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