ABCD is a rectangular loop made of uniform wi | vedclass

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(C) The total current $I = 20 	ext{ A}$ enters the loop and splits into two parallel branches,$ABC$ and $ADC$. Since the wire is uniform,the resistan

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$10^{-3} 	ext{ N m}^{-1}$

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(C) The total current $I = 20 	ext{ A}$ enters the loop and splits into two parallel branches,$ABC$ and $ADC$. Since the wire is uniform,the resistance of both branches is equal,so the current divides equally: $i_1 = i_2 = 10 	ext{ A}$.The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by a distance $r$ is given by:$\frac{F}{L} = \frac{\mu_0 i_1 i_2}{2 \pi r}$Given:$i_1 = 10 	ext{ A}$$i_2 = 10 	ext{ A}$$r = AD = BC = 2 	ext{ cm} = 2 	imes 10^{-2} 	ext{ m}$$\mu_0 = 4 \pi 	imes 10^{-7} 	ext{ T m/A}$Substituting the values:$\frac{F}{L} = \frac{(4 \pi 	imes 10^{-7}) 	imes 10 	imes 10}{2 \pi 	imes (2 	imes 10^{-2})}$$\frac{F}{L} = \frac{4 \pi 	imes 10^{-5}}{4 \pi 	imes 10^{-2}}$$\frac{F}{L} = 10^{-3} 	ext{ N m}^{-1}$.

$ABCD$ is a rectangular loop made of uniform wire. If $AD = BC = 2 \text{ cm}$,what is the magnetic force per unit length acting on wire $DC$ due to wire $AB$ if the ammeter reads $20 \text{ A}$? (The lengths of $AB$ and $DC$ are large in comparison with the other two sides).

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