A, B and C are three parallel conductors of e | vedclass

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(D) The force per unit length between two parallel conductors carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $f = \frac{\mu_0

Vedclass · Accepted Answer

$F_1=-F_2$

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(D) The force per unit length between two parallel conductors carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi r}$.$1$. Force $F_1$ exerted by $B$ on $A$:Since currents in $A$ and $B$ are in the same direction,the force is attractive (towards $B$).$F_1 = \frac{\mu_0 I \cdot I}{2 \pi x} \cdot L = \frac{\mu_0 I^2 L}{2 \pi x}$.$2$. Force $F_2$ exerted by $C$ on $A$:Since currents in $A$ and $C$ are in opposite directions,the force is repulsive (away from $C$).The distance between $A$ and $C$ is $2x$.$F_2 = \frac{\mu_0 I \cdot 2I}{2 \pi (2x)} \cdot L = \frac{\mu_0 I^2 L}{2 \pi x}$.Comparing the magnitudes,we see that $F_1 = F_2$. However,since the forces are in opposite directions (one is attractive towards $B$ and the other is repulsive away from $C$),we have $F_1 = -F_2$ in vector notation.

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