Magnetic fields at two points on the axis of | vedclass

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Question

We know, the magnetic field on the axis of a current carrying circular ring is given by

$B =\frac{\mu_{0}}{4 \pi} \frac{2 NIA }{\left( R ^{2}+ x ^{

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

We know, the magnetic field on the axis of a current carrying circular ring is given by

$B =\frac{\mu_{0}}{4 \pi} \frac{2 NIA }{\left( R ^{2}+ x ^{2}ight)^{3 / 2}}$

$	herefore \frac{ B _{1}}{ B _{2}}=\frac{8}{1}=\left[\frac{ R ^{2}+(0.2)^{2}}{ R ^{2}+(0.05)^{2}}ight]^{3 / 2}$

$4\left[ R ^{2}+(0.05)^{2}ight]=\left[ R ^{2}+(0.2)^{2}ight]$

$4 R ^{2}- R ^{2}=(0.2)^{2}-4 	imes(0.05)^{2}$

$4 R ^{2}- R ^{2}=(0.2)^{2}-(0.1)^{2}$

$3 R ^{2}=0.3 	imes 0.1$

$R ^{2}=(0.1)^{2} \Rightarrow R =0.1$

Magnetic fields at two points on the axis of a circular coil at a distance of $0.05\, m$ and $0.2\, m$ from the centre are in the ratio $8: 1.$ The radius of coil is .......... $m$

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