Question

Two infinitely long parallel wires carry currents of magnitude $I_1$ and $I_2$ are at a distance $4cm$ apart. The magnitude of the net magnetic field is found to reach a non-zero minimum value between the two wires and $1cm$ away from the first wire. The ratio of the two currents and their mutual direction is

• A. $\frac{I_2}{I_1}=9$, anti-parallel
• B. $\frac{I_2}{I_1}=9$, parallel
• C. $\frac{I_2}{I_1}=3$, anti-parallel
• D. $\frac{I_2}{I_1}3$, parallel

Answer 1

Answer: (A)

Let magnetic field is minimum at some point $P$, distant $x$ from first wire.

Net magnetic field at $P$ is
$B=\frac{{\mu }_0{I}_1}{2\pi x\times 10^{-2}}-\frac{{\mu }_0{I}_2}{2\pi \left(4x\right)\times 10^{-2}}$
$=\frac{{\mu }_0}{2\pi \times 10^{-2}}\left(\frac{I_1}{x}+\frac{I_2}{x-4}\right)$
For $B$ to be minimum,
$\frac{dB}{dx}=0\Rightarrow \frac{d}{dx}\left(\frac{I_1}{x}+\frac{I_2}{x-4}\right)=0$
$\Rightarrow -\frac{I_1}{x^2}-\frac{I_2}{{\left(x-4\right)}^2}=0\Rightarrow -\frac{I_1}{x^2}=\frac{I_2}{{\left(x-4\right)}^2}$
$\Rightarrow \frac{I_1}{I_2}=-{\left(\frac{x}{x-4}\right)}^2$
with $x$ = 1cm, we have
$\frac{I_1}{I_2}=-{\left(\frac{1}{1-4}\right)}^2=-\frac{1}{9}or\frac{I_2}{I_1}=-9$
Here, negative sign shows that they are anti-parallel