Question

Show that the magnetic field along the axis of a current carrying coil of radius r at a distance x from the centre of the coil is smaller by the fraction 3x^{2}/2r6{2} than the field at the centre of the coil carrying current.

Answer 1

Hi friend,
The magnetic field induction, at the center of circular coil of n turns ,radius r ,carrying current I is   $B_{0} = \frac{ u_{o} 2 \pi nI}{4 \pi r} = \frac{ u_{0}nI }{2r}$
Magnetic field induction at a point on the axis of the circular coil carrying current is given by $B_{0} = \frac{ u_{0}2 \pi nI r^{2} }{4 \pi ( r^{2} + x^{2} )^{3/2} } = \frac{ u_{0}nI r^{2} }{2 ( r^{2} + x^{2} )^{3/2} }$

$\frac{B}{ B_{0} } = \frac{ r^{3} }{ ( r^{2} + x^{2} )^{3/2} } = \frac{ r^{3} }{ r^{3} }[1+ \frac{ x^{2} }{ r^{2} } ] ^{-3/2} =[1- \frac{3 x^{2} }{2 r^{2} } +...]=[1- \frac{3 x^{2} }{2 r^{2} } ]$


Fractional decrease in magnetic field=[tex] \frac{B _{0}-B }{B _{0} } =1- \frac{B}{ B_{0} } =1-[1- \frac{3 x^{2} }{2 r^{2} } ]= \frac{3 x^{2} }{2 r^{2} } [/tex]
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