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(C) The magnetic moment $M$ of a circular coil is given by the formula $M = NIA$,where $N$ is the number of turns,$I$ is the current,and $A$ is the ar

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$N_A I_A = 4 N_B I_B$

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(C) The magnetic moment $M$ of a circular coil is given by the formula $M = NIA$,where $N$ is the number of turns,$I$ is the current,and $A$ is the area of the coil.Given that the magnetic moments of coils $A$ and $B$ are equal,we have $M_A = M_B$.Substituting the formula,we get $N_A I_A A_A = N_B I_B A_B$.The area of a circular coil is $A = \pi r^2$. Thus,$A_A = \pi (r_A)^2$ and $A_B = \pi (r_B)^2$.Substituting the values $r_A = 10 \ cm = 0.1 \ m$ and $r_B = 20 \ cm = 0.2 \ m$:$N_A I_A \pi (0.1)^2 = N_B I_B \pi (0.2)^2$$N_A I_A (0.01) = N_B I_B (0.04)$Dividing both sides by $0.01$,we get $N_A I_A = 4 N_B I_B$.

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 the number of turns and current of $A$ and $B$ respectively)

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