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Q21P

Expert-verifiedFound in: Page 858

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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?**

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

- Permeability of free space, ${\mu}_{0}=4\pi \times {10}^{-7}\frac{\text{Tm}}{\text{A}}$.
- ${d}_{1}=6.00\text{m}$.
- ${d}_{2}=4.00\text{m}$.
- Current carried by wire, $i=4.00\text{A}$.

**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.**

First of all, we have to find the r.

By using Pythagoras theorem, solve as:

$r=\sqrt{{\left(\frac{{d}_{1}}{2}\right)}^{2}+{\left({d}_{2}\right)}^{2}}$

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

$r=\sqrt{25}\phantom{\rule{0ex}{0ex}}r=5.00\text{m}$

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,

$sin\theta =\frac{{d}_{2}}{r}=\frac{4.00}{5.00}=0.8$

${B}_{p}=2\times 160\times 0.8$

${B}_{p}=256\text{nT}$

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