A small, charged bead can slide on a circular frictionless, insulating fixed wire frame. A point like dipole is fixed at the centre of circle, dipole moment is P→. Initially the bead is on the plane of symmetry of the dipole. Bead is released from rest. Ignore the effect of gravity. Select the correct options:

12mv2+q(pcosθ4πε0r2)=0 Or v=−2Qpcosθ4πε0mr2Circular motion of bead required a centripetal force Er=−∂V∂r=2pcosθ4πε0r3Note that QEr=mv2rThus wire frame does not exert any force on the bead to sustain circular motion. Bead will reach the point opposite its path executing a periodic motion.

A small, charged bead can slide on a circular frictionless, insulating fixed wire frame. A point like dipole is fixed at the centre of circle, dipole moment is P→. Initially the bead is on the plane of symmetry of the dipole. Bead is released from rest. Ignore the effect of gravity. Select the correct options:

12mv2+q(pcosθ4πε0r2)=0 Or v=−2Qpcosθ4πε0mr2Circular motion of bead required a centripetal force Er=−∂V∂r=2pcosθ4πε0r3Note that QEr=mv2rThus wire frame does not exert any force on the bead to sustain circular motion. Bead will reach the point opposite its path executing a periodic motion.

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A small, charged bead can slide on a circular frictionless, insulating fixed wire frame. A point like dipole is fixed at the centre of circle, dipole moment is $\vec{P}$ . Initially the bead is on the plane of symmetry of the dipole. Bead is released from rest. Ignore the effect of gravity. Select the correct options:

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Magnitude of velocity of bead as function of its angular position is $\sqrt{\frac{Q P \cos θ}{2 π ε_{0} m r^{2}}}$

Normal force exerted by the wire on bead is zero

Bead would move along a circular path until it reaches the position opposite to its starting position and then executes periodic motion

If the wire frame were not present bead executes circular motion and returns to initial point after tracing a complete circle

answer is A, B, C.

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Detailed Solution

$\frac{1}{2} m v^{2} + q (\frac{p \cos θ}{4 π ε_{0} r^{2}}) = 0 O r v = \sqrt{\frac{- 2 Q p \cos θ}{4 π ε_{0} m r^{2}}}$

Circular motion of bead required a centripetal force
$E_{r} = - \frac{\partial V}{\partial r} = \frac{2 p \cos θ}{4 π ε_{0} r^{3}}$
Note that $Q E_{r} = \frac{m v^{2}}{r}$
Thus wire frame does not exert any force on the bead to sustain circular motion. Bead will reach the point opposite its path executing a periodic motion.