Solution
System Analysis:
The ring rolls on the ceiling. Gravity acts downwards ($mg$). The particle P is fixed on the ring. For there to be “no force of interaction” between the ring and particle, the normal force/reaction force must be zero. This means the sum of external forces (Gravity + Magnetic) must provide exactly the required centripetal acceleration.
1. Condition at Top Point (Contact):
At the point of contact with the ceiling, the velocity of the particle is zero ($v=0$). Thus, magnetic force is zero. The acceleration is centripetal, directed downwards: $a = v^2/r$.
Forces: Gravity ($mg$, down). Interaction force ($N$). Condition $N=0$.
$$ mg = \frac{mv^2}{r} \implies r = \frac{v^2}{g} $$2. Condition at Bottom Point:
At the lowest point, the velocity of the particle is $2v$ (rolling motion). The acceleration is centripetal, directed upwards: $a = v^2/r$.
Forces: Gravity ($mg$, down). Magnetic Force ($F_m$).
For acceleration to be upwards, $F_m$ must be upwards and larger than gravity. With $B$ horizontal and perpendicular, $F_m = q(2v)B$.
$$ F_m – mg = \frac{mv^2}{r} $$Using $r = v^2/g$, we have $\frac{mv^2}{r} = mg$. So:
$$ 2qvB – mg = mg $$ $$ 2qvB = 2mg \implies q = \frac{mg}{Bv} $$