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(C) Let the distance between the wires be $2d$. The magnetic field due to a long wire at a distance $d$ is $B = \frac{\mu_0 I}{2\pi d}$.When currents

Vedclass · Accepted Answer

$3: 5$

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(C) Let the distance between the wires be $2d$. The magnetic field due to a long wire at a distance $d$ is $B = \frac{\mu_0 I}{2\pi d}$.When currents are in the same direction,the fields at the midpoint oppose each other: $B_1 - B_2 = \frac{\mu_0}{2\pi d} (I_1 - I_2) = 8 	imes 10^{-6} \ T$ (Equation $1$).When the direction of $I_2$ is reversed,the fields add up: $B_1 + B_2 = \frac{\mu_0}{2\pi d} (I_1 + I_2) = 3.2 	imes 10^{-5} \ T$ (Equation $2$).Dividing Equation $1$ by Equation $2$: $\frac{I_1 - I_2}{I_1 + I_2} = \frac{8 	imes 10^{-6}}{32 	imes 10^{-6}} = \frac{1}{4}$.Cross-multiplying gives $4I_1 - 4I_2 = I_1 + I_2$,which simplifies to $3I_1 = 5I_2$.Therefore,the ratio $\frac{I_2}{I_1} = \frac{3}{5}$.

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