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 _______ .

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TestHub · Accepted Answer

We know, the magnetic field on the axis of a current carrying circular ring is given by

$B = \frac{μ_{0}}{4 π} \frac{2 N I A}{{(R^{2} + x^{2})}^{3 / 2}}$

$∴ \frac{B_{1}}{B_{2}} = \frac{8}{1} = {[\frac{R^{2} + {(0.2)}^{2}}{R^{2} + {(0.05)}^{2}}]}^{3 / 2}$

$4 [R^{2} + {(0.05)}^{2}] = [R^{2} + {(0.2)}^{2}]$

$4 R^{2} - R^{2} = {(0.2)}^{2} - 4 \times {(0.05)}^{2}$

$4 R^{2} - R^{2} = {(0.2)}^{2} - {(0.1)}^{2}$

$3 R^{2} = 0.3 \times 0.1$

$R^{2} = {(0.1)}^{2} \Rightarrow R = 0.1$

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