  in  Fraunhofer's  diffrection  is  determined  by I  =   2 2 I0  (2 n   1)  I     =intensity  of  central  maxima I0 I0 I0 I0 22 22 0 61 n  =  order  of  maxima intensity  of  first  maxima  I   =  4 I0  I0       61 1 9 2 22 intensity  of  second  maxima  I   =  4 I0  I0 /a /a 0 /a /a 2 25 2 61 Angle   Diffraction  occurs  in  slit  is  always  fraunhofer  diffraction  as  diffraction  pattern  obtained  from  the  cracks  between the  fingers,  when  viewed  a  distant  tubelight  and  in  YDSE  experiment  are  fraunhofer  diffraction. GOLDEN  KEY  POINTS • The  width  of  central  maxima    ,  that  is,  more  for  red  colour  and  less  for  blue. i.e., wx   as blue  <  red w   <  w blue red • For  obtaining  the  fraunhofer  diffraction,  focal  length  of  second  lens  (L )  is  used. 2 wx  f  1/a width  will  be  more  for  narrow  slit • By  decreasing  linear  width  of  slit,  the  width  of  central  maxima  increase. RESOLVING  POWER  (R.P.) A  large  number  of  images  are  formed  as  a  consequence  of  light  diffraction  from  a  source.  If  two  sources  are separated  such  that  their  central  maxima  do  not  overlap,  their  images  can  be  distinguished  and  are  said  to  be resolved  R.P.  of  an  optical  instrument  is  its  ability  to  distinguish  two  neighbouring  points. Linear  R.P.  =  d/D here D  =  Observed  distance Angular  R.P.  =  d/ d  =  Distance  between  two  points (1) Microscope  :  In  reference  to  a  microscope,  the  minimum  distance  between   two  lines  at  which  they  are  just distinct  is  called  Resolving  limit  (RL)  and  it's  reciprocal  is  called  Resolving  power  (RP) NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Wave Optics\\Eng\\02_Diffraction of Light & Polarisation.p65 R.L.     and    R.P.  2µ sin      R.P.    1 sin   2µ   O   =  Wavelength  of  light  used  to  illuminate  the  object Objective µ  =  Refractive  index  of  the  medium  between  object  and  objective,   =  Half  angle  of  the  cone  of  light  from  the  point  object,  µsin  =  Numerical  aperture. (2) Telescope  :  Smallest  angular  separations  (d)  between  two  distant  object,  whose  images  are  separated  in  the telescope  is  called  resolving  limit.  So  resolving  limit  d  1.22   and  resolving  power a (RP)  =  1  a  R.P.  1 where  a  =  aperture  of  objective. d 1.22  E 51

JEE-Physics Example Light  of  wavelength  6000Å  is  incident  normally  on  a  slit  of  width  24  ×    10–5  cm.  Find  out  the  angular  position  of second  minimum  from  central  maximum  ? S o l . a  sin  =  2 given    =  6  ×    10–7  m,  a  =  24  ×    10–5  ×    10–2  m 2 2  6  107 1 sin=  a  =      =  30° 24  107 2 Example Light  of  wavelength  6328Å  is  incident  normally  on  a  slit  of  width  0.2  mm.  Calculate  the  angular  width  of  central maximum  on  a  screen  distance  9  m  ? S o l . given    =  6.328  ×    10–7  m,  a  =  0.2  ×    10–3  m w   =  2 2  6.328 107 6.328 103 180    =  radian    =      0.36° a 2 104 3.14 Example Light  of  wavelength  5000Å  is  incident  on  a  slit  of  width  0.1  mm.  Find  out  the  width  of  the  central  bright  line  on a  screen  distance  2m  from  the  slit  ? 2 f 2  2  5 107 S o l . wx  =  a =  104   =  20  mm Example The  fraunhofer  diffraction  pattern  of  a  single  slit  is  formed  at  the  focal  plane  of  a  lens  of  focal  length  1m.  The width  of  the  slit  is  0.3  mm.  If  the  third  minimum  is  formed  at  a  distance  of  5  mm  from  the  central  maximum  then calculate  the  wavelength  of  light. Sol. nf    =  ax n =  3 104  5 103   =  5000Å [  n  =  3] x   =   fn 3 1 na Example Find  the  half  angular  width  of  the  central  bright  maximum  in  the  Fraunhofer  diffraction  pattern  of  a  slit  of  width 12  ×    10–5  cm  when  the  slit  is  illuminated  by  monochromatic  light  of  wavelength  6000  Å. Sol.  sin      =  half  angular  width  of  the  central  maximum. a a  =  12  ×    10–5  cm,    =  6000  Å  =  6  ×    10–5  cm    sin     6 105  0.50      =  30° NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Wave Optics\\Eng\\02_Diffraction of Light & Polarisation.p65 a 12 105 Example Light  of  wavelength  6000  Å  is  incident  on  a  slit  of  width  0.30  mm.  The  screen  is  placed  2  m  from  the  slit.  Find (a)  the  position  of  the  first  dark  fringe  and  (b)  the  width  of  the  central  bright  fringe. S o l . The  first  fringe  is  on  either  side  of  the  central  bright  fringe. here n  =  1,  D  =  2  m,    =  6000  Å  =  6  ×    10–7  m  sin   x  a  =  0.30  mm  =  3  ×    10–4  m   a sin  = n ax  n DD x  nD x   1 6  10 7 2   4  103 m a  3  10 4  (a)    The  positive  and  negative  signs  corresponds  to  the  dark  fringes  on  either  side  of  the  central  bright  fringe. (b) The  width  of  the  central  bright  fringe  y  =  2x  =  2  ×    4  ×    10–3  =  8  ×    10–3  m  =  8  mm 52 E

JEE-Physics DIFFERENCE  BETWEEN  INTERFERENCE  AND  DIFFRACTION  : Diffraction Interference (1)         It  is  the  phenomenon  of  superposition (1) It  is  the  phenomenon  of  superposition         of  two  waves  coming  from  two  different of  two  waves  coming  from  two  different         coherent  sources. parts  of  the  same  wave  front. (2)         In  interference  pattern,  all  bright  lines (2) All  bright  lines  are  not  equally  bright         are  equally  bright  and  equally  spaced. and  equally  wide.  Brightness  and  width goes  on  decreasing  with  the  angle  of (3)         All  dark  lines  are  totally  dark diffraction. (4)       In  interference  bands  are  large  in  number (3) Dark  lines  are  not  perfectly  dark.  Their contrast  with  bright  lines  and  width  goes on  decreasing  with  angle  of  diffraction. (4) In  diffraction  bands  are  a  few  in  number. Example A  Slit  of  width  a  is  illuminated  by  monochromatic  light  of  wavelength  650nm  at  normal  incidence.  Calculate the  value  of  a  when  - (a) the  first  minimum  falls  at  an  angle  of  diffraction  of  30° (b) the  first  maximum  falls  at  an  angle  of  diffraction  of  30°. Sol. (a) for  first  minimum  sin  1  =                    a  =     =  650  10–9   =  650  10–9   =  1.3  ×  10–6m a sin  sin 30 0.5 1 3 3 3  650  10–9 (b) For  first  maximum    sin  1  =  2a         a  =  2 sin    =  2  0.5   =  1.95  ×  10–6m Example Red  light  of  wavelength  6500Å  from  a  distant  source  falls  on  a  slit  0.50  mm  wide.  What  is  the  distance  between the  first  two  dark  bands  on  each  side  of  the  central  bright  of  the  diffraction  pattern  observed  on  a  screen  placed 1.8  m.  from  the  slit. Sol. Given    =  6500Å  =  65  ×  10–8  m,  a  =  0.5  mm  =  0.5  ×  10–3  m.,  D  =  1.8  m. Required  distance  between  first  two  dark  bands  will  be  equal  to  width  of  central  maxima. NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Wave Optics\\Eng\\02_Diffraction of Light & Polarisation.p65 2D 2  6500  10 10  1.8   =  468  ×  10–5  m  =  4.68  mm Wx =  a   = 0.5  10 3 Example In  a  single  slit  diffraction experiment  first  minimum  for  1  =  660 nm  coincides with  first  maxima  for wa ve le ng th  2.  Calcu la te   . 2 S o l . For  minima  in  diffraction  pattern                d sin  =  n For  first minima                 d sin1  =  (1)1        sin1  =  1 d For  first  maxima                 dsin2  =  3 2           sin2  =  32 2 2d The  two  will  coincide  if,      1  =  2            or          sin1  =  sin2  1  32          =  2  2   ×  660  nm  =  440  nm. d 2d 3 1 3 2 E 53

JEE-Physics POLARISATION Experiments  on  interference  and  diffraction  have  shown  that  light  is  a  form  of  wave  motion.  These  effects  do not  tell  us  about  the  type  of  wave  motion  i.e.,  whether  the  light  waves  are  longitudinal  or  transverese. The  phenomenon  of  polarization  has  helped  to  establish  beyond  doubt  that  light  waves  are  transverse  waves. UNPOLARISED  LIGHT An  ordinary  beam  of  light  consists  of  a  large  number  of  waves  emitted  by  the  atoms  of  the  light  source.  Each  atom  produces  a  wave  with  its  own  orientation  of  electric  vector  E   so  all  direction  of  vibration  of  E   are  equally probable. ZY YX unpolarised light  unpolarised light propagating along . X-axis The  resultant  electromagnetic  wave  is  a  super  position  of  waves  produced  by  the  individual  atomic  sources NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Wave Optics\\Eng\\02_Diffraction of Light & Polarisation.p65 and  it  is  called  unpolarised  light.  In  ordinary  or  unpolarised  light,  the  vibrations  of  the  electric  vector  occur symmetrically  in  all  possible  directions  in  a  plane  perpendicular  to  the  direction  of  propagation  of  light. POL ARISATION The  phenomenon  of  restricting  the  vibration  of  light  (electric  vector)  in  a  particular  direction  perpendicular  to the  direction  of  propagation  of  wave  is  called  polarisation  of  light.  In  polarised  light,  the  vibration  of  the  electric vector  occur  in  a  plane  perpendicular  to  the  direction  of  propagation  of  light  and  are  confined  to  a  single direction  in  the  plane  (do  not  occur  symmetrically  in  all  possible  directions).    After  polarisation  the  vibrations become  asymmetrical  about  the  direction  of  propagation  of  light. POLARISER Tourmaline  crystal  :  When  light  is  passed  through  a  tomaline  crystal  cut  parallel  to  its  optic  axis,  the  vibrations of  the  light  carrying  out  of  the  tourmaline  crystal  are  confined  only  to  one  direction  in  a  plane  perpendicular  to the  direction  of  propagation  of  light.  The  emergent  light  from  the  crystal  is  said  to  be  plane  polarised  light. Nicol  Prism  :  A  nicol  prism  is  an  optical  device  which  can  be  used  for  the  production  and  detection  of  plane polarised  light.  It  was  invented  by  William  Nicol  in  1828. Polaroid    :  A  polaroid  is  a  thin  commercial  sheet  in  the  form  of  circular  disc  which  makes  use  of  the  property of  selective  absorption  to  produce  an  intense  beam  of  plane  polarised  light. PLANE  OF  POLARISATION  AND  PL ANE  OF  VIBR ATION  : The  plane  in  which  vibrations  of  light  vector      and  the  direction  of  propogation  lie  is  known  as  plane  of  vibration A  plane  normal  to  the  plane  of  vibration  and  in  which  no  vibration  takes  place  is  known  as  plane  of  polarisation A plane of vibration B E plane  polarised light unpolarised  F O' light G D plane of  C polarisation 54 E

