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(A) The magnetic field at the centre of a circular coil of radius $R$ is $B_{center} = \frac{\mu_0 I}{2R}$.The magnetic field at a point on the axis a

Vedclass · Accepted Answer

$17.89$

Vedclass · Answer

(A) The magnetic field at the centre of a circular coil of radius $R$ is $B_{center} = \frac{\mu_0 I}{2R}$.The magnetic field at a point on the axis at a distance $x$ from the centre is $B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.Given that $B_{center} = 6\sqrt{6} 	imes B_{axis}$,we have:$\frac{\mu_0 I}{2R} = 6\sqrt{6} 	imes \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.Simplifying,we get $1 = 6\sqrt{6} 	imes \frac{R^3}{(R^2 + x^2)^{3/2}}$.$(R^2 + x^2)^{3/2} = 6\sqrt{6} R^3$.Squaring both sides: $(R^2 + x^2)^3 = (6\sqrt{6})^2 R^6 = 216 	imes 6 	imes R^6 = 1296 R^6$.Taking the cube root: $R^2 + x^2 = (1296)^{1/3} R^2 = (6^4)^{1/3} R^2 = 6^{4/3} R^2$.$x^2 = (6^{4/3} - 1) R^2$.Using $R = 8 \ cm$ and $6^{4/3} \approx 10.9027$,$x^2 = (10.9027 - 1) 	imes 64 = 9.9027 	imes 64 = 633.77$.$x = \sqrt{633.77} \approx 25.17 \ cm$. Wait,re-evaluating the ratio: If $B_{center} = n 	imes B_{axis}$,then $(R^2+x^2)^{3/2} = n R^3$. For $n = 6\sqrt{6} = \sqrt{216}$,$(R^2+x^2)^3 = 216 R^6$,so $R^2+x^2 = (216)^{1/3} R^2 = 6 R^2$. Thus $x^2 = 5R^2$,$x = R\sqrt{5}$.$x = 8 	imes 2.236 = 17.888 \approx 17.89 \ cm$.

The magnetic induction at the centre of a current-carrying circular coil of radius $8 \ cm$ is $6 \sqrt{6}$ times the magnetic induction at a point on its axis. Then the distance of the point from the centre of the coil in $cm$ is $(\sqrt{5} = 2.236)$.

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