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 $20 \,mT$. If the direction of $I_{2}$ is reversed, the field becomes $50 \,mT$. 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}$