P1-Allens Made Physics Theory {PART-1}

JEE-Physics Substituting  the  given  expression  for  s,  we  have a   4 t v2 ...(ii) an  r Substituting  v  from  eq.  (i)  and  r =8m,  we  have an  1 t4 ...(iii) 2 The  instant  when  the  tangential  and  the  normal  accelerations  have  equal  magnitude,  can  be  obtained  by equating  their  expressions  given  in  eq.  (ii)  and  (iii). a   =  a      t  =  2s nT Substituting  the  above  value  of  t  in  eq.  (i),  we  obtain  v  =  8  m/s Example A  particle  is  moving  on  a  circular  path  of  radius  1.5  m  at  a  constant  angular  acceleration  of  2  rad/s2.  At  the instant  t  =  0,  angular  speed  is  60/  rpm.  What  are  its  angular  speed,  angular  displacement,  linear  velocity, tangential  acceleration  and  normal  acceleration  at  the  instant  t  =  2  s. Solution Initial  angular  speed  is  given  in  rpm  (revolution  per  minute).  It  is  expressed  in  rad/s  as 1 rpm  2 rad 60 s o   60   2 rad  2 rad/s    60 s At  the  instant  t  =  2  s,  angular  speed  2 and  angular  displacement  2 are  calculated  by  using  eq. 2  o  t Substituting  values  o  2 rad/s,    2 rad/s2,    t  2 s,  we  have 2  6   rad/s  2 1 t o  2 o  2 Substituting  values  o  0   rad,  o  2 rad/s,  2  6   rad/s  and  t  =2  s,  we  have 2  =  8  rad Linear  velocity  at  t  =  2  s,  can  be  calculated  by  using  eq. v  =  r 2 2 Substituting  r  1.5 m  and  2  6 rad/s,  we  have node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65 v   =  9  m/s 2 Tangential  acceleration  a   and  normal  acceleration  an can  be  calculated  by  using  eq.    and    respectively. a   =  r  Substituting  r  1.5 m  and    2 rad/s2,  we  have a   =  3  m/s2  a  =  2r n Substituting  2  =  6  rad/s  and  r  =  1.5  m,  we  have a   =  54  m/s2 n E 23

JEE-Physics Example A  particle  is  moving  in  a  circular  orbit  with  a  constant  tangential  acceleration.  After  2  s  from  the  beginning  of motion,  angle  between  the  total  acceleration  vector  and  the  radius  R  becomes  45°.  What  is  the  angular acceleration  of  the  particle?  a  a 45°   an center  Solution In  the  adjoining  figure  are  shown  the  total  acceleration  vector     and  its  components  the  tangential  accelerations  a a and  normal  accelerations  an are  shown.  These  two  components  are  always  mutually  perpendicular  to  each other  and  act  along  the  tangent  to  the  circle  and  radius  respectively.  Therefore,  if  the  total  acceleration  vector makes  an  angle  of  45°  with  the  radius,  both  the  tangential  and  the  normal  components  must  be  equal  in magnitude.  Now  from  eq.      and  ,  we  have a   a n  R  2R    2 ...(i) Since  angular  acceleration  is  uniform,  form  eq.,  we  have    o  t Substituting   =0    and  t  =  2  s,  we  have    =2 ...(ii) 0 From  eq.  (i)  and  (ii),  we  have    =  0.25  rad/s2 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65 24 E

JEE-Physics SOME WORKED OUT EXAMPLES Example#1 On  an  open  ground  a  motorist  follows  a  track  that  turns  to  his  left  by  an  angle  of  60°  after  every  500  m.  Starting from  a  given  turn,  specify  the  displacement  of  the  motorist  at  the  third,  sixth  and  eighth  turn.  Compare  the magnitude  of  displacement  with  the  total  path  length  covered  by  the  motorist  in  each  case. Solution D IV 60° C III 60° 60° 500m 60° E II B VIII V 60° 500m VI 60° 60° O 500m I VII A At  III  turn     Displacement =  OA  AB  BC  OC =  500  cos  60°  +  500  +  500  cos  60° 11 =  500  ×  +  500  +  500  ×  =  1000  m  from  O  to  C 22 Displacement 1000 2 Distance =  500  +  500  +  500  =  1500  m. So  Distance  1500 3 At  VI  turn  :   initial  and  final  positions  are  same  so  displacement         =  0  and  distance  =  500  ×    6  =  3000  m   Displacement  0  0 Distance 3000 At  VIII  turn  :  Displacement  =  2(500 ) cos  60    =  1000  ×  cos  30°  =  1000  3  500 3m  2  2 Displacement 500 3 3                 Distance  =  500  ×    8  =  4000  m    Dis tan ce 4000 8 Example#2 A drunkard walking in a narrow lane takes 5 steps forward and  3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1m long and requires 1s. Plot the x–t graph of his motion. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65 Determine graphically or otherwise how long the drunkard takes to fall in a pit 9m away from the start. x(m) x 9 Pit 7 5 4 2 5 8 13 16 21 t (sec)  E 25

JEE-Physics Solution from  x–t  graph      time  taken    =  21  s OR (5m  –  3m)  +  (5m  –  3m)  +  5m  =  9m       total  steps    =  21           time  =    21  s Example#3 A  man  walks  on  a  straight  road  from  his  home  to  a  market  2.5  km  away  with  a  speed  of 5  km/h.    On  reaching  the  market  he  instantly  turns  and  walks  back  with  a  speed  of  7.5  km/h.    What  is the (a)  magnitude  of  average  velocity  and (b)  average  speed  of  the  man,  over  the  interval  of  time  (i)  0  to  30  min.    (ii)  0  to  50  min  (iii)  0  to  40  min. Solution t1  dis tance 2.5 1 Time  taken  by  man  to  go  from  his  home  to  market,  h speed 5 2 2.5 1 h Time  taken  by  man  to  go  from  market  to  his  home, t2  7.5  3 115 h =  50  min.      To t al   t i m e   t a ken   =    t + t   2  2 36 1 (i) 0  to  30  min Average  velocity displacement  2.5 =5  km/h towards  market = time interval 30 Average  speed (ii) 0  to  50  min 60 Total  displacement = dis tan ce    2.5   =  5  km/h tim e  in terval 30 60 =  zero so  average  velocity  =  0 So,  average  speed 5 =  6  km/h =  50 / 60 Total  distance  travelled =  2.5  +  2.5 =  5  km. (iii) 0  to  40  min Distance  covered  in  30  min  (from  home  to  market)  =  2.5  km. 10 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65 Distance  covered  in  10  min  (from  market  to  home)  with  speed  7.5  km/h  =  7.5  ×  60   =  1.25  km So,  displacement  =  2.5  –  1.25  =  1.25  km  (towards  market) Distance  travelled  =  2.5  +  1.25  =  3.75  km Average  velocity  1.25   =  1.875  km/h.  (towards  market) 40 60 Average  speed   3.75 =  5.625  km/h. 40 60 Note  :  Moving  body  with  uniform  speed  may  have  variable  velocity.  e.g.  in  uniform  circular  motion  speed is  constant  but  velocity  is  non–uniform. 26 E

JEE-Physics Example#4 A  driver  takes  0.20  s  to  apply  the  brakes  after  he  sees  a  need  for  it.    This  is  called  the  reaction  time  of the  driver.    If  he  is  driving  a  car  at  a  speed  of  54  km/h  and  the  brakes  cause  a  deceleration  of  6.0m/s2, find  the  distance  travelled  by  the  car  after  he  sees  the  need  to  put  the  brakes  on  ? Solution Distance  covered  by  the  car  during  the  application  of  brakes  by  driver s  =  ut  =  54  5 (0.2)    =  15  x  0.2  =  3.0  m 1  18  After  applying  the  brakes;  v  =  0,        u  =  15  m/s,    a  =  6  m/s2 s   =  ? 2 U s i n g   v 2  =   u 2   –   2a s     0  =  ( 15) 2  –   2   ×   6   ×   s    12  s   =   2 25    s2  225  18.75 m 2 2 12 Distance  travelled  by  the  car  after  driver  sees  the  need  for  it      s  =  s   +  s   =  3  +  18.75  =  21.75  m. 12 Example#5 A  passenger  is  standing  d  distance  away  from  a  bus.  The  bus  begins  to  move  with  constant  acceleration a.  To  catch  the  bus,  the  passenger  runs  at  a  constant  speed  u  towards  the  bus.  What  must  be  the  minimum speed  of  the  passenger  so  that  he  may  catch  the  bus? Sol. Let  the  passenger  catch  the  bus  after  time  t. The  distance  travelled  by  the  bus, 1 ....(i) s   =  0  +    at2 and  the  distance  travelled  by  the  passenger 12 ....(ii) Now  the  passenger  will  catch  the  bus  if ....(iii) s   =  ut  +  0 2 d  +  s   =  s 12    d  +  1   at2  =  ut   1 at2  –  ut  +  d  =  0   t  [u  u2  2ad ] 2 2 a So  the  passenger  will  catch  the  bus  if  t  is  real,  i.e.,  u2    2  ad   u  2 ad So  the  minimum  speed  of  passenger  for  catching  the  bus  is  2ad . Example#6 If  a  body  travels  half  its  total  path  in  the  last  second  of  its  fall  from  rest,  find  :  (a)  The  time  and  (b)  height of  its  fall.  Explain  the  physically  unacceptable  solution  of  the  quadratic  time  equation.  (g  =  9.8  m/s2) Solution If  the  body  falls  a  height  h  in  time  t,  then 1 ...  (i) h=  gt2  [  u  =  0  as  the  body  starts  from  rest] 2 1 ...  (ii) Now,  as  the  distance  covered  in  (t  –  1)  second  is h'= g(t–1)2 2 So  from  Equations  (i)  and  (ii)  distance  travelled  in  the  last  second. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65 11 1 h  –  h'= gt2 –  g(t–1)2  i.e.,  h  –  h'= g(2t–1) 22 2 h But  according  to  given  problem  as  (h  –h’)  = 2 1 h   1  g 2t  1   or   1 gt2  g 2 t 1   [as  from  equation  (i)  h   1 gt2 ]    2   2   2   2   t2  –  4t  +  2  =  0  or  t  [4  (42  4  2)] / 2   t  2  2    t  =  0.59  s  or  3.41  s 0.59  s  is  physically  unacceptable  as  it  gives  the  total  time  t  taken  by  the  body  to  reach  ground  lesser  than one  sec  while  according  to  the  given  problem  time  of  motion  must  be  greater  than  1s. so      t=3.41s and    h= 1 ×  (9.8)  ×  (3.41)2=57  m 2 E 27

JEE-Physics Example#7 A  car  accelerates  from  rest  at  a  constant  rate   for  some  time,  after  which  it  decelerates  at  a  constant  rate  ,  to  come to  rest.  If  the  total  time  elapsed  is  t  evaluate  (a)  the  maximum  velocity  attained  and  (b)  the  total  distance  travelled. Solution (a)  Let  the  car  accelerates  for  time  t and  decelerates  for  time  t then  t  =  t +  t ....(i) 1  2  1  2 and  corresponding  velocity–time  graph  will  be  as  shown  in.  fig. From  the  graph  =  slope  of  line    OA  = v max   t = v max t1 1   vmax A v max v max O t1 B t2 t  and    –  slope of  line  AB =  t2   t =   2 v max v max    t          +     =  t    v  t    v  =  max max 1 1 t 1  t2  2   (b)  Total  distance    =  area  under  v–t  graph  =  2   ×  t  ×  v =  2   ×  t  ×    =       max  Note:  This  problem  can  also  be  solved  by  using  equations  of  motion  (v  =  u  +  at,  etc.). Example#8 Draw  displacement  time  and  acceleration  –  time  graph  for  the  given  velocity–time  graph v(ms-1) 10 05 t(s) 10 12 Solution 10 For  0  t      5    v  t        s  t2 and  a1=constant  5 =2  ms–2 1 for  whole  interval  s   =  Area  under  the  curve  =  ×      5  ×      10  =  25  m 12 For 5     t    10,      v  =  10ms–1    a  =0 for  whole  interval  s   =  area  under  the  curve  =    5  ×      10  =  50  m 2 For  10      t      12      v  linearly  decreases  with  time    a  =  –  10 =  –5  ms–2 3 2 1 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65 for  whole  interval  s =  Area  under  the  curve  =  ×    2  ×  10  =  10  m 32 s(m) a (ms—2) t 85 2 75 0   25 -5 5 10 12 t 28 E

