The fractional change in the magnetic field i | vedclass

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Question

$B _{	ext {axis }}=\frac{\mu_{0} i R ^{2}}{2\left( R ^{2}+ x ^{2}ight)^{3 / 2}}$

$B _{	ext {centre }}=\frac{\mu_{0} i }{2 R }$

$	herefore B

Vedclass · Accepted Answer

$\frac{3}{2} \frac{ r ^{2}}{ a ^{2}}$

Vedclass · Answer

$B _{	ext {axis }}=\frac{\mu_{0} i R ^{2}}{2\left( R ^{2}+ x ^{2}ight)^{3 / 2}}$

$B _{	ext {centre }}=\frac{\mu_{0} i }{2 R }$

$	herefore B _{	ext {centre }}=\frac{\mu_{0} i }{2 a }$

$	herefore B _{ axis }=\frac{\mu_{0} ia ^{2}}{2\left( a ^{2}+ r ^{2}ight)^{3 / 2}}$

$	herefore$ fractional change in magnetic field $=$

$\frac{\frac{\mu_{0} i }{2 a }-\frac{\mu_{0} ia ^{2}}{2\left( a ^{2}+ r ^{2}ight)^{3 / 2}}}{{\frac{\mu_{0} i }{2 a }}}=1-\frac{1}{\left[1+\left(\frac{ r ^{2}}{ a ^{2}}ight)ight]^{3 / 2}}$

$\approx 1-\left[1-\frac{3}{2} \frac{ r ^{2}}{ a ^{2}}ight]=\frac{3}{2} \frac{ r ^{2}}{ a ^{2}}$

Note : $\left(1+\frac{ r ^{2}}{ a ^{2}}ight)^{-3 / 2} \approx\left(1-\frac{3}{2} \frac{ r ^{2}}{ a ^{2}}ight)$

[True only if $r << a ]$

The fractional change in the magnetic field intensity at a distance $'r'$ from centre on the axis of current carrying coil of radius $'a'$ to the magnetic field intensity at the centre of the same coil is : (Take $r << a )$

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