JEE-Physics EXPERIMENTAL  DEMONSTR ATION  OF  POLARISATION  OF  LIGHT Take  two  tourmaline  crystals  cut  parallel  to  their  crystallographic  axis  (optic  axis). A PO Ordinary Polariser light First  hold  the  crystal  A  normally  to  the  path  of  a    beam  of  colour  light.  The  emergent  beam  will  be  slightly coloured.  Rotate  the  crystal  A  about  PO.  No  change  in  the  intensity  or  the  colour  of  the  emergent  beam  of  light. Take  another  crystal  B  and  hold  it  in  the  path  of  the  emergent  beam  of  so  that  its  axis  is  parallel  to  the  axis  of the  crystal  A.  The  beam  of  light  passes  through  both  the  crystals  and  outcoming  light  appears  coloured. AB PO Ordinary Polariser light Now,  rotate  the  crystal  B  about  the  axis  PO. It  will  be  seen  that  the  intensity  of  the  emergent    beam  decreases and  when  the  axes  of  both  the  crystals  are  at  right  angles  to  each  other  no  light  comes  out  of  the  crystal  B. AB PO Ordinary Polariser light If  the  crystal  B  is  further  rotated  light  reappears  and  intensity  becomes  maximum  again  when  their  axes  are parallel.  This  occurs  after  a  further  rotation  of  B  through  90°.  This  experiment  confirms  that  the  light  waves  are transverse  in  nature.  The  vibrations  in  light  waves  are  perpendicular  to  the  direction  of  propogation  of  the  the wave.    First  crystal  A  polarises  the  light  so  it  is  called  polariser.  Second  crystal  B,  analyses  the  light  whether  it  is polarised  or  not,  so  it  is  called  analyser. METHODS  OF  OBTAINING  PLANE  POLARISED  LIGHT • Polarisation  by  reflection   The  simplest  method  to  produce  plane  polarised  light  is    by    reflection.  This method  was  discovered  by  Malus  in      1808.  When  a  beam  of  ordinary  light  is  reflected  from  a  surface,  the reflected  light  is  partially  polarised.  The  degree  of  polarisation  of  the  polarised  light  in  the  reflected  beam  is greatest  when  it  is  incident  at  an  angle  called  polarising  angle  or  Brewster's  angle. NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Wave Optics\\Eng\\02_Diffraction of Light & Polarisation.p65 unpolarised  C 100% polarised  light perpendicular  to the plane of  A incidence p p B denser  r medium  partially   m polarised  D • Polarising  angle    :  Polarising  angle  is  that  angle  of  incidence  at  which  the  reflected  light  is  completely  plane polarisation. • Brewster's  Law  :  When  unpolarised  light  strikes  at  polarising  angle  P    on  an  interface  separating  a  rare medium  from  a  denser  medium  of  refractive  index    ,  such  that   =  tan  P    then  the  reflected  light  (light  in  rare medium)  is  completely  polarised.  Also  reflected  and  refracted  rays  are  normal  to  each  other.  This  relation  is known  as  Brewster's  law.  The  law  state  that  the  tangent  of  the  polarising  angle  of  incidence  of  a  transparent medium  is  equal  to  its  refractive  index  =  tan  P E 55

JEE-Physics In  case  of  polarisation  by  reflection  : (i) For  i  =  P  refracted  light  is  partially  polarised. (ii) For  i  =  P reflected  and  refracted  rays  are  perpendicular  to  each  other. (iii) For  i  <  P or i  >  P both  reflected  and  refracted  light  become  partially  polarised. sin p According  to  snell's  law   µ  =  sin r ............(i) sin p But  according  to  Brewster's  law    µ  =  tanp  =  cos p   ............(ii) sin p sin p From  equation  (i)  and  (ii)    sin r   =  cos p     sinr  =  cosp  sinr = sin (90° – p)  r = 90° – p     or p  +  r  =  90° Thus  reflected  and  refracted  rays  are  mutually  perpendicular By  Refraction In  this  method,  a  pile  of  glass  plates  is  formed  by    taking  20  to  30  microscope  slides  and  light  is  made  to  be incident  at  polarising  angle  57°.  According  Brewster  law,  the  reflected  light  will  be  plane  polarised  with  vibrations perpendicular  to  the  plane  of  incidence  and  the  transmitted  light  will  be  partially  polarised.    Since  in  one reflection  about  15%  of  the  light  with  vibration  perpendicular  to  plane  of  paper  is  reflected  therefore  after passing  through  a  number  of  plates  emerging  light  will  become  plane  polarised  with  vibrations  in  the  plane  of p a p e r. reflected light S 57° refracted light By  Dichroism NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Wave Optics\\Eng\\02_Diffraction of Light & Polarisation.p65 Some  crystals  such  as  tourmaline  and  sheets  of  iodosulphate  of  quinone  have  the  property  of  strongly  absorbing the  light  with  vibrations  perpendicular  of  a  specific  direction  (called  transmision  axis)  and  transmitting  the  light with  vibration  parallel  to  it.  This  selective  absorption  of  light  is  called  dichroism.  So  if  unpolarised  light  passes through  proper  thickness  of  these,  the  transmitted  light  will  plane  polarised  with  vibrations  parallel  to  transmission axis.  Polaroids  work  on  this  principle. Polaroid optic axis is  tourmaline S perpendicular  crystal to the plane of  Ordinary paper light transmission  axis 56 E

JEE-Physics By  scattering    : When  light  is  incident  on  small  particles  of  dust,  air  molecule  etc. (having  smaller  size  as  compared  to  the wavelength  of  light),  it  is  absorbed  by  the  electrons  and  is  re-radiated  in  all  directions.  The  phenomenon  is  called as  scattering.  Light  scattered  in  a  direction  at  right  angles  to  the  incident  light  is  always  plane-polarised. y Unpolarised light polarised polarised light light zx Unpolarised light Law  of  Malus When  a  completely  plane  polarised  light  beam      light  beamis  incident  analyser,  then  intensity    intensity  of  emergent light  varies  as  the  square  of  cosine  of  the  angle  between  the  planes  of  transmission  of  the  analyser  and  the polarizer.  I cos2I = I0 cos2 plane of polariser NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Wave Optics\\Eng\\02_Diffraction of Light & Polarisation.p65AAA Acos sin   A cos (i) If   =  0°  then  I  =  I  maximum  value  (Parallel  arrang ement) anplalaynseerof 0 (ii) If   =  90°  then  I  =  0  minimum  value  (Crossed  arrangement) If  plane  polarised  light  of  intensity  I (=  KA2)  is  incident  on  a  polaroid  and  its  vibrations  of  amplitude  A  make 0 angle    with  transmission  axis,  then  the  component  of  vibrations  parallel  to  transmission  axis  will  be  Acos    while perpendicular  to  it  will  be  A  sin  . Polaroid  will  pass  only  those  vibrations  which  are  parallel  to  transmission  axis  i.e.  Acos  ,    I0 A2   So  the  intensity  of  emergent  light I  =  K(Acos)2 =    KA2cos2 If  an  unpolarised  light  is  converted  into  plane  polarised  light  its  intensity  becomes  half. If  light  of  intensity  I ,  emerging  from  one  polaroid  called  polariser  is  incident  on  a  second  polaroid    (called 1 analyser)  the  intensity  of  light  emerging  from  the  second  polaroid  is I  =  I  cos2  =  angle  between  the  transmission  axis  of  the  two  polaroids. 2 1 E 57

JEE-Physics Optical  Activity When  plane  polarised  light  passes  through  certain  substances,  the  plane  of  polarisation  of  the  emergent  light  is rotated  about  the  direction  of  propagation  of  light  through  a  certain  angle.  This  phenomenon  is  optical  rotation. The  substance  which  rotate  the  plane  of  polarision  rotates  the  plane  of  polarisation  is  known  as  optical  active substance. Ex.  Sugar  solution,  sugar  crystal,  soldium  chlorate  etc. Optical  activity  of  a  substance  is  measured  with  the  help  of  polarimeter  in  terms  of  specific  rotation  which  is defined  as  the  rotation  produced  by  a  solution  of  length  10  cm  (1dm)  and  of  unit  concentration  (1g/cc)  for  a given  wave  length  of  light  at  a  given  temp. specific  rotation  [ ]    =  L   =  rotation  in  length  L  at  concentration tC C Types  of  optically  active  substances (a) Dextro  rotatory  substances Those  substances  which  rotate  the  plane  of  polarisation  in  clockwise  direction  are  called  dextro  rotatory of  right  handed  substances. (b) Laveo  rotatory  substances These  substances  which  rotate  the  plane  of  polarisation  in  the  anticlockwise  direction  are  called  laveo rotatory  or  left  handed  subsances. The  amount  of  optical  rotation  depends  upon  the  thickness  and  density  of  the  crystal  or  concentration  in case  of  solutions,  the  temperature  and  the  wavelength  of  light  used. Rotation  varies  inversely  as  the  square  of  the  wavelenth  of  light. APPLICATIONS  AND  USES  OF  POLARISATION NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Wave Optics\\Eng\\02_Diffraction of Light & Polarisation.p65  By  determining  the  polarising  angle  and  using  Brewster's  Law    =  tanP  refractive  index  of  dark  transparent substance  can  be  determined.  In  calculators  and  watches,  numbers  and  letters  are  formed  by  liquid  crystals  through  polarisation  of  light  called liquid  crystal  display  (L.C.D.)  In  CD  player  polarised  laser  beam  acts  as  needle  for  producing  sound  from  compact  disc.  It  has  also  been  used  in  recording  and  reproducing  three  dimensional  pictures.  Polarised    light  is  used  in  optical  stress  analysis  known  as  photoelasticity.  Polarisation  is  also  used  to  study  asymmetries  in  molecules  and  crystals  through  the  phenomenon  of  optical activity. Example Two  polaroids  are  crossed  to  each  other.  When  one  of  them  is  rotated  through  60°,  then  what  percentage  of  the incident  unpolarised  light  will  be  transmitted  by  the  polaroids  ? 58 E

JEE-Physics So l. Initially  the  polaroids  are  crossed  to  each  other,  that  is  the  angle  between  their  polarising  directions  is  90°.  When  one is  rotated  through  60°,  then  the  angle  between  their  polarising  directions  will  become  30°. Let  the  intensity  of  the  incident  unpolarised  light  =    I 0 1 Then  the  intensity  of  light  emerging  from  the  first  polaroid  is  I1  2 I0 This  light  is  plane  polarised  and  passes  through  the  second  polaroid. The  intensity  of  light  emerging  from  the  second  polaroid  is   I2  =  I1  cos2   =  the  angle  between  the  polarising  directions  of  the  two  polaroids. 1 and   =  30° so I2  I1 cos2 30  1 I0 cos2 30  I2 3 I1  2 I0 2  I0 8  transmission  percentage  =  I2 100  3 100  37.5% I0 8 Example At  what  angle  of  incidence  will  the  light  reflected  from  water  (  =  1.3)  be  completely  polarised  ? S o l .   =  1.3, From  Brewster's  law  tan  p  =    =  1.3    =  tan–1  1.3  =  53° Example If  light  beam  is  incident  at  polarising  angle  (56.3°)  on  air-glass  interface,  then  what  is  the  angle  of  refraction  in glass  ? Sol.  ip  +  r  =  90°   r  =  90°  –  i  =  90°  –  56.3°  =  33.7° p p p Example A  polariser  and  an  analyser  are  oriented  so  that  maximum  light  is  transmitted,  what  will  be  the  intensity  of outcoming  light  when  analyer  is  rotated  through  60°. Sol. According  to  Malus  Law  I  =  I  cos2    =  I  cos2  60°  = I0  1 2  I0 4 0 0  2  NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Wave Optics\\Eng\\02_Diffraction of Light & Polarisation.p65 Example A  300  mm  long  tube  containing  60  cm3  of  sugar  solution  produces  an  optical  rotations  of  10°  when  placed  in  a saccharimeter.  If  specific  rotation  of  sugar  is  60°,  calculate  the  quantity  of  sugar  contained  in  the  tube  solution. Sol.   =  300  mm  =  30  cm  =  3  decimetre,    =  10°,    T  60 ,  volume  of  solution  =  60  cm3     T C   C    10  1 g  cm–3 60  3 18   T   1 Quantity  of  sugar  contained  =  18 ×    60  =  3.33  g E 59

