A current-carrying closed loop in the form of a right-angle isosceles triangle ABC is placed in a uniform magn

Question

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:

Accepted Answer

1. **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} 	imes ec{B}) $ over a closed path.
2. **Force Components:** For the triangular loop ABC, the total force is the vector sum of the forces on its three arms:
   $$ ec{F}_{AB} + ec{F}_{BC} + ec{F}_{AC} = 0 $$
3. **Force on Arm AB:** The magnetic field is acting along the arm AB. Therefore, the angle $	heta$ 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	heta $ and $ \sin(0^{\circ}) = \sin(180^{\circ}) = 0 $, the force on arm AB is zero ($ ec{F}_{AB} = 0 $) .
4. **Calculation:** Substituting $ ec{F}_{AB} = 0 $ into the equilibrium equation:
   $$ 0 + ec{F}_{BC} + ec{F}_{AC} = 0 $$
   $$ ec{F}_{AC} = -\vec{F}_{BC} $$
5. **Conclusion:** Given that the force on arm BC is $\vec{F}$, the force on arm AC must be $-\vec{F}$.

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