The magnetic induction at the centre of a cur | vedclass

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(C) The magnetic field at any point on the axis of a current-carrying coil at a distance $x$ from the centre is given by:$B_{a} = \frac{\mu_{0} n i r^

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$20$

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(C) The magnetic field at any point on the axis of a current-carrying coil at a distance $x$ from the centre is given by:$B_{a} = \frac{\mu_{0} n i r^{2}}{2(r^{2} + x^{2})^{3/2}}$At the centre,$x = 0$,so the magnetic field is:$B_{c} = \frac{\mu_{0} n i}{2r}$Taking the ratio:$\frac{B_{c}}{B_{a}} = \frac{\mu_{0} n i}{2r} 	imes \frac{2(r^{2} + x^{2})^{3/2}}{\mu_{0} n i r^{2}} = \frac{(r^{2} + x^{2})^{3/2}}{r^{3}}$Given $B_{c} = 5\sqrt{5} B_{a}$,we have:$5\sqrt{5} = \frac{(r^{2} + x^{2})^{3/2}}{r^{3}}$Substituting $r = 10 \, cm$:$5\sqrt{5} = \frac{(100 + x^{2})^{3/2}}{1000}$$5000\sqrt{5} = (100 + x^{2})^{3/2}$Squaring both sides:$(5000)^{2} 	imes 5 = (100 + x^{2})^{3}$$125 	imes 10^{6} = (100 + x^{2})^{3}$Taking the cube root:$500 = 100 + x^{2}$$x^{2} = 400$$x = 20 \, cm$

The magnetic induction at the centre of a current-carrying circular coil of radius $10 \, cm$ is $5\sqrt{5}$ times the magnetic induction at a point on its axis. The distance of the point from the centre of the coil (in $cm$) is

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