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(A) The magnetic field at the centre of a circular coil carrying current $i$ and having radius $r$ is given by $B = \frac{\mu_0 i}{2r}$.Since the two

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$1 : \sqrt{2}$

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(A) The magnetic field at the centre of a circular coil carrying current $i$ and having radius $r$ is given by $B = \frac{\mu_0 i}{2r}$.Since the two coils are similar and carry the same current,the magnitude of the magnetic field produced by each coil at the centre is the same,let it be $B$.$B_1 = B_2 = B = \frac{\mu_0 i}{2r}$.Since the coils are mutually perpendicular,the magnetic field vectors $B_1$ and $B_2$ are perpendicular to each other.The resultant magnetic field $B_{net}$ at the centre is given by the vector sum of $B_1$ and $B_2$:$B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{B^2 + B^2} = \sqrt{2B^2} = B\sqrt{2}$.We need to find the ratio of the magnetic field due to one coil $(B)$ to the resultant magnetic field $(B_{net})$:Ratio $= \frac{B}{B_{net}} = \frac{B}{B\sqrt{2}} = \frac{1}{\sqrt{2}}$.Thus,the ratio is $1 : \sqrt{2}$.

Two similar coils are kept mutually perpendicular such that their centres coincide. At the centre,find the ratio of the magnetic field due to one coil and the resultant magnetic field by both coils,if the same current is flowing through them.

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