JEE-Physics EXERCISE-I CHECK YOUR GRASP 1 . Diffraction  and  interference  of  light  refers  to  :– (B)  wave  nature  of  light (A)  quantum  nature  of  light (D)  electromagnetic  nature  of  light (C)  transverse  nature  of  light 2 . The  phenomenon  of  diffraction  of  light  was  discovered  by  :– (A)  Huygens (B)  Newton (C)  Fresnel (D)  Grimaldi 3 . Sound  waves  shows  more  diffraction  as  compare  to  light  rays  :– (A)  wavelength  of  sound  waves  is  more  as  compare  to  light  rays (B)  wavelength  of  light  rays  is  more  as  compare  to  sound  waves (C)  wavelength  of  sound  waves  and  light  ray  is  same (D)  none  of  these 4 . The  conversation  going  on,  in  some  room,  can  be  heared  by  the  person  outside  the  room.  The  reason  for  it  is  :– (A)  interference  of  sound (B)  reflection  of  sound (C)  diffraction  of  sound (D)  refraction  of  sound 5 . Diffraction  initiated  from  obstacle,  depends  upon  the (A)  size  of  obstacle (B)  wave  length,  size  of  obstacle (C)  wave  length  and  distance  of  obstacle  from  screen (D)  size  of  obstacle  and  its  distance  from  screen 6 . Phenomenon  of  diffraction  occurs  :– (A)  only  in  case  of  light  and  sound  waves (B)  for  all  kinds  of  waves (C)  for  electro-magnetic  waves  and  not  for  matter  waves (D)  for  light  waves  only 7 . Diffraction  of  light  is  observed  only,  when  the  obstacle  size  is  :– (A)  very  large (B)  very  small (C)  of  the  same  order  that  of  wavelength  of  light (D)  any  size 8 . Which  of  the  following  ray  gives  more  distinct  diffraction  :– (A)  X-ray (B)  light  ray (C)   –ray (D)  Radio  wave 9 . All  fringes  of  diffraction  are  of  :– (A)  the  same  intensity (B)  unequal  width (C)  the  same  width (D)  full  darkness 1 0 . A  single  slit  of  width  d  is  placed  in  the  path  of  beam  of  wavelength   The  angular  width  of  the  principal maximum  obtained  is  :– d  2 2d (A)   (B)  d (C)  d (D)   1 1 . Direction  of  the  second  maximum  in  the  Fraunhofer  diffraction  pattern  at  a  single  slit  is  given  by  (a  is  the  width of  the  slit)  :– (A)  a sin    (B)  a cos   3 (C) a sin  (D)  a sin   3 2 2 2 1 2 . Angular  width  ()  of  central  maximum  of  a  diffraction  pattern  of  a  single  slit  does  not  depend  upon :– NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Wave Optics\\Eng\\02_Diffraction of Light & Polarisation.p65 (A)  distance  between  slit  and  source (B)  wavelength  of  light  used (C)  width  of  the  slit (D)  frequency  of  light  used 13 . Red  light  is  generally  used  to  observe  diffraction  pattern  from  single  slit.  If  green  light  is  used instead  of  red  light,  then  diffraction  pattern  :– (A)  will  be  more  clear (B)  will  be  contract (C)  will  be  expanded (D)  will  not  visualize 1 4 . Calculate  angular  width  of  central  maxima  if    =  6000  Å,  a  =  18  ×    10–5  cm  :– (A)  20° (B)  40° (C)  30° (D)  260° 1 5 . In  single  slit  Fraunhoffer  diffraction  which  type  of  wavefront  is  required  :– (A)  cylindrical (B)  spherical (C)  elliptical (D)  plane 1 6 . In  the  diffraction  pattern  of  a  single  slit  aperture,  the  width  of  the  central  fringe  compared  to  widths  of  the  other fringes,  is  :– (A)  equal (B)  less (C)  little  more (D)  double 1 7 . Central  fringe  obtained  in  diffraction  pattern  due  to  a  single  slit  :– (A)  is  of  minimum  intensity (B)  is  of  maximum  intensity (C)  intensity  does  not  depend  upon  slit  width (D)  none  of  the  above 60 E

JEE-Physics 1 8 . In  a  single  slit  diffraction  pattern,  if  the  light  source  is  used  of  less  wave  length  then  previous  one.  Then  width  of the  central  fringe  will  be  :– (A)  less (B)  increase (C)  unchanged (D)  none  of  the  above 1 9 . In  the  laboratory,  diffraction  of  light  by  a  single  slit  is  being  observed.  If  slit  is  made  slightly  narrow,  then diffraction  pattern  will  :– (A)  be  more  spreaded  than  before (B)  be  less  spreaded  than  before (C)  be  spreaded  as  before (D)  be  disappeared 2 0 . Find  the  half  angular  width  of  the  central  bright  maximum  in  the  Fraunhofer  diffraction  pattern  of  a  slit of  width  12  ×    10–5  cm  when  the  slit  is  illuminated  by  monochromatic  light  of  wavelength  6000  Å. (A)  40° (B)  45° (C)  (D)  60° 2 1 . In  a Fraunhofer's  diffraction by  a slit,  if  slit width  is a,  wave length  focal  length of  lens  is f,  linear  width of  central maxima  is  :– f fa 2f f (A)  a (B)   (C)  a (D)  2a 2 2 . In  a  Fraunhofer's  diffraction  obtained  by  a  single  slit  aperture,  the  value  of  path  difference  for  nth  order  of minima  is  :– (A)  n (B)  2n (2n 1)  (D)  (2n–1) (C)  2 2 3 . A  polariser  is  used  to  : (B)  Produce  polarised  light (A)  Reduce  intensity  of  light (D)  Produce  unpolarised  light (C)  Increase  intensity  of  light 2 4 . Light  waves  can  be  polarised  as  they  are  : (A)  Transverse (B)  Of  high  frequency (C)  Longitudinal (D)  Reflected 2 5 . Through  which  character  we  can  distinguish  the  light  waves  from  sound  waves  : (A)  Interference (B)  Refraction (C)  Polarisation (D)  Reflection 2 6 . The  angle  of  polarisation  for  any  medium  is  60°,  what  will  be  critical  angle  for  this  : (A)  sin–1  3 (B)  tan–1  3 (C)  cos–1  3 1 (D)  sin–1  3 2 7 . The  angle  of  incidence  at  which  reflected  light  is  totally  polarized  for  reflection  from  air  to  glass  (refractive  index  n) (A)  sin–1  (n) 1  1  (D)  tan–1  (n) (B)  sin–1   n  (C)  tan–1   n  2 8 . A  polaroid  is  placed  at  45°  to  an  incoming  light  of  intensity  I .  Now  the  intensity  of  light  passing  through  polaroid 0 after  polarisation  would  be  : NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Wave Optics\\Eng\\02_Diffraction of Light & Polarisation.p65 (A)  I (B)  I /2 (C)  I /4 (D)  Zero 0 0 0 2 9 . Plane  polarised  light  is  passed  through  a  polaroid.  On  viewing  through  the  polariod  we  find  that  when  the polariod  is  given  one  complete  rotation  about  the  direction  of  the  light,  one  of  the  following  is  observed. (A)  The  intensity  of  light  gradually  decreases  to  zero  and  remains  at  zero (B)  The  intensity  of  light  gradually  increases  to  a  maximum  and  remains  at  maximum (C)  There  is  no  change  in  intensity (D)  The  intensity  of  light  is  twice  maximum  and  twice  zero 3 0 . A  ray  of  light  is  incident  on  the  surface  of  a  glass  plate  at  an  angle  of  incidence  equal  to  Brewster's  angle  .  If  µ represents  the  refractive  index  of  glass  with  respect  to  air,  then  the  angle  between  reflected  and  refracted  rays is  : (A)  90  +   (B)sin–1  (µcos) (C)  90° (D)  90°  –  sin–1  (sin/µ) 3 1 . A  beam  of  light  strikes  a  glass  plate  at  an  angle  of  incident  60°  and  reflected  light  is  completely  polarised  than the  refractive  index  of  the  plate  is:- (A)    1.5 (B)  3 (C)  2 3 (D)    2 E 61

JEE-Physics 3 2 . Polarised  glass  is  used  in  sun  glasses  because  : (A)  It  reduces  the  light  intensity  to  half  an  account  of  polarisation (B)  It  is  fashionable (C)  It  has  good  colour (D)  It  is  cheaper 3 3 . When  unpolarized  light  beam  is  incident  from  air  onto  glass  (n=1.5)  at  the  polarizing  angle  : (A)  Reflected  beam  is  polarized  100  percent (B)  Reflected  and  refracted  beams  are  partially    polarized (C) The reason for (A) is that almost all the light is reflected (D) All of the above 3 4 . When  the  angle  of  incidence  on  a  material  is  60°,  the  reflected  light  is  completely  polarized.  The  velocity  of  the refracted  ray  inside  the  material  is  (in  ms–1)  : (A)  3  ×  108 3 (C)  3   ×  108 (D)  0.5  ×  108 (B)      ×  108  2 CHECK YOUR GRASP ANSWER-KEY EXERCISE-I Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Ans. B D A C B B C D B C D A B B D D B A A C Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 Ans. C A B A C D D B D C B A A C PREVIOUS YEARS QUESTIONS EXERCISE-II 1 . Electromagnetic  waves  are  transverse  in  nature  is  evident  by-                 [AIEEE  -  2002] (4)  diffraction (1)  polarization (2)  interference (3)  reflection 2 . When  an  unpolarized  light  of  intensity  I is  incident  on  a  polarizing  sheet,  the  intensity  of  the  light  which  does 0  not  get  transmitted  is- [AIEEE  -  2005] 1 1 (3)  zero (4)  I (1)  I (2)  I 0 20 40 3 . If  I is  the  intensity  of  the  principle  maximum  in  the  single  slit  diffraction  pattern,  then  what  will  be  its  intensity 0  when  the  slit  width  is  doubled- [AIEEE  -  2005] (1)  2I (2)  4I (3)  I (4)  I0 0 0 0 2 4 . Statement-1:  On  viewing  the  clear  blue  portion  of  the  sky  through  a  Calcite  Crystal,  the  intensity  of  transmitted NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Wave Optics\\Eng\\02_Diffraction of Light & Polarisation.p65 light  varies  as  the  crystal  is  rotated. [AIEEE  -  2011] Statement-1:  The  light  coming  from  the  sky  is  polarized  due  to  scattering  of  sun  light  by  particles  in  the atmosphere.  The  scattering  is  largest  for  blue  light. (1)  Statement-1  is  false,  statement-2  is  true (2)  Statement-1  is  true,  statement-2  is  false (3)  Statement-1  is  true,  statement-2  true;  statement-2  is  the  correct  explanation  of  statement-1 (4)  Statement-1  is  true,  statement-2  is  true;  statement  -2  is  not  correct  explanation  of  statement-1. 5 . A  beam  of  unpolarised  light  of  intensity  I0  is  passed  through  a  polaroid  A  and  then  through  another  polaroid B  which  is  oriented  so  that  its  principal  plane  makes  an  angle  of  45°  relative  to  that  of  A.  The  intensity  of  the emergent  light  is  :- (1)  I0 (2)  I0/2 (3)  I0/4 (4)  I0/8 PREVIOUS YEARS QUESTIONS ANSWER-KEY EXERCISE-II Que. 1 2 3 4 5 E Ans. 1 1 3 3 3 62



JEE-Physics ELECTROMAGNETIC INDUCTION MAGNETIC FLUX The magnetic flux () linked with a surface held in a magnetic field (B) is defined as the number of magnetic lines of force crossing that area (A). If  is the angle between the direction of the field and normal to the area, (area  vector) then   B.A = BA cos AB  FLUX LINKAGE If a coil has more than one turn, then the flux through the whole coil is the sum of the fluxes through the individual turns. If the magnetic field is uniform, the flux through one turn is  = BA cos If the coil has N turns, the total flux linkage  = NBA cos • Magnetic lines of force are imaginary, magnetic flux is a real scalar physical quantity with dimensions []  B  area  F  [L2 ]  B  F [ F = B I L sin]  sin  I L  I L   []  M L T2  [L2 ]  [M L2 T 2 A 1 ]    AL   SI UNIT of magnetic flux : or T–m2 (as tesla = Wb/m2) a m p e re  co ulo m b   [M L2T–2] corresponds to energy sec ond  joule  joule  second = weber (Wb) ampere coulomb CGS UNIT of magnetic flux : maxwell (Mx) 1Wb = 108 Mx AB P Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 • For a given area flux will be maximum : max = B A  AB P when magnetic field B is normal to the area min = 0  = 0° cos  = maximum = 1 max = B A For a given area flux will be minimum :  when magnetic field B is parallel to the area  = 90°  cos = minimum = 0 min = 0 E1

JEE-Physics Example A loop of area 0.06 m2 is placed in a magnetic field 1.2 T with its plane inclined 30° to the field direction. Find the flux linked with plane of loop. Solution Area of loop is 0.06 m2 , B = 1.2 T and  = 90° – 30° = 60° So, the flux linked with the loop is  = BA cos = 1.2 × 0.06 × cos = 1.2 × 0.06 × 1   = 0.036 Wb 2 Example A loop of wire is placed in a magnetic field   0.3 ˆj T . B Find the flux through the loop if area vector is   (2 ˆi  5 ˆj  3 kˆ)m 2 A Solution Flux linked with the surface      (0.3 ˆj)  (2 ˆi  5 ˆj  3 kˆ)m 2 = 1.5 Wb B A  Example At a given plane, horizontal and vertical components of earth's magnetic field B and B are along x and y axes HV respectively as shown in figure. What is the total flux of earth's magnetic field associated with an area S, if the area S is in (a) x-y plane (b) y-z plane and (c) z-x plane ? y y y BV BV BV S S x S BH x BH x z z BH (c) (b) z (a) Solution   ˆi B H  ˆj B V = constant, so   Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 B   B . S [ B = constant] (a) For area in x-y plane   S kˆ , xy  (ˆi B H  ˆj B V ).( kˆS )  0 S (b) For area S in y-z plane   S ˆi , yz  (ˆi B H  ˆj B V ).(ˆi S )  B H S S (c) For area S in z-x plane   S ˆj , zx  (ˆi B H  ˆj B V ).(ˆj S )  B V S S Negative sign implies that flux is directed vertically downwards. 2E

