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

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$\sqrt{2}: 1$

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(A) The magnetic field produced by a circular coil of radius $r$ carrying current $I$ at its centre is given by $B = \frac{\mu_0 I}{2r}$.Since the two coils are identical and carry the same current,the magnitude of the magnetic field due to each coil at the common centre is the same,i.e.,$B_1 = B_2 = B$.Since the planes of the coils are at right angles to each other,the magnetic field vectors $B_1$ and $B_2$ are mutually perpendicular.The resultant magnetic field $B_{	ext{net}}$ at the centre is given by the vector sum:$B_{	ext{net}} = \sqrt{B_1^2 + B_2^2}$Substituting $B_1 = B_2 = B$:$B_{	ext{net}} = \sqrt{B^2 + B^2} = \sqrt{2B^2} = \sqrt{2}B$The ratio of the magnitude of the resultant magnetic field to the magnetic field due to one of the coils is:$\frac{B_{	ext{net}}}{B} = \frac{\sqrt{2}B}{B} = \frac{\sqrt{2}}{1}$Therefore,the ratio is $\sqrt{2}: 1$.

Two identical coils having the same number of turns and carrying equal current have a common centre,and their planes are at right angles to each other. What is the ratio of the magnitude of the resultant magnetic field at the centre to the magnetic field due to one of the coils at the centre?

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