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Q.

Magnetic field 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 the coil is $n \times 10^{- 1} m .$ The value of ‘n’ is

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answer is 1.

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Detailed Solution

$\begin{array}{l} B = \frac{μ_{0} {NIR}^{2}}{2 {(r^{2} + x^{2})}^{3 / 2}} \Rightarrow \frac{B_{1}}{B_{2}} = {(\frac{r^{2} + x_{2}^{2}}{r^{2} + x_{1}^{2}})}^{3 / 2} \\ \Rightarrow {(\frac{2}{1})}^{3} = {(\frac{r^{2} + x_{2}^{2}}{r^{2} + x_{1}^{2}})}^{3 / 2} \end{array}$

$\Rightarrow r = 0.1 m$

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Magnetic field 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 the coil is n × 10−1m. The value of ‘n’ is