The magnetic field on the axis of a circular | vedclass

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Question

(A) The magnetic field on the axis of a circular loop at a distance $x$ from the center is given by $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.Given

Vedclass · Accepted Answer

$3.14 	imes 10^{-7}\,T$

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(A) The magnetic field on the axis of a circular loop at a distance $x$ from the center is given by $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.Given: $R = 100\,cm = 1\,m$,$x = 1\,m$,$I = \sqrt{2}\,A$,and $\mu_0 = 4\pi 	imes 10^{-7}\,T\cdot m/A$.Substituting the values:$B = \frac{4\pi 	imes 10^{-7} 	imes \sqrt{2} 	imes (1)^2}{2(1^2 + 1^2)^{3/2}}$$B = \frac{4\pi 	imes 10^{-7} 	imes \sqrt{2}}{2(2)^{3/2}}$Since $(2)^{3/2} = 2\sqrt{2}$,we have:$B = \frac{4\pi 	imes 10^{-7} 	imes \sqrt{2}}{2 	imes 2\sqrt{2}}$$B = \frac{4\pi 	imes 10^{-7}}{4} = \pi 	imes 10^{-7}\,T$$B = 3.14 	imes 10^{-7}\,T$.

The magnetic field on the axis of a circular loop of radius $R = 100\,cm$ carrying current $I = \sqrt{2}\,A$,at a point $x = 1\,m$ away from the centre of the loop is given by:

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