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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