What is the net force on the rectangular coil | vedclass

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(A) The force on a current-carrying conductor due to a long straight wire is given by $F = \frac{\mu_0 I_1 I_2 L}{2 \pi r}$.For the rectangular coil $

Vedclass · Accepted Answer

$25 	imes 10^{-7} 	ext{ N}$ moving towards wire

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(A) The force on a current-carrying conductor due to a long straight wire is given by $F = \frac{\mu_0 I_1 I_2 L}{2 \pi r}$.For the rectangular coil $ABCD$,the forces on the horizontal segments $BC$ and $AD$ are equal and opposite,so they cancel each other out.The force on segment $AB$ (at distance $r_1 = 2 	ext{ cm} = 0.02 	ext{ m}$) is attractive (towards the wire) because the currents are parallel:$F_{AB} = \frac{\mu_0 (2 	ext{ A})(1 	ext{ A})(0.15 	ext{ m})}{2 \pi (0.02 	ext{ m})} = \frac{2 	imes 10^{-7} 	imes 2 	imes 1 	imes 0.15}{0.02} = 30 	imes 10^{-7} 	ext{ N}$.The force on segment $CD$ (at distance $r_2 = 2 	ext{ cm} + 10 	ext{ cm} = 12 	ext{ cm} = 0.12 	ext{ m}$) is repulsive (away from the wire) because the currents are anti-parallel:$F_{CD} = \frac{\mu_0 (2 	ext{ A})(1 	ext{ A})(0.15 	ext{ m})}{2 \pi (0.12 	ext{ m})} = \frac{2 	imes 10^{-7} 	imes 2 	imes 1 	imes 0.15}{0.12} = 5 	imes 10^{-7} 	ext{ N}$.The net force is $F_{net} = F_{AB} - F_{CD} = 30 	imes 10^{-7} - 5 	imes 10^{-7} = 25 	imes 10^{-7} 	ext{ N}$.Since $F_{AB} > F_{CD}$,the net force is directed towards the wire.

What is the net force on the rectangular coil?

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