CHAPTER-28 Magnetic Field. Ch 28-2 What produces a Magnetic Field?  Electric field E produces a electric force F E on a stationary charge q  Magnetic.

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Presentation transcript:

CHAPTER-28 Magnetic Field

Ch 28-2 What produces a Magnetic Field?  Electric field E produces a electric force F E on a stationary charge q  Magnetic field B produces a magnetic force F B on a moving charge q  What produces a Magnetic Field? A Magnetic Monopole? Magnetic monopoles do not exists  Two ways to produce magnetic field B: Electromagnet: magnetic field due to a current Permanent magnet: net magnetic field due to intrinsic magnetic field of electrons in certain material.

Ch 28-3 The Definition of B  Electric field E: E field is tested by measuring force F E on a static charge q E=F E /q  Magnetic Field B is tested by measuring force F B on a moving charge q. If v is charge velocity then F B =q(vxB)=qvBsin   is angle between the v and B.  Direction of B: that direction of V for which F B =0  Direction of F B :  to plane of V and B

Ch 28-3 The Definition of B  Unit of B Tesla (T): B=F B /qvsin  SI unit of magnetic field B (T):Newtons/Coulomb.(m/s) = N/(C/s).m=N/A.m 1T=1N/A.m  Magnetic Field Lines:  B field line starts from N pole ans terminates at S pole  Like magnetic pole repel each other and opposite pole attract each other.

Ch 28-4 Crossed Fields  Crossed fields: a region with E and B field  to each other. If the net force due to these two fields on a charge particle is zero, particle travels undeviated Then F E =F B qE=qvB (  =90) E=vB or v=E/B

Ch 28-6 A Circulating Charged Particle  When a charge q moves in a B field then the magnetic force F B =qvBsin  =qvB (  =90)  For  =90 the particle moves in a circular orbit wit a radius given by : F R =F B  mv 2 /R=|q|vB (m is particle mass)  R=mv/qB or v=|q|RB/m  Particle period T=2  R/v  Frequency f=1/T=v/2  R  Angular frequency  =2  f  =2  f=v/R=|q|B/m If v has a component along B then particle trajectory is a helix (helical path)

Ch 28-8 Magnetic Force on a Current-Carrying Wire  A force is exerted by a magnetic field B on a charge +q moving with velocity v.  Direction of force is same for an electron (-q) moving with velocity v D ( in opposite direction to v).  For a wire carrying a current i, number of electrons passing a length of wire L in time t are q - given by :  q - =it=iL/v D & q - =-q and v D =-v  q - =iL/v D reduces to q=iL/v Then F B =q(vxB)= (iL/v)(vxB) F B = i(LxB) F B = iLB

Ch 28-9 Torque on a Current Loop A current carrying rectangular loop placed between the N and S poles of a magnet, no magnetic force on shorter sides but forces in opposite direction on longer sides

Ch 28-9 Torque on a Current Loop Same Magnitude of forces F 2 and F 4 on shorter sides F 2 =F 4 =ibBcos  Same Magnitude of forces F 1 and F 3 on longer sides F 1 =F 3 =iaBsin=iaB Net torque due to F1 and F3 is  ’ = iaB x (bsin  /2) + iaB x (bsin  /2) =iabBsin   ’ =iABsin  = i(AxB)=i 

Ch Magnetic Dipole Moment  Magnetic moment of a coil:   =NiA ( A is coil area); Then  =  x B =NiAB sin   Torque  exerted on an electric dipole p in an electric field E :  = p x E  Torque due to electric or magnetic field = vector product of dipole moment and field vector  Electric potential energy U(  ) of a electric dipole moment p in an E field U(  )= -p.E  Magnetic potential energy of a magnetic dipole  U(  ) = - . B = -  B cos  Max. value of U(  ) : for  = 180 Min. value of U(  ) for  =0 Work done by a magnetic field in rotating a a magnetic dipole from initial orientation  i to final orientation  f W=-  U= -(U f -U i ) W appl =-W

Suggested problems Chapter 28