Magnetic fields at two points on the axis of | vedclass

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(b) $B = \frac{{{\mu _0}}}{{4\pi }} 	imes \frac{{2\pi Ni{R^2}}}{{{{({R^2} + {x^2})}^{3/2}}}} $

$\Rightarrow B \propto \frac{1}{{{{({r^2} + {x^2})}

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$0.1$

Vedclass · Answer

(b) $B = \frac{{{\mu _0}}}{{4\pi }} 	imes \frac{{2\pi Ni{R^2}}}{{{{({R^2} + {x^2})}^{3/2}}}} $

$\Rightarrow B \propto \frac{1}{{{{({r^2} + {x^2})}^{3/2}}}}$
$ \Rightarrow \frac{8}{1} = \frac{{{{({R^2} + x_2^2)}^{3/2}}}}{{{{({R^2} + x_1^2)}^{3/2}}}} $

$\Rightarrow {\left( {\frac{8}{1}} ight)^{2/3}} = \frac{{{R^2} + 0.04}}{{{R^2} + 0.0025}}$
$ \Rightarrow \frac{4}{1} = \frac{{{R^2} + 0.04}}{{{R^2} + 0.0025}}$. On solving $R = 0.1\,m$

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 the coil is.....$m$

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