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(D) Let the distance between the two conductors be $2d$. The magnetic field at the midpoint due to a long straight wire carrying current $I$ at a dist

Vedclass · Accepted Answer

$\frac{3}{7}$

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(D) Let the distance between the two conductors be $2d$. The magnetic field at the midpoint due to a long straight wire carrying current $I$ at a distance $d$ is $B = \frac{\mu_0 I}{2 \pi d}$.When currents are in the same direction,the magnetic fields at the midpoint are in opposite directions. Thus,the net field is $B_1 = \frac{\mu_0}{2 \pi d} (I_1 - I_2) = 20 \mu T$.When the direction of $I_2$ is reversed,the magnetic fields at the midpoint are in the same direction. Thus,the net field is $B_2 = \frac{\mu_0}{2 \pi d} (I_1 + I_2) = 50 \mu T$.Dividing the two equations: $\frac{I_1 + I_2}{I_1 - I_2} = \frac{50}{20} = \frac{5}{2}$.Cross-multiplying: $2(I_1 + I_2) = 5(I_1 - I_2) \implies 2I_1 + 2I_2 = 5I_1 - 5I_2$.Rearranging terms: $7I_2 = 3I_1$,which gives $\frac{I_2}{I_1} = \frac{3}{7}$.

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