The magnetic moments associated with two closely wound circular coils $$A$$ and $$B$$ of radius $$r_A = 10$$ cm and $$r_B = 20$$ cm respectively are equal if: (Where $$N_A$$, $$I_A$$ and $$N_B$$, $$I_B$$ are number of turn and current of $$A$$ and $$B$$ respectively)

We need to find the condition for equal magnetic moments of two circular coils A and B.

Magnetic moment of a coil:

$$M = NIA = NI\pi r^2$$

For coil A: $$M_A = N_A I_A \pi r_A^2 = N_A I_A \pi (0.1)^2 = 0.01\pi N_A I_A$$

For coil B: $$M_B = N_B I_B \pi r_B^2 = N_B I_B \pi (0.2)^2 = 0.04\pi N_B I_B$$

Setting $$M_A = M_B$$:

$$0.01\pi N_A I_A = 0.04\pi N_B I_B$$

$$N_A I_A = 4 N_B I_B$$

The correct answer is Option 3: $$N_A I_A = 4N_B I_B$$.