JEE-Physics ELECTROMAGNETIC INDUCTION Michael Faraday demonstrated the reverse effect of Oersted experiment. He explained the possibility of producing emf across the ends of a conductor when the magnetic flux linked with the conductor changes. This was termed as electromagnetic induction. The discovery of this phenomenon brought about a revolution in the field of electric power generation. FAR A DAY'S EXPERIMENT Faraday performed various experiments to discover and understand the phenomenon of electromagnetic induction. Some of them are : v=0 • When the magnet is held stationary anywhere near or inside the coil, NS no deflection the galvanometer does not show any deflection. • When the N-pole of a strong bar magnet is moved towards the coil, v the galvanometer shows a deflection right to the zero mark. NS deflection to the right of zero mark • When the N-pole of a strong bar magnet is moved away from the coil, v the galvanometer shows a deflection left to the zero mark. NS deflection to the left of zero mark • If the above experiments are repeated by bringing the S-pole of the, v magnet towards or away from the coil, the direction of current in the coil is opposite to that obtained in the case of N-pole. SN • The deflection in galvanometer is more when the magnet moves deflection to faster and less when the magnet moves slow. the left of zero mark 2v NS more deflection to the right of zero mark Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 CONCLUSIONS Whenever there is a relative motion between the source of magnetic field (magnet) and the coil, an emf is induced in the coil. When the magnet and coil move towards each other then the flux linked with the coil increases and emf is induced. When the magnet and coil move away from each other the magnetic flux linked with the coil decreases, again an emf is induced. This emf lasts so long the flux is changing. Due to this emf an electric current start to flow and the galvanometer shows deflection. The deflection in galvanometer last as long the relative motion between the magnet and coil continues. Whenever relative motion between coil and magnet takes place an induced emf produced in coil. If coil is in closed circuit then current and charge is also induced in the circuit. This phenomenon is called electro magnetic induction. E3

JEE-Physics FAR A DAY'S L AWS OF ELECTROM AGNETIC INDUCTION Based on his experimental studies on the phenomenon of electromagnetic induction, Faraday proposed the following two laws. • First law Whenever the amount of magnetic flux linked with a closed circuit changes, an emf is induced in the circuit. The induced emf lasts so long as the change in magnetic flux continues. • Second law The magnitude of emf induced in a closed circuit is directly proportional to rate of change of magnetic flux linked with the circuit. If the change in magnetic flux in a time dt is = d then e  d dt LENZ'S LAW The Russian scientist H.F. Lenz in 1835 discovered a simple law giving the direction of the induced current produced in a circuit. Lenz's law states that the induced current produced in a circuit always flow in such a direction that it opposes the change or cause that produced it. If the coil has N number of turns and  is the magnetic flux linked with each turn of the coil then, the total magnetic flux linked with the coil at any time = N  e   d (N)  N d   N (2  1 ) dt dt t SN NS ACW v CW N S v NS rest rest (Coil face behaves as North pole (Coil face behaves as South pole to opposes the motion of magnet.) to opposes the motion of magnet.) e  () d , here negative sign indicates the concept of Lenz law. dt LENS'S L AW -  A CONSEQUENCE OF CONSERVATION OF ENERGY Copper coils are wound on a cylindrical cardboard and the two ends of the coil are NS connected to a sensitive galvanometer. When a bar magnet is moved towards the coil (fig.). The upper face of the coil near the magnet acquired north polarity. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 Consequently work has to be done to move the magnet further agains the force of G repulsion. When we withdraw the magnet away from the coil, its nearer face acquires south polarity. Now the workdone is against the force of attraction.When the magnet E is moved, the number of magnetic lines of force linking the coil changes, which causes an induced current of flow through the coil. The direction of the induced current, according to Lenz's law is always to oppose the motion of the magent. The workdone in moving the magnet is converted into electrical energy. This energy is dissipated as heat energy in the coil. Therefore, the induced current always flows in such a direction to oppose the cause. Thus it is proved that Lenz's law is the consequence of conservation of energy. 4

JEE-Physics GOLDEN KEY POINTS • Induced emf does not depends on nature of the coil and its resistance. • Magnitude of induced emf is directly proportional to the relative speed of coil magnet system, (e v). • Induced current is also depends on resistance of coil (or circuit). S S • Induced emf does not depends on resistance of circuit, It exist in  FR a<g a=g N N I cut open circuit also. metal loop ACW induced current = 0 but emf  0 • In all E.M.I. phenomenon, induced emf is non-zero induced parameter. • Induced charge in any coil (or circuit) does not depends on time in which change in flux occurs i.e. it is independent from rate of change of flux or relative speed of coil–magnet system. • Induced charge depends on change in flux through the coil and nature of the coil (or circuit) i.e. resistance. Example The radius of a coil decreases steadily at the rate of 10–2 m/s. A constant and uniform magnetic field of induction 10–3 Wb/m2 acts perpendicular to the plane of the coil. What will be the radius of the coil when the induced e.m.f. in the 1V Solution Induced emf e = d(BA )  Bd(r2 ) dr e 1  10 6 5 dt dt  2  103    102   cm GF JI= 2Br dt radius of coil r = 2B dr H Kdt Example The ends of a search coil having 20 turns, area of cross-section 1 cm2 and resistance 2 ohms are connected to a ballistic galvanometer of resistance 40 ohms. If the plane of search coil is inclined at 30° to the direction of a magnetic field of intensity 1.5 Wb/m2, coil is quickly pulled out of the field to a region of zero magnetic field, calculate the charge passed through the galvanometer. Solution The total flux linked with the coil having turns N and area A is  NBA 1  N (B . A ) = NBA cos = NBA cos(90° – 30°) = 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 NBA when the coil is pulled out, the flux becomes zero, 2 = 0 so change in flux is  = 2  NBA 20 1.5 10 4  0.357 104 C the charge flowed through the circuit is q = =  R 2R 2  42 Example When resistance of primary is changed according to figure then what is the direction of induced current in resistance ‘R’ of secondary? Solution L to N] 5 E

JEE-Physics Example The cross-sectional area of a closely-wound coil having 40 turns in 2.0 cm2 and its resistance is 16 ohm. The ends of the coil are connected to a B.G. of resistance 24 ohm. Calcualte the charge flowing through the B.G. when the coil is pulled quickly out of a region where field is 2.5 Wb/m2 to a region of zero magnetic field. Solution N = 40, B = 2.5 Wb/m2 , A = 2.0 cm × 10–4 m2 and R = 16 + 24 = 40 ohms the charge flowed through the circuit is NBA  40  2.5  (2.0 104 )  5.0 104 C q R 40 Example A current i = 3.36(1 + 2t) × 10–2 A increases at a steady rate in a long straight wire. A small circular loop of radius 10–3 m has its plane parallel to the wire and its centre is placed at a distance of 1m from the wire. The resistance of the loop is 8.4 × 10–4 . Find the magnitude and the direction of the induced current in the loop. Solution The field due to the wire at the centre of loop is B 0 I 2 10 7  I 1m  I 2d 1 So the flux linked with the loop wire wire loop  = BA = B × r2 = 2 I × 10–7 ×  × (10–3)2 = 2 × 10–13 So emf induced in the loop due to change of current | e| d  2 1013 d I dt dt  I = 3.36 (1 + 2t) × 10–2  d I  6.72 10 2 A/s dt And hence e= 2 × 10–13 × 6.72 × 10–2 = 13.44 × 10–15 V And the induced current in the loop i e 13.44 1015  16 1012 A R  8.4 104 Due to increase in current in the wire the flux linked with the loop will increase, so in accordance with Lenz's law Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 the direction of current induced in the loop will be inverse of that in wire, i.e., anticlockwise. Example A 10 ohm resistance coil has 1000 turns. It is placed in magnetic field of induction 5 × 10–4 tesla for 0.1 sec. If the area of cross-section is 1m2, then calculate the induced emf. Solution d NBA 1000  5 104 1 Magnetic flux through the coil  = NBA Induced emf =   = 5V dt t 0.1 Example A coil of mean area 500 cm2 and having 1000 turns is held perpendicular to a uniform field of 0.4 gauss. the coil is turned through 180° in 1/10 second. Calculate the average induced emf. 6E

JEE-Physics Solution  When the plane of coil is perpendicular to field, the angle between area A and field B is 0°. The flux linked with coil  = NBA cos 0 = NBA. When coil is turned through 180°, the flux linked  = NBA cos 12  = – NBA  change in flux  = 2 – 1 = – 2NBA the magnitude of the induced emf is    d  2NBA 2 1000  0.4 104  500 104  0.04 V dt dt  0.1 Example xxx R x x xx x K Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = B e–t is established inside the xx x xx x x rx x 0 x coil. If the key (K) is closed. Then calculate the electrical power developed right after closing the key. Solution Induced emf e  d  d (BA)  A dB  r 2 B0 d (et )   r 2 B0 et dt dt dt dt At t = 0, e = B e–0 . r2 = B r2 0 0 0 The electric power developed in the resistor R just at the instant of closing the key is P  e 2  B 2 2 r4 0 0 RR Example Two concentric coplanar circular loops made of wire, resistance per unit length 10–4 m–1, have diameters 0.2 m and 2 m. A time-varying potential difference (4 + 2.5 t) volt is applied to the larger loop. Calculate the current in the smaller loop. Solution The magnetic field at the centre O due to the current in the larger loop is B  0 I V=4+2.5t S r 2R R If  is the resistance per unit length, then I  potential difference  4  2.5 t  B  0 . 4  2.5 t resistance 2R . 2R 2R  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65  r << R, so the field B can be taken almost constant over the entire area of the smaller loop.  the flux linked with the smaller loop is   B  r2  0 . 4  2.5 t .r2 Induced emf e = d  0r2  2.5 2 R 2R  dt 4 R2  The corresponding current in the smaller loop is I' then I '  e  0r2  2.5  1  2.5 0r 2.5  4 107  0.1 R 4 R2    1.25A 2r  8  R 2 2 8   (1)2  (104 )2 E7

JEE-Physics Example (a) A current from A to B is increasing in magnitude then what is the direction of induced current in the loop and (b) if current is decreasing in magnitude then what is the direction of current in the loop. (c) If instead of current if an electron is moving in the same direction, what will happen ? Solution When current in the wire AB increases, the flux linked with the I=increasing (a) loop (which is out of the page) will increase, and hence the induced f=increasing Bout CW (b) current in the loop will be inverse, i.e., clockwise and will try to Ii = inverse decrease the flux linked with it, i.e., will repel the conductor AB Ii as shown in figure. I When current in the wire AB decreases, the flux linked with ACB the loop (which is out of the page) will decrease, and hence the induced current in the loop will be in anticlockwise direction and I=decreasing ACW will try to increase the flux linked with it, i.e., will attract the conductor AB as shown in figure. f=decreasing Bout Ii = direct Ii I B AC (c) If an electron moving from left to right, the flux with the loop (which is into the page) will first increase and then decrease as the electron passes by. So the induced current I in the loop will be first anticlockwise i and will change direction (i.e., will become clockwise) as the electron passes by. Example A magnet is moved in the direction indicated by an arrow between two coils AB and CD as shown in the figure. Suggest the direction of induced current in each coil. Solution For coil AB : Looking from the end A, the current in the coil AB will be in anticlokwise direction. For coil CD : Looking from the end D, direction of current in the coil CD will be anticlockwise. Note : as the N-pole of the magnet is moving away from the coil AB, the end B of the coil will behave as S-pole so as to oppose the motion of the magnet and the end C of the coil CD should behave as S-pole so as to repel the approaching magnet. Example  In a region of gravity free space, there exists a non-uniform magnetic field B  B0 x3kˆ . A uniform conductor AB of length L and mass m is placed with its end A at origin such that it extends along +x-axis. Find the initial acceleration of the centre of mass and that of end A when a current i flows from A to B. Solution Consider a small sectionof length dx at a ditance x from left end (or end A).  kˆ x dx B=-B0x3k dF  i dxˆi  i  The force on this section is B 0 ix 3 dxˆj A B B 0 x 3 = x=0 y Force on entire rod is given by, x    B 0iL4 ˆj   B 0iL4 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 dF  4 a cm 4m Fm  L B 0ix3dxˆj    Fm  ˆj 0 mass To find acceleration of a point on the rod, we first find angular acceleration of the rod about centre of mass. The  torque due to dF about centre of mass on the rod is    x  L  i              x  L  B 0 ix 3 dx kˆ d  2  dF  r F d  2  Net torque on the rod is  L B ikˆ  x 4  x3 L dx    B 0ikˆ  L L5   3B 0iL5 kˆ 0  2   5  8  40 0 8E

