Question

What is the net force on the rectangular coil?

Answer 1

What is the net force on the rectangular coil?

Answer 2

Given:

Length of the rectangular coil $=15 cm$

Breadth of the rectangular coil $=10 cm$

Current flowing in the coil $=1A$

Current flowing in the wire placed near the coil $=2A$

Distance of wire from the AB edge of the coil $\[ = 2cm = 2 \times {10^{ - 2}}m\]$

Distance of wire from the CD edge of the coil $\[ = 12cm = 12 \times {10^{ - 2}}m\]$

To Find: Net force on the rectangular coil

Solution:

The forces acting on each side of the rectangular coil are shown in the attachment.

Forces $\[{F_1},\,{F_2},\,{F_3}\,and\,{F_4}\]$ act on the sides AB, BC, CD and DA respectively.

Since the forces $\[{F_2}\,and\,{F_4}\,\]$ are equal forces acting in opposite directions, therefore, they will cancel each other.

The force acting between two current-carrying wires can be given as

$\[F = \frac{{{\mu _0}{I_1} \times {I_2}}}{{2\pi d}}\]$

Where, $\[{I_1}\,and\,{I_2}\]$ are the current flowing in the two wires, $d$ is the distance between the wires, $\[{{\mu _0}}\]$ is the permeability of free space ($\[{\mu _0} = 4\pi \times {10^{ - 7}}H/m\]$).

Therefore, the force $\[{F_1}\]$ can be given as:

$\[{F_1} = \frac{{4\pi \times {{10}^{ - 7}} \times 2 \times 1}}{{2\pi \times 2 \times {{10}^{ - 2}}}}\]$        

$\[{F_1} = 2 \times {10^{ - 5}}N\]$

Force $\[{F_3}\]$ can be given as:

$\[{F_3} = \frac{{4\pi \times {{10}^{ - 7}} \times 2 \times 1}}{{2\pi \times 12 \times {{10}^{ - 2}}}}\]$

$\[{F_3} = 0.33 \times {10^{ - 5}}N\]$

Thus, the net force on the rectangular coil can be given as:

$\[F = {F_1} - {F_3}\]$

$\[F = 2 \times {10^{ - 5}}N - 0.33 \times {10^{ - 5}}N\]$

$\[F = 1.67 \times {10^{ - 5}}N\]$

Hence, the net force on the rectangular coil is $\[1.67 \times {10^{ - 5}}N\]$.