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CHAPTER-28 Magnetic Field
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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.
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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
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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.
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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
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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)
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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
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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
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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
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Ch 28-10 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
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Suggested problems Chapter 28
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