JEE-Physics      3B0iL5 kˆ  9 B0iL3 kˆ dF     mL2  10 m CM  A CB As  , 40  12  x=0 x dx x=L L/2     Now, acceleration of end A is a A  a cm    rAC , where rAC is position vector of A with respect to centre of mass    B 0 iL4 ˆj   9 B 0 iL3 kˆ    L ˆi    B 0 iL4 ˆj  9 B 0 iL4 ˆj  B 0iL4 ˆj aA 4m  10 m  2  aA 4m 20 m 5m INDUCED ELECTRIC FIELD ×××× × When the magnetic field changes with time (let it increases with time) there is an induced E × electric field in the conductor caused by the changing magnetic flux. Important properties of induced electric field : ×E × ×× r  × ××× × (i) It is non conservative in nature. The line integral of E around a closed path is not E zero. When a charge q goes once around the loop, the total work done on it by the E × ×× electric field is equal to q times the emf. × Bin× Hence    e   d ...(i) dt E .d This equation is valid only if the path around which we integrate is stationary.  (ii) Due to of symmetry, the electric field E has the same magnitude at every point on the circle and it is tangential at each point (figure). (iii) Being nonconservative field, so the concept of potential has no meaning for such a field. (iv) This field is different from the conservative electrostatic field produced by stationary charges.  (v) The relation F  q E is still valid for this field. (vi) This field can vary with time. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 • For symmetrical situations E  d  A dB dt dt  = the length of closed loop in which electric field is to be calculated A = the area in which magnetic field is changing. Direction of induced electric field is the same as the direction of included current. Example The magnetic field at all points within the cylindrical region whose cross-section is x xx indicated in the figure start increasing at a constant rate  tesla . Find the xx xx sec ond x xx x x Rx x magnitude of electric field as a function of r, the distance from the geomatric x xx centreof the region. E9

JEE-Physics Solution For r < R : x xx x  E   A dB x R rx x dt xx  E (2r) = (r2)  E  r E  r xx 2 x E-r graph is straight line passing through origin. E At r = R, E  R xxx For r > R : 2 x xR r xx E  E   A dB R dt 2  Example E  R 2 1 E r E  1 2r Er r E (2r) = (R2)  R For the situation described in figure the magnetic field changes with time according to B = (2.00 t3 – 4.00 t2 + 0.8) T and r = 2R = 5.0 cm xxx r2 P2 2 xx xxx r1 P1 (a) Calculate the force on an electron located at P at t = 2.00 s xx x 2 x x (b) What are the magnetude and direction of the electric field at xx Rx x xx P when t = 3.00 s and r = 0.02m. x x 11 Bin Solution E   A dB E  R 2 d (2t3  4 t2  0.8)  R 2 (6 t2  8 t) dt 2r2 dt 2r2 (a) Force on electron at P is F = eE 2  at t = 2 s F 1.6 1019  (2.5 102 )2  [6(2)2  8(2)] I E  2  5 102 xx  1.6  2.5 1021  (24  16)  8 1021 N at t = 2s, r1 x x r2 4 x xx xx F = -e E Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 dB dt is positive so it is increasing.  direction of induced current and E are as shown in figure and hence force of electron having charge –e is right perpendicular to r downwards 2 (b) For r = 0.02 m and at t = 3s, E  r12 (6 t2  8 t)  0.02  [6(3)2  8(3)] = 0.3 V/m at t = 3sec, dB 1 2 r1 2 dt is positive so B is increasing and hence direction of E is same as in case (a) and it is left perpendicular to r1 upwards. 10 E

NO E.M.I. CASES horizontal JEE-Physics Condition of No EMI If : field lines No flux linkage through the coil =0 vertical wire or Flux linkage through the coil = constant =0 • If current I increases with respect to time no induced current in horizontal loop loop because no flux associated with it as plane of circular field lines of straight conductor is parallel to plane of loop. field lines in vertical plane • If current I increases with respect to time no induced parameter horizontal wire in solenoid because no flux associated with solenoid v vv v no relative motion NS (=constant) • no relative motion (=constant) NS   no relative motion no relative motion (=constant) (=constant) • Any rectangular coil or loop translates within the uniform B uniform transverse magnetic field its flux remains constant.  b rectangular loop V • Any coil or loop rotates about its geometrical axis in uniform transverse magnetic field its flux remains const. (=constant) • No flux associated for the coils or loops which are placed in mutually Buniform perpendicular planes. Hence If current of one either increase or  decrease, there is no effect on flux of other. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 (=constant) vertical I horizontal E 11

JEE-Physics METHODS OF PRODUCING INDUCED EMF (TYPES OF EMI) Emf can be induced in a closed loop by changing the magnetic flux linked with a circuit. The magnetic flux is  = BA cos  Magnetic flux can be changed by one of the following methods : (i) Changing the magnetic field B. (Static emi) (ii) Changing the area A of the coil and (dynamic emi)  (iii) Changing the relative orientation  of B and A (Periodic emi) Induced emf by changing the magnetic field B When there is a relative motion between the magnet and a closed loop, the magnetic lines of force passing through it changes, which results in change in magnetic flux. The changing magnetic flux produces induced emf in the loop. Induced emf by changing the area of the coil Q M M' P  B  dx v A U shaped frame of wire, PQRS is placed in a uniform magnetic field B N' perpendicular to the plane and vertically inward. A wire MN of length  is N S placed on this frame. The wire MN moves with a speed v in the direction shown. After time dt the wire reaches to the position M'N' and distance covered = dx. R The change in area A = Length × area = dx Change in the magnetic flux linked with the loop in the dt is d = B × A = B × dx induced emf e = d  B  dx = B v  v  dx  dt dt dt  If the resistance of circuit is R and the circuit is closed then the current through the circuit I e  I  Bv R R A magnetic force acts on the conductor in opposite direction of velocity is B2 2 x xx ax x R Fm  iB  v . x x Fm Fapplied xv x So, to move the conductor with a constant velocity v an equal and opposite B 2 2 v xxx bx x force F has to be applied in the conductor. F  Fm  R B2 2 v2 The rate at which work is done by the applied force is, Papplied  Fv  R and the rate at which energy is dissipated in the circuit is, Pdissipated  i2 R   B v 2 R  B2 2 v2 R   R  This is just equal to the rate at which work is done by the applied force. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 • In the figure shown, we can replaced the moving rod x x xa x x i e=Bv ab by a battery of emf Bv with the positive terminal at B xx a and the negative terminal at b. The resistance r of x xx v Rr the rod ab may be treated as the internal resistance of R x xx xx the battery. x x xb x x Hence, the current in the ciruit is i  e  Bv Rr R r 12 E

JEE-Physics Example Twelve wires of equal lengths are connected in the form of a skeleton cube which is moving with a velocity v in the directon of magnetic field  B . Find the e.m.f. in each arm of cube. Solution FHG IKJNo e.m.f.is v is   ed   .  v ) induced in any arm becuase parallel to B (B Example Wire PQ with negligible resistance slides on the three rails with 5 cm/sec. Calculate current in 10  resistance when switch S is connected to (a)position 1 (b)position 2 Solution e Bv 1  5  10 2  2  102 (a) For position 1 Induced current I = = = = 0.1 mA RR 10 e Bv(2) 1  5  10 2  4  10 2 = 0.2 mA (b) For position 2 Induced current I = = = RR 10 Example  B A copper wire of length 2m placed perpendicular to the plane of magnetic field  (2i  4j) T . If it moves with velocity (4i  6j  8k ) m/sec. Calculate dynamic emf across its ends. Solution Dynamic emf ed =  .    =  .     = i j k = i (32 – 0) – j (16 – 0) + k (12 – 16)  (v B)  (B v) (B  v) 2 4 0 468 = 32i  16j  4k , ed = (2k) · (32i  16j  4k) = –8 volt Example Two parallel rails with negligible resistance are 10.0 cm apart. x a xc ex They are connected by a 5.0  resistor. The circuit also contains x x 5.0  x two metal rods having resistance of 10.0 and 15.0  along the 4.0 m/s x 2.0 m/s rails (fig). The rods are pulled away from the resistor at xb fx constantspeeds 4.00 m/s and 2.00 m/s respectively. A uniform magnetic field of magnitude 0.01 T is applied perpendicular to x 10.0  x d 15.0  x the plane of the rails. Determine the current in the 5.0  resistor. Solution Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 Two conductors are moving in uniform magnetic field, so motional emf will induced across them. The rod ab will act as a source of emf e = Bv  = (0.01) (4.0) (0.1) = 4 × 10–3 V 1 and internal resistance r = 10.0  1 Similarly, rod ef will also act as source of emf e = (0.01) (2.0) (0.1) = 2 × 10–3 V 2 and internal resistance r = 15.0  a c e 2 e2 From right hand rule :V > V and V > V Also R = 5.0 , r2 ba ef f E eq  e1 r2  e2 r1 6 103  20 103  40 103  1.6 103 volt r1 R r1  r2  25 d e1 15 10 b r= 15 10  6  and I = E eq 1.6 103  1.6 103  8 103 amp from d to c eq 15 10  11 55 req  R 6 6 E 13

JEE-Physics MOTIONAL EMF FROM LORENTZ FORCE A conductor PQ is placed in a uniform magnetic field B, directed normal to – e(ve×– B) B P++++ B ––––Q the plane of paper outwards. PQ is moved with a velocity v, the free electrons e– of PQ also move with the same velocity. The electrons experience a magnetic v   Lorentz force, Fm  (v  B) . According to Fleming's left hand rule, this force acts in the direction PQ and hence the free electrons will move towards Q. A negative chagre accumulates at Q and a positive charge at P. An electric field E is setup in the conductor from P to Q. Force exerted by electric field on the  free electrons is, Fe  eE The accumulation of charge at the two ends continues till these two forces balance each other. so     e (    )     Fm Fe v B = – eE E  (v  B) The potential difference between the ends P and Q is V =  = (     .It is the magnetic force on the moving E. v B ). free electrons that maintains the potential difference and produces the emf  = B  v (for      ) B v  As this emf is produced due to the motion of a conductor, so it is called a motional emf. The concept of motional emf for a conductor can be generalized for any shape moving in any magnetic field  uniform or not. For an element d of conductor the contribution de to the emf is the magnitude d multiplied by     the component of v B parallel to d , that is de  (v B). d For any two points a and b the motional emf in the direction from b to a is, x x x x bx xxxx b cv c v a   x x x xxxx x e   (v B).d b a a v= v cos  x x x x xx xxxxx Motional emf in wire acb in a uniform magnetic field is the motional emf in an imaginary wire ab. Thus, e = e acb ab  = (length of ab) (v ) (B), v = the component of velocity perpendicular to both B and ab. From right hand rule   : b is at higher potential and a at lower potential. Hence, V = V – V = (ab) (v cos) (B) ba b a Direction of induced current or HP end of the rod find out with the help of Fleming right hand rule  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 Fore finger  In external field B direction. Thumb  In the direction of motion (  ) of conductor. v Middle finger It indicates HP end of conductor/direction of induced current. Left hand palm rule Fingers   In external field ( B ) direction. Palm  In direction of motion (  ) of conductor. v Thumb  It indicates HP end of conductor/direction of induced current in conductor. 14 E

