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(A) The torque (couple) on a current-carrying loop in a magnetic field is given by $C = NiAB \sin 	heta$.Since the current $i$,magnetic field $B$,and

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Couple on loop $A$ is more than on loop $B$

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(A) The torque (couple) on a current-carrying loop in a magnetic field is given by $C = NiAB \sin 	heta$.Since the current $i$,magnetic field $B$,and angle $	heta$ are the same for both,the torque is proportional to the product of the number of turns $N$ and the area $A$ of the loop: $C \propto N \cdot A$.Let $L$ be the length of the wire. For loop $A$ with $N_A = 1$ turn,$L = 2 \pi r_A$,so $r_A = L / (2 \pi)$. The area $A_A = \pi r_A^2 = \pi (L / 2 \pi)^2 = L^2 / (4 \pi)$.For loop $B$ with $N_B = 2$ turns,$L = 2 	imes (2 \pi r_B)$,so $r_B = L / (4 \pi)$. The area $A_B = \pi r_B^2 = \pi (L / 4 \pi)^2 = L^2 / (16 \pi)$.Now,calculate the ratio of the torques:$\frac{C_A}{C_B} = \frac{N_A A_A}{N_B A_B} = \frac{1 	imes (L^2 / 4 \pi)}{2 	imes (L^2 / 16 \pi)} = \frac{1/4}{2/16} = \frac{1/4}{1/8} = 2$.Thus,$C_A = 2 C_B$,which means the couple on loop $A$ is twice that on loop $B$,or the couple on loop $A$ is greater than on loop $B$.

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