A current of 10 A is flowing in two straight | vedclass

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(C) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $F = \frac{\mu_0}{4\p

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$4 	imes 10^{-3} \ N$

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(C) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $F = \frac{\mu_0}{4\pi} \frac{2 I_1 I_2}{r} \cdot l$.Given the initial force $F = 1 	imes 10^{-3} \ N$.When the current in both wires is doubled,the new currents become $I_1' = 2I_1$ and $I_2' = 2I_2$.The new force $F'$ is given by $F' = \frac{\mu_0}{4\pi} \frac{2(2I_1)(2I_2)}{r} \cdot l = 4 	imes \left( \frac{\mu_0}{4\pi} \frac{2 I_1 I_2}{r} \cdot l ight)$.Substituting the initial force value: $F' = 4 	imes (1 	imes 10^{-3} \ N) = 4 	imes 10^{-3} \ N$.

$A$ current of $10 \ A$ is flowing in two straight parallel wires in the same direction. The force of attraction between them is $1 \times 10^{-3} \ N$. If the current is doubled in both the wires,the force will be:

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