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(D) The magnetic field at the center of a circular coil of radius $R$ with $N$ turns carrying current $I$ is given by $B = \frac{\mu_0 NI}{2R}$.For th

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$\frac{\mu_0 NI}{R}$

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(D) The magnetic field at the center of a circular coil of radius $R$ with $N$ turns carrying current $I$ is given by $B = \frac{\mu_0 NI}{2R}$.For the first coil with current $I_1 = I$, the magnetic field is $B_1 = \frac{\mu_0 NI}{2R}$.For the second coil with current $I_2 = I\sqrt{3}$, the magnetic field is $B_2 = \frac{\mu_0 N(I\sqrt{3})}{2R} = \sqrt{3} \frac{\mu_0 NI}{2R}$.Since the coils are placed perpendicular to each other, the magnetic fields $B_1$ and $B_2$ are also perpendicular to each other $(	heta = 90^{\circ})$.The resultant magnetic field $B_{	ext{net}}$ is given by:$B_{	ext{net}} = \sqrt{B_1^2 + B_2^2}$Substituting the values:$B_{	ext{net}} = \sqrt{\left(\frac{\mu_0 NI}{2R}ight)^2 + \left(\sqrt{3} \frac{\mu_0 NI}{2R}ight)^2}$$B_{	ext{net}} = \frac{\mu_0 NI}{2R} \sqrt{1^2 + (\sqrt{3})^2}$$B_{	ext{net}} = \frac{\mu_0 NI}{2R} \sqrt{1 + 3} = \frac{\mu_0 NI}{2R} \sqrt{4} = 2 \left(\frac{\mu_0 NI}{2R}ight) = \frac{\mu_0 NI}{R}$.

Two identical coils of radius $R$ and number of turns $N$ are placed perpendicular to each other such that they have a common center. The currents through them are $I$ and $I\sqrt{3}$. The resultant intensity of the magnetic field at the center of the coils will be:

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