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| 1. |
Derivethe expressionfor magneticfieldat apointon theaxisof a circularcurrentloop . |
Answer» Solution : Let RBE theradius of acurrentloop, carryingcurrentI. LetR bea POINTON theaxisof aconductor.LetdBbe MAGNETICFIELD atP , dueto acurrentelement .idl. Fromthefigure ,`theta+ alpha= 90 ^(@) ,`so that` alpha= 90 ^(@)- theta ` and` cod alpha = cos(90 ^(@) - theta)=sin theta= (R^(2))/( (R^(2)+x^(2))^(1/2))` let`dB_(x)`be thehorizontalcomponentof dB . ApplyingBiot- Savart.s law , Wewrite`d vecB= ((mu_(0))/(4pi ))(I(d vecl xxvec r))/(r ^(3))` `i.e.,dB =((mu_(0))/(4pi))(Irdl)/(r^(3)) sintheta . ` where ` theta. = 90 ^(@)` `i.e.,dB =((mu _(0))/(4pi)) (i rdl )/(r^(3))=((mu_0)/(4pi))(idl)/(r^(2))` componentof dBalongthe horizontalis`dB_X =dBcos alpha ` or ` dB _x= dB sintheta= ((mu _(0))/(4pi))(idl)/(r^(2))sin theta `. Byusing(1)wewrite `dB_X =((mu_(0))/(4pi ))(Idl)/(r^(2))(R )/(r)= ((mu_(0))/(4pi )) (idlR )/((R^2+x^2)^(3/2))`.... (3) orintegrating `B_x = intdB_X= int ((mu_0)/(4pi))(IR )/((R^2+x^2)^(3/2))dl` `i.e.,B_(x) =((mu_(0))/(4pi)) (IR)/((R^2 +x^2)^(3/2))(2pi R )` Where` intdl= 2 piR or ` ` B_x =((mu_0)/(4 pi))((2 pi IR ^(2))/((R^(2)+x^2)^(3/2)))` tesl and ` vecb= B_(x)HATI= ((mu_(0))/(4pi )) ((2pi IR ^(2))/((R^(2)+x ^(2))^(3/2)))hati, dB _(y)=0 ` Fora circularloop, andatthecentre, `x = 0 , ` ` vec B _(0 )=((mu_0)/(4 pi))((2pi l)/(R ))hati ` fora circularconductorcontainingn turns, `B_x hati =((mu_0)/(4pi ))(2pi n IR ^2)/((R^2+x^2)^(3/2))hati ` and atthe centre , ` B_0 hati=((mu_0)/(4 pi))((2pi n l)/(R ))hati`. |
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