JEE-Physics Example#9 A  rocket  is  fired  upwards  vertically  with  a  net  acceleration  of  4  m/s2 and  initial  velocity  zero.  After  5  seconds  its fuel  is  finished  and  it  decelerates  with  g.  At  the  highest  point  its  velocity  becomes  zero.  Then  it  accelerates downwards  with  acceleration  g  and  return  back  to  ground.  Plot  velocity–time  and  displacement–time  graphs  for the  complete  journey.  Take  g  =  10  m/s2. Solution v (m/s) s(m) B A 70 50 A 20 BC t(s) O O5 7 10.7 C t(s) 10.7 57 In  the  graphs, v =  at   =  (4)  (5)  =  20  m/s v =  0  =  v –  gt A  OA  B  A  AB   tAB  vA 20  2s  t =  (5+2)s  =  7s g  OAB  10 Now, 1 s =  area  under  v–t  graph  between  0  to  7  s =  (7)  (20)  =  70  m OAB     2 Now, sOAB  sBC  1 g t 2   70    =  1   (10)  t2 2 B BC C 2   tBC  14  3.7 s  t   =  7+3.7  =  10.7s OABC  1 Also s   =  area  under  v–t  graph  between      OA  =  (5)  (20)  =  50  m OA  2 Example#10 At  the  height  of  500m,  a  particle  A  is  thrown  up  with  v  =  75  ms–1  and  particle  B  is  released  from  rest.  Draw, acceleration  –time,  velocity–time,  speed–time  and  displacement–time  graph  of  each  particle. For  particle  A  : A2 For  Particle  B B1 Time of flight                                                   A1 Time  of  flight u=0 1 500m   –500  =  +75  t  –    ×  10t2 2 500m 1 500=  2 (10)t2     t  =  10  s    t2  –  15t  –  100  =  0 Velocity  at  B B2    t  =  20  s node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65 A3 2 Time  taken  for  A A v  =    0  –  (10)  (10)  =  –  100  ms–1 12 v  =  0  =  75  –  10t     t  =  7.5  s Velocity  at  A , v  =  75  –  10  ×  20  =–  125  ms–1 3  1 Height  A A   =  75  ×  7.5  –    (10)  (7.5)2  =  281.25  m 21 2 E 29

JEE-Physics a(ms-2) t=20s a(ms-2) t=10s t(s) t(s) -10 t(s) -10 t(s) v(ms-1) v(ms-1) 75 -125 -100 t(s) speed t(s) (ms-1) speed (ms-1) displacement(m) 281.25 t(s)        t(s) displacement(m) t(s) –500 –500 Example#11   Two  ships  A  and  B  are  10  km  apart  on  a  line  running  south  to  north.  Ship  A  farther  north  is  streaming  west  at 20  km/h  and  ship  B  is  streaming  north  at  20  km/h.  What  is  their  distance  of  closest  approach  and  how  long  do they  take  to  reach  it? Solution Ships  A  and  B  are  moving  with  same  speed  20  km/h  in  the  directions  shown  in  figure.  It  is  a  two  dimensional,  two  body  problem  with  zero  acceleration.  Let  us  find  v BA          vA A N v BA   =  v B –  v A E  202  202   =  20  2 km/h vB Here,  vBA   B i.e.,  v BA   is  20  2 km/h  at  an  angle  of  45°  from  east AB=10km towards  north.  Thus,  the  given  problem  can  be  simplified  as  : vB=20km/h vBA= 20  2km/h                A node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65 C 450 vA=20km/h vBA  450 A  is  at  rest  and  B  is  moving  with  v BA   in  the  direction  shown  in  figure. B Therefore,  the  minimum  distance  between  the  two  is                    E 450  =  1 2 km s  =  AC  =  AB  sin  10  2    km  =  5  min  and  the  desired  time  is    t  =  B C 52 1 v BA   =    =    h  =  15  min 20 2 4 30

JEE-Physics Example#12 In  the  figure  shown,  the  two  projectile  are  fired  simultaneously.  Find  the  minimum  distance  between  them during  their  flight. 20 3 ms A 600 20ms-1 20m 300 B Solution 20 3 Taking  origin  at  A  and  x  axis  along  AB 20 vAB Velocity  of  A  w.r.t.  B 600 x    =  20 3 cos 60ˆi  sin 60ˆj  20 cos150ˆi  sin 150ˆj A B 20m 300 =  20 3  1 ˆi  3 ˆj  20  3 ˆi  1 ˆj  20 3ˆi  20ˆj dmin 900  2 2   2 2 tan   20  1   300 so  d min  sin   sin 30  1  d min  10m 20 3 3 20 2 Example  #13 A particle is dropped from the top of a high building of height 360 m. The distance travelled by the particle in ninth second is (g = 10 m/s2) (A) 85 m (B) 60 m (C) 40 m (D) can't be determined Solution Ans.  (C) Total   t i me taken  b y  par t i cl e  to  reac h  t he  grou nd   T  =  2H  2  360  6 2  8.484 s g 10 11 Distance travelled in 8 seconds =  gt2 =  (10) (8)2 = 320 m 22 Therefore distance travelled in ninth second = 360 – 320 = 40 m Example  #14 A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18 m away from the wall, the angle of projection of ball is node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65 3 2 1 3 (A) tan–1   2  (B) tan–1   3  (C) tan–1   2  (D) tan–1   4  Solution Ans.  (B) From equation of trajectory y = xtan  1  x  3 = 6 tan  1  1   tan = 2 R  4  3 y 3m  x 6m 18m E 31

JEE-Physics Example  #15 A particle moves in XY plane such that its position, velocity and acceleration are given by   xˆi  yˆj ;       v xˆi  v yˆj ;       a xˆi  a yˆj r v a which of the following condition is correct if the particle is speeding down? (A) xv  + yv  < 0 (B) xv  + yv  > 0 (C) a v  + a v  < 0 (D) a v  + a v  > 0 xy xy xx yy xx yy Solution Ans.  (C)  For speeding down  a.v  0   a v  +  av   < 0 xx yy Example  #16 A  particle  is  thrown  vertically  upwards  from  the  surface  of  the  earth.  Let  TP  be  the  time  taken  by  the  particle to  travel  from  a  point  P  above  the  earth  to  its  highest  point  and  back  to  the  point  P.  Similarly,  let  TQ  be  the time  taken  by  the  particle  to  travel  from  another  point  Q  above  the  earth  to  its  highest  point  and  back  to  the same  point  Q.  If  the  distance  between  the  points  P  and  Q  is  H,  the  expression  for  acceleration  due  to  gravity in  terms  of  TP,  TQ  and  H,  is  :- 6H 8H 2H H (A)   TP2  TQ2 (B)  TP2  TQ2 (C)  TP2  TQ2 (D)  TP2  TQ2 Solution Ans.  (B) Highest Time  taken  from  point  P  to  point  P                TP  2 2 h  H point g Q P h 2h H Time  taken  from  point  Q  to  point  Q            TQ  2 g   TP2  8(h  H) and  TQ2 8h    TP2  TQ2  8H g 8H g  g TP2  TQ2 g Example  #17 An  aeroplane  is  travelling  horizontally  at  a  height  of  2000  m  from  the  ground.    The  aeroplane,  when  at  a point  P,  drops  a  bomb  to  hit  a  stationary  target  Q  on  the  ground.  In  order  that  the  bomb  hits  the  target, what  angle    must  the  line  PQ  make  with  the  vertical  ?  [g  =  10ms–2] P Q node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65 (A)   15° (B)  30° (C)  90° (D)  45° Solution  : Ans.  (D) Let  t  be  the  time  taken  by  bomb  to  hit  the  target. 1 h  =  2000  =  gt2   t  =  20  sec 2 R  =  ut  =  (100)  (20)  =  2000  m  tan   R  2000  1    45 h 2000 32 E

JEE-Physics Example  #18 Some  informations  are  given  for  a    body  moving  in  a  straight  line.  The  body  starts  its  motion  at  t=0. Information  I  :  The  velocity  of  a  body  at  the  end  of  4s  is  16  m/s Information  II  :  The  velocity  of  a  body  at  the  end  of  12s  is  48  m/s Information  III  :  The  velocity  of  a  body  at  the  end  of  22s  is  88  m/s The  body  is  certainly  moving  with (A)  Uniform  velocity (B)  Uniform  speed (C)  Uniform  acceleration (D)  Data  insufficient  for  generalization Solution Ans.  (D) Here  average  acceleration  =  16  0  48  16  88  48  4 4  0 12  4 22 12 But  we  can't  say  certainly  that  body  have  uniform  acceleration. Example  #19 A  large  number  of  particles  are  moving  each  with  speed  v  having  directions  of  motion  randomly  distributed. What  is  the  average  relative  velocity  between  any  two  particles  averaged  over  all  the  pairs? (A)  v (B)  (/4)v (C)  (4/)v (D)  Zero Solution  : Ans.  (C)  Relative  velocity  ,  vr  v1  v2   where  v   =  v   =  v 12 If  angle  between  them  be  ,  then  vr  v2  v2  2v2 cos   2v2 (1  cos )  2v sin     2  Hence,  average  relative  velocity    2  d  4v vr 2  2v sin 0 2  d 0 Example  #20 A ball is projected as shown in figure. The ball will return to point : y(vertical) wind gcot u gravity node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65  x (horizontal) O (A) O (B) left to point O (C) right to point O (D) none of these Solution  : Ans.  (A) ax  g cot   1  ux  Ini t i al veloc i t y  & ac c eler a t i on  are op posi te  to ea c h ot her. Here  a y g tan  uy Ball will return to point O. E 33

JEE-Physics Example  #21 Throughout  a  time  interval,  while  the  speed  of  a  particle  increases  as  it  moves  along  the  x-axis,  its  velocity  and acceleration  might  be (A) positive and positive respectively. (B) positive and negative respectively. (C) negative and negative respectively. (D) negative and positive respectively. Solution  : Ans. (A, C) Speed increases if both velocity & acceleration have same signs. Example  #22 u Three point particles A, B and C are projected from same A point with same speed at t=0 as shown in figure. 53° B For this situation select correct statement(s). u (A) All of them reach the ground at same time. 37° (B) All of them reach the ground at different time. Cu Hg (C) All of them reach the ground with same speed. (D) All of them have same horizontal displacement when they reach the ground. Solution  : Ans. (B, C) Ver t i cal  co mpo nent o f i nit ial  vel oci t ie s are  dif ferent reach  t he  gro und  at di fferent  t im e. Example  #23 A projectile is thrown with speed u into air from a point on the horizontal ground at an angle  with horizontal. If the air exerts a constant horizontal resistive force on the projectile then select correct alternative(s). (A) At the farthest point, the velocity is horiozntal. (B) The time for ascent equals the time for descent. (C) The path of the projectile may be parabolic. (D) The path of the projectile may be a straight line. Solution  : Ans.  (C,D) Here total acceleration a =  g2  a 2  = constant, so path may be parabolic or straight line. x Example  #24  to  26 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65 A river of width 'd' with straight parallel banks flows due North with speed u. A boat, whose speed is v relative yd  yv to water, starts from A and crosses the river. If the boat is steered due West and u varies with y as u =  d2 then answer the following questions. B W SN yu Ax E 2 4 . The time taken by boat to cross the river is d d d 2d (A)  (B) (C)  (D)  2v v 2v v 34 E

