Question

The value of the line integral of $\overrightarrow{\mathrm{B}}$around the closed path in FIGURE EX29.22 is $3.77\times {10}^{-5}\mathrm{Tm}$. What is${\mathrm{I}}_{3.}$?

Short answer

The current${\mathrm{I}}_3=7\mathrm{A}.$

Step-by-step solution

Given information

We have given,

line integral =$3.77\times {10}^{-5}\mathrm{Tm}$

We have to find the current${\mathrm{I}}_{3.}$

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

$$\begin{gathered} \mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3 \\ \mathrm{I}=2\mathrm{A}-6\mathrm{A}+{\mathrm{I}}_3 \\ \mathrm{I}=-4\mathrm{A}+{\mathrm{I}}_3.......................\left(1\right) \end{gathered}$$

since

$$\begin{gathered} \oint \overrightarrow{B}.\overrightarrow{dS}={\mu }_{0}I \\ \oint \overrightarrow{B}.\overrightarrow{dS}=(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A})\mathrm{I} \\ \mathrm{I}=\frac{\oint \overrightarrow{B}.\overrightarrow{dS}}{4\mathrm{\pi }\times {10}^{-7}}=\frac{3.77\times {10}^{-5}\mathrm{T}.\mathrm{m}}{4\times 3.14\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}} \\ \mathrm{I}=3\mathrm{A} \end{gathered}$$

using equation (1),

$$\begin{gathered} \mathrm{I}=-4\mathrm{A}+{\mathrm{I}}_3=3\mathrm{A} \\ {\mathrm{I}}_3=7\mathrm{A}. \end{gathered}$$