Question

$A$, $B$ and $C$ are parallel conductors of equal lengths carrying currents $I$, $I$ and $2I$ respectively. Distance between $A$ and $B$ is $x$. Distance between $B$ and $C$ is also $x$. $F_1$ is the force exerted by $B$ on $A$. $F_2$ is the force exerted by $C$ on $A$. Choose the correct answer.

• A. $F_1=2F_2$
• B. $F_2=2F_1$
• C. $F_1=F_2$
• D. $F_1=-F_2$

Answer 1

The correct option is D $F_1=-F_2$

According to Right hand thumb rule ,the direction of magnetic field due to current in wire B , in the region of wire A will be normal outward to the plane of paper and the force between wire A and B is given by ,

$F_1=K\frac{I\times I\times l}{x}$ (where $l=$length of each wire)

Now , by Fleming's left hand rule , the direction of $F_1$ will be perpendicularly towards wire B .

Again , according to Right hand thumb rule ,the direction of magnetic field due to current in wire C , in the region of wire A will be normal inward to the plane of paper and the force between wire A and C is given by ,

$F_2=K\frac{I\times 2I\times l}{2x}$ (where $l=$length of each wire)

or $F_2=K\frac{I\times I\times l}{x}$

Now , by Fleming's left hand rule , the direction of $F_2$ will be perpendicularly outwards from wire A .

Hence , it is clear that $F_1=-F_2$ ,where -ive shows that $F_1$ and $F_2$ are in opposite directions .