JEE-Physics 2 5 . Absolute velocity of boat when it reaches the opposite bank is 4 (B) v, towards West 4 (D) v, towards East (A)  v, towards East (C)  v, towards West 3 3 2 6 . Equation of trajectory of the boat is x2 y2 x2 x3 y2 y3 (A)  y  (B)  x  (C)  y  2d  3d2 (D)  x  2d  3d2 2d 2d Solution  : 24. Ans.  (B) dd Time taken =   vy v 25. Ans.  (B) At y=d, u =0 so absolute velocity of boat = v towards West. 26. Ans.  (D) y d  y dy dx y d  y v For boat (w.r.t. ground)    vy = v, vx = u =  v   v  and   d2 dt dt d2 dx y d  y x y yd  y2  y2 y3    dy d2  dx  d2 dy  x  2d  3d2 00 Example  #27 As shown in the figure there is a particle of mass 3 kg, is projected with speed 10 m/s at an angle 30° with horizontal (take g = 10 m/s2) then match the following 10 m/s A 30° B Column  I Column  II (A)  Average velocity (in m/s) during half of the time of flight, is 1 (P) 2 (B) The time (in sec) after which the angle between velocity (Q) 5 vector and initial velocity vector becomes /2, is 13 2 (C) Horizontal range (in m), is (R) 53 (D) Change in linear momentum (in N-s) when particle is at (S) At an angle of tan –1  1 highest point, is  from horizontal  2 3  (T) 2 Solution  : Ans. (A) (Q,S); (B)  (T); (C)  (R); (D)  (R) node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65  1 0 sin 3 0   0  2 75  25  5  2  42 v avx 2  v avy 2  10 cos 302 For (A) :    vav    13  m/s For (B) : Angle with horizontal    tan 1  v avy   tan 1  5 / 2  tan 1 1  u  v avx   5 3   2 3  60° 30° 60° gt By using        We have   u  sin 30  t  10 2  2 v u at gt 101 / u2 sin 2 100  3 / 2 30° For (C) : Horizontal range(R) =  g  10  5 3  m v For (D) : C h ang e  i n   l i n e ar  m o m e n t u m     =   m u  =    3   ×   1 0s i n 30°  =   5 3   N- s y E 35

JEE-Physics Example  #28 A particle is moving along a straight line along x-axis with an initial velocity of 2 m/s towards positive x-axis. A constant acceleration of 0.5 m/s2 towards negative x-axis starts acting on particle at t=0. Find velocity (in m/s) of particle at t = 2s. Solution  : Ans.  1 v = u + at  v= 2 + (–0.5) (2) = 1 m/s Example  #29 B (72/5) km/hr In the given figure points A and C are on the horizontal u D ground  &  A  and  B  are  in  same  vertical  plane. Simultaneously bullets are fired from A, B and C and  (54/5) km/hr they collide at D. The bullet at B is fired horizontally with A C ground 72 speed of   km/hr and the bullet at C is projected 5 54 vertically upward at velocity of  5  km/hr. Find velocity of the bullet projected from A in m/s. Solution  : Ans.  5  72 5  2  54 5  2  5 18   5 18  For collision  u  u 2  u 2      42  32  5 m/s B C Example  #30 Two stones A and B are projected simultaneously as shown in figure. It has been observed that both the stones reach the ground at the same place after 7 sec of their projection. Determine difference in their vertical components of initial velocities in m/s. (g = 9.8 m/s2) u2 B 2 49m u1  A Solution  : Ans.  7 In time of flight i.e. 7 s, the vertical displacement of A is zero and that of B is 49 m so for relative motion of B node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65 w.r.t.  A   ( u si n2–u1si n1)×  7 = 49   u si n2  – u s i n1 = 7  m/s 2 2 1 Example  #31  a.vˆ A particle moves with a tangential acceleration  a t   where    5ˆi   m/s2 . If the speed of the particle is zero a at x=0, then find v (in m/s) at x = 4.9 m. a y at v x 36 E

JEE-Physics Solution  : Ans.  7    4.9 v2 As  vdv  a.dr  adx  5dx  dx v   5 4.9  v2 = 49  v = 7 m/s vdv  5 0 02 Example  #32 A body is thrown up with a speed 49 m/s. It travels 5 m in the last second of its upward journey. If the same body is thrown up with a velocity 98 m/s, how much distance (in m) will it travel in the last second. (g = 10 m/s2) Solution  : Ans.  5 In last second of upward journey, all bodies travel same distance (= g/2 = 5m) Example  #33 A particle is moving in a circle of radius R in such a way that at any instant the normal and the tangential component of its acceleration are equal. If its speed at t=0 is v  then the time it takes to complete the first revolution is 0 R 1  e   . Find the value of ( v 0 Solution  : Ans.  3 v2 v t v t v0 v0 2 R t v0 R v0 v0 R 1  v0 1  v0 0 dt  ds R R 0 0 dv   dv  1    1    v   ds   1 v0 t dt v2 R  v  dt R    dt t t   2R  R n 1  v0 t  t  2  n 1  v0 t 1  v0 t  e 2   t  R 1  e 2 R  0 R R v0    1,  2      1  2  3 Example  #34     Find  the  relation  between  acceleration  of  blocks  a1,  a2  and  a3. x2 x3 x4 x1 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Kinametics\\English\\theory.p65 2 13 Solution x1  +  x2  +  x3  +  x4  =     x1  x2  x3  x4  0   a1  +  a2  +  a2  +  a3  =  0   a1  +  2a2  +  a3  =  0 Example#35 Two  moving  particles  P  and  Q  are  10  m  apart  at  any  instant.  Velocity  of  P  is  8  m/s  at  30,  from  line  joining  the P   and  Q  and  velocity  of  Q  is  6m/s  at  30.Calculate  the  angular  velocity  of  P  w.r.t.  Q Solution 37 8 sin 30  (6 sin 30 ) PQ    =  10 =  0.7  rad/s. E



JEE-Physics LOGIC GATES INTRODUCTION :  A logic gate is a digital circuit which is based on certain logical relationship between the input and the output voltages of the circuit.  The logic gates are built using the semiconductor diodes and transistors.  Each logic gate is represented by its characteristic symbol.  The operation of a logic gate is indicated in a table, known as truth table. This table contains all possible combinations of inputs and the corresponding outputs.  A logic gate is also represented by a Boolean algebraic expression. Boolean algebra is a method of writing logical equations showing how an output depends upon the combination of inputs. Boolean algebra was invented by George Boole. BASIC LOGIC GATES (2) AND gate, and (3) NOT gate There are three basic logic gates. They are (1) OR gate  The OR gate :- The output of an OR gate attains the state 1 if one or more inputs attain the state 1. Logic symbol of OR gate A Y=A+B B The Boolean expression of OR gate is Y = A + B, read as Y equals A 'OR' B. Truth table of a two-input OR gate A BY 0 00 0 11 1 01 1 11  The AND gate :- The output of an AND gate attains the state 1 if and only if all the inputs are in state 1. A Y\" Y Output Logic symbol of AND gate Input A BY B 0 00 0 10 The Boolean expression of AND gate is Y = A.B 1 00 1 11 It is read as Y equals A 'AND' B Truth table of a two-input AND gate  The NOT gate : The output of a NOT gate attains the state 1 if and only if the input does not attain the state 1. Logic symbol of NOT gate A Y E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\2. Logic Gates.p65 The Boolean expression is Y = A , read as Y equals NOT A. Truth table of NOT gate AY 01 10 COMBINATION OF GATES : The three basis gates (OR, AND and NOT) when connected in various combinations give us logic gates such as NAND, NOR gates, which are the universal building blocks of digital circuits. E 29

JEE-Physics  The NAND gate : A Y\" Y Logic symbol of NAND gate Input Output B The Boolean expression of NAND gate isY = A.B A BY 0 01 0 11 Truth table of a NAND gate 1 0 1 1 10  The NOR gate : Logic symbol of NOR gate A Y B The Boolean expression of NOR gate is Y  A  B Truth table of a NOR gate A BY 0 01 0 10 1 00 1 10 UNIVERSAL GATES : The NAND or NOR gate is the universal building block of all digital circuits. Repeated use of NAND gates (or NOR gates) gives other gates. Therefore, any digital system can be achieved entirely from NAND or NOR gates. We shall show how the repeated use of NAND (and NOR) gates will gives us different gates.  The NOT gate from a NAND gate :- When all the inputs of a NAND gate are connected together, as shown in the figure, we obtain a NOT gate Truth table of a single input NAND gate A Y\" A A B= (A) Y 00 1 11 0  The AND gate from a NAND gates :- If a NAND gate is followed by a NOT gate (i.e., a single input NAND gate), the resulting circuit is an AND gate as shown in figure and truth table given show how an AND gate has been obtained from NAND gates. Truth table A B Y' Y E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\2. Logic Gates.p65 A 00 1 0 Y 01 1 0 B 10 1 0 11 0 1  The OR gate from NAND gates :- If we invert the inputs A and B and then apply them to the NAND gate, the resulting circuit is an OR gate. Truth table A AB A B Y A 00 1 1 0 Y 01 1 0 1 B 10 0 1 1 B 11 0 0 1 30 E

JEE-Physics  The NOT gate from NOR gates :- When all the inputs of a NOR gate are connected together as shown in the figure, we obtain a NOT gate AY  The AND gate from NOR gates :- If we invert the inputs A and B and then apply them to the NOR gate, the resulting circuit is an AND gate. A A Y B B  The OR gate from NOR gate :- If a NOR gate is followed by a single input NOR gate (NOT gate), the resulting circuit is an OR gate. A Y B XOR AND XNOR GATES :  The Exclusive - OR gate (XOR gate):- The output of a two-input XOR gate attains the state 1 if one and only one input attains the state 1. Logic symbol of XOR gate A Y B The Boolean expression of XOR gate is Y  A. B  A. B or Y = A B Truth table of a XOR gate A BY 0 00 0 11 1 01 1 10  Exclusive - NOR gate (XNOR gate):- The output is in state 1 when its both inputs are the same that is, both 0 or both 1. E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\2. Logic Gates.p65 Logic symbol of XNOR gate A Y B The Boolean expression of XNOR gate is Y  A.B  A.B or Y  A  B or A  B Truth table of a XNOR gate A BY 0 01 0 10 1 00 1 11 E 31

JEE-Physics LAWS OF BOOLEAN ALGEBR A Basic OR, AND, and NOT operations are given below : OR AND NOT A+0=A A. 0 = 0 A+ A =1 A+1=1 A. 1 = A A. A =0 A+A=A A.A=A A .A=A Boolean algebra obeys commutative, associative and distributive laws as given below :  Commutative laws : A+B=B+A; A.B = B.A  Associative laws : A + (B + C) = (A + B) + C A. (B . C) = (A. B) . C  Distributive laws : A. (B + C) = A.B + A.C  Some other useful identities : (i) A + AB = A (ii) A . (A + B) = A (iii) A + ( A B) = A + B (iv) A. ( A + B) = A.B (v) A +(B.C) = (A + B). (A + C) (vi) ( A + B).(A + C) = A .C + B.A +B.C  De Morgan's theorem : E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\2. Logic Gates.p65 First theorem : A  B  A.B Second theorem : A.B  A  B 32 E

JEE-Physics SUMMARY OF LOGIC GATES Names Symbol Boolean Truth table Electrical Circuit diagram Expression analogue (Practical Realisation) OR A Y Y=A+B A BY A A D1 B 0 00 B 0 11 B Y 1 01 D2 R 1 11 AND A Y = A. B AD D2 B Y\" Y A BY 0 00 0 10 AB Y 1 00 1 11 R CCC NOT Y= A AY VCC 01 or A Y 10 RC Inverter RB A A Y NOR A Y Y A B A BY VCC RC Y (OR +NOT) B 0 01 A D1 RB 0 10 AB B R1 1 00 D2 1 10 A D1 B NAND A Y\" Y Y  A.B A BY VCC 0 01 D2 Y (AND+NOT) B 0 11 A 1 01 B 1 10 RB R1 VCC E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\2. Logic Gates.p65 XOR A Y A BY B 0 00 (Exclusive Y  A B 0 11 OR) A or 1 01 B 1 10 XNOR Y  A.B  AB A BY (Exclusive Y=AB 0 01 NOR) 0 10 Y or 1 00 1 11 Y  A.B  A.B or 33 Y  A B E

