Question

Two parallel long wires carry currents $i_1$ and $i_2$ with $i_1>i_2$. When the currents are in the same direction, the magnetic field midway between the wires is $10\mu T$. When the direction of $i_2$ is reversed, it becomes $40\mu T$. Then, the ratio of $i_1/i_2$ is

• A. $3:4$
• B. $5:3$
• C. $7:11$
• D. $11:7$

Answer 1

The correct option is B $5:3$

When the current in the wires is in same direction. Magnetic field at mid point $O$ due to $I$ and $II$ wires are respectively
$B_I=\frac{{\mu }_0}{4\pi }\frac{2i_1}{x}\otimes ;B_{II}=\frac{{\mu }_0}{4\pi }\frac{2i_2}{x}\oslash$
So, the net magnetic field at $O$
$B_{net}=\frac{{\mu }_0}{4\pi }\times \frac{2}{x}(i_1-i_2)$
$\Rightarrow 10\times {10}^{-6}=\frac{{\mu }_0}{4\pi }\frac{2}{x}(i_1-i_2)...\left(i\right)$
when the direction of $i_2$ is reversed
$B_I=\frac{{\mu }_0}{4\pi }\frac{2i_1}{x}\otimes ;B_{II}=\frac{{\mu }_0}{4\pi }\frac{2i_2}{x}\otimes$
so, net magnetic field at $O$
$B_{net}=\frac{{\mu }_0}{4\pi }\times \frac{2}{x}(i_1+i_2)$
$\Rightarrow 40\times {10}^{-6}=\frac{{\mu }_0}{4\pi }\frac{2}{x}(i_1+i_2)...\left(ii\right)$
Dividing EQ. (ii) by (i) we get
$\frac{i_1+i_2}{i_1-i_2}=\frac{4}{1}\Rightarrow \frac{i_1}{i_2}=\frac{5}{3}$