Question

In unit-vector notation, what is the magnetic field at pointPin Fig. 29-86 if$\mathit{i}\mathbf{=}\mathbf{10}\mathbf{ }\mathbf{\text{A}}$and$\mathit{a}\mathbf{}\mathbf{=}\mathbf{}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{ }\mathbf{\text{cm}}$? (Note that the wires are notlong.)

Short answer

The net magnetic field at point P is $\overrightarrow{B}=-\left(2.0\times {10}^{-4} \text{T}\right)\overrightarrow{\mathrm{k}}$.

Step-by-step solution

Step 1: Identification of the given data

The current through wire is,$i=10 \text{A}$

The length of wire is,$a=10 \text{cm}$

Understanding the concept

For this problem, we will use the equation for magnetic field due to finite wire of length L, carrying currenti, at a point P above the right end of the wire, and at distanceR from the wire. Direction of the magnetic field will depend on the direction of the current;we can determine this by right hand thumb rule.

Formula:

Magnetic field due to finite wire of length L, carrying currenti, at a point P above the right end of the wire, and at distance R from the wire is

$B_p=\frac{{\mu }_{0}i}{4\pi R}·\left(\frac{L}{\sqrt{L^2+R^2}}\right)$

Calculate the net magnetic field at point P

We break the wire loop in eight parts as shown in the figure.We can find the magnetic field due to each part at point P

Due to part 1 magnetic field is $B_1$

$$\begin{gathered} B_1=\frac{{\mu }_{0}i}{\frac{4\pi a}{4}}·\frac{\frac{a}{4}}{{\left({\left(\frac{a}{4}\right)}^2+{\left(\frac{a}{4}\right)}^2\right)}^{\frac{1}{2}}} \\ {B}_1=-\frac{{\mu }_{0}i}{\sqrt{2}a\pi }\overrightarrow{\mathrm{k}} \end{gathered}$$

Due to part 2 magnetic field is $B_2$

$$\begin{gathered} B_2=\frac{{\mu }_{0}i}{\frac{4\pi ·a}{4}}·\frac{\frac{3a}{4}}{{\left({\left(\frac{3a}{4}\right)}^2+{\left(\frac{a}{4}\right)}^2\right)}^{\frac{1}{2}}} \\ {B}_2=-\frac{3{\mu }_{0}i}{\sqrt{10}a\pi }\overrightarrow{\mathrm{k}} \end{gathered}$$

Due to part 3 magnetic field is $B_3$

$$\begin{gathered} B_3=\frac{{\mu }_{0}i}{\frac{4\pi ·3a}{4}}·\frac{\frac{a}{4}}{{\left({\left(\frac{3a}{4}\right)}^2+{\left(\frac{a}{4}\right)}^2\right)}^{\frac{1}{2}}} \\ {B}_3=-\frac{{\mu }_{0}i}{3\sqrt{10}a\pi }\overrightarrow{\mathrm{k}} \end{gathered}$$

Due to part 4 magnetic field is $B_4$

$$\begin{gathered} B_4=\frac{{\mu }_{0}i}{\frac{4\pi ·3a}{4}}·\frac{\frac{3a}{4}}{{\left({\left(\frac{3a}{4}\right)}^2+{\left(\frac{3a}{4}\right)}^2\right)}^{\frac{1}{2}}} \\ {B}_4=-\frac{\sqrt{2}{\mu }_{0}i}{6a\pi }\overrightarrow{\mathrm{k}} \end{gathered}$$

Due to part 5 magnetic field is $B_5$

$$\begin{gathered} B_5=\frac{{\mu }_{0}i}{\frac{4\pi ·3a}{4}}·\frac{\frac{3a}{4}}{{\left({\left(\frac{3a}{4}\right)}^2+{\left(\frac{3a}{4}\right)}^2\right)}^{\frac{1}{2}}} \\ {B}_5=-\frac{\sqrt{2}{\mu }_{0}i}{6a\pi }\overrightarrow{\mathrm{k}} \end{gathered}$$

Due to part 6 magnetic field is $B_6$

$$\begin{gathered} B_6=\frac{{\mu }_{0}i}{\frac{4\pi ·3a}{4}}·\frac{\frac{a}{4}}{{\left({\left(\frac{3a}{4}\right)}^2+{\left(\frac{a}{4}\right)}^2\right)}^{\frac{1}{2}}} \\ {B}_6=-\frac{{\mu }_{0}i}{3\sqrt{10}a\pi }\overrightarrow{\mathrm{k}} \end{gathered}$$

The net magnetic field at point Pthen becomes

$B=B_1+B_2+B_3+B_4+B_5+B_6+B_7+B_8$

$\begin{array}{rcl}\overrightarrow{B}& =& -\overrightarrow{k}\frac{{\mu }_{0}i}{a\pi }·\left(\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{10}}+\frac{1}{3\sqrt{10}}+\frac{\sqrt{2}}{6}+\frac{\sqrt{2}}{6}+\frac{1}{3\sqrt{10}}+\frac{3}{\sqrt{10}}+\frac{1}{\sqrt{2}}\right)\\ \overrightarrow{B}& =& -\overrightarrow{k}\frac{2{\mu }_{0}i}{a\pi }·\left(\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{10}}+\frac{1}{3\sqrt{10}}+\frac{\sqrt{2}}{6}\right)\\ \overrightarrow{B}& =& -\overrightarrow{k}\frac{8{\mu }_{0}i}{a4\pi }·\left(\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{10}}+\frac{1}{3\sqrt{10}}+\frac{\sqrt{2}}{6}\right)\end{array}$

Substitute all the value in the above equation.

$\begin{array}{rcl}\overrightarrow{B}& =& \frac{-\overrightarrow{k}\left(4\pi \times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A} \times 10 \text{A}\right)}{4\pi \times 10\times {10}^{-2} \text{m}}\left(\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{10}}+\frac{1}{3\sqrt{10}}+\frac{\sqrt{2}}{6}\right)\\ \overrightarrow{B}& =& -\overrightarrow{k}·{10}^{-4}\left(\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{10}}+\frac{1}{3\sqrt{10}}+\frac{\sqrt{2}}{6}\right) \text{T}\\ \overrightarrow{B}& =& -\left(2.0\times {10}^{-4} \text{T}\right)\overrightarrow{k}\end{array}$

Hence the net magnetic field at pointP is$\overrightarrow{B}=-\left(2.0\times {10}^{-4} \text{T}\right)\overrightarrow{k}$