Question

A circular loop of radius $R$ is bent along a diameter and given a shape as shown in the figure. One of the semicircles $\text{KNM}$ lies in the $\text{x-z}$ plane and the other one $\text{KLM}$ in the $\text{y-z}$ plane with their centres at the origin. Current $I$ is flowing through each of the semicircles as shown in figure. If an external uniform magnetic field $B\hat{j}$ ​is applied, determine the forces $F_1$ ​ and $F_2$ ​ on the semicircles $\text{KLM}$ and $\text{KNM}$ due to the field and the net force $F$ on the loop.

• A. $2IBR,IBR,4IBR$
• B. $IBR,2IBR,4IBR$
• C. $2IBR,2IBR,0$
• D. $IBR,2IBR,4IBR$

Answer 1

The correct option is D $IBR,2IBR,4IBR$

Force on a current carrying conductor due to a magnetic field is $$\overrightarrow{F}=i(\overrightarrow{l}\times \overrightarrow{B})$$Here, the displacement vector $\overrightarrow{l}$ is same for the both parts. i.e, $\overrightarrow{l}=2R(-\hat{k})$ and $\overrightarrow{B}=B\hat{j}$. So,

Force on $\text{KLM:}$

${\overrightarrow{F}}_{KLM}=I\left[2R\right(-\hat{k})\times B\hat{j}]=2IBR\hat{i}$

Thus, $|{\overrightarrow{F}}_{KLM}|=F_1=2IBR$

Force on $\text{KNM:}$

${\overrightarrow{F}}_{KNM}=I\left[2R\right(-\hat{k})\times B\hat{j}]=2IBR\hat{i}$

Thus, $|{\overrightarrow{F}}_{KNM}|=F_2=2IBR$

Net force:

$F=F_1+F_2=2IBR+2IBR=4IBR$

Hence, option (d) is the correct answer.