Question 5.12: A rectangular coil ABCD of size 20 cm × 10 cm is placed 1 cm......
A rectangular coil ABCD of size 20 cm × 10 cm is placed 1 cm away from a long current-carrying conductor with its longer sides parallel to the conductor. A current of 100 A is flowing through the long conductor while the coil is carrying a current of 10 A. Calculate the resultant force developed on the coil.
Force will be developed between the longer sides of the coil and the conductor as they are parallel to each other. There will be force of attraction with one coil side and force of repulsion with the other. The difference between the two forces will be the resultant force acting on the coil. There will be no force developed on the sides which are perpendicular to the conductor as shown in Fig. 5.32. Force experienced per unit length by a current-carrying conductor due to flux density created by the other current-carrying conductor is
\mathrm{F}_1=\mathrm{BI}_2=\frac{\mu_0 \mathrm{I}_1 \times \mathrm{I}_2}{2 \pi \mathrm{r}_1} \mathrm{~N}Force of attraction between the conductor and side AB
\mathrm{F}_1=\frac{\mu_0 \mathrm{I}_1, \mathrm{I}_2 l}{2 \pi \mathrm{r}_1}Force of repulsion between the conductor and side CD
\mathrm{F}_2=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2 l}{2 \pi \mathrm{r}_2}The net force =\mathrm{F}_1-\mathrm{F}_2=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2 l}{2 \pi}\left[\frac{1}{\mathrm{r}_1}-\frac{1}{\mathrm{r}_2}\right]
Substituting the values
\begin{aligned} \mathrm{F}_1-\mathrm{F}_2 & =\frac{4 \pi \times 10^{-7} \times 100 \times 10 \times 0.2}{2 \pi}\left[\frac{1}{0.01}-\frac{1}{0.11}\right] \\ & =3.63 \times 10^{-3} \mathrm{~N} \end{aligned}