The magnetic field at the center of a current | vedclass

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(A) The magnetic field at the center of a current-carrying loop is given by $B_{C} = \frac{\mu_{0}I}{2r}$.The magnetic field at a point on the axis at

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(A) The magnetic field at the center of a current-carrying loop is given by $B_{C} = \frac{\mu_{0}I}{2r}$.The magnetic field at a point on the axis at a distance $x$ from the center is given by $B_{A} = \frac{\mu_{0}Ir^{2}}{2(x^{2} + r^{2})^{3/2}}$.Given that $B_{C} = 5\sqrt{5} B_{A}$,we substitute the expressions:$\frac{\mu_{0}I}{2r} = 5\sqrt{5} \left( \frac{\mu_{0}Ir^{2}}{2(x^{2} + r^{2})^{3/2}} ight)$.Simplifying the equation:$\frac{1}{r} = \frac{5\sqrt{5}r^{2}}{(x^{2} + r^{2})^{3/2}}$.$(x^{2} + r^{2})^{3/2} = 5\sqrt{5}r^{3} = (5^{1/2})^{3}r^{3} = (5^{1/2}r)^{3}$.Taking the cube root on both sides:$(x^{2} + r^{2})^{1/2} = 5^{1/2}r$.Squaring both sides:$x^{2} + r^{2} = 5r^{2}$.$x^{2} = 4r^{2} \Rightarrow x = 2r$.Given $r = 0.1 \ m$,we get $x = 2 	imes 0.1 \ m = 0.2 \ m$.

The magnetic field at the center of a current-carrying loop of radius $0.1 \ m$ is $5\sqrt{5}$ times that at a point along its axis. The distance of this point from the center of the loop is: (in $m$)

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