Area of the coil initially, $$A = \pi R^{2}$$

Initial magnetic moment, $$\mu = NIA = NI(\pi R^{2})$$

Length of the coil initially, L = L= $$N(2\pi R)$$

Radius of new coil formed,$${R}’ = \frac{R}{2}$$

Let the number of turns in new coil be$${N}’$$

But the length of the new coil formed must be equal to that of the original coil $${L}’ = L$$

Therefore, $${N}’ * 2\pi \frac{R}{2} = N * 2\pi R$$ $$\Rightarrow {N}’ = 2N$$

Thus area of new coil $${A}’ = {A}’ = \frac{\pi R^{2}}{4}$$

Therefore, the magnetic moment of new coil, $${\mu }’ = {N}’I{A}’$$ $$\Rightarrow 2NI * \frac{\pi R^{2}}{4}$$ $$\Rightarrow \frac{NI\pi R^{2}}{2}$$ $$\Rightarrow \frac{\mu }{{\mu }’} = \frac{1}{2}$$

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