Question

The value of the line integral of $\overrightarrow{\mathrm{B}}$around the closed path in FIGURE EX29.23 is $1.38\times {10}^{-5}\mathrm{T}.\mathrm{m}$. What are the direction (in or out of the page) and magnitude of $I_3$?

Short answer

The current ${\mathrm{I}}_3=23\mathrm{A}$and it is points into the paper.

Step-by-step solution

Given information

We have given,

line integral =$1.38\times {10}^{-5}\mathrm{T}.\mathrm{m}$

We have to find the current ${\mathrm{I}}_3$and its flow of direction.

Simplify

Using ampere's law, for any closed path is given by,

$\oint \overrightarrow{B}.\overrightarrow{dS}={\mu }_{0}I$

Where I is a total enclosed current in the magnetic field curve.

then total current will be

$$\begin{gathered} \mathrm{I}={\mathrm{I}}_2+{\mathrm{I}}_3 \\ \mathrm{I}=-12\mathrm{A}+{\mathrm{I}}_3.......................\left(1\right) \end{gathered}$$

since other currents are out of closed curve.

since,

$$\begin{gathered} \oint \overrightarrow{B}.\overrightarrow{dS}={\mu }_{0}I \\ \oint \overrightarrow{B}.\overrightarrow{dS}=(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A})\mathrm{I} \\ \mathrm{I}=\frac{\oint \overrightarrow{B}.\overrightarrow{dS}}{4\mathrm{\pi }\times {10}^{-7}}=\frac{1.38\times {10}^{-5}\mathrm{T}.\mathrm{m}}{4\times 3.14\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}} \\ \mathrm{I}=11\mathrm{A} \end{gathered}$$

using equation (1),

localid="1649421657359" $$\begin{gathered} \mathrm{I}=11\mathrm{A}=-12\mathrm{A}+{\mathrm{I}}_3 \\ {\mathrm{I}}_3=23\mathrm{A}. \end{gathered}$$

the positive sine indicates that current will be into the paper.