A closed loop PQRS carrying a current is placedin auniform magnetic field. If the magnetic force on the segments PS,SR and RQ are `F_(1),F_(2)"and" F_(3)` respectively and are in the plane of the p[aper and along the directions shown in the figure, the force on the segment QP is

Options

A. `sqrt((F_(3)-F_(1))^(2)-F_(2)^(2))`
B. `F_(3)-F_(1)-F_(2)`
C. `F_(3)-F_(1)-F_(2)`
D. `sqrt((F_(3)-F_(1))^(2)+F_(2)^(2))`

Answer

D

Explanation

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

`F_("horizontal")=(F_(3)-F_(1))`
`F_("vertical")=F_(2)`
Resultant of `vecF_(1), vecF_(2) and vecF_(3)` is `vecF`
`"where "F=sqrt((F_(3)-F_(1))^(2)+F_(2)^(2))`
Since total force = 0, hence force on QP is equal
to `vecF` in magnitude but opposite direction.
`F_(QP)=sqrt((F_(3)-F_(1))^(2)+F_(2)^(2))`