JEE-Physics Example An aircraft with a wing span of 40 m flies with a speed of 1080 kmh–1 in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of earth's magnetic field is 1.75 × 10–5 T. Find the emf that develops between the tips of the wings. Solution The metallic part between the wing-tips can be treated as a single conductor cutting flux-lines due to vertical component of earth's magnetic field. So emf is induced between the tips of its wings. v Here l = 40 m, B = 1.75 × 10–5 T, v  1080 km h 1  1080 1000 ms1 = 300 ms–1  V 3600 BV   = B  v = 1.75 × 10–5 × 40 × 300 = 0.21 V V Example A rod PQ of length L moves with a uniform velocity v parallel to a long straight wire carrying a current i, the end P remaining at a distance r from the wire. Calculate the emf induced across the rod. Take v = 5.0 m/s, i = 100 amp, r = 1.0 cm and L = 19 cm. Solution The rod PQ is moving in the magnetic field produced by the current-carrying P dx v Q long wire. The field is not uniform throughout the length of the rod (changing i with distance). Let us consider a small element of length dx at distance x from wire. if magnetic field at the position of dx is B then emf induced x d= B v dx  0 i v dx 2 x  emf  is induced in the entire length of the rod PQ is    d  Q 0 i v dx P 2 x Now x = r at P, and x = r + L at Q. hence 0 i v rL dx 0 iv rL 0 i v 0 i v r  L 2 r x 2  2 2 r    lo g e x r  [loge (r  L)  loge r]  log Putting the given values :  = (2 × 10–7) (100) (5.0) log 1.0  19 = 10–4 log 20 Wb/s = 3 × 10–4 volt e 1.0 e Example xxxxxx A horizontal magnetic field B is produced across a narrow gap between square xxxxxx iron pole-pieces as shown. A closed square wire loop of side , mass m and xxxxxx resistance R is allowed of fall with the top of the loop in the field. Show that the xxxxxx xxxxxx loop attains a terminal velocity given by v  Rmg while it is between the poles of xxxxxx B2 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 the magnet. mg Solution As the loop falls under gravity, the flux passing through it decreases and so an induced emf is set up in it. Then a force F which opposes its fall. When this force becomes equal to the gravity force mg, the loop attains a terminal velocity v. The induced emf e = B v , and the induced current is i e Bv xxxxxx  xxxxxx RR xxxxxx x x x Fx x The force experienced by the loop due to this current is F  B  i  B2 v 2 xxxxxx R xxxxxx When v is the terminal (constant) velocity F = mg or B2 v 2 or R mg mg  mg v  B2 2 R E 15

JEE-Physics Example Figure shows a rectangular conducting loop of resistance R, width L, and d B length b being pulled at constant speed v through a region of width d in b v which a uniform magnetic field B is set up by an electromagnet.Let L = 40 mm, b = 10 cm, d = 15 cm, R = 1.6 , L B = 2.0 T and v = 1.0 m/s x (i) Plot the flux  through the loop as a function of the position x of the right side of the loop. (ii) Plot the induced emf as a function of the positioin of the loop. (iii) Plot the rate of production of thermal energy in the loop as a function of the position of the loop. Solution (i) When the loop is not in the field : coil out coil entering coil in coil leaving coil out The flux linked with the loop  = 0 8 When the loop is entirely in the field : Magnitic flux linked with the loop (mWb) 4  = B L b = 2 × 40 × 10–3 × 10–1 = 8 mWb 0 x(in cm) When the loop is entering the field : 0 5 10 15 20 25 30 The flux linked with the loop  = B L x fig. (i) When the loop is leaving the field : The flux  = B L [b – (x – d)] (ii) Induced emf is e   d   d dx   d v coil out coil entering coil in coil leaving coil out dt dx dt dx = – slope of the curve of figure (i) × v 80 The emf for 0 to 10 cm : 40 e = – (8  0) 103 1  80 mV 0 x(in cm) (10  0) 102 (mV) –40 The emf for 10 to 15 cm : e = 0 × 1 = 0 –80 The emf for 15 to 25 cm : 0 5 10 15 20 25 30 fig. (ii) e = – (0  8) 103 1  80 mV (25  15) 102 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 e2 coil out coil entering coil in coil leaving coil out (iii) The rate of thermal energy production is P  P(mW) R for 0 to 10 cm : P = (80 103 )2  4 mW 4 1.6 0 for 10 to 15 cm : P = 0 0 5 10 15 20 25 30 for 15 to 25 cm : P = (80 103 )2  4 mW fig. (iii) 1.6 16 E

JEE-Physics Example Two long parallel wires of zero resistance are connected to each other by a battery of 1.0 V. The separation between the wires is 0.5 m. A metallic bar, which is perpendicular to the wires and of resistance 10 , moves on these wires. When a magneticfield of 0.02 testa is acting perpendicular to the plane containing the bar and the wires. Find the steady-state veclocity of the bar. If the mass of the bar is 0.002 kg then find its velocity as a function of time. Solution The current in the 10 bar is 1.0 V x xB=x0.0x2Tx x I   0.1 A 10  xxxxxx 1.0V I F x 0.5m xxx x  xxxxxx The current carrying bar is placed in the magnetic field B (0.2 T) perpendicular to the plane of paper and directed downwards. xxxxxx The magnetic force of the bar is F = B I  = 0.02 × 0.5 × 0.1 0 = 1 × 10–3 N  The moving bar cuts the lines of force of B . If v be the instantaneous velocity of the bar, then the emf induced in the bar is  = Bv = 0.02 × 0.5 × v = 0.01 v volt. By Lenz's law,  will oppose the motion of the bar which will ultimately attain a steady velocity. In this state, the induced emf  will be equal to the applied emf (1.0 volt).  0.01 v = 1.0 or v  1.0  100 ms–1 0.01 Again, a magnetic force F acts on the bar. If m be the mass of the bar, the acceleration of the rod is dv F  dv  F .dt Integrating,  dv   F dt  v  F tC (constant) dt m m m m If at t = 0, v = 0 then C = 0.  v Ft But F = 1 × 10–3 N, m = 0.002 kg  v  1 103 t = 0.5 t m 0.002 Example B xxxxx F xxxxx In figure, a rod closing the circuit moves along a U-shaped wire at a xxxxx constant speed v under the action of the force F. The circuit is in a xxxxx uniform magnetic field perpendicualr to its plane. Calculate F if the rate generation of heat is P.  Solution The emf induced across the ends of the rod,  = Bv Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 Current in the circuit,   Bv Magnetic force on the conductor, F' = IB, towards left I RR  acceleration is zero F' = F  BI = F or I  F  P =  I = B v× F = Fv  F  P B B v Example xx x ax x The diagram shows a wire ab of length  and resistance R sliding on a smooth + x vx x pair of rails with a velocity v towards right. A uniform magnetic field of induction S x bx x B acts normal to the plane containing the rails and the wire inwards. S is a – current source providing a constant I in the circuit. xx Determine the potential difference between a and b. E 17

JEE-Physics Solution The wire ab which is moving with a velocity v is equivalent to an emf source of value B v  with its positive terminal towards a.  Potential difference V – V = Bv – IR ab Example A thin semicircular conduting ring of radius R is falling with its plane vertical NB v  M Q in a horizontal magnetic induction B (fig.). At the position MNQ, the speed of M the ring is v. What is the potential difference developed across the ring at the M' NB position MNQ ? N' Solution v Let semiconductor ring falls through an infinitesimally small distance dx from its initial position MNQ to M'Q'N' in time dt (fig). Q decrease in area of the ring inside the magnetic field, Q'dx 2R dA = – MQQ'M' = – M'Q' × QQ' = –2R dx  change in magnetic flux linked with the ring, d = B × dA = B × (–R dx) = – 2BR dx The potential difference developed across the ring, e   d   2 B R dx   2B R v dt dt  the speed with which the ring is falling v  dx dt INDUCED E.M.F. DUE TO ROTATION OF A CONDUCTOR ROD IN A UNIFORM M AGNETIC FIELD Let a conducting rod is rotating in a magnetic field around an axis passing through its one end, normal to its plane. Consider an small element dx at a distance x from axis of rotation. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 Suppose velocity of this small element = v So, according to Lorent's formula induced e.m.f. across this small element d = B v. dx  This small element dx is at distance x from O (axis of rotation)  Linear velocity of this element dx is v =  x substitute of value of v in eqn (i) d = B  x dx Every element of conducting rod is normal to magnetic field and moving in perpendicular direction to the field   dx   B  x2   2 d  B x    0 0  So, net induced e.m.f. across conducting rod or   1 B  2   1 B  2  f  2 [f = frequency of rotation] 2 2 = B f ( 2) area traversed by the rod A =  2 or   B A f 18 E

JEE-Physics Example A wheel with 10 metallic spokes each 0.5 m long is rotated with angular speed of 120 revolutions per minute in a plane normal to the earth's mangetic field. If the earth's magnetic field at the given place is 0.4 gauss, find the e.m.f. induced between the axle and the rim of the wheel. Solution  = 2n = 2  120  4 , B = 0.4 G = 4 × 10–5 T, length of each spoke = 0.5 m 60 induced emf e  1 B 2  1  4 105  4  0.5 2  6.28 105 volt 22 As all the ten spokes are connected with their one end at the axle and the other end at the rim, so they are connected in parallel and hence emf across each spoke is same. The number of spokes is immaterial. Example A horizontal copper disc of diameter 20 cm, makes 10 revolutions/sec about a vertical axis passing through its centre. A uniform magnetic field of 100 gauss acts perpendicular to the plane of the disc. Calculate the potential difference its centre and rim in volts. Solution B = 100 gauss = 100 × 10–4 Wb/m2 = 10–2, r = 10 cm = 0.10 m, frequency of rotaion = 10 rot/sec The emf induced between centre and rim   1 B 2  1 B r2 ( r = ) A 22   = 2f = 2 × 3.14 × 10 = 62.8 s–1    1 10  62.8  (0.1)2 = 3.14 × 10–3 V = 3.14 mV. 2 Example A circular copper disc 10 cm in radius rotates at 20  rad s–1 about an axis through its centre and perpendicular to the disc. A uniform magnetic field of 0.2 T acts perpendicular to the disc. (a) Calculate the potential difference developed between the axis of the disc and the rim. (b) What is the induced current, if the resistance of 2 is connected in between axis and rim of the disc. Solution Here B = 0.2 T radius of the circular disc, r = 10 cm = 0.1 m resistance of the disc, R = 2 angular speed of rotation of the disc,  = 20  rad s–1 (a) If e is the induced e.m.f. produced between the axis of the disc and its rim, then Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 e = 1 Br2  1  0.2  20  (0.1)2 = 0.0628 V 22 (b) Induced current I e 0.0628  0.0314 A  R2 SELF INDUCTION When the current through the coil changes, the magnetic flux linked with the coil also changes. Due to this change of flux a current induced in the coil itself according to lenz concept it opposes the change in magnetic flux. This phenomenon is called self induction and a factor by virtue of coil shows opposition for change in magnetic flux called cofficient of self inductance of coil. Considering this coil circuit in two cases. E 19

JEE-Physics Case - I Current through the coil is constant (N,) If I  B   (constant)  No EMI I flux lines +– total flux of coil (N)  current through the coil () E N  I  NLI L  N  NBA  total K Rh II I where L = coefficient of self inductance of coil S I unit of L : 1 Wb =1 Henry = 1 N.m J Dimensions : [M1L2T-2 A–2] amp A2 = 1 A2 Note : L is constant of coil it does not depends on current flow through the coil. Case - II Current through the coil changes w.r.t. time dI dB d If dt  dt  dt  Static EMI  N = LI d dI d – N = – L , (– N ) called total self induced emf of coil 'e ' dt dt dt S es   L dI V. s dt S.I. unit of L   A SELF-INDUCTANCE OF A PLANE COIL Total magnetic flux linked with N turns,  = NBA = N  0NI  A  µ0N2 A  µ0N2 × r² = µ0 N 2 r  But  = L I L  µ0 N2r  2r  2r 2r 2 2 Example A soft iron core is introduced in an inductor. What is the effect on the self-inductance of the inductor? Solution Since soft iron has a large relative permeability therefore the magnetic flux and consequently the self-induc- tance is considerably increased. SELF-INDUCTANCE OF A SOLENOID Current flowing through it= Let cross-sectional area of solenoid=A, Length of the solenoid =, then   NBA  N µ0 N  A  µ0N2 A    But   LI  L  µ0N2 A or L= 0 r N 2 A Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65  m  If no iron or similar material is nearby, then the value of self-inductance depends only on the geometrical factors (length, cross-sectional area, number of turns). Example The current in a solenoid of 240 turns, having a length of 12 cm and a radius of 2 cm, changes at the rate of 0.8 As–1. Find the emf induced in it. Solution | | L dI  0N2 A . dI  4 107  (240)2    (0.02)2  0.8  6  104 V dt  dt 0.12 20 E

