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(A) The force per unit length between two long parallel conductors carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by $F = \fra

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$F/4$ and unchanged in direction

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(A) The force per unit length between two long parallel conductors carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by $F = \frac{\mu_0}{4\pi} \frac{2 I_1 I_2}{d}$.Initially,$I_1 = 2\,	ext{A}$ and $I_2 = 2\,	ext{A}$,so $F = \frac{\mu_0}{4\pi} \frac{2 	imes 2 	imes 2}{d} = \frac{\mu_0}{4\pi} \frac{8}{d}$.When the current in each is changed to $1\,	ext{A}$,the new force $F^{\prime}$ is $F^{\prime} = \frac{\mu_0}{4\pi} \frac{2 	imes 1 	imes 1}{d} = \frac{\mu_0}{4\pi} \frac{2}{d}$.Comparing the two,$F^{\prime} = F/4$.Since both currents are reversed,the relative direction of the currents remains the same (both are still parallel). Therefore,the nature of the force (attractive or repulsive) remains unchanged,and the direction of the force on each conductor remains unchanged.

In the given figure,$X$ and $Y$ are two long straight parallel conductors each carrying a current of $2\,\text{A}$. The force on each conductor is $F$ newtons. When the current in each is changed to $1\,\text{A}$ and reversed in direction,the force on each is now

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