question_answer

A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ 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                     [AIPMPT (S) 2008]

A)       \[{{F}_{3}}-{{F}_{1}}-{{F}_{2}}\]

B) \[\sqrt{{{({{F}_{3}}-{{F}_{1}})}^{2}}+F_{2}^{2}}\]

C) \[\sqrt{{{({{F}_{3}}-{{F}_{1}})}^{2}}-F_{2}^{2}}\]

D) \[{{F}_{3}}-{{F}_{1}}+{{F}_{2}}\]

Correct Answer: B

Solution :

The FBD of the loop is as shown

Therefore, force on QP will be equal and opposite to sum of forces on other sides.

Thus,    \[{{F}_{QP}}=\sqrt{{{({{F}_{3}}-{{F}_{1}})}^{2}}+F_{2}^{2}}\]

Alternative:

\[{{F}_{4}}\sin \theta ={{F}_{2}}\]

\[{{F}_{4}}\cos \theta =({{F}_{3}}-{{F}_{1}})\]

\[\therefore \]      \[{{F}_{4}}=\sqrt{{{({{F}_{3}}-{{F}_{1}})}^{2}}+F_{2}^{2}}\]