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(C) The magnetic field at the centre of a circular coil is given by the formula: $B = \frac{\mu_{0} n i}{2 r}$.Here,$n$ is the number of turns,$i$ is

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become $4$ times

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(C) The magnetic field at the centre of a circular coil is given by the formula: $B = \frac{\mu_{0} n i}{2 r}$.Here,$n$ is the number of turns,$i$ is the current,and $r$ is the radius of the coil.From the formula,we can see that $B \propto \frac{n}{r}$.Let the initial magnetic field be $B = k \frac{n}{r}$.Given that the new number of turns $n' = 2n$ and the new radius $r' = \frac{r}{2}$.The new magnetic field $B'$ is given by $B' = k \frac{n'}{r'} = k \frac{2n}{r/2} = 4 \left( k \frac{n}{r} \right) = 4B$.Therefore,the magnetic field becomes $4$ times the original value.

If the radius of a coil is halved and the number of turns is doubled,then the magnetic field at the centre of the coil,for the same current,will

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