Question

Figure 29-49 shows two very long straight wires (in cross section) that each carry a current of$\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{A}$ directly out of the page. Distance ${\mathbf{d}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$ and distance ${\mathbf{d}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$. What is the magnitude of the net magnetic field at point P, which lies on a perpendicular bisector to the wires?

Short answer

Magnitude of the net magnetic field at point P is $256\text{nT}$.

Step-by-step solution

Given

1. Permeability of free space, ${\mu }_0=4\pi \times {10}^{-7}\frac{\text{Tm}}{\text{A}}$.
2. $d_1=6.00\text{m}$.
3. $d_2=4.00\text{m}$.
4. Current carried by wire,$i=4.00\text{A}$.

Understanding the concept

By using the concept of magnetic field due to long wire carrying current and component of magnetic field, determine the net magnetic field at point P.

Formula:

Magnetic field due to the wire at point P is

$B=\frac{{\mu }_{0}i}{2\pi r}$

Here, $i=\mathrm{current}$, ${\mu }_0=$permeability of free space,$r=$distance of wire.

Calculate the magnitude of the net magnetic field at point P.

First of all, we have to find the r.

By using Pythagoras theorem, solve as:

$r=\sqrt{{\left(\frac{d_1}{2}\right)}^2+{\left(d_2\right)}^2}$

$r=\sqrt{{\left(\frac{6}{2}\right)}^2+{\left(4\right)}^2}$

$$\begin{gathered} r=\sqrt{25} \\ r=5.00\text{m} \end{gathered}$$

Now, the magnetic field is given by

$B=\frac{{\mu }_{0}i}{2\pi r}$

$B=\frac{4\pi \times {10}^{-7}\times 4.0}{2\times 3.14\times 5.00}$

$B=1.6\times {10}^{-7}\text{T}$

$B=160\text{nT}$

The component of the magnetic field cancels with each other, so the magnetic field at point P is

$B_p=2Bsin\theta$

Here,

$sin\theta =\frac{d_2}{r}=\frac{4.00}{5.00}=0.8$

$B_p=2\times 160\times 0.8$

$B_p=256\text{nT}$