Question

Two long parallel straight conductors carry currents $I_1$ and $I_2\left(I_1>I_2\right)$. When the currents are in the same direction, the magnetic field at a point midway between the wires is $20mT$. If the direction of $I_2$ is reversed, the field becomes $50mT$. The ratio of the currents $\frac{I_1}{I_2}$ is

• A. $\frac{5}{2}$
• B. $\frac{7}{3}$
• C. $\frac{4}{3}$
• D. $\frac{5}{3}$

Answer 1

Answer: (B)

When the currents $I_1$ and $I_2$ are in the same direction, the magnetic field at point $P$ is
$B_1=\frac{{\mu }_0{I}_1}{2\pi (r/2)}-\frac{{\mu }_0{I}_2}{2\pi (r/2)}$
$20\times 10^{-3}=\frac{{\mu }_0}{\pi r}\left(I_1-I_2\right)\dots (i)$
When the direction of $I_2$ is reversed, the magnetic field at same point $P$ is $B_2=\frac{{\mu }_0}{2\pi }\frac{I_1}{(r/2)}+\frac{{\mu }_0}{2\pi }\frac{I_2}{(r/2)}$
$50\times 10^{-3}=\frac{{\mu }_0}{\pi r}\left(I_1+I_2\right)\dots (ii)$
Divide (i) by (ii), we get
$\frac{2}{5}=\frac{I_1-I_2}{I_1+I_2},2\left(I_1+I_2\right)=5\left(I_1-I_2\right)$
$\frac{2}{5}\left[\frac{I_1}{I_2}+1\right]=\left[\frac{I_1}{I_2}-1\right]$
$\therefore \frac{I_1}{I_2}=\frac{7}{3}$