Question

A square loop of side 'a' is placed at a distance 'a' away from a long wire carrying a current $I_1$ if the loop carries a current $I_2$ then the nature of the force on the loop and its magnitude is

• A. $\frac{{\mu }_0{l}_1{l}_2}{2\pi a}$, attractive
  
• B. $\frac{{\mu }_0{l}_1{l}_2}{4\pi a}$, attractive
• C. $\frac{{\mu }_0{l}_1{l}_2}{4\pi a}$, repulsive
• D. $\frac{{\mu }_0{l}_1{l}_2}{4\pi a}$, replusion

Answer 1

The correct option is B $\frac{{\mu }_0{l}_1{l}_2}{4\pi a}$, attractive

Given: Current in wire $=I_1$

Current in loop $=I_2$

Solution: Let distance between wire and AB is $r_1.$

Let distance between wire and CD is $r_2.$

So, $r_1=a$ and $r_2=2a$

Using figure we can conclude that, BC and DA sides of square are collinear and at same distance from wire but force experienced in equal in magnitude but have opposite nature, so

$F_{BC}+F_{DA}=0$ $...\left(1\right)$

Now, force on AB:

$F_{AB}=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{r_1}$

$==>F_{AB}=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{a}$ $\left(attractive\right)$

Now, force on CD:

$F_{CD}=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{r_2}$

$==>F_{CD}=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{2a}$ $\left(repulsive\right)$

Net force is:

$F=F_{AB}+F_{BC}+F_{CD}+F_{DA}$

Also, $F_{AB}>F_{CD}$ also both have opposite nature.

So, on using $e{q}^n\left(1\right),$ we get

$==>F=\frac{{\mu }_o}{2\pi }\frac{I_1{I}_2}{a}-\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{2a}$

$==>F=\frac{{\mu }_o}{2\pi }\frac{I_1{I}_2}{a}(1-1/2)$

$==>F=\frac{{\mu }_o}{4\pi }\frac{I_1{I}_2}{a}$ $\left(attractive\right)$

hence,

The correct opt : B