Question

Magnetic field in a region is given by $\overrightarrow{B}=B_{0}x\hat{k}$. Two loops each of side of $a$, separated by a distance $d$ is placed in this magnetic region in the $xy-$plane with one of its sides on $x$-axis as shown in figure. If $F_1$ is the force on loop $1$ and $F_2$ be the force on loop $2$, then

• A. $F_1=F_2=0$
• B. $F_1>F_2$
• C. $F_2>F_1$
• D. $F1=F_2\ne 0$

Answer 1

The correct option is D $F1=F_2\ne 0$

Force on a current carrying conductor of length $l$ & current $i$ in a magnetic field is given by,

$\overrightarrow{F}=i(\overrightarrow{l}\times \overrightarrow{B})$


Let us, find force on loop $1$ by each side individually,

As we can see, side $\text{AB}$ and $\text{CD}$ having same length, carries equal amount of current in opposite direction. Also, both will experiences same amount of magnetic field. Thus, they will experience equal amount of force in opposite direction. Therefore, sum of $F_{AB}$ and $F_{CD}$ will be zero.

Now, $F_{BC}=I\left(l\right(-\hat{j})\times B_{0}x\hat{k})=I\left(a\right(-\hat{j})\times B_{0}a\hat{k})$

$=I{a}^2{B}_0(-\hat{i})[\because l=x=a]$

$F_{AD}=I(l\hat{j}\times B_{0}x\hat{k})=0[\because x=0]$

Net force on the loop $1$ is,

$F_{net}=(F_{AB}+F_{CD})+F_{BC}+F_{DA}$

$=0+I{a}^2{B}_0(-\hat{i})+0$

$(F_1{)}_{net}=I{a}^2{B}_0(-\hat{i})$

Similarly, force on loop $2$ by each side individually,

Similar as side $\text{AB}$ and $\text{CD}$ in loop $1$, here side $\text{PQ}$ and $\text{RS}$ will experience equal force in opposite directions. Hence, sum of $F_{PQ}$ and $F_{RS}$ will be zero.

$F_{SP}=I(l\hat{j}\times B_{0}x\hat{k})=I\left(\right(a\hat{j})\times B_0(a+d\left)\hat{k}\right)$

$=I(a^2+ad)B_0\left(\hat{i}\right)$

$F_{QR}=I\left(l\right(-\hat{j})\times B_{0}x\hat{k})=I\left(a\right(-\hat{j})\times B_0(2a+d\left)\hat{k}\right)$

$=I(2a^2+ad)B_0(-\hat{i})$

Net force on the loop $2$ is,

$F_{net}=(F_{PQ}+F_{RS})+F_{SP}+F_{QR}$

$=0+I(a^2+ad)B_0\left(\hat{i}\right)+I(2a^2+ad)B_0(-\hat{i})$

$(F_2{)}_{net}=I{a}^2{B}_0(-\hat{i})$

Hence, $F_1=F_2\ne 0$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.

Why this Question ? To understand the concept of force on a current carrying conductor in a variable magnetic field.