Question

Determine the magnitude and direction of the force between two parallel wires 30 m long and 6.0 cm apart, each carrying 30A in the same direction. Determine the magnitude of the force between two parallel wires.

Answer 1

Concepts and reason

The concepts used in this problem are magnetic field due to a current carrying wire and the magnetic force. First, derive to expression for force between two wires using the concept of magnetic field due to a current carrying wire and the magnetic force. Then, calculate the force by substituting the values.

Fundamentals

Magnetic field due to a current carrying wire:

The expression for magnetic field due to a straight current carrying wire is:

$B = \frac{{{\mu _{\rm{o}}}I}}{{2\pi d}}$

Here, ${\mu _{\rm{o}}}$ is the magnetic permeability, $I$ is the current flowing in the wire and $d$ is the distance of point from the wire.

Magnetic force:

The magnetic force is perpendicular to the magnetic field and the direction of current. The magnitude of magnetic force is given by:

$\begin{array}{c}\\{{\vec F}_{{\rm{mag}}}} = L\left( {\vec I \times \vec B} \right)\\\\{F_{{\rm{mag}}}} = ILB\sin \theta \\\end{array}$

Here, $I$ is the current, $v$ is the velocity, $B$ is the magnetic field and $\theta$ is the angle between current and the magnetic field.

The diagram for the given problem is:

The force on wire 2 is:

$F = {I_2}LB\sin \theta$

Substitute ${90^{\rm{o}}}$ for $\theta$ ,

$F = {I_2}LB\sin {90^{\rm{o}}}$

$F = {I_2}LB$ …… (1)

The magnetic field at wire 2 due to current in wire 1 is:

$B = \frac{{{\mu _{\rm{o}}}{I_1}}}{{2\pi d}}$ …… (2)

Substitute equation (2) in equation (1).

$F = {I_2}L\left( {\frac{{{\mu _{\rm{o}}}{I_1}}}{{2\pi d}}} \right)$

$F = \frac{{{\mu _{\rm{o}}}L{I_1}{I_2}}}{{2\pi d}}$ …… (3)

The expression of force is:

$F = \frac{{{\mu _{\rm{o}}}L{I_1}{I_2}}}{{2\pi d}}$

Substitute $30{\rm{ A}}$ for ${I_1}$ , $30{\rm{ A}}$ for ${I_2}$ , $30{\rm{ m}}$ for $L$ , $6{\rm{ cm}}$ for $d$ and $4\pi \times {10^{ - 7}}{\rm{ H/m}}$ for ${\mu _{\rm{o}}}$ .

$\begin{array}{c}\\F = \frac{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ H/m}}} \right)\left( {30{\rm{ m}}} \right)\left( {30{\rm{ A}}} \right)\left( {30{\rm{ A}}} \right)}}{{2\pi \left( {6{\rm{ cm}}} \right)}}\\\\ = \frac{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ H/m}}} \right)\left( {30{\rm{ m}}} \right)\left( {30{\rm{ A}}} \right)\left( {30{\rm{ A}}} \right)}}{{2\pi \left( {6{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)}}\\\\ = 0.09{\rm{ N}}\\\end{array}$

Ans:

The force between two wires is ${\bf{0}}{\bf{.09 N}}$ .