JEE-Physics NUMBER SYSTEMS Decimal Number system The base of this system is 10 and in this system 10 numbers [0,1,2,3,4,5,6,7,8,9] are used. Ex. 1396, 210.75 are decimal numbers. Binary Number System The base of this system is 2 and in this system 2 numbers (0 and 1) are used. Ex. 1001, 1101.011 are Binary numbers. Binary to decimal conversion We can write any decimal number in following form 2365.75 = 2000 + 300 + 60 + 5 + 0.7 + 0.05 11 = 2 × 1000 + 3 × 100 + 6 × 10 + 5 × 1 + 7 × +5× 10 100 = 2 × 103 + 3 × 102 + 6 × 101 + 5× 100 +7 × 10–1+ 5 × 10–2 Similarly we can write any binary number in following form 10101.11 = 1 × 24 + 0 × 23 + 1 × 22 + 0 × 21 + 1 × 20 + 1× 2–1 + 1 × 2–2 11 = 1 × 16 + 0 × 8 + 1 × 4 + 0 × 2 + 1 × 1 + 1 × + 1 × 24 11 = 16 + 4 + 1 + + = 21.75 24 Ex .1 Conver t binar y number 1011.01 into decimal number. 1011.01= 1 × 23 + 0 × 22 + 1 × 21 + 1 × 20 + 0 × 2–1 + 1 × 2–2 1 = 8 + 2 + 1 + 4 = 11.25 Ex .2 Conver t binar y number 1000101.101 into decimal number. 1000101.101  1× 26+0× 25+0× 24+0× 23+1× 22+0× 21+1× 20+1× 2–1+0× 2–2+1× 2–3 11 = 64 + 4 + 1 + + = 69 + 0.5 + 0.125 = 69.625 28 Question for Practise : Convert the following binary numbers into decimal numbers – (a) 101 (b) 110.001 (c) 11111 (d) 1011.11 Ans. : (a) 5 (b) 6.125 (c) 31 (d) 11.75 DECIMAL TO BINARY CONVERSION You should remember this table for decimal to binary conversion 2–3 2–2 2–1 20 21 22 23 24 25 26 27 28 29 210 0.125 0.25 0.5 1 2 4 8 16 32 64 128 256 512 1024 E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\2. Logic Gates.p65 Ex.3 Convert the decimal number 25 into its binary equivalent Sol. 25=16+8+1  24+ 23 + 20 = 1 × 24 + 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 so (25)10  (11001)2 Ex.4 Convert 69 into its binary equivalent Sol. 69 =64 + 4 + 1 = 1 × 26 + 1× 22 + 1 × 20  (69)10  (1000101)2 Ex.5 Convert 13.5 into its binary equivalent 13.5 = 8 + 4 + 1 + 0.5 = 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 + 1 × 21  (13.5)10 = (1101.1)2 Question for Practise : Convert the following decimal numbers into binary numbers (a) 6 (b) 65 (c) 106 (d) 268 (e) 8.125 Ans. (a) 110 (b) 1000001 (c) 1101010 (d) 100001100 (e) 1000.001 34 E

JEE-Physics MAGNETIC EFFECT OF CURRENT The branch of physics which deals with the magnetism due to electric current or moving charge (i.e. electric current is equivalent to the charges or electrons in motion) is called electromagnetism. ORESTED'S DISCOVERY SN The relation between electricity and magnetism was discovered by Orested in 1820. Orested showed that the electric current through the conducting wire deflects the North magnetic needle held below the wire. • When the direction of current in conductor is reversed then deflection of –+ SI magnetic needle is also reversed • On increasing the current in conductor or bringing the needle closer to the N N conductor the deflection of magnetic needle increases. Oersted discovered a magnetic field around a conductor carrying electric Oersted's experiment. Current in current. Other related facts are as follows: the wire deflects the compass needle (a) A magnet at rest produces a magnetic field around it while an electric charge at rest produce an electric field around it. (b) A current carrying conductor has a magnetic field and not an electric field around it. On the other hand, a charge moving with a uniform velocity has an electric as well as a magnetic field around it. (c) An electric field cannot be produced without a charge whereas a magnetic field can be produced without a magnet. (d) No poles are produced in a coil carrying current but such a coil shows north and south polarities. (e) All oscillating or an accelerated charge produces E.M. waves also in additions to electric and magnetic fields. • Current Element dB A very small element ab of length d of a thin conductor carrying P I r current I called current element. Current element is a vector quantity whose magnitude is equal to the product of current and I length of small element having the direction of the flow of current. a d b  d NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 • Biot – Savart's Law With the help of experimental results, Biot and Savart arrived at a mathematical expression that gives the magnetic field at some point in space in terms of the current that produces the field. That expression is based  on the following experimental observations for the magnetic field dB at a point P associated with a length  element d  of a wire carrying a steady current I. dB I, dB  d, dB  sinand dB  1  Id sin   dB = 0 Id sin  r2 dB  r2 4 r2 Vector form of Biot-Savar t's law   0 Id sin  nˆ n =unit vector perpendicular to the plane of  dB ( Id  ) and ( r ) 4 r2   0 Id     ×  = (Id) (r)sin nˆ ] dB 4  r r [ Id  r3 E1

JEE-Physics GOLDEN KEY POINTS •   0 Id    field vector  is always perpendicular to the plane of dB  r (dB) According to r 3 , direction of magnetic 4 vectors  Id   and  where plane of  Id   and  is the plane of wire. ( r ), (r ) • Magnetic field on the axis of current carrying conductor is always zero (=0° or  = 180°) • Magnetic field on the perimeter of circular loop or coil is always minimum. MAGNETIC FIELD LINES (By Michal Faraday) In order to visualise a magnetic field graphically, Michal faraday introduced the concept of field lines. Field lines of magnetic field are imaginary lines which represents direction of magnetic field continuously. GOLDEN KEY POINTS • Magnetic field lines are closed curves. • Tangent drawn at any point on field line represents direction of the field at that point. • Field lines never intersects to each other. • At any place crowded lines represent stronger field while distant lines represents weaker field. • In any region, if field lines are equidistant and straight the field is uniform otherwise not. Non-uniform Field Uniform Field Magnitude is Direction is Both magnitude Both magnitude not constant not constant and direction are and direction are not constant constant • Magnetic field lines emanate from or enters in the surface of a magnetic material at any angle. • Magnetic field lines exist inside every magnetised material. • Magnetic field lines can be mapped by using iron dust or using compass needle. RIGHT HAND THUMB RULE This rule gives the pattern of magnetic field lines due to current carrying wire. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 (i) Straight current (ii) Circular current Thumb  In the direction of current Curling fingers  In the direction of current, Curling fingers  Gives field line pattern Thumb  Gives field line pattern Case I : wire in the plane of the paper Case I : wire in the plane of the paper I ACW CW Magnetic field lines BB I Towards observer or Away from the observer perpendicular or perpendicular out-wards inwards 2 E

JEE-Physics Case II : Wire is  to the plane of the paper. Case II : Wire is  to the plane of the paper ACWconcentric & CWconcentric & ACW CW circular field lines circular field lines SN NS B I Towards observer Away from the observer GOLDEN KEY POINTS • When current is straight, field is circular • When current is circular, field is straight (along axis) • When wire is in the plane of paper, the field is perpendicular to the plane of the paper. • When wire is perpendicular to the plane of paper, the field is in the plane of the paper. APPLICATION OF BIOT-SAVART LAW : A • Magnetic field surrounding a thin straight current carrying conductor id r P  AB is a straight conductor carrying current i from B to A. At a point P, whose a   perpendicular distance from AB is OP =a, the direction of field is perpendicular   to the plane of paper, inwards (represented by a cross) =a tan dl=a sec2 d...(i) O =90°– & r=asec B • By Biot-Savart’s law   0 id sin   (due to a current element id at point P) dB 4 r2  B=  dB  0 id sin  (due to wire AB)  B= 0i  cos d 4 r2 4 Taking limits of integration as –2 2 to 1 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 0i 1 0i sin  1 0i 2 4 a cos d      B  4 a 2 4 a sin 1  sin 2 (inwards) Example Magnetic field due to infinite length wire at point 'P' Solution  I 2 = 90° M 90° P B = 0 I [sin90° + sin90°] B = 0I P 4 d 1 = 90° P 2d d  E3

JEE-Physics Example Magnetic field due to semi infinite length wire at point 'P' I Sol. B= 0 I [sin + sin90°] 90° 2=90° P B= 0 I [sin + 1] P 4 d M 1= P 4 d Ld Example Magnetic field due to special semi infinite length wire at point 'P' Solution B = 0I [sin0° + sin90°] B = 0I P 4 d P 4 d Example I 90° 2=90° M 1=° P d N Magnetic field due to special finite length wire at point 'P' 90° 2=° P M d 1=0° Solution B= 0 I [sin0° + sin] ;B = 0 I sin NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 P 4 d P 4 d Example If point ‘P’ lies out side the line of wire then magnetic field at point ‘P’ : d 2  P  I 1 Solution BP  0 I sin 90  1   sin 90  2   0I (cos 1  cos 2 ) 4 d 4 d 4 E

JEE-Physics • Magnetic field due to a loop of current Magnetic field lines due to a loop of wire are shown in the figure B i i i The direction of magnetic field on the axis of current loop can be determined by right hand thumb rule. If fingers of right hand are curled in the direction of current, the stretched thumb is in the direction of magnetic field. • Calculation of magnetic field z-axis Consider a current loop placed in y-z plane carrying current i in id y-axis anticlockwise sense as seen from positive x-axis. Due to a small Ri  r= R2+x2 current element id shown in the figure, the magnetic field at P is given by dB  0 id sin 900 . O 4 r2 i   dB x  P x-axis   id The angle between id and r is 900 because is along y- axis, while x-z plane.  perpendicular  lies in The direction of dB is r  to  as shown. The vector dB can be resolved into two components, r dB cos along z-axis and dB sin along x-axis. For any two diametrically opposite current elements, the components along x-axis add up, while the other two components cancel out. Therefore, the field at P is due to x-component of field only. Hence, we have 0 id 0 id R 0iR d  d  2R 4 r2 4 r2 r 4 r 3    B= sin   dB sin   =  B= z-axis = 0 i  2R 2 dBcos dB 4 R2  x2 3/2      0 i  2R 2 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 B= 4 r3 r  R2  x2 P x-axis dBsin (a) At the centre, x=0, B = 0i centre 2R (b) At points very close to centre, x<<R  B= 0i   x2  3 / 2 = 0i   3x2  2 1 R2  2 1 2R2  (c) At points far off from the centre, x>>R  B= 0 2R 2 4 x3 (d) The result in point (c) is also expressed as B = 0 2M 4 x3 where M=   R 2 , is called magnetic dipole moment. E5

