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 - Physics MCQ Physics | ExamTest.live

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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 $F_{3} - F_{1} - F_{2}$
C $(F_{3} - F_{1})^{2} + F_{2}^{2}$
D $F_{1} + F_{2} + F_{3}$

Solution & Step-by-step Explanation

The net magnetic force on any closed current-carrying loop in a uniform magnetic field is zero. F_{n e t} = F_{PS} + F_{SR} + F_{RQ} + F_{QP} = 0 F_{1} + F_{2} + F_{3} + F_{QP} = 0 F_{QP} = - (F_{1} + F_{2} + F_{3}) If F_{1} and F_{3} are collinear (opposite sides) and F_{2} is perpendicular:The magnitude will be (F_{3} - F_{1})^{2} + F_{2}^{2} .

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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:

F_{3} - F_{1} + F_{2}

F_{3} - F_{1} - F_{2}

(F_{3} - F_{1})^{2} + F_{2}^{2}

F_{1} + F_{2} + F_{3}

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Magnetism Magnetic Force Current Carrying Loop

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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:

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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 F1​, F2​, and F3​ respectively and are in the plane of the paper and along the directions shown, the force on the segment QP is:

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Discussion

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: