The radius of the current carrying coil is 'R'. If fractional decreases in field value with respect to a center of the coil for a nearby axial point is $1 %$ then find an axial position of that point.

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Solution

magnetic field around circular current carrying coil is $\frac{μ_{0 i R^{2}}}{{2 (x^{2} + r^{2})}^{\frac{3}{2}}}$
Magnetic field at the center $= \frac{μ_{0 i}}{2 R}$
$\frac{μ_{0 i R^{2}}}{{2 (x^{2} + R^{2})}^{\frac{3}{2}}} = \frac{1}{2 \sqrt{2}} \frac{μ_{0 i}}{2 R}$
$8 R^{6} = {(x^{2} + R^{2})}^{3}$
$2 R^{2} = (x^{2} + R^{2})$
$x = \pm \sqrt{{2 R}^{2} - R^{2}} = \pm R$

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The radius of the current carrying coil is 'R'. If fractional decreases in field value with respect to a center of the coil for a nearby axial point is 1% then find an axial position of that point.

magnetic field around circular current carrying coil is μ0iR22(x2+r2)32Magnetic field at the center=μ0i2Rμ0iR22(x2+R2)32=12√2μ0i2R 8R6=(x2+R2)3 2R2=(x2+R2) x=±√2R2−R2=±R

The radius of the current carrying coil is 'R'. If fractional decreases in field value with respect to a center of the coil for a nearby axial point is $1 %$ then find an axial position of that point.