Question

The value of the line integral of B?  around the closed path in the figure (Figure 1) is 11.33 × 10^?6 Tm.What is I3?

Answer 1

Concepts and reason

The concepts used to solve this problem are ampere’s law.

First find the enclosed current in the loop by summing individual currents in it.

Finally find the current ${I_3}$ by rearranging ampere’s law expression.

Fundamentals

Ampere’s law for any closed loop states,

“The line integral of magnetic field and the infinitesimal element of the curve over a closed path are permeability times the current surrounded in the loop”.

The expression for ampere’s law is,

$\oint {\vec B \cdot dl} = {\mu _0}{I_{enclosed}}$

Here, $\vec B$ is the magnetic field, $dl$ is the infinitesimal element length, ${\mu _0}$ is the permeability of free space, and ${I_{enclosed}}$ is the enclosed current.

In a closed loop the total current enclosed will be equal to the addition of the individual currents in the loop.

The expression for total currents in the loop is,

${I_{enclosed}} = {I_1} - {I_2} + {I_3}$

In the loop the direction of current ${I_2}$ is opposite to the direction of current ${I_1}$ and ${I_3}$ . So the current ${I_2}$ indicates negative sign.

Substitute $2.0\,{\rm{A}}$ for ${I_1}$ and $6.0\,{\rm{A}}$ for ${I_2}$ .

$\begin{array}{c}\\{I_{enclosed}} = 2.0\,{\rm{A}} - 6.0\,{\rm{A}} + {I_3}\\\\ = {I_3} - 4.0\,{\rm{A}}\\\end{array}$

The expression for ampere’s law is,

$\oint {\vec B \cdot dl} = {\mu _0}{I_{enclosed}}$

Substitute ${I_3} - 4.0\,{\rm{A}}$ for ${I_{enclosed}}$ in the above expression.

$\oint {\vec B \cdot dl} = {\mu _0}\left( {{I_3} - 4.0\,{\rm{A}}} \right)$

By rearranging the expression,

$\begin{array}{c}\\\oint {\vec B \cdot dl} = {\mu _0}\left( {{I_3} - 4.0\,{\rm{A}}} \right)\\\\\left( {{I_3} - 4.0\,{\rm{A}}} \right) = \frac{{\oint {\vec B \cdot dl} }}{{{\mu _0}}}\\\\{I_3} = \frac{{\oint {\vec B \cdot dl} }}{{{\mu _0}}} + 4.0\,{\rm{A}}\\\end{array}$

Substitute $11.33 \times {10^{ - 6}}\,{\rm{T}} \cdot {\rm{m}}$ for $\oint {\vec B \cdot dl}$ and $4\pi \times {10^{ - 7}}\,{\rm{T}} \cdot {\rm{m/A}}$ for ${\mu _0}$ .

$\begin{array}{c}\\{I_3} = \frac{{11.33 \times {{10}^{ - 6}}\,{\rm{T}} \cdot {\rm{m}}}}{{4\pi \times {{10}^{ - 7}}\,{\rm{T}} \cdot {\rm{m/A}}}} + 4.0\,{\rm{A}}\\\\{\rm{ = 9}}{\rm{.02}}\,{\rm{A}} + 4.0\,{\rm{A}}\\\\ = 13.0\,{\rm{A}}\\\end{array}$

Ans:

Thus, the value of the current ${I_3}$ is $13.0\,A$ .