JEE-Physics MUTUAL INDUCTION (N1,1B1,L1) air gap (N2,2B2,L2) Whenever the current passing through primary coil I A1 MI A2 or circuit change then magnetic flux neighbouring primary (S.I.) secondary coil or circuit will also change. Acc. to secondary (M.I.) Lenz for opposition of flux change, so an emf induced in the neighbouring coil or circuit. + – () G E (1) Rh (2) 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 2<1 and 2 = B1A2 (=0°). Case - I When current through primary is constant Total flux of secondary is directly proportional to current flow through the primary coil N 2  I  N 2 = MI , M  N 2 2  N2B1A2  (T )s where M : is coefficient of mutual induction. 2 2 1 I1 I1 Ip 1 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 = MI –N d2 = –M dI1 ,  N 2 d  1 2 dt dt dt  called total mutual induced emf of secondary coil e . m • 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 : • Number of turns (N , N ). • Cofficient of self inductance (L , L ). 12 12 • Area of cross section. • Magnetic permeabibility of medium (r). • Distance between two coils (As d  = M  ). • Orientation between two coils. • Coupling factor 'K' between two coils. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 DIFFERENT COEFFICIENT OF MUTUAL INDUCTANCE • In terms of their number of turns • In terms of their coefficient of self inductances • In terms of their nos of turns (N , N ) 12 ( a ) Two co–axial solenoids (M ) S1S2 (N1) Coefficient of mutual inductance between two solenoids A M s1s2 = N2B1A = N2  0 N 1 I1  A  M s1s2  0N1N2 A  I1 I1       (N1) S  E 21

JEE-Physics (b) Two plane concentric coils (M ) C1C2 N1 M c1c2 = N2B1A2 where B = 0N1I1 , A = r22 I1 1 2 N2 r2 O r1 2r1 (r1>>r2) I M c1c2 = N2  0 N 1 I1  (r22) M  0 N1N 2 r22 I1  2r1  2 r1   c1 c2 Two concentric loop : Two concentric square loops : A square and a circular loop M  r22 (r >> r) r2 O M  b2 tiny M  r2 tiny r1 1 2 r1 a a a r b a In terms of L  and L : For two magnetically coupled coils :- 12 M  K L1L 2 here 'K' is coupling factor between two coils and its range 0   K  1 • For ideal coupling K = 1  M max  L1L2 (where M is geometrical mean of L and L ) max 12 • For real coupling (0 < K < 1) M  K L1L 2 • Value of coupling factor 'K' decided from fashion of coupling. • Different fashion of coupling PS plane are parallel: 0° fashion P S air gap K=1 P S M = L1L2 (max) 0<K<1 S ideal coupling M = K L1L2 (coaxial fashion) no flux coupling no flux coupling d (zero overlapping) P (zero overlapping) normal coupling if d KM K=0, M=0 K=0, M=0 'K' also defined as K = s = mag. flux linked with sec ondary (s) p mag. flux linked with p rimary (p) INDUCTANCE IN SERIES AND PAR ALLEL Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 Two coil are connected in series : Coils are lying close together (M) If M = 0, L = L + L If M  0 L  L1  L2  2M 12 (a) When current in both is in the same direction Then L = (L + M) + (L + M) 12 (b) When current flow in two coils are mutually in opposite directions. L = L + L – 2M 12 Two coils are connected in parallel : 11 1 L  L1L2 11 1   L1  L2  (a) If M = 0 then L L1 L2 or (b) If M  0 then L (L1  M) (L2  M) 22 E

JEE-Physics Example On a cylindrical rod two coils are wound one above the other. What is the coefficient of mutual induction if the inductance of each coil is 0.1 H ? Solution One coil is wound over the other and coupling is tight,so K = 1, M  L1L 2  0.1  0.1  0.1H Example How does the mutual inductance of a pair of coils change when : (i) the distance between the coils is increased ? (ii) the number of turns in each coil is decreased ? (iii) a thin iron rod is placed between the two coils, other factors remaining the same ? Justify your answer in each case . Solution (i) The mutual inductance of two coils, decreases when the distance between them is increased. This is because the flux passing from one coil to another decreases. (ii) Mutual inductance M  0 N1 N2 A i.e., M  N N Clearly, when the number of turns N and N in the  1 2 1 2 two coils is decreased, the mutual inductance decreases. (iii) When an iron rod is placed between th two coils the mutual inductance increases, because M  permeability () Example A coil is wound on an iron core and looped back on itself so that the core has two sets of closely would wires in series carrying current in the opposite sense. What do you expect about its self-inductance ? Will it be larger or small ? Solution As the two sets of wire carry currents in opposite directions, their induced emf's also act in opposite directions. These induced emf's tend to cancel each other, making the self-inductance of the coil very small. This situation is similar to two coils connected in series and producing fluxes in opposite directions. Therefore, their equivalent inductance must be L = L + L – 2M = L + L – 2L = 0 eq Example Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 A solenoid has 2000 turns wound over a length of 0.3 m. The area of cross-section is 1.2 × 10–3 m2. Around its central section a coil of 300 turns is closely would. If an initial current of 2A is reversed in 0.25 s, find the emf induced in the coil. Solution M  0N1N2 A  4 107  2000  300 1.2 103  3 103 H  0.3   M dI  3 103  2  2   48 103 V = 48 mV dt  0.25  E 23

JEE-Physics ENERGY STORED IN INDUCTOR The energy of a capacitor is stored in the electric field between its plates. Similary, an inductor has the capability of storing energy in its magnetic field.An increasing current in an inductor causes an emf between its terminals. Power P = The work done per unit time  dW  ei   L di  i  L i di dt dt  dt here i = instanteneous current and L = inductance of the coil From dW = – dU (energy stored) so dW   dU  dU  Li di dU = Li di dt dt dt dt The total energy U supplied while the current increases from zero to final value i is, I 1 L (i2 )0I 1 L I2 2 2  L idi 0 U  U  the energy stored in the magnetic field of an inductor when a current I is  1 L I2 . 2 The source of this energy is the external source of emf that supplies the current. • After the current has reached its final steady state value I, di  0 and no more energy is input to the inductor. dt • When the current decreases from i to zero, the inductor acts as a source that supplies a total amount of energy 1 Li2 to the external circuit. If we interrupt the circuit suddenly by opening a switch the current decreases very 2 rapidly, the induced emf is very large and the energy may be dissipated in an arc the switch. MAGNETIC ENERGY PER UNIT VOLUME OR ENERGY DENSITY • The energy is an inductor is actually stored in the magnetic field within the coil. For a long solenoid its magnetic field can be assumed completely within the solenoid. The energy U stored in the solenoid when a current I is, U  1 L I2  1 n2 V ) I2 (L = 0 n2 V) (V = Volume = A) 2 2 (0 The energy per unit volume uU  1 0 n2 I2  (0 n I)2 B2 1 B2 V 2 2 0  (B = 0 n I)  u  2 0 20 Example Figure shows an inductor L a resistor R connected in paralled to a battery Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 through a switch. The resistance of R is same as that of the coil that makes L. Two identical bulb are put in each arm of the circuit. (a) Which of two bulbs lights up earlier when S is closed? (b) Will the bulbs be equally bright after some time? Solution (i) When switch is closed induced e.m.f. in inductor i.e. back e.m.f. delays the glowing of lamp P so lamp Q light up earlier. (ii) Yes. At steady state inductive effect becomes meaningless so both lamps become equally bright after some time. 24 E

JEE-Physics PERIODIC EMI  Let a coil intially placed perpendicular to uniform magnetic field. Now this coil starts rotation about an axis that the flux linked with the coil change due to change in B A  oriantation of area vector A with respect to magnetic field B  Angle in between area vector A and magnetic field B is  then Example A coil of 160 turns of cross-sectional area 250 cm2 rotates at an angular velocity of 300 radian/sec about an axis parallel to the plane of the coil in a uniform magnetic field of 0.6 Wb/m2. What is the maximum emf induced in the coil ? If the coil is connected to a resistance of 2 ohm, what is the maximum torque that has to be delivered to maintain its motion. Solution The instantaneous induced emf is  = NBA  sin t  NBA    = N B A  = 160 × 0.6 × (250 × 10–4) × 300 = 720 volt max The maximum current through the coil is i max  max 720 = 360 amp. R  2 The torque on a current-carrying coil placed in a magnetic field is  = BINAsin  = BINA sint  maximum torque = B I N A = 0.6 × 360 × 160 × (250 × 10–4) = 864 newton meter. By Lenz's law, this torque opposes the rotaion of the coil. Hence to maintain the rotation an equal torque must be inserted in the opposite direction. Therefore the required torque is 864 N-m. Example A very small circular loop of area 5 × 10–4 m2, resistance 2 ohm and negligible inductance is initially coplanar and concentric with a much larger fixed circular loop of radius 0.1 m. A constant current of 1 ampere is passed in the bigger loop and the smaller loop is rotated with angular velocity  rad/s about a diameter. Calculate (a) the flux linked with the smaller loop (b) induced emf and induced current in the smaller loop as a function of time. Solution (a) The field at the centre of larger loop B1  0 I  2 107 = 2 × 10–6 Wb/m2 2R 0.1 is initially along the normal to the area of smaller loop. Now as the smaller loop (and hence normal to its Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65  plane) is rotating at angular velocity , with respect to B so the flux linked with the smaller loop at time t is, 2 = B A cos  = (2 × 10–6) (5 × 10–4) cos t 1 2 i.e., 2 =  × 10–9 cos t Wb  (b) The induced emf in the smaller loop e2   d2   d ( 109 cos t) IA a dt dt b =  × 10–9  sin t volt (c) The induced current in the smaller loop is, I2  e2  1  109 sin t ampere R 2 E 25

JEE-Physics TRANSFORMER Laminated Load Core Working principle Mutual induction Primary winding Transformer has basic two section AC mains Secondary ( a ) Shell : Consist of primary and secondary coil of copper. winding The effective resistance between primary and secondary coil is infinite because electric circuit between two is open (R = ) ps ( b ) Core : Which is between two coil and magnetically coupled two coils. Two coils of transformer would on the same core. The alternating current passing through the primary creats a continuously changing flux through the core. This changing flux induces alternating emf in secondary. Work It regulates AC voltage and transfer the electrical electrical power without change in freqency of input supply. (The alternating current changes itself.) Special Points • It can't work with D.C. supply, and if a battery is connected to its primary, then output is across scondary is always zero ie. No working of transformer. • It can't called 'Amplifier' as it has no power gain like transistor. • It has no moving part, hence there are no mech. losses in transformer. Types : According to voltage regulation it has two – (i) Step up transformer : N > N (ii) Step down transformer N < N SP SP 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 = I R , I = line current R = total line resistance, LL L L I = power to be transmission power losses = I2L R L L line voltage 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%) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 (a) No flux leakage s = p  ds = dp dt dt e = e = e induced emf per turn of each coil is also same. sp total induced emf for secondary E = N e total induced emf for primary E = N e ss pp Es Ns or p where n : turn ratio , p : transformation ratio Ep = Np = n 26 E

JEE-Physics (b) No load condition V =E and E = V VS  N S from (i) and (ii) VS  NS =n or p PP SS VP N P VP NP (c) No power loss P =P and V I = V I VS  IP valid only for ideal transformer out in SS PP VP IS from equation (iii) and (iv) VS  IP  N S  n or p VP IS N P Note : Generally transformers deals in ideal condition i.e. P = P , if other information are not given. in out 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  VS . IS  100 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 : Kota Jaipur Raja Park Sub-Station Jaipur Area Station Step up transformer House Kota thermal 200 V to Step up Step down Step down Power Station transformer transformer Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 11 kV transformer 135 kV 11 kV 11kV to 135 kV to 11kV to 220kV Example 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. (b) 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. E 27

JEE-Physics Solution The diagram shows the network : For sub-station, P = 800 kW = 800 × 103 watts V = 220 V P 800 103 40 S = V = = × 103 A. 220 11 Primary current (p) in sub-station transformer will be given by 220  40 103 4000 × P = 220 × S, P = 11  4000 = 200 A (a) Hence transmission line current = 200 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. Power 1400 kW 1400  1000 (c) Voltage in secondary at power plant has characteristics = Current = 200 A = 200 = 7000 V Step-up transformer at power plant has characteristics 440 - 7000 V. Example 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? Solution ES  NS NS  NP  ES  4000  230  400 EP NP EP E = 2300 V ; N = 4000, E = 230 V 2300 p PS Example 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. Solution E = 240 V, E = 24 V, E I = 24 W Current in primary coil I= ESIS  24  0.1A PS SS EP 240 P LOSSES OF TRANSFORMER (a) Copper or joule heating losses W h e r e : There losses occurs in both coils of shell part Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 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. (b) Flux leakage losses W h e r e : There losses occurs in between both the coil of shell part. C a u s e : Due to air gap between both the coils. Remmady : 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 28 E

