You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Two Parallel Currents

A closed PQ...

Question

A closed $P Q R S$ carrying a current is placed in uniform magnetic field. If the magnetic forces on segments $P S, S R$ and $R Q$ are $F_{1}, F_{2}$ and $F_{3}$ respectively and are in the plane of the paper and along the directions shown, then force on the segment $Q P$ is -

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\sqrt{(F_{3} - F_{1})^{2} - F_{2}^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\sqrt{(F_{3} - F_{1})^{2} + F_{2}^{2}}$
From above daigram
$F^{'}$ in y axis
$F^{'} = F_{3} - F_{1}$
and so resultantant of two forces
$F^{'} = F_{3} - F_{1}$ and $F_{2}$ will be
$F_{n e t} = \sqrt{(F_{3} - F_{1})^{2} + F_{2}^{2}}$

Suggest Corrections

Similar questions

Q. A closed loop

P Q R S

carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments

P S, S R

and

R S

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

Q P

Q. 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}, a n d F_{3}

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

Q. If

F_{1}

F_{2}

makes an angle of

30^{\circ}

and

45^{\circ}

respectively with

F_{3}

as shown in the figure, and magnitude of

F_{3}

5 N

, then the magnitude of

F_{1}

and

F_{2}

respectively are (Given,

_{1} +_{2} =_{3}

)

Q. If

F_{1}

F_{2}

makes an angle of

60^{\circ}

and

37^{\circ}

respectively with

F_{3}

as shown in the figure, and magnitude of

F_{3}

2 N

, then the magnitude of

F_{1}

and

F_{2}

respectively are (Given,

_{1} +_{2} =_{3}

)

The forces F₁, F_2, and F₃ are acting on a particle of mass m, such that F₂ and F₃ are mutually perpendicular and under the effect of F₁, F_2, and F₃ , the particle remains stationary. What will be the acceleration of the particle, if the force F₁ is removed?

Join BYJU'S Learning Program

A closed PQRS carrying a current is placed in 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, then force on the segment QP is -

The correct option is A √(F3−F1)2+F22 From above daigram F′ in y axisF′=F3−F1and so resultantant of two forcesF′=F3−F1 and F2 will be Fnet=√(F3−F1)2+F22

A closed $P Q R S$ carrying a current is placed in uniform magnetic field. If the magnetic forces on segments $P S, S R$ and $R Q$ are $F_{1}, F_{2}$ and $F_{3}$ respectively and are in the plane of the paper and along the directions shown, then force on the segment $Q P$ is -

The correct option is A $\sqrt{(F_{3} - F_{1})^{2} + F_{2}^{2}}$
From above daigram
$F^{'}$ in y axis
$F^{'} = F_{3} - F_{1}$
and so resultantant of two forces
$F^{'} = F_{3} - F_{1}$ and $F_{2}$ will be
$F_{n e t} = \sqrt{(F_{3} - F_{1})^{2} + F_{2}^{2}}$