JEE-Physics Example Find the magnetic field at the centre of a current carrying conductor bent in the form of an arc subtending angle  at its centre. Radius of the arc is R. Solution y-axis Let the arc lie in x-y plane with its centre at the origin. id i  Consider a small current element id as shown. The field due to this element at the centre is id sin 900  R2 id and R are perpendicular  dB= 0 R 4  d  Now d  Rd  dB  0 iRd 0 i d R 4 R2  dB = 4 R O The direction of field is outward perpendicular to plane of paper x-axis Total magnetic field B=  dB  B  0i    0i  B= 0i  4 R 4 R d  0 4 R 0 Example Find the magnetic field at the centre of a current carrying conductor bent in the form of an arc subtending angle 1 and 2 at the centre. Solution Magnetic field at the centre of arc abc and adc wire of circuit loop b B abc  0 I11 and B adc  0 I22  B abc  I11  angle = arc length  1 1 1 4 r 4 r B adc I2 2 2 = I1 radius a 2 2  V = IR = IR  I1  R2  I1  2  R    R  ) d I2 R1 I2 1 A 11 22 I2 c B abc =  2   1  B 1  1  Badc  1   2   B 2 1     HELMHOLT'Z COILS ARR ANGEMENT R This arrangement is used to produce uniform magnetic field (N, I, R) (N, I, R) of short range. It consists :- ACW • Two identical co–axial coils (N, I, R same) ACW • Placed at distance (center to center) equal to radius ('R') of coils O1 • Planes of both coils are parallel to each other. O2 • Current direction is same in both coils (observed from same side) M otherwise this arrangement is not called \"Helmholtz coil arrangement\". (uniformmagnetic field NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 of short range BHC) P SQ O1R/2 M O2 R/2 Example A pair of stationary and infinitely long bent wires are placed in the x-y plane IQ as shown in fig. The wires carry currents of 10 ampere each as shown. The segments L and M are along the x-axis. The segments P and Q are parallel to RO S the y-axis such that OS = OR = 0.02 m. Find the magnitude and direction of LM the magnetic induction at the origin O. PI 6E

JEE-Physics Solution As point O is along the length of segments L and M so the field at O due to these segments will be zero. Further,   as the point O is near one end of a long wire, BR  BP  BQ = 0 I (kˆ) + 0 I (kˆ) [as RO = SO = d] 4 d 4 d so,   0  2I  (kˆ) Substituting the given data,  10–7 × 2 10 (kˆ) = 10–4 W b (kˆ) BR 4  d  BR  0.02 m2 B = 10–4 T and in (+z) direction. Example Calculate the field at the centre of a semi-circular wire of radius R in situations depicted in figure (i), (ii) and (iii) if the straight wire is of infinite length. bI Ia Ia R R a c bI O R OI I OI I bI Rc c (ii) R (i) (iii) Solution The magnetic field due to a straight current carring wire of infinite length, for a point at a distance R from one of its ends is zero if the point is along its length and 0I if the point is on a line perpendicular to its length while at the 4 R centre of a semicircular coil is 0 I   4R so net magnetic field at the centre of semicircular wire is BR  Ba  Bb  Bc (i)  =0+ 0 I  + 0 = 0 I ( into the page) BR 4R 4R (ii)  = 0 I  + 0 I  + 0 I  = 0 I [ + 2]  (out of the page) BR 4 R 4R 4 R 4 R (iii)  = 0 I  + 0 I  + 0 I  = 0 I [ – 2] (in to the page) BR 4 R 4R 4 R 4 R NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 Example b Calculate the magnetic induction at the point O, if the current carrying wire is in the shape shown in figure. The radius of the curved part of the wire is a and a cI d linear parts are assumed to be very long and parallel. e Solution Magnetic induction at the point O due to circular portion of the wire B= µ0 I = 0 i  3  (out of the page) (= 3 ) 1 4 R 4 a 2 2 Magnetic induction at O due to wire cd will be zero since O lies on the line cd itself when extended backward. Magnetic induction at O due to infinitely long straight wire ae is  B = µ0i [sin 1  sin 2 ] where r = a, 1 = 0, 2 =   B= µ0 i sin 0  sin    = µ0 i 2 4 r 2 4 a   2  4 a 2 Because both the fields are in same direction i.e. perpendicular to plane of paper and directed upwards, hence the resultant magnetic induction at O is B=B +B = 0 i  3  1   12 4 a  2  E7

JEE-Physics Example Bi In the frame work of wires shown in figure, a current i is allowed to flow. Calculate the magnetic induction at the centre O. If angle  is equal to 90°, A R2 D then what will be the value of magnetic induction at O ? C O Solution B= 0i   i R1 Magnetic induction at O due to the segment BC is 1 4 R 2 E Similarly, the magnetic induction at O due to circular segment AED is B = 0 i (2  )  2 4 R 1 Magnetic field due to segments AB and CD is zero, because point 'O' lies on axis of these parts. Hence resultant magnetic induction at O is B=B +B = 0i    2    ,  12 4  R2 R1     B= 0 i    3  = 0 i  1  3  If  = 90° = , then 4  R 2R1  8  R2 R1   2 2    2 Example Two concentric circular coils X and Y of radii 16 cm and 10 cm respectively lie in the same vertical plane containing the north-south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and in Y clockwise, for an observer looking at the coils facing the west. What is the magnitude and direction of the magnetic field at their common centre (i) Due to coil X alone ? (ii) Due to coil Y alone ? (iii) Due to both the coils ? Solution According to the figure the magnitude of the magnetic field at the centre of coil X is Bx = 0 x N x  2  107 × 16 20 N 2 rx 2 0.16 = 4 × 10–4 T Ix Coil X Since the current in coil X is anticlockwise, the direction of Bx Coil Y is towards the east as shown in figure. W E Iy By Bx The magnitude of magnetic field at the centre of the coil Y is given by BY = 0 Y N Y 4   10 7 18 25 × 10–4 T S 2 = × =9 rY 2 2  since the current in coil Y is clockwise, the direction of field BY is towards the west (see fig.). Since the two fields are collinear and oppositely directed. The magnitude of the resultant field = difference between the two fields and its direction is that of the bigger field. Hence the net magnetic field at the common centre NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 is 5  × 10–4 T and is directed towards the west. Example ZI A long wire bent as shown in the figure carries current I. If the radius of the semi-circular portion is \"a\" then find the magnetic induction at the centre C. O a Y Solution C X II  Due to semi circular part  0 I ˆi B1 4a due to parallel parts of currents   2  0 I (kˆ) , B = B =  = 0 I (ˆi ) + 0 I (kˆ) B2 4 a net C B1  B2 4a 2 a magnitude of resultant field B = B 2  B 2 = 0I 2  4 1 2 4 a 8E

JEE-Physics Example A piece of wire carrying a current of 6 A is bent in the form of a circular arc of radius 10.0 cm, and it subtends an angle of 120° at the centre. Find the magnetic field due to this piece of wire at the centre. Solution Magnetic field at centre of arc B = µ0 I ,  = 120° = 2 I 4 R 3 rad B = 0I  2 = 0I 4 107  6 100 120° 4R 3 6R = T = 12.57 µT 6 10 Example Y An infinitely long conductor as shown in fig. carrying a current I with a semicircular loop on X-Y plane and two straight parts, one parallel to x-axis and another coinciding with I Z-axis. What is the magnetic field induction at the centre C of the semi-circular loop. IC r Solution X O The magnetic field induction at C due to current through straight part of the I conductor parallel to X-axis is Z B= 0 r I 2 sin   sin 0  = 0 I acting along + Z direction. i.e.  = 0 I kˆ 1 4 / 2  2 r 2 r B1 The magnetic field induction at C due to current through the semi-circular loop in X-Y plane is B = 0 r I 2   = 0 I acting along + Z-direction i.e.  = 0 I kˆ 2 4 / 2r B2 2r The magnetic field induction at C due to current through the straight part of the conductor coinciding with Z-axis is 0 I sin   0 I  0 I 4 / 2  2 r i.e. B3 2 r  B3 =  sin 0  ˆi r 2 = acting along (–X)-axis Total magnetic field induction at C is    0 I kˆ + 0 I kˆ – 0 I ˆi = 0 I 1   kˆ  ˆi  B  B1  B2  B3 = 2 r 2 2 r 2 r Example A conductor carrying a current i is bented as shown in figure. Find the magnitude of magnetic field at the origin Solution    Field at O due to part 1   0i ˆi   1  0i  kˆ y B1 4 r Field at O due to part 2 B2 4  2r   i 2i B3 1  Wire 3 passes through origin when it is extended backwards 0 r x 3   0 i  ˆi kˆ  O i B0  B1  B2  B3   z =   NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p654r   2  0.04m 0.05m Example 0.03m A conductor of length 0.04 m is tangentially connected to a circular 4A  loop of radius 0.03 m perpendicular to its plane. Find the magnetic field induction at the centre of the loop if 4 ampere current is passed through the conductor as shown in fig. Solution Magnetic field induction at the centre of the loop due to the straight current-carrying conductor, B= 0 I sin 1  sin 2  = 4 107  4 sin 0  0.04  = 1.07 × 10–5T 4 r 4  0.03 0.05  Magnetic fields due to the two halves of the loop are equal in magnitude and opposite in direction. So, the magnetic field induction due to the loop at the centre of the loop is zero. So, the magnetic field induction at the centre of the loop is 1.07 × 10–5T. E9

JEE-Physics Example P Figure shows a right-angled isosceles triangle PQR having its base equal to a. A current of I ampere is passing downwards along a thin straight wire cutting the 90° r plane of the paper normally as shown at Q. Likewise a similar wire carries an r equal current passing normally upwards at R. Find the magnitude and direction Q 45° a R of the magnetic induction at P. Assume the wires to be infinitely long. Solution a Let r = PQ = PR and a2 = r2 + r2 = 2r2 or r = 2 Magnetic induction at P due to conductor at Q is B = 0 I = 20I = 0I (along PR) 1 2r 2a 2a Magnetic induction at P due to conductor at R is B = 0I (along PQ) 2 2a Now, resultant of these two is B = B12  B 2 =  0I  2   0I 2 = 2 0I = 0 I 2  2a   2a  2a a  The direction of B is towards the mid-point of the line QR. AMPERE'S CIRCUITAL LAW I5 I4 I3 ACW Ampere's circuital law state that line integral of the magnetic field around circulation I1 I2 Positive any closed path in free space or vacuum is equal to 0 times of net  I=(I1I2+I3) Negative current or total current which crossing through the area bounded by the  closed path. Mathematically  B . d  0I This law independent of size and shape of the closed path. Any current outside the closed path is not included in writing the right hand side of law Note : • This law suitable for infinite long and symmetrical current distribution. • Radius of cross section of thick cylinderical conductor and current density must be given to apply this law. MAGNETOMOTIVE FORCE (M.M.F.)   .    0 I , where   .   0 I    .   I B d B  0H , d H d  0H The line integral of magnetising field around any closed path is equal to net current crossing through the area NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65  bounded by the closed path, also called 'magnetomotive force'. Magnetomotive force (M.M.F.) = H. d APPLICATION OF AMPERE'S CIRCUITAL LAW • Magnetic field due to infinite long thin current carrying straight conductor  Consider a circle of radius 'r'. Let XY be the small element of length d. B and d are in the same direction because direction of along the tangent of the circle. By A.C.L.   ,  B d cos   0 I . d  0  I  B I (where  = 0°)  B d cos 0  0 I  B  d  0 I (where  d  2r ) ACW rOy B x d  0I I 2 r B (2r) = 0I  B 10 E

• Magnetic field due to infinite long solid cylinderical conductor JEE-Physics • For a point inside the cylinder r < R, Current from area R2 is = I I R so current from area r2 is = I (r2 ) = Ir2 12 3 R 2 R2 axis By Ampere circuital law for circular path 1 of radius r Cross-sectional Bin (2r) = 0I' = 0 Ir2  Bin = 0 Ir Bin  r view R2 2R 2 • For a point on the axis of the cylinder (r = 0); B axis  0 I • For a point on the surface of cylinder (r = R) 1 23 By Ampere circuital law for circular path 2 of radius R R Bs (2  R) = 0I  Bs = 0 I (it is maximum) 2R • For a point outside the cylinder (r > R) :- B By Ampere circuital law for circular path 3 of radius r Bmax Br Bout 1r 0 I 1 r< R 2 r  Bout  r Bout (2  r) = 0I  B out  r= 0 r=R r>R Magnetic field outside the cylinderical conductor does not depend upon nature (thick/thin or solid/hollow) of the conductor as well as its radius of cross section. • Magnetic field due to infinite long hollow cylinderical conductor E • For a point at a distance r such that r < a < b B1  0 ba 12 3 • For a point at a distance r such that a < r < b Axis B2(2r) =0 I'   B2(2r)  r2  a2  Side View = 0I  b2  a2   I  a12 3 B2  0 I  r2  a2  b 2r  b2  a2  Cross sectional   view NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 • For a point at a distance r such that r>b > a, B3 (2r) = 0I   B3  0I 2 r • For a point at the axis of cylinder r=0 Baxis = 0 Magnetic field at specific positions for thin hollow cylinderical conductor I At point 1 B1 = 0 O 12 3 R At point 2 B2 = 0 I (maximum) [outer surface] and 2 R B2 = 0 (minimum) [inner surface] B 0 I Bout 1 2 r r At point 3 B3 = (for the point on axis Baxis =0) r=0 r=R r 11

