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

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