Question

$A,B$ and $C$ are parallel conductors of equal lengths carrying currents $i$ 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 relation among the following.

• A. $|{\overrightarrow{F}}_1|=2|{\overrightarrow{F}}_2|$
• B. $|{\overrightarrow{F}}_2|=2|{\overrightarrow{F}}_1|$
• C. $|{\overrightarrow{F}}_1|=3|{\overrightarrow{F}}_2|$
• D. $|{\overrightarrow{F}}_1|=|{\overrightarrow{F}}_2|$

Answer 1

The correct option is D $|{\overrightarrow{F}}_1|=|{\overrightarrow{F}}_2|$

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 normally outwards to the plane of paper and the force between wire $A$ and $B$ is given by,

$F_1=\frac{{\mu }_0{i}^{2}l}{2\pi x}[l=\text{length of wire}]$

The direction of $F_1$ ​ will be perpendicular towards wire $B$ .

Similarly, the direction of magnetic field due to current in wire $C$, in the region of wire $A$ will be normally inwards to the plane of paper and the force between wire $A$ and $C$ is given by,

$F_2=\frac{{\mu }_0\left(i\right)\left(2i\right)l}{2\pi \left(2x\right)}=\frac{{\mu }_0{i}^{2}l}{2\pi x}[l=\text{length}]$

The direction of $F_2$​ will be leftwards from wire $A$.

Hence, it is clear that $\overrightarrow{F_1}=-\overrightarrow{F_2}$ ​ ,where $-ve$ shows that $F_1$ ​ and $F_2$ ​ are in opposite directions.


Hence, option (d) is the correct answer.