Question

The figure below shows, in cross section, several conductorsthat carry currents through the plane of the figure. The currentshave the magnitudes I₁ = 2.0 A, I₂ = 6.5 A, and I₃ = 2.4 A. and the directions shown. Four paths,labeled a through d, are shown. What is the lineintegral dl for each paths? Each integralinvolves going around the path in the counterclockwisedirection.

path a
1 T · m

path b
2 T · m

path c
3 T · m

path d
4 T · m

Answer 1

Concepts and reason

The main concept required to solve the problem is Ampere’s circuital law.

For the line integral, calculate the net current flowing through the loop and apply Ampere’s circuital law.

Fundamentals

Ampere’ s circuital law states that the line integral of the magnetic field around a closed loop is permeability times the algebraic sum of the current passing through the loop.

Consider a loop C such that current I passes through the surface S enclosed by C. The line integral of magnetic field around the closed curve is,

$\oint {\vec B} .d\vec l = {\mu ^\circ }I$

Here, ${\mu ^\circ }$ is the permeability of the free space, B is the magnetic field vector and dl is the infinitesimal element of the curve.

(1)

Calculate the integral of magnetic field vector around path a.

The line integral of magnetic field around the closed curve is,

$\oint {\vec B} .d\vec l = {\mu ^\circ }I$

Substitute $0{\rm{ A}}$ for I and $4\pi \times {10^{ - 7}}{\rm{ H/m}}$ for ${\mu ^\circ }$ as follows:

$\begin{array}{c}\\\oint {\vec B} .d\vec l = \left( {4\pi \times {{10}^{ - 7}}{\rm{ H/m}}} \right)\left( {0{\rm{ A}}} \right)\\\\ = 0{\rm{ T}} \cdot {\rm{ m}}\\\end{array}$

(2)

Calculate the integral of magnetic field vector around path b.

The line integral of magnetic field around the closed curve is,

$\oint {\vec B} .d\vec l = {\mu ^\circ }I$

Substitute $- 2.0{\rm{ A}}$ for I and $4\pi \times {10^{ - 7}}{\rm{ H/m}}$ for ${\mu ^\circ }$ as follows:

$\begin{array}{c}\\\oint {\vec B} .d\vec l = - \left( {4\pi \times {{10}^{ - 7}}{\rm{ H/m}}} \right)\left( {2.0{\rm{ A}}} \right)\\\\ = - 2.5 \times {10^{ - 6}}{\rm{ T}} \cdot {\rm{ m}}\\\end{array}$

The direction of path is taken along counter clockwise direction. Hence the integral points outward from Right hand thumb rule, while the current through loop b is inwards. Therefore, a negative sign is involved here.

(3)

Calculate the integral of magnetic field vector around path c.

The line integral of magnetic field around the closed curve is,

$\oint {\vec B} .d\vec l = {\mu ^\circ }I$

The net current in the loop c is,

$I = {I_1} + {I_2}$

Here, ${I_1}$ and ${I_2}$ are the current through two wires passing through loop c.

So, the line integral is given as follows:

$\oint {\vec B} .d\vec l = {\mu ^\circ }\left( {{I_1} + {I_2}} \right)$

Substitute $- 2.0{\rm{ A}}$ for ${I_1}$ , and ${\rm{6}}{\rm{.5 A}}$ for ${I_2}$ , $4\pi \times {10^{ - 7}}{\rm{ H/m}}$ for ${\mu ^\circ }$ as follows:

$\begin{array}{c}\\\oint {\vec B} .d\vec l = \left( {4\pi \times {{10}^{ - 7}}{\rm{ H/m}}} \right)\left( { - 2.0{\rm{ A}} + 6.5{\rm{ A}}} \right)\\\\ = 5.65 \times {10^{ - 6}}{\rm{ T}} \cdot {\rm{ m}}\\\end{array}$

The direction of path is taken along counter clockwise direction. Hence the integral points outward from Right hand thumb rule, while the current ${I_1}$ is inwards. Therefore, a negative sign is involved here.

(4)

Calculate the integral of magnetic field vector around path c.

The line integral of magnetic field around the closed curve is,

$\oint {\vec B} .d\vec l = {\mu ^\circ }I$

The net current in the loop c is,

$I = {I_1} + {I_2} + {I_3}$

Here, ${I_1}$ , ${I_2}$ and ${I_3}$ are the current through the wires passing through loop d.

So, the line integral is given as follows:

$\oint {\vec B} .d\vec l = {\mu ^\circ }\left( {{I_1} + {I_2} + {I_3}} \right)$

Substitute $- 2.0{\rm{ A}}$ for ${I_1}$ , ${\rm{6}}{\rm{.5 A}}$ for ${I_2}$ and ${\rm{2}}{\rm{.4 A}}$ for ${I_3}$ , $4\pi \times {10^{ - 7}}{\rm{ H/m}}$ for ${\mu ^\circ }$ as follows:

$\begin{array}{c}\\\oint {\vec B} .d\vec l = \left( {4\pi \times {{10}^{ - 7}}{\rm{ H/m}}} \right)\left( { - 2.0{\rm{ A}} + 6.5{\rm{ A + 2}}{\rm{.4 A}}} \right)\\\\ = 8.67{\rm{ 1}}{{\rm{0}}^{ - 6}}{\rm{ T}} \cdot {\rm{ m}}\\\end{array}$

The direction of path is taken along counter clockwise direction. Hence the integral points outward from Right hand thumb rule, while the current ${I_1}$ is inwards. Therefore, a negative sign is involved here.

Ans: Part 1

The value of line integral around path a is $0{\rm{ T}} \cdot {\rm{ m}}$ .

Part 2

The value of line integral around path b is $- 2.5 \times {10^{ - 6}}{\rm{ T}} \cdot {\rm{ m}}$ .

Part 3

The value of line integral around path c is $5.65 \times {10^{ - 6}}{\rm{ T}} \cdot {\rm{ m}}$ .

Part 4

The value of line integral around path d is $8.67 \times {10^{ - 6}}{\rm{ T}} \cdot {\rm{ m}}$ .