Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A current-carrying closed loop in the form of a right-angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F, the force on the arm AC is:

-F

√2F

-√2F

Step-by-Step Solution

Net Force on a Closed Loop: The net magnetic force on any closed current-carrying loop placed in a uniform magnetic field is always zero ( $ec{F}_{net} = 0$ ). This is derived from the integral of the force vector $I(d ec{l} \times ec{B})$ over a closed path.
Force Components: For the triangular loop ABC, the total force is the vector sum of the forces on its three arms: $ec{F}_{AB} + ec{F}_{BC} + ec{F}_{AC} = 0$
Force on Arm AB: The magnetic field is acting along the arm AB. Therefore, the angle $\theta$ between the current vector $I ec{l}$ and the magnetic field $\vec{B}$ is either $0^{\circ}$ or $180^{\circ}$ . Since the magnetic force magnitude is $F = IlB \sin\theta$ and $\sin(0^{\circ}) = \sin(180^{\circ}) = 0$ , the force on arm AB is zero ( $ec{F}_{AB} = 0$ ) .
Calculation: Substituting $ec{F}_{AB} = 0$ into the equilibrium equation: $0 + ec{F}_{BC} + ec{F}_{AC} = 0$ $ec{F}_{AC} = -\vec{F}_{BC}$
Conclusion: Given that the force on arm BC is $\vec{F}$ , the force on arm AC must be $-\vec{F}$ .

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