Question

On a printed circuit board, a relatively long straight conductorand a conducting rectangular loop lie in the same plane, as shownin Figure P31.9. Taking h = 0.500 mm, w = 1.50 mm, and L = 2.50 mm, find their mutual inductance.
pH

Figure P31.9

Listed location of solution for your convenience.

Answer 1

Concepts and reason

The given problem can be solved by using the expression of magnetic field due to straight current carrying conductor, the concept of mutual inductance.

First, use the expression of magnetic field to calculate the magnetic flux. Then, find the mutual inductance.

Fundamentals

The expression for magnetic field due to straight current carrying conductor can be expressed as follows,

$B = \frac{{{\mu _0}I}}{{2\pi r}}$

Here, ${\mu _0}$ is the absolute permeability, $I$ is the current and $r$ is the distance.

The expression for magnetic flux can be expressed as follows,

${\varphi _B} = \int {\vec B \cdot d\vec A}$

Here, $B$ is the magnetic field and $A$ is the area.

The expression for the mutual inductance can be expressed as follows,

$M = \frac{{N\varphi }}{I}$

Consider the given diagram as follows,

The magnetic field due to the current carrying conductor can be expressed as follows,

$B = \frac{{{\mu _0}I}}{{2\pi x}}$

Here, $x$ is the distance from the wire.

The small area of the rectangular loop can be calculated as follows,

$d\vec A = Ldx$

Calculate the magnetic flux.

${\varphi _B} = \int {\vec B \cdot d\vec A}$

Substitute $\frac{{{\mu _0}I}}{{2\pi x}}$ for $\vec B$ and $Ldx$ for $d\vec A$ in the expression of magnetic flux.

$\begin{array}{c}\\{\varphi _B} = \int\limits_h^{h + w} {\left( {\frac{{{\mu _0}I}}{{2\pi x}}} \right)\left( {Ldx} \right)} \\\\ = \frac{{L{\mu _0}I}}{{2\pi }}\ln \left( {\frac{{h + w}}{h}} \right)\\\end{array}$

The expression for the mutual inductance can be expressed as follows,

$M = \frac{{N{\varphi _B}}}{I}$

Substitute $\frac{{L{\mu _0}I}}{{2\pi }}\ln \left( {\frac{{h + w}}{h}} \right)$ for ${\varphi _B}$ and $1$ for $N$ .

$\begin{array}{c}\\M = \frac{{\left( 1 \right)\left( {\frac{{L{\mu _0}I}}{{2\pi }}\ln \left( {\frac{{h + w}}{h}} \right)} \right)}}{I}\\\\ = \frac{{L{\mu _0}}}{{2\pi }}\ln \left( {\frac{{h + w}}{h}} \right)\\\end{array}$

Substitute $4\pi \times {10^{ - 7}}{\rm{ H}} \cdot {{\rm{m}}^{ - 1}}$ for ${\mu _0}$ , $0.500{\rm{ mm}}$ for $h$ , $1.50{\rm{ mm}}$ for $w$ and $2.50{\rm{ mm}}$ for $L$ .

$\begin{array}{c}\\M = \frac{{\left( {2.50{\rm{ mm}}} \right)\left( {4\pi \times {{10}^{ - 7}}{\rm{ H}} \cdot {{\rm{m}}^{ - 1}}} \right)}}{{2\pi }}\ln \left( {\frac{{0.500{\rm{ mm}} + 1.50{\rm{ mm}}}}{{0.500{\rm{ mm}}}}} \right)\\\\ = 693 \times {10^{ - 12}}{\rm{ H}}\\\end{array}$

Convert ${\rm{H}}$ to ${\rm{pH}}$ .

$\begin{array}{c}\\M = \left( {693 \times {{10}^{ - 12}}{\rm{ H}}} \right)\left( {\frac{{1{\rm{ pH}}}}{{{{10}^{ - 12}}{\rm{ H}}}}} \right)\\\\ = 693{\rm{ pH}}\\\end{array}$

Ans:

The value of mutual induction is $693{\rm{ pH}}$ .