A closed loop `PQRS` carrying a current is place in a unifrom 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

Options

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

C

Explanation

The resultant of the forces `F_3 and F_1` is abs `(F_3-F_1)`. The force `(F_3-F_1) and F_2 ` are pendicular to each other. `Therefore` The resultant force`=sqrt(F_2^(2)+(F_3-F_1)^(2))therefore` The force on QP will be equal and opposite to the sum of the forces acting on the other 3 sides. `therefore F_(QP)=sqrt(F_2^(2)+(F_3-F_1)^(2))`.