Question

The value of the line integral around the closed path in the figure(Figure 1) is 1.96×10^-5 Tm . What is the magnitude of I3?

Answer 1

Concepts and reason

The concepts required to solve the given problem are Ampere’s circuital law, and right hand rule.

Initially, use right hand rule to determine the direction of unknown current enclosed by the loop. Then, determine the net current enclosed by the loop using direction of each current inside the loop. Finally, determine the magnitude of unknown current by using Ampere’s circuital law.

Fundamentals

According to right hand rule, if the thumb of right hand is along the direction of current in a wire then the curl of remaining figures gives the direction of magnetic field around the wire.

Ampere’s circuital law states that the line integral of magnetic field B along a closed path is equal to ${\mu _0}$ times the current ${I_{{\rm{enc}}}}$enclosed by the closed path.

$\oint {\vec B \cdot d\vec l = {\mu _0}{I_{{\rm{enc}}}}}$

Here, ${\mu _0}$is the permeability of free space.

The direction of unknown current ${I_3}$is into the page because the direction of magnetic field is in clockwise direction.

The net current enclosed by the closed path is,

${I_{{\rm{enc}}}} = {I_3} - {I_2}$

The Ampere’s circuital law is given by following expression.

$\oint {\vec B \cdot d\vec l = {\mu _0}{I_{{\rm{enc}}}}}$

Substitute $1.96 \times {10^{ - 5}}\,{\rm{T}} \cdot {\rm{m}}$for $\oint {\vec B.d\vec l,}$$4\pi \times {10^7}\,{\rm{T}}{\rm{.m}}/{\rm{A}}$for ${\mu _0}$, and ${I_3} - {I_2}$for ${I_{{\rm{enc}}}}$in the above equation to solve for ${I_3}.$

$\begin{array}{c}\\1.96 \times {10^{ - 5}}\,{\rm{T}} \cdot {\rm{m}} = \left( {4\pi \times {{10}^{ - 7}}\,{\rm{T}}{\rm{.m/A}}} \right)\left( {{I_3} - {I_2}} \right)\\\\{I_3} = \frac{{1.96 \times {{10}^{ - 5}}\,{\rm{T}} \cdot {\rm{m}}}}{{\left( {4\pi \times {{10}^{ - 7}}\,{\rm{T}}{\rm{.m/A}}} \right)}} + {I_2}\\\\ = 15.6\,{\rm{A}} + {I_2}\\\end{array}$

Substitute 12 A for ${I_2}$in the above equation.

$\begin{array}{c}\\{I_3} = 15.6\,{\rm{A}} + 12\,{\rm{A}}\\\\{\rm{ = 27}}{\rm{.6}}\,{\rm{A}}\\\end{array}$

Ans:

The magnitude of unknown current is 27.6 A.