# What Is The Net Force On The Square Coil

Force on side BC and AD are equal but opposite, so their net will be zero.

$$F_{AB} = 10^{-7} * \frac{2*2*1}{2*10^{-2}} * 15 * 10^{-2}$$ $$\Rightarrow 3 * 10^{-6} N$$ $$F_{CD} = 10^{-7} * \frac{2*2*1}{12*10^{-2}} * 15 * 10^{-2}$$ $$\Rightarrow 0.5 * 10^{-6} N$$ $$\Rightarrow F_{net} = F_{AB} – F_{CD}$$ $$\Rightarrow 3 * 10^{-6} – 0.5 * 10^{-6}$$ $$\Rightarrow 2.5 * 10^{-6} N$$

The Net Force On The Square Coil is $$25 * 10^{-7}N$$

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