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(A) For wire $A$ bent into a circle of radius $r$,the circumference is $2 \pi r = 40 \ cm$. Thus,$r = \frac{40}{2 \pi} \ cm$. The magnetic field at th

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$3: 4$

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(A) For wire $A$ bent into a circle of radius $r$,the circumference is $2 \pi r = 40 \ cm$. Thus,$r = \frac{40}{2 \pi} \ cm$. The magnetic field at the center is $B_1 = \frac{\mu_0 i_1}{2 r}$.For wire $B$ bent into an arc of radius $r$,the length of the arc is $l = r 	heta = 30 \ cm$. Thus,the angle subtended is $	heta = \frac{30}{r} = \frac{30}{40 / 2 \pi} = \frac{3}{2} \pi$ radians. The magnetic field at the center is $B_2 = \frac{\mu_0 i_2 	heta}{4 \pi r}$.Given $B_1 = B_2$,we have $\frac{\mu_0 i_1}{2 r} = \frac{\mu_0 i_2 	heta}{4 \pi r}$.Substituting $	heta = \frac{3}{2} \pi$,we get $\frac{i_1}{2} = \frac{i_2 (3/2 \pi)}{4 \pi} = \frac{3 i_2}{8}$.Therefore,$\frac{i_1}{i_2} = \frac{3}{8} 	imes 2 = \frac{3}{4}$.

Two wires $A$ and $B$ are of lengths $40 \ cm$ and $30 \ cm$. $A$ is bent into a circle of radius $r$ and $B$ into an arc of radius $r$. $A$ current $i_1$ is passed through $A$ and $i_2$ through $B$. To have the same magnetic inductions at the centre,the ratio of $i_1: i_2$ is

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