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Question

(b) $\frac{{{B_{centre}}}}{{{B_{axis}}}} = {\left( {1 + \frac{{{x^2}}}{{{R^2}}}} \right)^{3/2}},$

also ${B_{axis}} = \frac{1}{8}{B_{centre}}$
$ \R

Vedclass · Accepted Answer

$R\sqrt 3 $

Vedclass · Answer

(b) $\frac{{{B_{centre}}}}{{{B_{axis}}}} = {\left( {1 + \frac{{{x^2}}}{{{R^2}}}} \right)^{3/2}},$

also ${B_{axis}} = \frac{1}{8}{B_{centre}}$
$ \Rightarrow \frac{8}{1} = {\left( {1 + \frac{{{x^2}}}{{{R^2}}}} \right)^{3/2}} \Rightarrow 2 = {\left( {1 + \frac{{{x^2}}}{{{R^2}}}} \right)^{1/2}}$
$ \Rightarrow 4 = 1 + \frac{{{x^2}}}{{{R^2}}} \Rightarrow 3 = \frac{{{x^2}}}{{{R^2}}} \Rightarrow {x^2} = 3{R^2}$$ \Rightarrow x = \sqrt 3 R$

A circular current carrying coil has a radius $R$. The distance from the centre of the coil on the axis where the magnetic induction will be $\frac{1}{8}^{th}$ to its value at the centre of the coil, is

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