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(A) For wire $A$, the magnetic field at the center of a circular loop is $B_1 = \frac{\mu_0 i_1}{2r}$.Given the length of wire $A$ is $40 \ cm$, the c

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$3: 4$

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(A) For wire $A$, the magnetic field at the center of a circular loop is $B_1 = \frac{\mu_0 i_1}{2r}$.Given the length of wire $A$ is $40 \ cm$, the circumference is $2\pi r = 40 \ cm$, so $r = \frac{40}{2\pi}$.For wire $B$, the magnetic field at the center of an arc of radius $r$ subtending an angle $	heta$ is $B_2 = \frac{\mu_0 i_2 	heta}{4\pi r}$.The length of wire $B$ is $30 \ cm$, so the arc length $s = r	heta = 30 \ cm$.Thus, $	heta = \frac{30}{r} = \frac{30}{40/2\pi} = \frac{3}{2}\pi$.Setting $B_1 = B_2$:$\frac{\mu_0 i_1}{2r} = \frac{\mu_0 i_2 	heta}{4\pi r}$$\frac{i_1}{2} = \frac{i_2 	heta}{4\pi} = \frac{i_2 (3\pi/2)}{4\pi} = \frac{3i_2}{8}$$\frac{i_1}{i_2} = \frac{3}{8} 	imes 2 = \frac{3}{4}$.Therefore, the ratio $i_1: i_2$ is $3: 4$.

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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