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Two circular coils X and Y having equal number of turns and carrying equal current in same sense are placed co-axially, such that they subtend the same solid angle at O. Let B be the magnetic field at “O” due to smaller coil x, then

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Magnetic field at ‘O’ due to larger coil= $2 B$

Magnetic field at ‘O’ due to larger coil $= \frac{B}{2}$

Total magnetic field at ‘O’ is $\frac{3 B}{2}$

Total magnetic field at ‘O’ is $\frac{\sqrt{5} B}{2}$

answer is B, C.

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Detailed Solution

Due to coil 1 at ’O’ field is $B = \frac{μ_{0} n i r^{2}}{2 {(r^{2} + a^{2})}^{\frac{3}{2}}}$

Due to coil 2 at 'O' field is
$\begin{array}{l} B^{1} = \frac{μ_{0} n i {(2 r)}^{2}}{2 {(2 r)}^{2} + {(2 a^{2})}^{3 / 2}} \\ B^{1} = \frac{B}{2} \end{array}$

$T o t a l f i e l d a t' o' i s B + \frac{B}{2} = \frac{3 B}{2}$

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