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

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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 current-carrying loop of radius $R$ is given by $B = \frac{\mu_0 I}{2R}$.Since the three rings are mutually perpendicular and centered at the origin,their magnetic field vectors will be directed 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,$\vec{B}_x = \frac{\mu_0 I}{2R} \hat{i}$,$\vec{B}_y = \frac{\mu_0 I}{2R} \hat{j}$,and $\vec{B}_z = \frac{\mu_0 I}{2R} \hat{k}$.The resultant magnetic field $\vec{B}$ at the origin is the vector sum: $\vec{B} = \vec{B}_x + \vec{B}_y + \vec{B}_z = \frac{\mu_0 I}{2R} (\hat{i} + \hat{j} + \hat{k})$.The magnitude of the resultant magnetic field is $|\vec{B}| = \sqrt{(\frac{\mu_0 I}{2R})^2 + (\frac{\mu_0 I}{2R})^2 + (\frac{\mu_0 I}{2R})^2} = \sqrt{3 \left(\frac{\mu_0 I}{2R}\right)^2} = \sqrt{3} \frac{\mu_0 I}{2R}$.

Three rings,each having equal radius $R$,are placed mutually perpendicular to each other,and each has 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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