JEE-Physics Magnetic field due to an infinite plane sheet of current -z dB dBsin P x P y  dBcos dBcos   dB x-axis r dBsin xx An infinite sheet of current lies in x-z plane, carrying current along-z axis. The field at any point P on y is along a line parallel to x-z plane. We can take a rectangular amperian loop as shown. If you traverse the loop in clockwise direction, inward current will be positive. By Ampere circuital law,  =  0 enclosed ....(i) a B.d  B PQRS B Let  represents current per unit length. P Q b The current enclosed is given by  =enclosed a x-axis      b Now,  B.d   B.d  R  B.d   B.d   B.d PQRS PQ QR RS SR S Now,     as B  d  B.d   B.d  0 QR   SP   0 B.d  B.d 2  B  a as B d 2   Also, =  2B  a  0a  B  PQ RS Example A long straight solid conductor of radius 5 cm carries a current of 3A, which is uniformly distributed over its circular cross-section. Find the magnetic field induction at a distance 4 cm from the axis of the conductor. Relative permeability of the conductor = 1000. Solution Imagine a circular path of radius r whose centre lies on the axis of I=3A solid conductor such that the point P lies on it. the current threading this closed path I'= I  r 2 = Ir 2 R 2 R2 r Magnetic field B acts tangential to the amperian circular path at P and is same in NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 magnitude at every point on circular path. PB Using Ampere circuital law  0r  Ir2  0r I r R=5cm B.d    2R 2 = µ µ I'  B (2r) =  R2  B = I 0r B= 4 107  1000  3  0.04 = 9.6 × 10–3 T. 2  0.05 2 Example A current I flows along a thin walled tube of radius R with a long longitudinal slit of width b (<<R). What is the magnetic field induction at a distance r (< R) ? Solution Using principle of superposition, field due to strip in place of slit + field due to tube with slit B = 0 so B = field due to strip in place of slit= 0  I  b E 2r (2R  b) 12

JEE-Physics MAGNETIC FIELD DUE TO SOLENOID It is a coil which has length and used to produce uniform magnetic field of long range. It consists a conducting wire which is tightly wound over a cylinderical frame in the form of helix. All the adjacent turns are electrically insulated to each other. The magnetic field at a point on the axis of a solenoid can be obtained by superposition of field due to large number of identical circular turns having their centres on the axis of solenoid. Magnetic field due to a long solenoid A solenoid is a tightly wound helical coil of wire. If length of solenoid is large, as compared to its radius, then in the central region of the solenoid, a reasonably uniform magnetic field is present. Figure shows a part of long solenoid with number of turns/length n.We can find the field by using Ampere circuital law. Consider a rectangular loop ABCD. For this loop   B.d  0ienc ABCD           Now  B.d   B.d   B.d   B.d   B.d  B  a ABCD AB BC CD DA a A       D B This is because  B.d   B.d  0, B  d . b C AB CD And,   (  B outside the solenoid is negligible  B.d  0 DA Now, ienc = n  a  i s  B  a  0 n  a  i  B  0ni (Magnetic field lines) Finite length solenoid : Its length and diameter are comparable.  2 r B 1 (uniform) E PM E'  End-I End-II NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 By the concept of BSL magnetic field at the axial point 'P' obtained as : BP  0nI (cos 1  cos 2 ) 2 Angle 1 and 2 both measured in same sense from the axis of the solenoid to end vectors. Infinite length solenoid : Its length very large as compared to its diameter i.e. ends of solenoid tends to infinity. ( a) Magnetic field at axial point which is well inside the solenoid 1  0° and 2  180° B  0 n I [cos 0° – cos 180°]  0nI [(1) – (–1)]  0nI 2 2 (b) Magnetic field at both axial end points of solenoid 1 = 90°and 2  180°  B  0 n I [cos 90° – cos 180°]  0nI [(0) – (–1)]  0nI 2 22 E 13

JEE-Physics Example The length of solenoid is 0.1m. and its diameter is very small. A wire is wound over it in two layers. The number of turns in inner layer is 50 and that of outer layer is 40. The strength of current flowing in two layers in opposite direction is 3A. Then find magnetic induction at the middle of the solenoid. Solution Direction of magnetic field due to both layers is opposite, as direction of current is opposite so Bnet = B1 – B2 = 0n1I1 – 0n2I2 = 0 N1 I – 0 N2 I ( I1 = I2 = I)   = 0 I (N1 – N2) = 4  10–7  3 (50 – 40) = 12 × 10–5 T  0 1 Example A closely wound, solenoid 80 cm. long has 5 layers of winding of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0A. Estimate the magnetic field (a) Inside the solenoid (b) Axial end points of the solenoid Solution (a) Magnetic field inside the solenoid Bin = 0nI = 0 N I, (N=400× 5= 2000) 4  107 2000  8 = 8 × 10–3 T  = (80  102 ) (b) Magnetic field at axial end points of solenoid Bends = 0 n I 8   103 = 4 × 10–3 T = 22 Example A straight long solenoid is produced magnetic field 'B' at its centre. If it cut into two equal parts and same number of turns wound on one part in double layer. Find magnetic field produced by new solenoid at its centre. Solution Magnetic field produced by a long solenoid is B = 0nI, where n = N/  Same number of turns wound over half length N   0 N I    / 2  I     Magnetic field produced by new solenoid is B' = 0 = 2 =2B NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 Example 30° 60° Find out magnetic field at axial point ‘P’ of solenoid shown in figure P (where turn density ‘n’ and current through it is I) end Solution 30° 60° Magnetic field at point ‘P’ due to finite length solenoid P BP = 0 n I [cos 1 – cos 2], where 1 = 30° (CW), 2 E 0nI [cos 30°–cos 120°] end 2 2 = (180°–60°) = 120° (CW) = = 0 n I  3    1  = 0nI ( 3 + 1) 2  2  2  4  14

JEE-Physics Example  A uniform magnetic field B  B 0kˆ exists in a region. A current carrying wire is placed in x-y plane as shown. Find the force acting on part AB of the wire. y B=B0k a a a i aB Aa x Solution The conductor consists of 5 straight sections viz AC, CD, DE, EF and FB as shown. As the field is uniform, force on the sections is given by       F  i ˆi  B . Thus, FAC  i aˆi  B 0kˆ  B 0iaˆj   FDE  i FEF  i       FCD  i aˆj  B 0kˆ  B 0iaˆi , aˆi  B 0kˆ  B 0iaˆj , aˆj  B 0kˆ  B 0iaˆi        F  FAC  FCD  FDE  FEF  FFB  3B0iaˆj FFB  i aˆj  B 0kˆ  B 0iaˆj Net force, CURRENT CARRYING CONDUCTOR IN MAGNETIC FIELD When a current carrying conductor placed in magnetic field, a magnetic force exerts on each free electron which are present inside the conductor. The resultant of these forces on all the free electrons is called magnetic force on conductor. • Magnetic force on current element  Through experiments Ampere established that when current element I d is placed in magnetic field B , it   experiences a magnetic force dFm  I (d  B ) Id  x  (Cw)  B dFm (external) • Current element in a magnetic field does not experience any force if the current in it is parallel or NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 anti–parallel with the field  = 0° or 180° dF = 0 (min.) m • Current element in a magnetic field experiences maximum force if the current in it is perpendicular with the field  = 90° dF = BId (max.) m • Magnetic force on current element is always perpendicular to the current element vector and magnetic   field vector. dFm  I d  and dFm  B (always) • Total magnetic force on straight current carrying conductor in uniform magnetic field given as   f   f d      dFm  B, Fm  I(L  B) Fm = =I× i i  I f i LL N Where  = f L d , vector sum of all length elements from initial to final point, which is in accordance with the i law of vector addition and | L | = length of the condutor. E 15

JEE-Physics  • Total magnetic force on arbitrary shape current carrying conductor in uniform magnetic field B is  f  f d     dFm I B, Fm  I(L  B)  × (L = ab) i i  Initial I Final point a L point b  f d   Where L =i , vector sum of all length elements from initial to final point or displacement between free ends of an arbitrary conducter from initial to final point. GOLDEN KEY POINT • A current carrying closed loop (or coil) of any shape placed in uniform magnetic field then no net magnetic force act on it (Torque may or may not be zero)  f   I  L= i d =0 or  d =0  i=f So net magnetic force acting on a current carrying closed loop Fm  0 (always) • When a current carrying closed loop (or coil) of any shape placed in non uniform magnetic field then net magnetic force is always acts on it (Torque may or may not be zero) Example  A wire bent as shown in fig carries a current i and is placed in a uniform field of magnetic induction B that emerges from the plane of the figure. Calculate the force acting on the wire. RR ii Solution  The total force on the whole wire is F = I| L |B = I(R + 2R + R)B = 4RIB m R NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 Example  A square of side 2.0 m is placed in a uniform magnetic field B  2.0 T in a direction perpendicular to the plane of the square inwards. Equal current i = 3.0 A is flowing in the directions shown in figure. Find the magnitude of magnetic force on the loop. xx  C x Bx xx D xx xx xA x xx x Ex Solution  Net force on the loop = 3 ( FAD )     Force on wire ACD = Force on AD = Force on AED  Fnet = 3(i) (AD) (B) = (3) (3.0) (2 2 )(2.0) N = 36 2 N. Direction of this force is towards EC. 16 E

JEE-Physics Example A metal rod of mass 10 gm and length 25 cm is suspended on two springs as shown in figure. The springs are extended by 4 cm. When a 20 ampere current passes through the rod it rises by 1 cm. Determine the magnetic field assuming acceleration due to gravity to be 10 m/s2. x k xx k F=bil x TT x Mg I x x S o l . Let tension in each spring is = T 0 Initially the rod will be in equilibrium if 2T = Mg then T = kx ...(i) 0 00 Now when the current I is passed through the rod it will experience a force F = BIL vertically up; so in this situation for its equilibrium, 2T + BIL = Mg with T = kx...(ii) (x = 4 – 1 = 3cm) So from eq. (i) and eq.(ii) T  Mg  BIL  x  1  BIL T0 Mg x0 Mg  B = Mgx0  x = 10 103 10  3 102 = 1.5 × 10–2T 20  25 102  4  102 IL x 0 Example Two conducting rails are connected to a source of e.m.f. and form an incline as shown in fig. A bar of mass 50 g slides without friction down the incline through a vertical magnetic field B. If the length of the bar is 50 cm and a current of 2.5 A is provided by the battery, for what value of B will the bar slide at a constant velocity ? [g = 10 m/s2] I B v=constI Solution Fmcos R Force on current carrying wire F = BIL The rod will move down the plane with constant velocity only if  mgsin F cos  = mg sin   BIL cos  = mg sin  Fm mg  NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65  or, B= mg tan  = 50 103 10  3 = 0.3 T IL 2.5  50 102 4 Example A wire PQ of mass 10g is at rest on two parallel metal rails. The separation between the rails is 4.9 cm. A magnetic field of 0.80 tesla is applied perpendicular to the plane of the rails, directed downwards. The resistance of the circuit is slowly decreased. When the resistance decreases to below 20 ohm, the wire PQ begins to slide on the rails. Calculate the coefficient of friction between the wire and the rails. P 6V 4.9cm Q E 17

