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(A) The magnetic field at the center of a circular ring of radius $R$ carrying current $I$ is given by $B_0 = \frac{\mu_0 I}{2R}$.Since the three ring

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$\sqrt{3} \frac{\mu_0 I}{2R}$

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(A) The magnetic field at the center of a circular ring of radius $R$ carrying current $I$ is given by $B_0 = \frac{\mu_0 I}{2R}$.Since the three rings are mutually perpendicular,their magnetic field vectors at the origin will be along the $x$,$y$,and $z$ axes respectively.Let the magnetic fields due to the rings in the $yz$,$zx$,and $xy$ planes be $\vec{B}_x$,$\vec{B}_y$,and $\vec{B}_z$ respectively.Thus,$\vec{B}_x = B_0 \hat{i}$,$\vec{B}_y = B_0 \hat{j}$,and $\vec{B}_z = B_0 \hat{k}$.The resultant magnetic field $\vec{B}$ at the origin is $\vec{B} = \vec{B}_x + \vec{B}_y + \vec{B}_z = B_0 \hat{i} + B_0 \hat{j} + B_0 \hat{k}$.The magnitude of the resultant magnetic field is $B = \sqrt{B_0^2 + B_0^2 + B_0^2} = \sqrt{3 B_0^2} = \sqrt{3} B_0$.Substituting $B_0 = \frac{\mu_0 I}{2R}$,we get $B = \sqrt{3} \frac{\mu_0 I}{2R}$.

Three rings,each having equal radius $R$,are placed mutually perpendicular to each other and each having its centre at the origin of the coordinate system. If current $I$ is flowing through each ring,then the magnitude of the magnetic field at the common centre is

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