Thank you for reporting, we will resolve it shortly
Q.
What is the net force on the square coil?
Solution:
Force on side $B C$ and $A D$ are equal but opposite so their net will be zero.
But $F_{A B} =10^{-7} \times \frac{2 \times 2 \times 1}{2 \times 10^{-2}} \times 15 \times 10^{-2}$
$=3 \times 10^{-6} N $
and $ F_{C D} =10^{-7} \times \frac{2 \times 2 \times 1}{\left(12 \times 10^{-2}\right)} \times 15 \times 10^{-2} $
$=0.5 \times 10^{-6} N $
$ \Rightarrow F_{\text {net }} =F_{ AB }-F_{ CD }=2.5 \times 10^{-6} N $
$=25 \times 10^{-7} N $, towards the wire.