JEE-Physics Solution Wire PQ begins to slide when magnetic force is just equal to the force of friction, i.e.,    µ mg = i B sin  ( = 90°) so i= E = 6 = 0.3 A so µ = i B = 0.3  4.9  102 0.8  = 0.12 R 20 mg 10  103 9.8  Example A wire abcdef with each side of length '' bent as shown in figure and carrying a current I is placed in a uniform magnetic field B parallel to +y direction.What is the force experienced by the wire. Z c   ba d e Y X fB Solution   Magnetic force on wire abcdef in uniform magnetic field is Fm = I (L B) ,  L is displacement between free ends of the conductor from initial to   =(B) ˆj ; L   (ˆi  ˆj) = BI (kˆ) = BI, along +z direction L = () ˆi and B final point. B F = I  = BIL m MAGNETIC FORCE BETWEEN TWO PAR ALLEL CURRENT CARRYING CONDUCTORS Like currents unlike currents 12 12 I1 I1 Repulsion B2 B1 dF12 B2 B1 dF21 dF12 dF21 d d The net magnetic force acts on a current carrying conductor due to its own field is zero. So consider two infinite long parallel conductors separated by distence 'd' carrying currents I and I . 1 02I1 Magnetic field at each point on conductor (ii) due to current I is 2d [uniform field for conductor (2)] 1 B= 1 Magnetic field at each point on conductor (i) due to curent I is B= 0 I2 [Uniform field for conductor (1)] NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 2 2 2d consider a small element of length 'd' on each conductor. These elements are right angle to the external magnetic field, so magnetic force experienced by elements of each conductor given as dF = B I d =  0 I2  I d ...(i) (Where I d  B ) 12 2 1  2 d  12   1 dF = B I d =  0 I1  I d ... (ii) (Where I d  B ) 21 1 2  2 d  21   2 Where dF is magnetic force on element of conductor (i), due field of conductor (i) and dF is magnetic force on 12 21 element of conductor (ii), due to field of conductor (i). dF12 dF21 0 I1 I2 d d 2d Magnetic force per unit length of each conductor is = = f= 0 I1I2 N/m (in S.I.) f = 2 I1I2 dyne/cm (In C.G.S.) 2d d 18 E

JEE-Physics Definition of ampere : 12 Magnetic force/unit length for both infinite length conductor gives as 0 I1 I2 (4 107 ) (1) (1) 1A 1A 2d = 2 (1) f= = 2 × 10–7 N/m 'Ampere' is the current which, when passed through each of two parallel infinite long straight conductors placed in free space at a distance of 1 m from each 1m other, produces between them a force of 2 × 10–7 N/m L • Force scale f= 0 I1 I2 is applicable when at least one conductor must be of infinite I1 I2 2d length so it behaves like source of uniform magnetic field for other conductor. Magnetic force on conductor 'LN' is F = f ×   FLN   0 I1 I2   d LN  2d  (source)   N • Equillibrium of free wire Case I : Upper wire is free : Consider a long horizontal wire which is rigidly fixed another wire is placed directly above and parallel to fixed wire. (Stable (free) I2 fm Finite length (m,) Equilibrium) g Infinite length h (Source) + (fixed) I1 – Magnetic force per unit length of free wire f = 0 I1I2 , and it is repulsive in nature because currents are unlike. m 2h Free wire may remains suspended if the magnetic force per unit length is equal to weight of its unit length At balanced condition f = W'. Weight per unit length of free wire = 0 I1 I2  m g (stable equillibrium condition) 2 h  m If free wire is slightly pluked and released then it will executes S.H.M. in vertical plane. h The time period of motion is T  2 g Case II : Lower wire is free : Consider a long horizontal wire which is rigidly fixed. Another wire is placed directly below and parallel to the fixed wire. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 – I1 (Source) Infinite length (fixed) + d (depht) fm g Finite length (m,) (Unstable (free) I2 Equilibrium) Magnetic force per unit length of free wire is f= 0 I1 I2 , and it is attractive in nature because currents are like. 2 d m Free wire may remains suspended if the magnetic force per unit length is equal to weight of its unit length At balanced condition f = W' m Weight per unit length of free wire 0 I1 I2  m g (unstable equillibrium condition) 2 d  E 19

JEE-Physics Example Two horizontal parallel straight conductors, each 20 cm long, are arranged one vertically above the other and carry equal currents in opposite directions. The lower conductor is fixed while the other is free to move in guides remaining parallel to the lower. If the upper conductor weights 1.20 g, what is the approximate current that will maintain the conductors at a distance 0.75 cm apart. Solution Fm P iQ In equilibrium magnetic force F will balance weight mg. d mg m 0 i 2  So mg = F  mg = 2 d m 2 m g d 2 1.2 103  9.8  0.75 102 Ri S i = 0   = 4  107  20 102 = 2205 = 47 A MAGNETIC DIPOLE MOMENT A magnetic dipole consists of a pair of magnetic poles of equal and opposite strength separated by small distance. Ex. Magnetic needle, bar magnet, solenoid, coil or loop. • Magnetic moment of Bar magnet Neutral point A Magnetic -m +m axis S N Cross sectional 0 view The magnetic moment of a bar magnet is defined as a vector quantity having magnitude equal to the product of pole strength (m) with effective length () and directed along the axis of the magnet from south pole to north pole. M= m It is an axial vector S.I. unit : - A.m2 GOLDEN KEY POINTS • Attractive property : A bar magnet attracts certain magnetic substances (eg. Iron dust). The attracting power of the bar magnet is maximum at two points near the ends called poles. So the attracting power of a bar magnet at its poles called 'pole strength' • The 'pole strength' of north and south pole of a bar magnet is conventionally represented by +m and –m respectively. • The 'pole strength' is a scalar quantity with S.I. unit A–m. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 • The 'pole strength' of bar magnet is directly proportional to its area of cross section. m  A • The attracting power of a bar magnet at its centre point is zero, so it is called 'neutral point'. • Magnetic poles are always exists in pairs i.e. mono pole does not exist in magnetism. So Gauss law in  magnetism given as  B    0 ds • Effective length or magnetic length :– It is distance between two poles along the axis of a bar magnet. As pole are not exactly at the ends, the effective length () is less than the geometrical length ( ) of the bar magnet. 0  ~ 0  91 0 • Inverse square law (Coulomb law) : The magnetic force between two isolated magnetic poles of strength m and m lying at a distance 'r' is directly proportional to the product of pole strength and inversely proportional 12 to the square of distance between their centres. The magnetic force between the poles can be attractive or repulsive according to the nature of the poles. 20 E

JEE-Physics Fm  m1m2 m1m2 µ0 (S.I.) r2 4 1 Fm=k where k 1 (C.G.S.) r2 Fm  Inverse square law of Coulomb in magnetism is applicable only for two long bar magents becasue isolated poles cannot exist. • If a magnet is cut into two equal parts along the length then pole strength is reduced to half and length remains unchanged. New magnetic dipole moment M'=m'()= m    M . 22 The new magnetic dipole moment of each part becomes half of original value. -m  m' = (m/2), × '=  m' = m '=/2 +m SN SN SN SM N SN M'= m× 2=2M M' = (m/2) × '= M/2 M= m×  • If a magnet is cut into two equal parts transverse to the length then pole strength remains unchanged and length is reduced to half. New magnetic dipole moment M'm     M .  2  2 The new magnetic dipole moment of each part becomes half of original value. • The magnetic dipole moment of a magnet is equal to product of pole strength and distance between poles. M=m   N N S S N S  • As magnetic moment is a vector, in case of two magnets having magnetic moments M1 and M2 with angle  between them, the resulting magnetic moment. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65  2 2 1/2 tan    M2 sin   1 2 M  M  M  2M1M 2 cos  with  M  M2 cos     1 Example The force between two magnetic poles in air is 9.604 mN. If one pole is 10 times stronger than the other, calculate the pole strength of each if distance between two poles is 0.1 m? Solution Force between poles F  µ0 m1m2 or 9.604  10 3  10 7  m  10m or m2= 96.04 N2T –2m=9.8N/T 4 r2 0.1  0.1 So strength of other pole is 9.8 × 10 = 98 N/T Example A steel wire of length L has a magnetic moment M. It is then bent into a semicircular arc. What is the new magnetic moment? E 21

JEE-Physics Solution If m is the pole strength then M=m.L  m  M L If it is bent into a semicircular arc then L= r  rL  So new magnetic moment M '  m 2r  M 2  L  2M L  Example Two identical bar magnets each of length L and pole strength m are placed at right angles to each other with the north pole of one touching the south pole of other. Evaluate the magnetic moment of the system. Solution S M2 MR M1 =M2 = mL  MR  M 2  M 2  2M1M 2 cos   2 mL M1  1 2 2 M1 and tan   M sin 90  1 i.e. = tan–11 = 45° NS N M  M cos 90 M2 MAGNETIC MOMENT OF CURRENT CARRYING COIL ( LOOP) Current carrying coil (or loop) behaves like magnetic dipole. The face of coil in which current appears to flow anti clock wise acts as north pole while face of coil in which current appears to flow clock wise acts as south pole. • A loop of geometrical area 'A', carries a current 'I' then magentic moment of coil M = I A • A coil of turns 'N', geometrical area 'A', carries a current 'I' then magentic moment M = N I A Magnetic moment of current carrying coil is an axial vector M  NIA where A is a area vector perpendicu- lar to the plane of the coil and along its axis. SI UNIT : A-m2 or J/T ACW CW M M M  NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 Direction of M find out by right hand thumb rule  • Curling fingers  In the direction of current • Thumb  Gives the direction of M For a current carrying coil, its magnetic moment and magnetic field vectors both are parallel axial vectors. Example Find the magnitude of magnetic moment of the current carrying loop ABCDEFA. Each side of the loop is 10 cm long and current in the loop is i = 2.0 A CD B E A F 22 E

JEE-Physics Solution By assuming two equal and opposite currents in BE, two current carrying loops (ABEFA and BCDEB) are formed. Their magnetic moments are equal in magnitude but perpendicular to each other. CD B E A F Hence, M = M2  M2 = 2M where M = iA = (2.0)(0.1)(0.1) = 0.02 A-m2 net    Mnet = ( 2 )(0.02) A-m2 = 0.028 A-m2 Example The wire loop PQRSP formed by joining two semicircular wires of radii R and R carries a current I as 12 shown in fig. What is the magnetic induction at the centre O and magnetic moment of the loop in cases (A) and (B) ? (A) (B) R2 I R R2 S R O 1Q SR O P R1 I Q P Solution As the point O is along the length of the straight wires, so the field at O due to them will be zero and hence. (A)   0  I   I  i.e.,  0 I  1  1   &  × I 1 R 2   1 R 2  = 1 I R 2  R 2   B 4  R1  B 4  R1 R2  M  NIS =1  2 2 2 1 2 2 1  R   2 (B) Following as in case (A), in this situation,   0 I  1  1   and,   1 I R 2  R 2   B 4  R1 R2  M 2 2 1   MAGNETIC DIPOLE IN MAGNETIC FIELD Torque on magnet ic dipole NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 N Fm=mB CW S  +m  B=(uniform)  CW Fm=mB-m (a) Bar magnet  = force × perpendicular distance between force couple  = (mB) (sin) , where M = m  = 90°   = MB (maximum)    M B  = MBsin Vector form  = 0° or 180° = 0 (minimum) (b) Coil or Loop I M CW       M B   N I  A  B   O B(uniform)  = 90°   = BINA (maximum) Plane of coil  = BINAsin  = 0° or 180° = 0 (minimum) E 23

