Question

In Fig. 29-43, two long straight wires at separation $\mathit{d}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$carry currents ${\mathbf{i}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{61}\mathbf{}\mathbf{\text{mA}}$and ${\mathit{i}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{00}{\mathit{i}}_{\mathbf{1}}$out of the page. (a) Where on the x axis is the net magnetic field equal to zero? (b) If the two currents are doubled, is the zero-field point shifted toward wire 1, shifted toward wire 2, or unchanged?

Short answer

1. The magnetic field is zero for $x=4.0\text{cm}$.
2. There is no change in the zero point if the current is doubled.

Step-by-step solution

Given

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

Determine the formulas:

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.

$\mathit{B}\mathbf{=}\frac{\mathbf{\mu I}}{\mathbf{2}\mathbf{\pi R}}$

(a) Calculate where on the x axis is the net magnetic field equal to zero

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

$B_1=\frac{{\mu }_0{I}_1}{2\pi x}$

Now for wire 2 it is as follows

$B_2=\frac{{\mu }_0{I}_2}{2\pi \left(d-x\right)}$

As net magnetic field is zero

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

Substitute the values and solve as:

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

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

(b) Find out if the zero-field point shifted toward wire 1, shifted toward wire 2, or unchanged with the two currents are doubled

Now in second part, the current is doubled

So

$\frac{{\mu }_0{I}_1}{2\pi x}=\frac{{\mu }_0{I}_2}{2\pi \left(d-x\right)}$

Substitute the values and solve as:

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

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