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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

Option: 1

F_3-F_1-F_2


Option: 2

\sqrt{\left(\mathrm{F}_3-\mathrm{F}_1\right)^2+\mathrm{F}_2^2}


Option: 3

\sqrt{F_3-F_1+F_2}


Option: 4

None


Answers (1)

best_answer

According to the figure the magnitude of the force on the segment QM is F_3-F_1 and PM is F_2.

Therefore, the magnitude of the force on segment PQ is \sqrt{\left(\mathrm{F}_3-\mathrm{F}_1\right)^2+\mathrm{F}_2^2}

Posted by

Irshad Anwar

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