JEE-Physics GOLDEN KEY POINTS • Torque on dipole is an axial vector and it is directed along axis of rotation of dipole.  • Tendency of torque on dipole is try to align the M in the direction of B or tries to makes the axis of dipole  parallel to B or makes the plane of coil (or loop) perpendicular to B . • Dipole in uniform magnetic field Fnet=0 (no translatory motion)  may or may not be zero (decides by ) • Dipole in non uniform magnetic field Fnet0 (translatory motion)  may or may not be zero (decides by ) • When a current carrying coil (or loop) is placed in longitudenal magnetic field then maxmium torque acts on it. M   = 90°  B   max = MB = BINA • When a current carrying coil (or loop) is placed in transverse magnetic field the no torque acts on it.   MB M anti B    = 0°  min = 0 or  = 180° WORK DONE IN ROTATING A MAGNETIC DIPOLE Work done in rotating a dipole in a uniform magnetic field through small angle 'd' dW = .d = MBsind So work done in rotating a dipole from angular position 1 to 2 with respect to the Magnetic field direction 2 dW = 2 MB sin d 1 1  W = = MB(cos 1  cos 2 ) • If magnetic dipole is rotated from field direction i.e. 1 = 0° to position 2= then work done is W = MB (1 – cos) = 2MB sin2 /2  in one rotation  = 0° or 360° W = 0 in 1/4 rotation  = 90°  W = MB in half rotation  = 180°  W = 2MB in 3/4 rotation  = 270°  W = MB • Work done to rotate a dipole in a magnetic field is stored in the form of potential energy of magnetic dipole. POTENTIAL ENERGY OF MAGNETIC DIPOLE NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 The potential energy of dipole defined as work done in rotating the dipole from a direction perpendicular to the  given direction. U = W – W  U = MB (1 – cos) – MB = MB cos In vector form U  90°  M .B GOLDEN KEY POINTS  • When M and B are parallel ( = 0°), the dipole has minimum potential energy and it is in stable equillibrium. U   MB (minimum)  • When M and B are anti parallel ( = 180°), the dipole has maximum potential energy and it is in unstable equillibrium. U  MB (maximum)  • When M and B are prpendicular to each other ( = 90°), the dipole has potential energy U=0 and in this situation maximum torque acts on it hence no equillibrium. 24 E

JEE-Physics Example A circular coil of 25 turns and radius 6.0 cm, carrying a current of 10 A, is suspended vertically in a uniform magnetic field of magnitude 1.2 T. The field lines run horizontally in the plane of the coil. Calculate the force and torque on coil due to the magnetic field. In which direction should a balancing torque be applied to prevent the coil from turning ? Solution  Magnetic  d    force F= I  B m   B For coil or close loop  d  0 so Fm  0  I The torque  on a coil of any shape having N turns and I current  in a magnetic field B is given by  = NIABsin  = 25 × 10 ×  × 6 × 6 × 10–4 × 1.2 × sin90° = 3.39 N  The direction of  is vertically upwards. To prevent the coil from turning, an equal and opposite torque must be applied. Example A uniform magnetic field of 5000 gauss is established along the positive z-direction. A rectangular loop of side 20 cm and 5 cm carries a current of 10 A is suspended in this magnetic field. What is the torque on the loop in the different cases shown in the following figures ? What is the force in each case ? Which case corresponds to stable equilibrium ? Z ZZ  Y  Y  Y BI B B I O O I X (a) O X (b) X (c) ZZ Z  Y  Y  Y B B B O O O 30° I (f) (e) X (d) NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 Solution (a) Torque on loop, =BIA sin Here, =90°; B=5000 gauss=5000× 10–4 tesla = 0.5 tesla   = 10 ampere, A=20 × 5cm2=100 × 10–4=10–2 m2 Now,   = 0.5 × 10× 10–2=5 × 10–2 Nm It is directed along –y-axis (b) Same as (a). (c) =5 × 10–2 Nm along –x-direction (d)  =5×10–2 N m at an angle of 240° with +x direction. (e)  is zero. [Angle between plane of loop and direction of magnetic field is 90°] (f)  is zero. Resultant force is zero in each case. Case (e) corresponds to stable equilibrium. Example A circular coil of 100 turns and having a radius of 0.05 m carries a current of 0.1 A. Calculate the work required to turn the coil in an external field of 1.5 T through 180° about an axis perpendicular to the magnetic field ? The plane of coil is initially at right angles to magnetic field. Solution Work done W = MB (cos1–cos2) =NAB (cos1–cos2)  W = Nr2B ( c os  – c o s  ) = 100 × 0.1 × 3.14 × (0.05)2 × 1.5 (cos 0° –cos) = 0.2355J 1 2 E 25

JEE-Physics Example A bar magnet of magnetic moment 1.5 JT–1 lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required to turn the magnet so as to align its magnetic moment. (i) Normal to the field direction? (ii) Opposite to the field direction? (b) What is the torque on the magnet in case (i) and (ii)? Solution Here, M = 1.5 JT–1, B = 0.22 T. (a) P.E. with magnetic moment aligned to field = – MB P.E. with magnetic moment normal to field = 0 P.E. with magnetic moment antiparallel to field = + MB (i) Work done = increase in P.E. = 0 – (–MB) = MB = 1.5 × 0.22 = 0.33 J. (ii) Work done = increase i n P.E. = MB – (–MB) = 2MB = 2 × 1.5 × 0.22 = 0.66 J. (b) We have  = MBsin (i)  = MB sin = 1.5 × 0.22 × 1 = 0.33 J. (=90° sin=1) This torque will tend to align M with B. (ii)  = MB sin = 1.5 × 0.22 × 0=0 ( =180°  sin=0) Example A short bar magnet of magnetic moment 0.32 J/T is placed in uniform field of 0.15 T. If the bar is free to rotate in plane of field then which orientation would correspond to its (i) stable and (ii) unstable equilibrium? What is potential energy of magnet in each case? Solution (i) If M is parallel to B then =0°. So potential energy U = Umin = – MB Umin = –MB = –0.32 × 0.15 J= –4.8 × 10–2 J (stable equilibrium) (ii) If M is antiparallel to B then  = ° So potential energy U = Umax = + MB = + 0.32 × 0.15 = 4.8 × 10–2 J (unstable equilibrium.) ATOMIC MAGNETISM An atomic orbital electron, which doing bounded uniform circular motion around nucleus. A current constitues with this orbital motion and hence orbit behaves like current carrying loop. Due to this magnetism produces at nucleus position. This phenomenon called as 'atomic magnetism. Bohr's postulates : v (i) mv2 kze2 L = mvr = n  h  , where n = 1, 2,3 ....... ACW r = r2 (ii)  2  e– +Ze Basic elements of atomic magnetism : r (a) Orbital current :- I  ef  e  ev  e Fe T 2r 2 I(current) (b) Magnetic induction at nucleus position :- As circular orbit behaves like current carrying loop, so magnetic induction at nucleus position B = 0I NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 N 2r BN  0 ef  0 e  0 e v  0 e 2r 2Tr 4 r2 4 r (c) Magnetic moment of circular orbit :- Magnetic dipole moment of circular orbit  M = IA where A is r 2 e r 2 evr er2 area of circular orbit. M  ef   T2 2 • Relation between magnetic moment and angular momentum of orbital electron Magnetic moment M = evr m eL (angular momentum L = mvr) I(current)    r eL 2m 2m e Vector form   2m  M M  For orbital electron its M and L both are antiparallel axial vectors. 26 E

JEE-Physics BOHR MAGNETON (B) According of Bohr's theory, angular momentum of orbital electron is given by nh , where n = 1, 2, and h is plank's L = 2 moment of orbital 3 ........ is given by M constant. Magnetic electron = eL = n eh 4 m 2m eh • If n = 1 then M = 4m , which is Bohr magneton denoted by B • Definition of B : Bohr magneton can be defined as the magnetic moment of orbital electron which revolves in first orbit of an atom. eh 1.6  10 19  6.6  10 34 • B = 4 m = 4  3.14  9.1  10 31 = 0.923 × 10–23 A.m.2 • Basic elements of atomic magnetism for first orbit of H-atom (n=1, z = 1) (a) Accurate form :- (v = 2.18 × 106 m/sec, f = 6.6 × 1015 cy/sec. r = 0.529Å) • Orbital current I = 0.96 mA • Magnetic induction at nucleus position B = 12.8 T N • Magnetic moment of orbital electron M = 0.923 × 10–23 A.m2 (b) Simple form :- (v  2 × 106 m/sec, f  6 × 1015cy/sec, r  0.5Å) • Orbital current I  1mA • Magnetic induction at nucleus position B 4T N • Magnetic moment of orbital electron M = µ A.m2 B A NONCONDUCTING CHARGED BODY IS ROTATED WITH SOME ANGUL AR SPEED. q In this case the ratio of magnetic moment and angular momentum is constant which is equal to 2m here q = charge and m = the mass of the body. Example In case of a ring, of mass m, radius R and charge q distributed on it circumference. Angular momentum L = I = (mR2)() ... (i) Magnetic moment M = iA = (qf) (R2) HGF IJKM = (q)  (R2) = q R2 ...(ii) +++ + + + + + +q+++ 2 2 +R +++ ++ + + NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 + ++  f =  From Eqs. (i) and (ii) Mq + + + +  2  L 2m Although this expression is derived for simple case of a ring, it holds good for other bodies also. For example, for b gqL q I a disc or a sphere. M =  M , where L = I  2m 2m Rigid body Ring Disc Solid sphere Spherical shell Moment of inertia (I) mR2 mR2 2 2 2 mR2 mR2 5 3 qI qR 2 qR 2 qR 2 qR 2 Magnetic moment= 2 4 5 3 2m E 27

JEE-Physics FORCE ON A CHARGED PARTICLE IN A MAGNETIC FIELD     Force experienced by a current element d  in magnetic field B is given by dF = Id × B ....(i)  Now if the current element d is due to the motion of charge particles, each particle having a charge a cross-section A,   q moving with velocity  through d  = nqA v  d   nqdV   [with volume dV=A d] v v From eqn (i) we can write  q (    ) dF =ndV v B ndV=the total number of charged particles in volume dV  (n=number of charged particles per unit volume), force on a charged particle  = 1 dF =q  n dV F (v B) GOLDEN KEY POINT • The force   and the field  F is always perpendicular to both the velocity v B • A charged particle at rest in a steady magnetic field does not experience any force. If the charged particle is at rest then    , so   v 0 vB 0 • A moving charged particle does not experience any force in a magnetic field if its motion is parallel or antiparallel to the field. q=0° F F qv B B B O v O 90° v vq q q  =180° F=vBsin Fmax=qvB F=0 (A) (B) (C) • If the particle is moving perpendicular to the field. In this situation all the three vectors   and  F, v B are mutually perpendicular to each other. Then sin  = max = 1, i.e.,  = 90°, The force will be maximum Fmax = q v B • Work done by force due to magnetic field in motion of a charged particle is always zero. When a charged particle move in a magnetic field, then force acts on it is always perpendicular to displacement, so the work done,W =  F.ds   Fds cos 90  0 (as  = 90°), And as by work-energy theorem W =  KE, the kinetic energy   1 m v 2  remains unchanged and  2  hence speed of charged particle v remains constant. NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-09\\Magnetic Effect of current & Magnetism\\Eng\\01.Theory.p65 However, in this situation the force changes the direction of motion, so the direction of velocity  of the v charged particle changes continuously. MOTION OF A CHARGED PARTICLE IN A MAGNETIC FIELD Motion of a charged particle when it is moving collinear with the field magnetic field is not affected by the field (i.e. if motion is just along or opposite to magnetic field) ( F = 0) Only the following two cases are possible : • Case  : When the charged particle is moving perpendicular to the field. The angle between  and  is =90°. So the force will be maximum (= qvB) and always perpendicular B v to motion (and also field); Hence the charged particle will move along a circular path (with its plane perpendicular to the field). Centripetal force is provided by the force qvB, So mv2  qvB  r mv r qB Angular frequency of circular motion, called cyclotron or gyro-frequency. = v  qB r m E 28