JEE-Physics (c) Iron losses W h e r e : There losses occurs in core part. Types : (i) Hysteresis losses (ii) Eddy currents losses (i) Hysteresis losses C a u s e : 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. Remmady : 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 example soft iron (low retentivity and low coercivity) EDDY CURRENTS (or Focalt's currents) (F <<<F ) 12 • Eddy currents are basically the induced currents set up inside the body of conductor whenever the magnetic flux linked with it changes. • Eddy currents tend to follow the path of least resistance inside a conductor. So they from irregularly shaped loops. However, their directions are not random, but guided by Lenz's law. • Eddy currents have both undesirable effects and practically useful applications. Applications of eddy currents : (i) Induction furnace (ii) Electromagnetic damping (iii) Electric brakes (iv) Speedometers (v) Induction motor (vi) Electromagnetic shielding (vii) Inductothermy (viii) Energy meters GOLDEN KEY POINTS • These currents are produced only in closed path within the entire volume and on the surface of metal body. Therefore their measurement is impossible. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 • Circulation plane of these currents is always perpendicular to the external field direction. • Generally resistance of metal bodies is low so magnitude of these currents is very high. • These currents heat up the metal body and some time body will melt out (Application : Induction furnace) • Due to these induced currents a strong eddy force (or torque) acts on metal body which always apposes the translatory (or rotatory) motion of metal body, according to lenz. • Transformer Cause : Transformer core is always present in the effect of alternating magnetic field (B = B si nt). Due to this 0 eddy currents are produced in its volume, so a part of magnetic energy of core is wasted as heat. R e m m a d y : To minimise these losses transformer core should be laminated. with the help of lamination process, circulation path of eddy current is greatly reduced & net resistance of system is greatly increased. So these currents become E 29

JEE-Physics R-L DC CIRCUIT R L Current Growth I switch ( i ) emf equation E  IR  L dI +– dt E (ii) Current at any instant When key is closed the current in circuit increases exponentially with respect to time. The current in  t  circuit at any 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) I  I 0 (i i i ) Just after the closing of the key inductance behaves like open circuit and current in circuit is zero. + – () Open circuit, t = 0, I = 0 Inductor provide infinite resistence ( i v ) Some time after closing of the key inductance behaves like simple R connecting wire (short circuit) and current in circuit is constant. + – () I0 E Short circuit, t  , I  I , Inductor provide zero resistence  R 0 (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 () L   It is a time in which current increases up to 63% or 0.63 times of peak current value. R sec. (vi) Half life (T) It is a time in which current increases upto 50% or 0.50 times of peak current value. I=I (1 – e–t/), t = T, I = I0  I0 =I 1  eT/ = 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 (1 – e–T/)  e–T/ = 0 2 20 2 T log e = loge2 T  0.693   T L  e  0.693 R sec. (vii) Rate of growth of current at any instant :–  dI   E (et/ )  t=0   dI   E  dI   dt  L  dt max L t =    dt   0 Note : Maximum or initial value of rate of growth of current does not depends upon resistance of coil. 30 E

JEE-Physics Current Decay R L0 (i) Emf equation IR  L dI  0 21 dt (ii) Current at any instant +– E 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 = I , source emf E is cut off from circuit 0 I  I0 (et/ ) Just after opening of key t=0 E  I = I0 = R Some time after opening of key t   I  0 (iii) 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  e  t /  t=0   dI  = E t  dI  0 dt   L  dt max. L dt  Graph for R–L circuit :– Current Growth :– I I0 line dI t=0, rate of growth I0 exp. growth dt maximum ( a ) 0.63I0 (b) growth t=0 =0 t rate exp. decrease t=0 Current decay :– t= rate=0 t Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\02 EMI.p65 I t=0, I = I0 =(max) dI dt t=0, rate of decay maximum ( a ) 0.37I0 exp. decrease decay t=0 t= t (b) rate exp. decrease t=0 t= rate=0 t E 31



JEE-Physics ELECTROMAGNETIC WAVES INTRODUCTION A changing electric field produces a changing magnetic field and vice versa which gives rise to a transverse wave known as electromagnetic waves. The time varying electric field and magnetic field mutually perpendicualr to each other are also perpendicualr to the direction of propogation. Thus the electromagnetic waves consist of sinusoidally Y time varying electric and magnetic field acting at right Ey angles to each other as well as at right angles to the c direction of propogation. X Bz Z HISTORY OF ELECTROMAGNETIC WAVES  In the year 1865, Maxwell predicted the electromagnetic waves theoretically. According to him, an accelerated charge sets up a magnetic field in its neighborhood.  In 1887, Hertz produced and detected electromagnetic waves experimentally at wavelength of about 6m.  Seven year later, J.C. Bose became successful in producing electromagnetic waves of wavelength in the range 5mm to 25mm.  In 1896, Marconi discovered that if one of the spark gap terminals is connected to an antenna and the other terminal is earthed, the electromagnetic waves radiated could go upto several kilometers.  The antenna and the earth wires are connected to the two plates of a capacitor which radiates radio frequency waves. These waves could be received at a large distance by making use of an antenna earth system as d e t e c to r.  Using these arrangements; in 1899 Marconi first established wireless communication across the English channel i.e., across a distance of about 50 km. CONCEPT OF DISPLACEMENT CURRENT When a capacitor is allowed to charge in an electric circuit, the current flows through connecting wires. As capacitor charges, charge accumulates on the two plates of capacitor and as a result, a changing electric field is produced across between the two plate of the capacitor. +– According to maxwell changing electric field intensity is equivalent to E a current through capacitor known as displacement curent (I ). If + q +– d I=Ic + I=Id – I=Ic and – q be the charge on the left and right plates of the capacitor + – respectively at any instant if  be the surface charge density of plate of capacitor the electric field between the plate is given by + – q +– E  0 0 A Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\04 EMW Theory.p65 If charge on the plates of the capacitor increases by dq in time dt then dq = I dt change in electric field is dE = d q Idt dE I = = 0 A 0 A dt 0 A I = 0 dE d (EA) = 0 dE ( E = EA) Id  0 dE A dt = 0 d t dt dt The conduction current is the current due to the flow of charges in a conductor and is denoted as Ic and displacement current is the current due to changing electric field between the plate of the capacitor and denoted as I so the total current I is sum of I and I i.e. I = I + I d cd cd Ampere's circuital law can be written as z  B.dz  = 0 (I + I)  B.d = 0 (I + 0 dE ) c d c dt E 109

JEE-Physics MA XWELL'S EQUATION There are four maxwel' equation given below (1) Gauss law in electrostatics : z  q ...(i) E.ds = 0 ...(ii) z  ...(iii) (2) Gauss law in magnetism : B.ds = 0 ... (iv) z(3)  dB Faraday's law of electromagnetic induction : emf = E. d =– dt L O  dE dt z NM PQB.d = 0 (4) Maxwell - Ampere's circuital law : Ic  0 HERTZ EXPERIMENT (Practical production of EM waves) Input  In 1888, Hertz demonstrated the production of electromagnetic waves by oscillating charge. His experimental apparatus is shown schematically in fig.  An induction coil is connected to two spherical electrodes with a Induction narrow gap between them. It acts as a transmitter. The coil provides coil short voltage surges to the spheres making one positive and the other negative. A spark is generated between the spheres when the voltage between them reaches the breakdown voltage for air. Transmitter As the air in the gap is ionised, it conducts more rapidly and the q -q discharge between the spheres becomes oscillatory.  The above experimental arrangement is equivalent to an LC circuit, Receiver where the inductance is that of the loop and the capacitance is due to the spherical electrodes.  Electromagnetic waves are radiated at very high frequency ( 100 MHz) as a result of oscillation of free charges in the loop.  Hertz was able to detect these waves using a single loop of wire with its own spark gap (the receiver).  Sparks were induced across the gap of the receiving electrodes when the frequency of the receiver was adjusted to match that of the transmitter. PROPERTIES OF ELECTROMAGNETIC WAVES  The electric and magnetic fields satisfy the following wave equations, which can be obtained from Maxwell's third and fourth equations. 2E  00 2E and 2B  00 2 B x 2 t 2 x 2 t2  Electromagnetic waves travel through vacuum with the speed of light c, where Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\04 EMW Theory.p65 c  1  3 108 m / s 0 0  The electric and magnetic fields of an electromagnetic wave are prependicular to each other and also perpendicular to the direction of wave propagation. Hence, these are trnasverse waves.   The instantaneous magnitudes of E and B in an electromagnetic wave are related by the expression E c B  Electromagnetic waves carry energy. The rate of flow of energy crossing a unit area is described by the  Where  1  S EB Poynting vector S . 0 110 E

JEE-Physics  Electromagnetic waves carry momentum and hence can exert pressure(P) on surfaces,which is known as  radiation pressure. For an electromagnetic wave with Poynting vector S , incident upon a perfectly absorbing surface S P c and if incident upon a perfectly reflecting surface P  2S c  The electric and magnetic fields of a sinusoidal plane electromagnetic wave propagating in the positive x-direction can also be written as E = E sin(kx – t) and B = B sin(kx – t) m m where  is the angular frequency of the wave and k is wave number which are given by   2f and k  2   The intensity of a sinusoidal plane electro-magnetic wave is defined as the average value of Poynting vector S av  EmBm  E 2  c B 2 20 m 20 m taken over one cycle. 20 c  The fundamental sources of electromagnetic waves are accelerating electric charges. For examples radio waves emitted by an antenna aries from the continuous oscillations (and hence acceleration) of charges within the antenna structure.  Electromagnetic waves obey the principle of superposition.  The electric vector of an electromagnetic field is responsible for all optical effects. For this reason electric vector is also called a light vector. TR ANSVERSE NATURE OF ELECTROMAGNETIC WAVES Maxwell showed that a changing electric field produces a changing magnetic field and vice-versa. This alternate production of time 'varying electric and magnetic fields gives rise to the propagation of electromagnetic waves.  The variation of electric field ( E ) and magnetic field ( B ) are mutually perpendicular to each other as well as to the direction of the propagation of the wave i.e., the electromagnetic waves are transverse in nature. Proof : Consider a plane electromagnetic wave travelling along Y B plane wave F A front X-direction with its wave front in the Y–Z plane and ABCD is E its portion at time t. The values of electric field and magnetic C field to the left of ABCD will depend on x and t (and not on O y and z as the wave under consideration is a plane wave ZG X propagating in x direction. direction of Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\04 EMW Theory.p65 According to Gauss' law, the total electric flux across the D propagation parallelopiped' ABCDOEFG is zero because it does not z  enclose any charge. i.e. E.dS  0 z z z z z zor E.dS  E.dS  E.dS  E.dS  E.dS  E.dS  0   ...(i) ABCD EFOG ADGE BCOF OCDG FBAE  since electric field E does not depend on y and z, so the contribution to the electric flux coming from the faces normal to y and z axes cancel out in pairs. z z   i.e. E.dS  E.dS  0 ... (ii) OCDG FBAE E 111

JEE-Physics z z   ... (iii) and E.dS  E.dS  0 ADGE BCOF Using equation (ii) and (iii) in equation (i), we get z z   E.dS  E.dS  0 ...(iv) ABCD EFOG z z z zNow  (  E.dS  E x.dS cos0  ExdS  Ex dS E x is parallel to dS ) ABCD ABCD ABCD ABCD = E × area of face ABCD = E S ... (v) xx (   E'x is antiparallel to dS ) z z zand E'.dS  E'x dS cos180  E'x dS EFOG EFOG EFOG = E'x × area of faceEFOG = E ' S ... (vi) x where, E and E'x are the x-components of electric field on the faces ABCD and EFOG respectively. x Substituting the values of equations (v) and (vi) in equation (iv), we get E S – E 'S = 0 or S(E – E ') = 0 xx xx  S0  E – E' = 0 or E 'x = E x xx This equation shows that the value of the x-component of electric field does not change with time. In other Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No.-10\\EMI & AC\\English\\04 EMW Theory.p65 words, electric field along x-axis is static. Since the static electric field cannot propagate the wave, hence the electric field parallel to the direction of the propagation of the wave is zero. i.e. E ' = E = 0 xx It means, electric field is perpendicular to the direction of propagation of the wave. similarly, it can be proved that the magnetic field is perpendicular to the direction of the propagationof the wave. Since both electric and magnetic fields are perpendicualr to the direction of the propagation of the wave, so electromagnetic wave is transverse in nature. GOLDEN KEY POINTS • When a capacitor is connected across the battery through the connecting wires there is flow of conduction current, while thorugh the gap between the plates of capacitor,there is flow of displacement current. • Maxwell's equation are mathematical formulation of Gauss's law in electrostatics (I) Gauss's law in electromagnetism (II) foradays law of electromagnetic induction (III) and Ampere's circuital law (IV) • Frequency of electromagnetic waves is its inherent characterstic, when an electromagnetic wave travels from one medium to another, its wavelength changes but frequency remains unchanged. • Ozone layer absorbs the ultra-violet rays from the sun and these prevents them from producing harmful effect on living organisms on the earth. Further it traps the infra-red rays and prevents them from escaping the surface of earth. It helps to keeps the earth's at atmosphere warm 112 E