Question

Two infinitely long parallel wires carry currents of magnitude $I_1$ and $I_2$ are at a distance $4\, cm$ apart. The magnitude of the net magnetic field is found to reach a non-zero minimum value between the two wires and $1\, cm$ 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
$B=\frac{\mu_{0}I_{1}}{2\pi x\times10^{-2}}-\frac{\mu_{0}I_{2}}{2\pi\left(4x\right)\times10^{-2}}$
$=\frac{\mu_{0}}{2\pi\times10^{-2}}\left(\frac{I_{1}}{x}+\frac{I_{2}}{x-4}\right)$
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