Question

In Figure, two long straight wires are perpendicular to the page and separated by distance ${\mathit{d}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{}\mathbf{\text{cm}}$. Wire 1 carries $\mathbf{6}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\text{A}}$into the page. What are (a) magnitude and (b) direction (into or out of the page) of the current in wire 2 if the net magnetic field due to the two currents is zero at point P located at distance ${\mathit{d}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\text{cm}}$from wire 2?

Short answer

1. The value of the current is $4.3\text{A}$.
2. The direction of the current is out of the page.

Step-by-step solution

Given

1. Wire separation is$d_1=0.75\text{cm}=7.5\times {10}^{-3}\text{m}$
2. Wire 1 carries current$6.5\text{A}$
3. $d_2=1.50\text{cm}=1.50\times {10}^{-2}\text{m}$
   

Understanding the concept

Use the formula for magnetic field at a certain point away from the semi-infinite wire as follows:

$\mathit{B}\mathbf{=}\frac{\mathit{\mu }\mathit{I}}{\mathbf{4}\mathit{\pi }\mathit{R}}$

Given the net magnetic field is zero which means that the magnetic field by wire 1 must cancel with magnetic field by wire 2.

$\mathit{B}\mathbf{=}\frac{{\mathit{\mu }}_{\mathbf{0}}\mathbf{\times }\mathit{I}}{\mathbf{4}\mathit{\pi }\mathit{R}}$

(a) Calculate the magnitude of the current in wire 2

Magnetic field at a certain point from the wire is as follows

$B=\frac{{\mu }_{0}I}{4\pi R}$

So, for wire 1 magnetic field at point p is as follows

$B_1=\frac{{\mu }_0{I}_1}{4\pi \left(d_1+d_2\right)}$

So, for wire 2 magnetic field at point p is as follows

$B_2=\frac{{\mu }_0{I}_2}{4\pi \left(d_2\right)}$

The net magnetic field is zero means

$$\begin{gathered} B_1=B_2 \\ \frac{{\mu }_0{I}_1}{4\pi \left(d_1+d_2\right)}=\frac{{\mu }_0{I}_2}{4\pi d_2} \end{gathered}$$

Substitute the values and solve:

$$\begin{gathered} \frac{{\mu }_0\times 6.5}{4\pi \left(7.5\times {10}^{-3}+1.50\times {10}^{-2}\right)}=\frac{{\mu }_0\times I_2}{4\pi \times 1.50\times {10}^{-2}} \\ {I}_2=4.3\text{A} \end{gathered}$$

hence the current is, $4.3\text{A}$.

(b) Calculate the magnitude of the current in wire 2

Direction of current $I_2$is out of the page or opposite $I_1$ to create an opposite magnetic field. That is out of the page.