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(D) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 n I}{2r}$,where $n$ is the number of turns,$I$ is the current,and

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$1: 4$

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(D) The magnetic field at the centre of a circular coil is given by $B = \frac{\mu_0 n I}{2r}$,where $n$ is the number of turns,$I$ is the current,and $r$ is the radius of the coil.Let $L$ be the total length of the wire. For a coil with $n$ turns and radius $r$,$L = n(2\pi r)$,which implies $r = \frac{L}{2\pi n}$.Substituting $r$ into the magnetic field formula: $B = \frac{\mu_0 n I}{2(L / 2\pi n)} = \frac{\mu_0 n^2 I \pi}{L}$.Since the length of the wire $L$ and the current $I$ are constant,$B \propto n^2$.Therefore,the ratio of the magnetic fields is $\frac{B_1}{B_2} = \left( \frac{n_1}{n_2} \right)^2$.Given $n_1 = 5$ and $n_2 = 10$,the ratio is $\frac{B_1}{B_2} = \left( \frac{5}{10} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.

$A$ wire is first bent in the form of a circular coil of $5$ turns and the same wire is then bent in the form of another circular coil of $10$ turns. If the same current is passed in both the coils,then the ratio of the magnetic fields at their centres is

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