Two straight parallel wires,both carrying 10 | vedclass

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Question

(C) The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by distance $a$ is given by $F = \frac{\mu_0}{4\p

Vedclass · Accepted Answer

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

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(C) The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by distance $a$ is given by $F = \frac{\mu_0}{4\pi} \frac{2 i_1 i_2}{a}$.Given,$F = 1 	imes 10^{-3} \ N$ for $i_1 = 10 \ A$ and $i_2 = 10 \ A$.When both currents are doubled,the new currents are $i_1' = 2 i_1$ and $i_2' = 2 i_2$.The new force $F'$ is given by $F' = \frac{\mu_0}{4\pi} \frac{2 (2 i_1) (2 i_2)}{a} = 4 	imes \left( \frac{\mu_0}{4\pi} \frac{2 i_1 i_2}{a} ight) = 4 	imes F$.Substituting the value of $F$,we get $F' = 4 	imes 1 	imes 10^{-3} \ N = 4 	imes 10^{-3} \ N$.

Two straight parallel wires,both carrying $10 \ A$ in the same direction,attract each other with a force of $1 \times 10^{-3} \ N$. If both currents are doubled,the force of attraction will be:

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