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Obtain an expression for magnetic field in terms of magnetic dipole moment associated with circular current loop. ""(or) Show that circular current loop can be associated with a magnetic dipole. |
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Answer» Solution :We KNOW that the magnetic field at a point on the axis of a circular loop. `""B=(mu_(0)/(4PI))(2piIR^(2))/(x^(2)+R^(2))^(3/2)` R - radius of the loop. For a distance `"x>>R", x^(2)+R^(2)=x^(2)` and `""(x^(2)+R^(2))^(3/2)=x^(3)` HENCE, `""B=(mu_(0)/(4pi))(2piIR^(2))/(x^(3))"where", piR^(2)=A` (area of circular loop) i.e., `""B=(mu_(0)/(4pi))(2IA)/(x^(3))"where", m=IA=` magnetic moment Hence,`""vec(B)=(mu_(0)/(4pi))(2vec(m))/(x^(3))"...(1)"` Hence circular loop acts as a magnetic dipole. Note : `*` As an analogous to the electric field intensity `""vec(E)=(1/(4piepsilon_(0)))(2vec(p)_(e))/(x^(3)) "...(2)"` `*` Comparing (1) and (2) we note that by substituting `mu_(0) " as " 1/epsilon_(0)` and `vec(B)` by `vec(E), vec(m) " by " vec(p)_(e)`, one can relate the magnetic and electric fields due to the corresponding dipoles. |
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