Solution:
Solution
Let the length of the wire be $$L$$ and radius of the coil be $$r$$
$$\therefore$$ $$L = 2\pi r$$
Magnetic field at the centre of coil $$B = \dfrac{\mu_o I}{2r}$$ ........(1)
Now the same wire is bent into circular loops of n turns.
$$\therefore$$ $$L =n 2\pi r' $$
$$2\pi r = 2n\pi r'$$ $$\implies r' = \dfrac{r}{n}$$ ........(2)
Magnetic field at the centre of circular loops of n turns $$B' = \dfrac{\mu_o I}{4\pi r'} \times 2\pi n = \dfrac{\mu_o I}{2r'}n$$
Using equation (2), we get $$B' = \dfrac{\mu_o I}{2 \frac{r}{n}}n = \dfrac{\mu_o I}{2 r} n^2$$
$$\implies$$ $$B' = n^2 B$$