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(A) The magnetic field at the center of a circular coil carrying current $I$ with radius $R$ is given by $B = \frac{\mu_0 I}{2R}$.Since the two coils

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$\mu_0$

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(A) The magnetic field at the center of a circular coil carrying current $I$ with radius $R$ is given by $B = \frac{\mu_0 I}{2R}$.Since the two coils are identical and their planes are perpendicular,the magnetic fields produced by them at the center,$B_1$ and $B_2$,will be equal in magnitude: $B_1 = B_2 = \frac{\mu_0 I}{2R}$.Because the planes are perpendicular,the magnetic field vectors $B_1$ and $B_2$ are also perpendicular to each other.The net magnetic field $B_{net}$ is given by the vector sum: $B_{net} = \sqrt{B_1^2 + B_2^2}$.Substituting $B_1 = B_2 = B$,we get $B_{net} = \sqrt{B^2 + B^2} = \sqrt{2} B$.Substituting the value of $B$: $B_{net} = \sqrt{2} \left( \frac{\mu_0 I}{2R} ight) = \frac{\mu_0 I}{\sqrt{2} R}$.Given $I = \sqrt{2} 	ext{ A}$ and $R = 1 	ext{ m}$,we have $B_{net} = \frac{\mu_0 	imes \sqrt{2}}{\sqrt{2} 	imes 1} = \mu_0$.

Two identical current-carrying coils with the same center are placed with their planes perpendicular to each other. If current $I = \sqrt{2} \text{ A}$ and the radius of the coil is $R = 1 \text{ m}$,then the magnetic field at the center is equal to (where $\mu_0$ is the permeability of free space).

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