Question

A closed loop $PQRS$ carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments $PS,SR$ and $RS$ are $F_1,F_2$ and $F_3$ respectively and are in the plane of the paper and along the directions shown, the force on the segment $QP$ is

• A. $F_3-F_1+F_2$
• B. $F_3-F_1-F_2$
• C. $\sqrt{{(F_3-F_1)}^2-F_2^2}$
• D. $\sqrt{{(F_3-F_1)}^2+F_2^2}$

Answer 1

The correct option is D $\sqrt{{(F_3-F_1)}^2+F_2^2}$

Total force on the current carrying closed loop should be zero, if placed in a uniform magnetic field.

So, net force, $\overrightarrow{F}=I[\oint d\overrightarrow{l}\times \overrightarrow{B}]=0$

$\therefore {\overrightarrow{F}}_1+{\overrightarrow{F}}_3+{\overrightarrow{F}}_2+{\overrightarrow{F}}_{PQ}=0$

$\Rightarrow F_1(-\hat{i})+F_3\hat{i}+F_2(-\hat{j})+{\overrightarrow{F}}_{PQ}=0$

$\Rightarrow {\overrightarrow{F}}_{PQ}=F_1\hat{i}-F_3\hat{i}+F_2\hat{j}$

$\Rightarrow {\overrightarrow{F}}_{PQ}=(F_1-F_3)\hat{i}+F_2\hat{j}$

$\therefore \left|{\overrightarrow{F}}_{PQ}\right|=\sqrt{(F_1-F_3{)}^2+F_2^2}$

Hence, option (c) is the correct answer.