91Ó°ÊÓ

1. Separation distance is$d=16\text{cm}$
2. Current$i_1=3.61\text{mA}$
3. Current$i_2=3.0i_1\text{mA}$

Use the equation of magnetic field by long straight wire carrying current to solve this problem. By equating this equation for both wires, find the position of point of zero magnetic field.

$\mathit{B}\mathbf{=}\frac{\mathbf{μ\pm õ}}{\mathbf{2}\mathbf{Ï€¸é}}$

As both wires carry current in the same direction, the magnetic field can cancel in the region between them.

Let us assume that x is the distance from wire 1 where magnetic field is zero. So, magnetic field is as follows

$B_1=\frac{{\mathrm{μ}}_0{I}_1}{2\mathrm{π}x}$

Now for wire 2 it is as follows

$B_2=\frac{{\mathrm{μ}}_0{I}_2}{2\mathrm{π}\left(d-x\right)}$

As net magnetic field is zero

$$\begin{gathered} B_1=B_2 \\ \frac{{\mathrm{μ}}_0{I}_1}{2\mathrm{π}x}=\frac{{\mathrm{μ}}_0{I}_2}{2\mathrm{π}\left(d-x\right)} \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} \frac{3.61\text{mA}}{x}=\frac{3.0×3.61\text{mA}}{16\text{cm}-x} \\ \left(16\text{cm}-x\right)3.61=3×3.61x \\ 16\text{cm}-x=3x \\ x=4.0\text{cm} \end{gathered}$$

So, magnetic field is zero at $4.0\text{cm}$from wire 1.

Now in second part, the current is doubled

So

$\frac{{\mathrm{μ}}_0{I}_1}{2\mathrm{π}x}=\frac{{\mathrm{μ}}_0{I}_2}{2\mathrm{π}\left(d-x\right)}$

Substitute the values and solve as:

$$\begin{gathered} \frac{3.61×2}{x}=\frac{3×3.61×2}{d-x} \\ \frac{1}{x}=\frac{3}{16\text{cm}-x} \\ x=4.0\text{cm} \end{gathered}$$

So increase in current does not affect the position of zero potential point.