The magnetic induction at the centre of a cur | vedclass

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(C) The magnetic field at the centre of a circular coil is $B_0 = \frac{\mu_0 I}{2R}$.The magnetic field at a distance $x$ from the centre on the axis

Vedclass · Accepted Answer

$20$

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(C) The magnetic field at the centre of a circular coil is $B_0 = \frac{\mu_0 I}{2R}$.The magnetic field at a distance $x$ from the centre on the axis is $B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.We are given that $B_0 = 5\sqrt{5} \cdot B_{axis}$.Substituting the expressions,we get: $\frac{\mu_0 I}{2R} = 5\sqrt{5} \cdot \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.Simplifying,we have: $1 = \frac{5\sqrt{5} R^3}{(R^2 + x^2)^{3/2}}$.Taking the power of $2/3$ on both sides: $1 = \frac{(5\sqrt{5})^{2/3} R^2}{R^2 + x^2}$.Since $5\sqrt{5} = 5^{3/2}$,then $(5^{3/2})^{2/3} = 5^1 = 5$.So,$1 = \frac{5 R^2}{R^2 + x^2}$.$R^2 + x^2 = 5R^2 \implies x^2 = 4R^2 \implies x = 2R$.Given $R = 10 \ cm$,therefore $x = 2 	imes 10 \ cm = 20 \ cm$.

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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